Some time ago, I recall that in the thread "What's Your Biggest Hurdle?" Alphanumeric posted this question about holonomy groups and orbifolds.
The question caught my interest, and although I am not at all familiar with the applications of orbifolds to string theory, I am fairly well versed in manifolds, quotient groups and quotient spaces, symmetries, etc., and so I thought I'd take on the question from a mathematical perspective.
Review of what I (think I) know so far:
Let's start by generalizing to T^n, the torus formed by R^n mod a lattice. First, note that the lattice is not necessarily rectangular, but may have a structure like that of a simple triclinic crystal (in n dimensions, of course). Therefore, the lattice itself can be considered as a linear transformation of the fundamental lattice of integers in n dimensions, i.e. all n-vectors whose components are integers. The torus under consideration is then topologically R^n/Z^n.
To form an orbifold, you mod R^n by a finite discrete symmetry group G, with the restriction that G must send the lattice to itself (athough it will not necessarily fix any single point of the lattice), and then consider the action of G on the torus. There are some restrictions on what G can look like, as for instance there is no five-fold symmetry on a lattice in the plane R^2, so G cannot by(Z/5)^2 in two dimensions. It is probably well known, but not known to me, exactly what structures are possible for G. However, this is not necessary for solving the holonomy problem.
The holonomy group around a singluar point is studied by parallel transporting vectors around small loops near the point and studying any anomalies. If a vector undergoes a direction shift of 2pi/k when carried around a loop, this can be regarded as the effect of a rotation matrix applied to the vector, and these matrices form a group. In the simplest case, this group is isomorphic to Z/k.
End of review. Starting here, these are my thoughts on the problem.
G will produce various stationary points in the torus, some of which may have been stationary in R^n, but some will only be stationary points in the torus via the identification that occurs when you mod R^n by the lattice. Now fix attention on any single stationary point P. Neither the topology nor the geometry of the torus is affected if the lattice is slid around in R^n, as this merely amounts to repositioning the origin of coordinates in the torus. Therefore, the chosen symmetry point can be regarded as at the center of a lattice cell. Then for any studies of the local properties of the orbifold near this point, it is effectively far from the edges of the lattice cell. This frees one from considering any complications which may arise from identifications at the lattice cell edges, and reduces the problem to one in R^n instead of T^n.
Now consider a point Q nearby P and a set of small loop paths starting and ending at Q. In 2 dimensions, the idea of the loop going around P is well defined, but in higher dimensions this no longer makes any sense. Therefore, it is more difficult to see what the loop path is doing when it detects a parallel transport anomaly (i.e. the effect of a holonomy element). It is easiest to proceed by analogy with the 2 dimensional case.
In 2 dimensions, each stationary point (which can always be thought of as NOT on the lattice cell edge) must be a point identified with itself under R^2/G, which means that it must be at the point of a cone, with R^2 wrapped several times around the cone. There are two ways to look at the cone: as a seamless whole, with R^2 multiply covering it, or as a sector taken from R^2 with the radial edges joined. The latter case is the useful point of view, because it introduces a seam in the cone, which is a half-line terminating at P. Note that the actual location of the edge is arbitrary, but its existence is required, much like a branch cut in complex contour integration.
In higher dimensions, R^n is a multiple covering of R^n/G, and so the edge has dimension n-1, and the boundary of this edge has dimension n-2. The point P can then no longer be uniquely defined as the endpoint of its edge, because the dimension is wrong. Therefore, P can only be the common intersection of the boundaries of several edges. Now think about how each of these edges arises. If there are several edges, there must be several independent foldings of R^n, similarly to how one could fold a piece of paper both horizontal and vertical directions (although the analogy is not perfect). Each folding is associated with a k-fold covering of R^n/G by R^n, which in turn will be generated by an subgroup of G that is of order k. For each edge, its boundary is the set of points fixed by this subgroup.
Since we are dealing with Abelian groups here, G must split as a product of single-generator groups of various orders, so that G is homomorphic to product(Z/k)^l where the product is over k and l depends on k. Therefore, each edge is associated with a subgroup of G with a single generator of order k. This subgroup fixes points on the n-2 dimensional boundary B of its associated edge and rotates points by 2pi/k in the 2-dimensional plane O orthogonal to the boundary. Looking at the inverse image of this edge in R^n, one sees that it becomes a set of k radial half-planes of dimension n-1 fanning out from an n-2 dimensional plane and cutting R^n into k sectors. Crossing this edge is equivalent to having the path jump to the identified point on the other side of the sector, which entails an angular change in direction of 2pi/k for the parallel transported vector along the loop.
In R^n, each factor group's O plane will lie in the B plane of every other factor group, because if not, the rotations would interfere with each other and G would not be Abelian. NOTE: There is one exception to the previous paragraph. Although for k>2, the action of Z/k must be a rotation on one or more 2 dimensional subspaces of R^n, this is not so for k=2. Here a reflection through a fixed plane of dimension n-1 could also represent Z/2, as well as the rotation by pi which fits the pattern of the other Z/k groups. Since G is Abelian, you cannot have mixtures of reflections and rotations in the same subgroup, so for instance the group of symmetries on the square is not allowed, because it is not Abelian. Therefore, each factor group of G is either reflection or one of the rotation groups. Also, note that all the rotation groups fix an n-2 dimensional subspace, i.e. the subspaces have codimension 2, and if the intersections of these fixed subspaces is to be a point, then n/2 such subspaces are required. Therefore, R^n must have even n in order to give fixed points. If R^n is odd dimensioned, then an odd number of the factor groups of G MUST be Z/2 acting as reflection, so orientation cannot be preserved.
Now consider the loop through Q again. This loop may return to Q without crossing any of the edges, or may cross each one independently of crossing any other one. The situation is very similar to the way light bounces off a corner reflector. From a single light source, you will get 7 images, which together with the original source, put one point in each octant of R^3. The octant of the image depends on which mirrors the light struck on its path, but all choices are open. In the present case, the orbifold will look like the Cartesian product of a bunch of sectors of R^2 (each with its own angular width 2pi/k, which may or may not be different from that of any other sector), together with (if needed) one or more copies of the half-plane in R^2. These half planes come from the reflection action of the Z/2 subgroups.
Any path that crosses an edge associated with a group element that belongs to a subgroup of order k will find its parallel transported vector has been rotated by 2pi/k, which is the generator of the finite group of rotations homomorphic to Z/k. Since the path may wander around and encounter each edge an arbitrary number of times and in any order, the total set of changes in orientation for the parallel transported vector is an arbitrary combination of the generator elements of each factor group of G. That is, it is G itself.
Another Note: Once again, the reflection acts differently. Reflection does not identify the opposite edges (an angle pi apart) on the R^2 half plane. Instead, it reflects paths back from the edge, since a path crossing the edge at angle 0 would pass into the other half-plane, and be reflected back into the original half-plane still near 0, not near pi. This means that if you try to parallel transport a vector across this kind of edge, instead of the path reappearing elsewhere with its vector rotated, the path and vector bounce off the edge, just like light off a mirror, with vector's orientation to the path unchanged. Since you cannot cross the boundary, you only get the identity element of the Z/2 group. Holonomy doesn't make much sense in this scenario, and so the case of reflection should probably be exluded on physical grounds. Of course that leaves you with the theorem that for odd dimension n, there is no Abelian group G that will produce stationary points. Speaking of which, if you choose too small a group G, you won't get stationary points anyway, but larger dimensional singularities in the orbifold (creases, etc.). The case of point singularities seems to arise only if G is chosen to make it so.
Last minute addition: These orbifolds appear to be a kind of approximation to the Calabi-Yau manifolds, which raises an interesting idea for me. The holonomy in neighborhoods away from the singular points is trivial because the torus is flat, while in the Calabi-Yau manifolds, the curvature is nonzero and so their is nontrivial holonomy everywhere. Perhaps you could start with an orbifold, get the holonomy near the singular points, and then spread this out as a kind of "holonomy density" from which you could infer a local curvature. This might give a manifold that is topologically closer to the behavior of a true Calabi-Yau manifold (i.e. nonsingular), but has a fairly simple "patchwork approximation" formula for local curvature. Just a thought.
There, that's my contribution for the day. And remember, these are things I've never thought seriously about before, and so I'm FIGURING IT OUT AS I GO ALONG. It could be riddled with errors and misconceptions; therefore, all comments are welcome. Anyway, I hope it is of interest to a nonzero number of readers.
--Stuart Anderson
QUOTE (mr_homm+Nov 25 2007, 02:38 AM)
It is probably well known, but not known to me, exactly what structures are possible for G. However, this is not necessary for solving the holonomy problem.
Though specific to the T^6 torus of string theory this paper lists the various Z_p and Z_p x Z_q discrete groups which act on the lattice crystalographically on page 34 (as numbered on the pages themselves)
I don't know if you're familiar with Dynkin diagrams and root systems, but that table also lists the lattice layout. SU(2)^6 has a Dynkin diagram of just 6 isolated dots, which means they are all at right angles to one another. SU(3)^3 has 3 pairs of dots joined by a single line, so there's 3 pair of vectors which are at 90 degrees to any other pair and 120 degrees from their partner. It's a pretty nice way to write down a lattice with a particular symmetry.
Though specific to the T^6 torus of string theory this paper lists the various Z_p and Z_p x Z_q discrete groups which act on the lattice crystalographically on page 34 (as numbered on the pages themselves)
I don't know if you're familiar with Dynkin diagrams and root systems, but that table also lists the lattice layout. SU(2)^6 has a Dynkin diagram of just 6 isolated dots, which means they are all at right angles to one another. SU(3)^3 has 3 pairs of dots joined by a single line, so there's 3 pair of vectors which are at 90 degrees to any other pair and 120 degrees from their partner. It's a pretty nice way to write down a lattice with a particular symmetry.
QUOTE (mr_homm+Nov 25 2007, 02:38 AM)
The holonomy group around a singluar point is studied by parallel transporting vectors around small loops near the point and studying any anomalies. If a vector undergoes a direction shift of 2pi/k when carried around a loop, this can be regarded as the effect of a rotation matrix applied to the vector
It definitely receives a shift of 2pi/k, but despite the path taken around the singularity being almost arbitrary, to then compare the different results, a straight line transportation is done, which seems a little odd to me. Suddenly it's not arbitrary.
I can see where it all comes from, I think I'm just missing something slight and it's making me feel a bit uneasy with it. It's things like this which could be explained in a few seconds by someone who knows about it, but which books can't help with since they aren't interactive.
I can see where it all comes from, I think I'm just missing something slight and it's making me feel a bit uneasy with it. It's things like this which could be explained in a few seconds by someone who knows about it, but which books can't help with since they aren't interactive.
QUOTE (mr_homm+Nov 25 2007, 02:38 AM)
In 2 dimensions, each stationary point (which can always be thought of as NOT on the lattice cell edge) must be a point identified with itself under R^2/G, which means that it must be at the point of a cone, with R^2 wrapped several times around the cone. There are two ways to look at the cone: as a seamless whole, with R^2 multiply covering it, or as a sector taken from R^2 with the radial edges joined. The latter case is the useful point of view, because it introduces a seam in the cone, which is a half-line terminating at P. Note that the actual location of the edge is arbitrary, but its existence is required, much like a branch cut in complex contour integration.
This is precisely the example I had in my head. Nakahara gives a very short mention of orbifolds and that's one of the examples he gives. I even made myself a folded paper version 
QUOTE (mr_homm+Nov 25 2007, 02:38 AM)
In higher dimensions, R^n is a multiple covering of R^n/G, and so the edge has dimension n-1, and the boundary of this edge has dimension n-2. The point P can then no longer be uniquely defined as the endpoint of its edge, because the dimension is wrong. Therefore, P can only be the common intersection of the boundaries of several edges. Now think about how each of these edges arises. If there are several edges, there must be several independent foldings of R^n, similarly to how one could fold a piece of paper both horizontal and vertical directions (although the analogy is not perfect). Each folding is associated with a k-fold covering of R^n/G by R^n, which in turn will be generated by an subgroup of G that is of order k. For each edge, its boundary is the set of points fixed by this subgroup.
I see what you're getting at. It's going to be the intersection of all the various subspaces which are invariant under the various elements of G.
I'm not going to continue quoting each paragraph, but suffice to say I agree with what you're saying and you certainly explain it in a way which helps with my understanding
I'm not going to continue quoting each paragraph, but suffice to say I agree with what you're saying and you certainly explain it in a way which helps with my understanding
QUOTE (mr_homm+Nov 25 2007, 02:38 AM)
. Holonomy doesn't make much sense in this scenario, and so the case of reflection should probably be exluded on physical grounds.
Z_2 orbifolds are common in string constructions but mostly as a matter of model building and seeing principles in action. As that list in that pdf I linked to shows, it's not considered a viable model in general, but I have seen papers talking about it. Z_2 x Z_p are common though. The majority of the work I've been doing recently has been on such constructions.
QUOTE (mr_homm+Nov 25 2007, 02:38 AM)
Last minute addition: These orbifolds appear to be a kind of approximation to the Calabi-Yau manifolds, which raises an interesting idea for me. The holonomy in neighborhoods away from the singular points is trivial because the torus is flat, while in the Calabi-Yau manifolds, the curvature is nonzero and so their is nontrivial holonomy everywhere. Perhaps you could start with an orbifold, get the holonomy near the singular points, and then spread this out as a kind of "holonomy density" from which you could infer a local curvature. This might give a manifold that is topologically closer to the behavior of a true Calabi-Yau manifold (i.e. nonsingular), but has a fairly simple "patchwork approximation" formula for local curvature. Just a thought.
You're right, they are. These orbifold constructions are singular limits of Calabi Yau spaces. They are everywhere flat except at the singularities. If you blow up the singularities into smooth 'blobs' you have a Calabi Yau manifold (orbifolds are not manifolds in general). A great deal of work is being done in such things because orbifolds are easier to describe but not as general as Calabi Yaus. I've spoken to some people who consider 'conifolds' which seem to be what these singular points are close up (as you said with the R^2/Z_p example). They introduce a perturbation to the singularity, making it non-zero size and see how it evolves in time.
Excellent post I wait to see if any of the cranks are going to try and feign understanding. 
Hey Neo', your cue!
QUOTE (mr_homm+Nov 25 2007, 02:38 AM)
There, that's my contribution for the day. And remember, these are things I've never thought seriously about before, and so I'm FIGURING IT OUT AS I GO ALONG. It could be riddled with errors and misconceptions; therefore, all comments are welcome. Anyway, I hope it is of interest to a nonzero number of readers.
Considering you've never thought much about them before you have certainly grasped it damn quickly!
Excellent post I wait to see if any of the cranks are going to try and fain understanding.
Excellent post
I wait to see if any of the cranks are going to try and fain understanding.
QUOTE (AlphaNumeric+Nov 25 2007, 08:20 AM)
Excellent post
I wait to see if any of the cranks are going to try and feign understanding. Hey Neo', your cue!
Hi Alphanumeric,
I haven't read the whole thing yet, but the table of groups did hold some surprises. Z/7 is a rather bizarre symmetry for a lattice. I would not have thought it was possible on a 6 dimensional torus, but these things are rather hard to picture, and sometimes one's intuition does let one down!
I haven't read the whole thing yet, but the table of groups did hold some surprises. Z/7 is a rather bizarre symmetry for a lattice. I would not have thought it was possible on a 6 dimensional torus, but these things are rather hard to picture, and sometimes one's intuition does let one down!
I don't know if you're familiar with Dynkin diagrams and root systems, but that table also lists the lattice layout. SU(2)^6 has a Dynkin diagram of just 6 isolated dots, which means they are all at right angles to one another. SU(3)^3 has 3 pairs of dots joined by a single line, so there's 3 pair of vectors which are at 90 degrees to any other pair and 120 degrees from their partner. It's a pretty nice way to write down a lattice with a particular symmetry.
My graduate work was mainly in algebraic topology (loved homology groups, hated homotopy groups, never used holonomy groups), but I also did a year course on Lie groups, which included the whole classification scheme for the semi-simple Lie groups via roots and Dynkin diagrams. That was 20 years ago, so a lot of the detail has gone kind of vague since I haven't touched that stuff since grad school, but the general outline I still recall. Roots were used mostly for classification, and weights for describing different representations of the Lie Groups, i.e. their actions on different spaces.
So let's see if I'm connecting up these concepts properly: You choose a Lie group, which represents a basic physical symmetry (right?), and its roots form a lattice, which you use to create a torus of appropriate dimension. The torus then in a sense encodes some information about the roots and therefore the Lie group. At the same time, the lattice places restrictions on the values of k in Z/k that will preserve the lattice and therefore can be used to create orbifolds. The Lie group determines the dimension of the torus and the menu of Z/k choices, and the particular choice made then finally determines the exact orbifold. So the orbifold encodes information about the underlying Lie group and the choice of modding group. Does that sound right?
It definitely receives a shift of 2pi/k, but despite the path taken around the singularity being almost arbitrary, to then compare the different results, a straight line transportation is done, which seems a little odd to me. Suddenly it's not arbitrary.
This seems more than just a little odd to me too! In fact, it seems absurd to introduce the straight line transport at all. It may not actually be necessary, but just be a particular way that one person decided to do it, that got passed along. I've seen that happen a lot in mathematics, although eventually someone notices and it gets sorted out. My feeling about it right now is that you can get the holonomy group without using the straight line transport at all, because you can always compare the original test vector with what comes back after you parallel transport it around a loop path in the orbifold, and see how much it has rotated. Do you have a reference where they discuss how to get the holonomy group and mention this straight line transport? I'd like to take a look and try to see what's going on there.
Here is how I would think of doing it without the straight line transport: Start with a singular point P and a starting point Q at a distance d from P, with d chosen "sufficiently small." Consider ALL loop paths starting and ending at Q, confined to a ball around P of radius D>d. On each loop path, parallel transport an entire frame around the path (frame = complete orthonormal set of vectors). Compare the parallel transported frame with the original frame for each path. It is known that any two frame orientations are related by a UNIQUE element of O_n, the orthogonal group of the appropriate dimension, so each path becomes associated with a group element from O_n. Concatenated paths associate to the product of these group elements (because it is a composition of rotations or reflections), and concatenated paths are already included among all paths in the ball, so you get closure, hence a subgroup of O_n. You know already that it is a discrete group, since the rotations are by fixed finite increments, depending on how often the path crosses each "seam" on the orbifold. This is clearly the holonomy group, is it not? But the straight transport seems unnecessary in this approach.
This seems more than just a little odd to me too! In fact, it seems absurd to introduce the straight line transport at all. It may not actually be necessary, but just be a particular way that one person decided to do it, that got passed along. I've seen that happen a lot in mathematics, although eventually someone notices and it gets sorted out. My feeling about it right now is that you can get the holonomy group without using the straight line transport at all, because you can always compare the original test vector with what comes back after you parallel transport it around a loop path in the orbifold, and see how much it has rotated. Do you have a reference where they discuss how to get the holonomy group and mention this straight line transport? I'd like to take a look and try to see what's going on there.
Here is how I would think of doing it without the straight line transport: Start with a singular point P and a starting point Q at a distance d from P, with d chosen "sufficiently small." Consider ALL loop paths starting and ending at Q, confined to a ball around P of radius D>d. On each loop path, parallel transport an entire frame around the path (frame = complete orthonormal set of vectors). Compare the parallel transported frame with the original frame for each path. It is known that any two frame orientations are related by a UNIQUE element of O_n, the orthogonal group of the appropriate dimension, so each path becomes associated with a group element from O_n. Concatenated paths associate to the product of these group elements (because it is a composition of rotations or reflections), and concatenated paths are already included among all paths in the ball, so you get closure, hence a subgroup of O_n. You know already that it is a discrete group, since the rotations are by fixed finite increments, depending on how often the path crosses each "seam" on the orbifold. This is clearly the holonomy group, is it not? But the straight transport seems unnecessary in this approach.
I can see where it all comes from, I think I'm just missing something slight and it's making me feel a bit uneasy with it. It's things like this which could be explained in a few seconds by someone who knows about it, but which books can't help with since they aren't interactive.
In my experience, books have to make a choice between being hard-core technical or "friendly." In the former case, they assume you know nearly everything already, and in the latter case they reduce everything to the level of "Mr. Sun shines because of thing called fusion, which is like a really hot fire..." In neither case do they provide much wisdom about how or why the theory came to be constructed as it presently exists. Interactive would be great, but there's such a limited availability of "face time" with real experts, alas for the rest of us.
Yes, and actually this bothered me at first, because it seemed that the entire n-2 dimensional fixed subspace of each group element should be singular. And so it should, if the group G factors into subgroups that act independently on 2 dimensional subspaces of R^n. A group G like that would be for instance (Z/3)^3 on R^6, which would rotate each of 3 2-d subspaces independently by an angle 0 or 2pi/3 or -2pi/3. This is possible basically because T^n is the n-fold Cartesian product of T^1, which is rather unusual (spheres don't behave this way, for instance). The fixed point of each 2-d rotation on its own private 2-d subspace becomes n-2 dimensional via the Cartesian product. This appears to happen whenever G splits into subgroups with separate actions on the separate Cartesian factors of T^n. So these would not be good orbifolds, because the singularity is a bunch of n-2 dimensional planes rather than a bunch of points.
More generally, it seems that G must not factor into subgroups that each act on a separate invriant subspace of T^n (in the sense that G maps these spaces to themselves, not the stronger sense that G fixes points within them) in any way at all. I mean, T may also be possibly a Cartesian product of two things that don't look like lower dimensional tori, and the action of G should not split that way either. It looks like it's OK for G to factor, as long as the factors don't act on separate Cartesian components of T^n.
So what DOES work? One of the examples in your references is Z/3, acting on each 2-d Cartesian factor in a coordinated way, rather than independently, so that vectors are rotated by the same angle in each 2-d factor torus. There is obviously no possibility of separate factors operating on separate subspaces, since Z/3 does not factor, and in fact, this action of Z/3 has just a single fixed point in R^6. This expands to 27 fixed points in T^6 according to your references, because (I suspect) the interaction with each torus factor splits the fixed point into 3 copies, which happens 3 times, for a total of 27.
Now I'm a very visual thinker, so this situation has just given me an idea about how to represent all this in a nice clean way. To start with, the groups you are working with all do seem to be defined by actions on each Cartesian factor torus, but these actions are coordinated instead of independent. Nevertheless, let's start with an independent action, for instance (Z/3)^3 with each factor acting independently on one Cartesian factor torus. Consider just one of these group factors Z/3 acting on its own Cartesian factor T^2. Following the idea in my previous post, zoom in on a particular stationary point and locate that point at the center of the lattice cell, and confine attention to a small disk around the point, so that the periodicity across the lattice boundary is irrelevant, and you are looking at Z/3 acting on R^2.
Reducing mod Z/3 splits R^2 into 3 sectors of angle 2pi/3 each, and a circle around the origin would pass through these factors one at a time. Cut this circle at one of the sector boundaries and unwrap it to represent these three sectors as three line segments adjacent in a straight line. Moving from one segment to the next is a way of visually representing moving from one sector to the next. Passing out of the rightmost segment, you will of course pass into the leftmost segment, since the actual sectors wrap around a circle. It doesn't geometrically look like the original "pie chart" image of sectors, but this will turn out not to matter, since all we want to do is know when a path has moved from one sector to another (because that's what causes the rotation of the parallel transported vector).
Now do this for the other two factors of (Z/3)^3 on the other two Cartesian factor tori. Since the whole group and T^6 are products of these subgroups and Cartesian factors, you can represent the various regions in T^6 as the Cartesian product of 3 copies of these line segment diagrams. Of course, the product of 2 copies of the 3-line-segment diagram will form a 3 by 3 grid of squares, and all 3 copies will form a 3 by 3 by 3 box of cubes, like a Rubik's cube, with opposite faces identified. So this cube is basically a 3-torus, but it is discrete, since each cube represent a whole Cartesian product of sectors, so moving around within a cube is basically meaningless. It may as well be a 3 by 3 by 3 grid of points, i.e. a 27 point discrete 3-torus.
What's so great about this picture? Well, since all 27 points are identified, and since following a path from point to point indicates crossings from one sector to another, every path through the picture encodes all the direction changes a parallel transported vector will undergo as its path carries it through the actual space T^6/G. Since all points are identified, any path, not just closed paths, represent closed paths in T^6/G. Further, the effect on the parellel transported vector is PATH INDEPENDENT on this diagram! This is because each step in each of the 3 directions on this diagram represents a 2pi/3 direction change in one of the 3 T^2 components of T^6. A path that wiggles around or circles back just undoes some of the effect of its own earlier segments. Equivalently, a loop path on this diagram (including paths that leave one edge and wrap back from the other edge, just so long as the starting and ending point are the same) represent NO direction change for the parallel transported vector, since the same number of transitions (mod 3) must be made up/down, right/left and front/back in order to return to the starting point, and this causes all the angle changes to either cancel out or sum to 2pi. (This is very similar to the situation in electrostatics where the fact that every loop integral of E is zero is equivalent to path independence of line integrals between fixed starting and ending points.)
That's what independently acting subgroups look like. Now what about Z/3 acting by IDENTICAL 2pi/3 rotations on each T^2 factor? You get the same exact 3 by 3 by 3 grid of points, but now not every point is identified. The coordinated action shows that points are identified only if they belong to sectors that are shifted by 2pi/3 in all three Cartesian factors. This means that in the grid diagram, points are identified along space diagonals. For instance, if the points are given coordinates (0,0,0) through (2,2,2), then the points (0,0,0), (1,1,1), and (2,2,2) represent regions in T^6 that are identified mod G. Because the grid wraps back at the edges like T^3, all the other diagonals parallel to the main space diagonal are also identified. This means that closed loop paths in T^6 are represented by paths in the grid that terminate in a point equivalent to their starting point, i.e. on the same diagonal. Since path independence of the effect on the parallel transported vector is still true, this means that there are as many effects on this vector as there are distinct displacements along a fixed diagonal.
By periodicity, there are 3 possibilities: stay home, (displacement 0,0,0), move along the diagonal 1 step (displacement 1,1,1), or 2 steps (2,2,2). Of course 3 steps brings you back to your starting point, so it is equivalent to (0,0,0). These steps obviously form a group under vector addition with periodic boundaries, and if one of the points is taken to be (0,0,0) itself, then the displacements are precisely the set of all points equivalent to (0,0,0). But by definition, two points are equivalent if the action of an element of G maps the regions they represent to each other. Therefore, the group structure of vector addition on the equivalent points is identical to the structure of G. But the displacement written as a vector, can also be read as a list of the angles that the parallel transported vector will be turned through by a closed path in T^6/G crossing from the region at the start of the displacement vector to the region at its end. But the group of these angle rotations is the holonomy group, so you get holonomy = vector addition on identified points of grid = G.
THAT'S what's so great about this picture.
Oh, and it lets you visualize the relevant aspects of a continuous 6-dimensional structure as a discrete 3-dimensional structure, which saves wear and tear on the old visual cortex. It should be straightforward to sketch the grid for any choice of G if you know its action on T^6. It's just a matter of coloring in the points that are equivalent to (0,0,0) and then connecting the dots to get the displacements.
And now on to other points....
Yes, and actually this bothered me at first, because it seemed that the entire n-2 dimensional fixed subspace of each group element should be singular. And so it should, if the group G factors into subgroups that act independently on 2 dimensional subspaces of R^n. A group G like that would be for instance (Z/3)^3 on R^6, which would rotate each of 3 2-d subspaces independently by an angle 0 or 2pi/3 or -2pi/3. This is possible basically because T^n is the n-fold Cartesian product of T^1, which is rather unusual (spheres don't behave this way, for instance). The fixed point of each 2-d rotation on its own private 2-d subspace becomes n-2 dimensional via the Cartesian product. This appears to happen whenever G splits into subgroups with separate actions on the separate Cartesian factors of T^n. So these would not be good orbifolds, because the singularity is a bunch of n-2 dimensional planes rather than a bunch of points.
More generally, it seems that G must not factor into subgroups that each act on a separate invriant subspace of T^n (in the sense that G maps these spaces to themselves, not the stronger sense that G fixes points within them) in any way at all. I mean, T may also be possibly a Cartesian product of two things that don't look like lower dimensional tori, and the action of G should not split that way either. It looks like it's OK for G to factor, as long as the factors don't act on separate Cartesian components of T^n.
So what DOES work? One of the examples in your references is Z/3, acting on each 2-d Cartesian factor in a coordinated way, rather than independently, so that vectors are rotated by the same angle in each 2-d factor torus. There is obviously no possibility of separate factors operating on separate subspaces, since Z/3 does not factor, and in fact, this action of Z/3 has just a single fixed point in R^6. This expands to 27 fixed points in T^6 according to your references, because (I suspect) the interaction with each torus factor splits the fixed point into 3 copies, which happens 3 times, for a total of 27.
Now I'm a very visual thinker, so this situation has just given me an idea about how to represent all this in a nice clean way. To start with, the groups you are working with all do seem to be defined by actions on each Cartesian factor torus, but these actions are coordinated instead of independent. Nevertheless, let's start with an independent action, for instance (Z/3)^3 with each factor acting independently on one Cartesian factor torus. Consider just one of these group factors Z/3 acting on its own Cartesian factor T^2. Following the idea in my previous post, zoom in on a particular stationary point and locate that point at the center of the lattice cell, and confine attention to a small disk around the point, so that the periodicity across the lattice boundary is irrelevant, and you are looking at Z/3 acting on R^2.
Reducing mod Z/3 splits R^2 into 3 sectors of angle 2pi/3 each, and a circle around the origin would pass through these factors one at a time. Cut this circle at one of the sector boundaries and unwrap it to represent these three sectors as three line segments adjacent in a straight line. Moving from one segment to the next is a way of visually representing moving from one sector to the next. Passing out of the rightmost segment, you will of course pass into the leftmost segment, since the actual sectors wrap around a circle. It doesn't geometrically look like the original "pie chart" image of sectors, but this will turn out not to matter, since all we want to do is know when a path has moved from one sector to another (because that's what causes the rotation of the parallel transported vector).
Now do this for the other two factors of (Z/3)^3 on the other two Cartesian factor tori. Since the whole group and T^6 are products of these subgroups and Cartesian factors, you can represent the various regions in T^6 as the Cartesian product of 3 copies of these line segment diagrams. Of course, the product of 2 copies of the 3-line-segment diagram will form a 3 by 3 grid of squares, and all 3 copies will form a 3 by 3 by 3 box of cubes, like a Rubik's cube, with opposite faces identified. So this cube is basically a 3-torus, but it is discrete, since each cube represent a whole Cartesian product of sectors, so moving around within a cube is basically meaningless. It may as well be a 3 by 3 by 3 grid of points, i.e. a 27 point discrete 3-torus.
What's so great about this picture? Well, since all 27 points are identified, and since following a path from point to point indicates crossings from one sector to another, every path through the picture encodes all the direction changes a parallel transported vector will undergo as its path carries it through the actual space T^6/G. Since all points are identified, any path, not just closed paths, represent closed paths in T^6/G. Further, the effect on the parellel transported vector is PATH INDEPENDENT on this diagram! This is because each step in each of the 3 directions on this diagram represents a 2pi/3 direction change in one of the 3 T^2 components of T^6. A path that wiggles around or circles back just undoes some of the effect of its own earlier segments. Equivalently, a loop path on this diagram (including paths that leave one edge and wrap back from the other edge, just so long as the starting and ending point are the same) represent NO direction change for the parallel transported vector, since the same number of transitions (mod 3) must be made up/down, right/left and front/back in order to return to the starting point, and this causes all the angle changes to either cancel out or sum to 2pi. (This is very similar to the situation in electrostatics where the fact that every loop integral of E is zero is equivalent to path independence of line integrals between fixed starting and ending points.)
That's what independently acting subgroups look like. Now what about Z/3 acting by IDENTICAL 2pi/3 rotations on each T^2 factor? You get the same exact 3 by 3 by 3 grid of points, but now not every point is identified. The coordinated action shows that points are identified only if they belong to sectors that are shifted by 2pi/3 in all three Cartesian factors. This means that in the grid diagram, points are identified along space diagonals. For instance, if the points are given coordinates (0,0,0) through (2,2,2), then the points (0,0,0), (1,1,1), and (2,2,2) represent regions in T^6 that are identified mod G. Because the grid wraps back at the edges like T^3, all the other diagonals parallel to the main space diagonal are also identified. This means that closed loop paths in T^6 are represented by paths in the grid that terminate in a point equivalent to their starting point, i.e. on the same diagonal. Since path independence of the effect on the parallel transported vector is still true, this means that there are as many effects on this vector as there are distinct displacements along a fixed diagonal.
By periodicity, there are 3 possibilities: stay home, (displacement 0,0,0), move along the diagonal 1 step (displacement 1,1,1), or 2 steps (2,2,2). Of course 3 steps brings you back to your starting point, so it is equivalent to (0,0,0). These steps obviously form a group under vector addition with periodic boundaries, and if one of the points is taken to be (0,0,0) itself, then the displacements are precisely the set of all points equivalent to (0,0,0). But by definition, two points are equivalent if the action of an element of G maps the regions they represent to each other. Therefore, the group structure of vector addition on the equivalent points is identical to the structure of G. But the displacement written as a vector, can also be read as a list of the angles that the parallel transported vector will be turned through by a closed path in T^6/G crossing from the region at the start of the displacement vector to the region at its end. But the group of these angle rotations is the holonomy group, so you get holonomy = vector addition on identified points of grid = G.
THAT'S what's so great about this picture. Oh, and it lets you visualize the relevant aspects of a continuous 6-dimensional structure as a discrete 3-dimensional structure, which saves wear and tear on the old visual cortex. It should be straightforward to sketch the grid for any choice of G if you know its action on T^6. It's just a matter of coloring in the points that are equivalent to (0,0,0) and then connecting the dots to get the displacements.
And now on to other points....
Z_2 orbifolds are common in string constructions but mostly as a matter of model building and seeing principles in action. As that list in that pdf I linked to shows, it's not considered a viable model in general, but I have seen papers talking about it. Z_2 x Z_p are common though. The majority of the work I've been doing recently has been on such constructions.
But what is the action of Z/2 on the torus? Z/2 can in general act by reversing the sign of k coordinates while preserving the sign of the remaining n-k coordinates. But if k is odd, then Z/2 is not orientation preserving, and you get the weird behavior that I mentioned before where the orbifold has a reflective boundary instead of a wraparound. It seems puzzling to me, not what you expect from an orbifold, but perhaps there was a flaw in my reasoning and it won't really happen that way.
Thanks! But it's not really that special, because I have known about groups, manifolds, and quotient spaces for many years, it was just a matter of applying them to this situation. I'm rather proud of my grid diagram idea in this post however, because that was a genuine creative inspiration that I had while I was typing this post. Always a good feeling to know that the old brain hasn't completely rotted away from teaching undergrad physics and engineering for 20 years.
--Stuart Anderson
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Though specific to the T^6 torus of string theory this paper lists the various Z_p and Z_p x Z_q discrete groups which act on the lattice crystalographically on page 34 (as numbered on the pages themselves)
I haven't read the whole thing yet, but the table of groups did hold some surprises. Z/7 is a rather bizarre symmetry for a lattice. I would not have thought it was possible on a 6 dimensional torus, but these things are rather hard to picture, and sometimes one's intuition does let one down!
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| QUOTE |
| Though specific to the T^6 torus of string theory this paper lists the various Z_p and Z_p x Z_q discrete groups which act on the lattice crystalographically on page 34 (as numbered on the pages themselves) |
I haven't read the whole thing yet, but the table of groups did hold some surprises. Z/7 is a rather bizarre symmetry for a lattice. I would not have thought it was possible on a 6 dimensional torus, but these things are rather hard to picture, and sometimes one's intuition does let one down!
I don't know if you're familiar with Dynkin diagrams and root systems, but that table also lists the lattice layout. SU(2)^6 has a Dynkin diagram of just 6 isolated dots, which means they are all at right angles to one another. SU(3)^3 has 3 pairs of dots joined by a single line, so there's 3 pair of vectors which are at 90 degrees to any other pair and 120 degrees from their partner. It's a pretty nice way to write down a lattice with a particular symmetry.
My graduate work was mainly in algebraic topology (loved homology groups, hated homotopy groups, never used holonomy groups), but I also did a year course on Lie groups, which included the whole classification scheme for the semi-simple Lie groups via roots and Dynkin diagrams. That was 20 years ago, so a lot of the detail has gone kind of vague since I haven't touched that stuff since grad school, but the general outline I still recall. Roots were used mostly for classification, and weights for describing different representations of the Lie Groups, i.e. their actions on different spaces.
So let's see if I'm connecting up these concepts properly: You choose a Lie group, which represents a basic physical symmetry (right?), and its roots form a lattice, which you use to create a torus of appropriate dimension. The torus then in a sense encodes some information about the roots and therefore the Lie group. At the same time, the lattice places restrictions on the values of k in Z/k that will preserve the lattice and therefore can be used to create orbifolds. The Lie group determines the dimension of the torus and the menu of Z/k choices, and the particular choice made then finally determines the exact orbifold. So the orbifold encodes information about the underlying Lie group and the choice of modding group. Does that sound right?
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It definitely receives a shift of 2pi/k, but despite the path taken around the singularity being almost arbitrary, to then compare the different results, a straight line transportation is done, which seems a little odd to me. Suddenly it's not arbitrary.
This seems more than just a little odd to me too! In fact, it seems absurd to introduce the straight line transport at all. It may not actually be necessary, but just be a particular way that one person decided to do it, that got passed along. I've seen that happen a lot in mathematics, although eventually someone notices and it gets sorted out. My feeling about it right now is that you can get the holonomy group without using the straight line transport at all, because you can always compare the original test vector with what comes back after you parallel transport it around a loop path in the orbifold, and see how much it has rotated. Do you have a reference where they discuss how to get the holonomy group and mention this straight line transport? I'd like to take a look and try to see what's going on there.
Here is how I would think of doing it without the straight line transport: Start with a singular point P and a starting point Q at a distance d from P, with d chosen "sufficiently small." Consider ALL loop paths starting and ending at Q, confined to a ball around P of radius D>d. On each loop path, parallel transport an entire frame around the path (frame = complete orthonormal set of vectors). Compare the parallel transported frame with the original frame for each path. It is known that any two frame orientations are related by a UNIQUE element of O_n, the orthogonal group of the appropriate dimension, so each path becomes associated with a group element from O_n. Concatenated paths associate to the product of these group elements (because it is a composition of rotations or reflections), and concatenated paths are already included among all paths in the ball, so you get closure, hence a subgroup of O_n. You know already that it is a discrete group, since the rotations are by fixed finite increments, depending on how often the path crosses each "seam" on the orbifold. This is clearly the holonomy group, is it not? But the straight transport seems unnecessary in this approach.
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| QUOTE |
It definitely receives a shift of 2pi/k, but despite the path taken around the singularity being almost arbitrary, to then compare the different results, a straight line transportation is done, which seems a little odd to me. Suddenly it's not arbitrary. |
This seems more than just a little odd to me too! In fact, it seems absurd to introduce the straight line transport at all. It may not actually be necessary, but just be a particular way that one person decided to do it, that got passed along. I've seen that happen a lot in mathematics, although eventually someone notices and it gets sorted out. My feeling about it right now is that you can get the holonomy group without using the straight line transport at all, because you can always compare the original test vector with what comes back after you parallel transport it around a loop path in the orbifold, and see how much it has rotated. Do you have a reference where they discuss how to get the holonomy group and mention this straight line transport? I'd like to take a look and try to see what's going on there.
Here is how I would think of doing it without the straight line transport: Start with a singular point P and a starting point Q at a distance d from P, with d chosen "sufficiently small." Consider ALL loop paths starting and ending at Q, confined to a ball around P of radius D>d. On each loop path, parallel transport an entire frame around the path (frame = complete orthonormal set of vectors). Compare the parallel transported frame with the original frame for each path. It is known that any two frame orientations are related by a UNIQUE element of O_n, the orthogonal group of the appropriate dimension, so each path becomes associated with a group element from O_n. Concatenated paths associate to the product of these group elements (because it is a composition of rotations or reflections), and concatenated paths are already included among all paths in the ball, so you get closure, hence a subgroup of O_n. You know already that it is a discrete group, since the rotations are by fixed finite increments, depending on how often the path crosses each "seam" on the orbifold. This is clearly the holonomy group, is it not? But the straight transport seems unnecessary in this approach.
I can see where it all comes from, I think I'm just missing something slight and it's making me feel a bit uneasy with it. It's things like this which could be explained in a few seconds by someone who knows about it, but which books can't help with since they aren't interactive.
In my experience, books have to make a choice between being hard-core technical or "friendly." In the former case, they assume you know nearly everything already, and in the latter case they reduce everything to the level of "Mr. Sun shines because of thing called fusion, which is like a really hot fire..." In neither case do they provide much wisdom about how or why the theory came to be constructed as it presently exists. Interactive would be great, but there's such a limited availability of "face time" with real experts, alas for the rest of us.
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I see what you're getting at. It's going to be the intersection of all the various subspaces which are invariant under the various elements of G.
Yes, and actually this bothered me at first, because it seemed that the entire n-2 dimensional fixed subspace of each group element should be singular. And so it should, if the group G factors into subgroups that act independently on 2 dimensional subspaces of R^n. A group G like that would be for instance (Z/3)^3 on R^6, which would rotate each of 3 2-d subspaces independently by an angle 0 or 2pi/3 or -2pi/3. This is possible basically because T^n is the n-fold Cartesian product of T^1, which is rather unusual (spheres don't behave this way, for instance). The fixed point of each 2-d rotation on its own private 2-d subspace becomes n-2 dimensional via the Cartesian product. This appears to happen whenever G splits into subgroups with separate actions on the separate Cartesian factors of T^n. So these would not be good orbifolds, because the singularity is a bunch of n-2 dimensional planes rather than a bunch of points.
More generally, it seems that G must not factor into subgroups that each act on a separate invriant subspace of T^n (in the sense that G maps these spaces to themselves, not the stronger sense that G fixes points within them) in any way at all. I mean, T may also be possibly a Cartesian product of two things that don't look like lower dimensional tori, and the action of G should not split that way either. It looks like it's OK for G to factor, as long as the factors don't act on separate Cartesian components of T^n.
So what DOES work? One of the examples in your references is Z/3, acting on each 2-d Cartesian factor in a coordinated way, rather than independently, so that vectors are rotated by the same angle in each 2-d factor torus. There is obviously no possibility of separate factors operating on separate subspaces, since Z/3 does not factor, and in fact, this action of Z/3 has just a single fixed point in R^6. This expands to 27 fixed points in T^6 according to your references, because (I suspect) the interaction with each torus factor splits the fixed point into 3 copies, which happens 3 times, for a total of 27.
Now I'm a very visual thinker, so this situation has just given me an idea about how to represent all this in a nice clean way. To start with, the groups you are working with all do seem to be defined by actions on each Cartesian factor torus, but these actions are coordinated instead of independent. Nevertheless, let's start with an independent action, for instance (Z/3)^3 with each factor acting independently on one Cartesian factor torus. Consider just one of these group factors Z/3 acting on its own Cartesian factor T^2. Following the idea in my previous post, zoom in on a particular stationary point and locate that point at the center of the lattice cell, and confine attention to a small disk around the point, so that the periodicity across the lattice boundary is irrelevant, and you are looking at Z/3 acting on R^2.
Reducing mod Z/3 splits R^2 into 3 sectors of angle 2pi/3 each, and a circle around the origin would pass through these factors one at a time. Cut this circle at one of the sector boundaries and unwrap it to represent these three sectors as three line segments adjacent in a straight line. Moving from one segment to the next is a way of visually representing moving from one sector to the next. Passing out of the rightmost segment, you will of course pass into the leftmost segment, since the actual sectors wrap around a circle. It doesn't geometrically look like the original "pie chart" image of sectors, but this will turn out not to matter, since all we want to do is know when a path has moved from one sector to another (because that's what causes the rotation of the parallel transported vector).
Now do this for the other two factors of (Z/3)^3 on the other two Cartesian factor tori. Since the whole group and T^6 are products of these subgroups and Cartesian factors, you can represent the various regions in T^6 as the Cartesian product of 3 copies of these line segment diagrams. Of course, the product of 2 copies of the 3-line-segment diagram will form a 3 by 3 grid of squares, and all 3 copies will form a 3 by 3 by 3 box of cubes, like a Rubik's cube, with opposite faces identified. So this cube is basically a 3-torus, but it is discrete, since each cube represent a whole Cartesian product of sectors, so moving around within a cube is basically meaningless. It may as well be a 3 by 3 by 3 grid of points, i.e. a 27 point discrete 3-torus.
What's so great about this picture? Well, since all 27 points are identified, and since following a path from point to point indicates crossings from one sector to another, every path through the picture encodes all the direction changes a parallel transported vector will undergo as its path carries it through the actual space T^6/G. Since all points are identified, any path, not just closed paths, represent closed paths in T^6/G. Further, the effect on the parellel transported vector is PATH INDEPENDENT on this diagram! This is because each step in each of the 3 directions on this diagram represents a 2pi/3 direction change in one of the 3 T^2 components of T^6. A path that wiggles around or circles back just undoes some of the effect of its own earlier segments. Equivalently, a loop path on this diagram (including paths that leave one edge and wrap back from the other edge, just so long as the starting and ending point are the same) represent NO direction change for the parallel transported vector, since the same number of transitions (mod 3) must be made up/down, right/left and front/back in order to return to the starting point, and this causes all the angle changes to either cancel out or sum to 2pi. (This is very similar to the situation in electrostatics where the fact that every loop integral of E is zero is equivalent to path independence of line integrals between fixed starting and ending points.)
That's what independently acting subgroups look like. Now what about Z/3 acting by IDENTICAL 2pi/3 rotations on each T^2 factor? You get the same exact 3 by 3 by 3 grid of points, but now not every point is identified. The coordinated action shows that points are identified only if they belong to sectors that are shifted by 2pi/3 in all three Cartesian factors. This means that in the grid diagram, points are identified along space diagonals. For instance, if the points are given coordinates (0,0,0) through (2,2,2), then the points (0,0,0), (1,1,1), and (2,2,2) represent regions in T^6 that are identified mod G. Because the grid wraps back at the edges like T^3, all the other diagonals parallel to the main space diagonal are also identified. This means that closed loop paths in T^6 are represented by paths in the grid that terminate in a point equivalent to their starting point, i.e. on the same diagonal. Since path independence of the effect on the parallel transported vector is still true, this means that there are as many effects on this vector as there are distinct displacements along a fixed diagonal.
By periodicity, there are 3 possibilities: stay home, (displacement 0,0,0), move along the diagonal 1 step (displacement 1,1,1), or 2 steps (2,2,2). Of course 3 steps brings you back to your starting point, so it is equivalent to (0,0,0). These steps obviously form a group under vector addition with periodic boundaries, and if one of the points is taken to be (0,0,0) itself, then the displacements are precisely the set of all points equivalent to (0,0,0). But by definition, two points are equivalent if the action of an element of G maps the regions they represent to each other. Therefore, the group structure of vector addition on the equivalent points is identical to the structure of G. But the displacement written as a vector, can also be read as a list of the angles that the parallel transported vector will be turned through by a closed path in T^6/G crossing from the region at the start of the displacement vector to the region at its end. But the group of these angle rotations is the holonomy group, so you get holonomy = vector addition on identified points of grid = G.
THAT'S what's so great about this picture.
Oh, and it lets you visualize the relevant aspects of a continuous 6-dimensional structure as a discrete 3-dimensional structure, which saves wear and tear on the old visual cortex. It should be straightforward to sketch the grid for any choice of G if you know its action on T^6. It's just a matter of coloring in the points that are equivalent to (0,0,0) and then connecting the dots to get the displacements.And now on to other points....
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| QUOTE |
| I see what you're getting at. It's going to be the intersection of all the various subspaces which are invariant under the various elements of G. |
Yes, and actually this bothered me at first, because it seemed that the entire n-2 dimensional fixed subspace of each group element should be singular. And so it should, if the group G factors into subgroups that act independently on 2 dimensional subspaces of R^n. A group G like that would be for instance (Z/3)^3 on R^6, which would rotate each of 3 2-d subspaces independently by an angle 0 or 2pi/3 or -2pi/3. This is possible basically because T^n is the n-fold Cartesian product of T^1, which is rather unusual (spheres don't behave this way, for instance). The fixed point of each 2-d rotation on its own private 2-d subspace becomes n-2 dimensional via the Cartesian product. This appears to happen whenever G splits into subgroups with separate actions on the separate Cartesian factors of T^n. So these would not be good orbifolds, because the singularity is a bunch of n-2 dimensional planes rather than a bunch of points.
More generally, it seems that G must not factor into subgroups that each act on a separate invriant subspace of T^n (in the sense that G maps these spaces to themselves, not the stronger sense that G fixes points within them) in any way at all. I mean, T may also be possibly a Cartesian product of two things that don't look like lower dimensional tori, and the action of G should not split that way either. It looks like it's OK for G to factor, as long as the factors don't act on separate Cartesian components of T^n.
So what DOES work? One of the examples in your references is Z/3, acting on each 2-d Cartesian factor in a coordinated way, rather than independently, so that vectors are rotated by the same angle in each 2-d factor torus. There is obviously no possibility of separate factors operating on separate subspaces, since Z/3 does not factor, and in fact, this action of Z/3 has just a single fixed point in R^6. This expands to 27 fixed points in T^6 according to your references, because (I suspect) the interaction with each torus factor splits the fixed point into 3 copies, which happens 3 times, for a total of 27.
Now I'm a very visual thinker, so this situation has just given me an idea about how to represent all this in a nice clean way. To start with, the groups you are working with all do seem to be defined by actions on each Cartesian factor torus, but these actions are coordinated instead of independent. Nevertheless, let's start with an independent action, for instance (Z/3)^3 with each factor acting independently on one Cartesian factor torus. Consider just one of these group factors Z/3 acting on its own Cartesian factor T^2. Following the idea in my previous post, zoom in on a particular stationary point and locate that point at the center of the lattice cell, and confine attention to a small disk around the point, so that the periodicity across the lattice boundary is irrelevant, and you are looking at Z/3 acting on R^2.
Reducing mod Z/3 splits R^2 into 3 sectors of angle 2pi/3 each, and a circle around the origin would pass through these factors one at a time. Cut this circle at one of the sector boundaries and unwrap it to represent these three sectors as three line segments adjacent in a straight line. Moving from one segment to the next is a way of visually representing moving from one sector to the next. Passing out of the rightmost segment, you will of course pass into the leftmost segment, since the actual sectors wrap around a circle. It doesn't geometrically look like the original "pie chart" image of sectors, but this will turn out not to matter, since all we want to do is know when a path has moved from one sector to another (because that's what causes the rotation of the parallel transported vector).
Now do this for the other two factors of (Z/3)^3 on the other two Cartesian factor tori. Since the whole group and T^6 are products of these subgroups and Cartesian factors, you can represent the various regions in T^6 as the Cartesian product of 3 copies of these line segment diagrams. Of course, the product of 2 copies of the 3-line-segment diagram will form a 3 by 3 grid of squares, and all 3 copies will form a 3 by 3 by 3 box of cubes, like a Rubik's cube, with opposite faces identified. So this cube is basically a 3-torus, but it is discrete, since each cube represent a whole Cartesian product of sectors, so moving around within a cube is basically meaningless. It may as well be a 3 by 3 by 3 grid of points, i.e. a 27 point discrete 3-torus.
What's so great about this picture? Well, since all 27 points are identified, and since following a path from point to point indicates crossings from one sector to another, every path through the picture encodes all the direction changes a parallel transported vector will undergo as its path carries it through the actual space T^6/G. Since all points are identified, any path, not just closed paths, represent closed paths in T^6/G. Further, the effect on the parellel transported vector is PATH INDEPENDENT on this diagram! This is because each step in each of the 3 directions on this diagram represents a 2pi/3 direction change in one of the 3 T^2 components of T^6. A path that wiggles around or circles back just undoes some of the effect of its own earlier segments. Equivalently, a loop path on this diagram (including paths that leave one edge and wrap back from the other edge, just so long as the starting and ending point are the same) represent NO direction change for the parallel transported vector, since the same number of transitions (mod 3) must be made up/down, right/left and front/back in order to return to the starting point, and this causes all the angle changes to either cancel out or sum to 2pi. (This is very similar to the situation in electrostatics where the fact that every loop integral of E is zero is equivalent to path independence of line integrals between fixed starting and ending points.)
That's what independently acting subgroups look like. Now what about Z/3 acting by IDENTICAL 2pi/3 rotations on each T^2 factor? You get the same exact 3 by 3 by 3 grid of points, but now not every point is identified. The coordinated action shows that points are identified only if they belong to sectors that are shifted by 2pi/3 in all three Cartesian factors. This means that in the grid diagram, points are identified along space diagonals. For instance, if the points are given coordinates (0,0,0) through (2,2,2), then the points (0,0,0), (1,1,1), and (2,2,2) represent regions in T^6 that are identified mod G. Because the grid wraps back at the edges like T^3, all the other diagonals parallel to the main space diagonal are also identified. This means that closed loop paths in T^6 are represented by paths in the grid that terminate in a point equivalent to their starting point, i.e. on the same diagonal. Since path independence of the effect on the parallel transported vector is still true, this means that there are as many effects on this vector as there are distinct displacements along a fixed diagonal.
By periodicity, there are 3 possibilities: stay home, (displacement 0,0,0), move along the diagonal 1 step (displacement 1,1,1), or 2 steps (2,2,2). Of course 3 steps brings you back to your starting point, so it is equivalent to (0,0,0). These steps obviously form a group under vector addition with periodic boundaries, and if one of the points is taken to be (0,0,0) itself, then the displacements are precisely the set of all points equivalent to (0,0,0). But by definition, two points are equivalent if the action of an element of G maps the regions they represent to each other. Therefore, the group structure of vector addition on the equivalent points is identical to the structure of G. But the displacement written as a vector, can also be read as a list of the angles that the parallel transported vector will be turned through by a closed path in T^6/G crossing from the region at the start of the displacement vector to the region at its end. But the group of these angle rotations is the holonomy group, so you get holonomy = vector addition on identified points of grid = G.
THAT'S what's so great about this picture.
Oh, and it lets you visualize the relevant aspects of a continuous 6-dimensional structure as a discrete 3-dimensional structure, which saves wear and tear on the old visual cortex. It should be straightforward to sketch the grid for any choice of G if you know its action on T^6. It's just a matter of coloring in the points that are equivalent to (0,0,0) and then connecting the dots to get the displacements.And now on to other points....
Z_2 orbifolds are common in string constructions but mostly as a matter of model building and seeing principles in action. As that list in that pdf I linked to shows, it's not considered a viable model in general, but I have seen papers talking about it. Z_2 x Z_p are common though. The majority of the work I've been doing recently has been on such constructions.
But what is the action of Z/2 on the torus? Z/2 can in general act by reversing the sign of k coordinates while preserving the sign of the remaining n-k coordinates. But if k is odd, then Z/2 is not orientation preserving, and you get the weird behavior that I mentioned before where the orbifold has a reflective boundary instead of a wraparound. It seems puzzling to me, not what you expect from an orbifold, but perhaps there was a flaw in my reasoning and it won't really happen that way.
QUOTE
Considering you've never thought much about them before you have certainly grasped it damn quickly!
Thanks! But it's not really that special, because I have known about groups, manifolds, and quotient spaces for many years, it was just a matter of applying them to this situation. I'm rather proud of my grid diagram idea in this post however, because that was a genuine creative inspiration that I had while I was typing this post. Always a good feeling to know that the old brain hasn't completely rotted away from teaching undergrad physics and engineering for 20 years.
--Stuart Anderson
QUOTE (mr_homm+Nov 26 2007, 07:25 AM)
Z/7 is a rather bizarre symmetry for a lattice. I would not have thought it was possible on a 6 dimensional torus, but these things are rather hard to picture, and sometimes one's intuition does let one down!
I had exactly the same thing. I could see things like Z_12, it's just a sort of 'enhanced' symmetry in 6 dimensions but 7 doesn't have any common factors with 6 so it's not something you'd expect at just a glance. That's why I think those Dynkin diagrams are useful, they give a natural way to construct lattice roots and it turns out that one of them has a natural Z_7 symmetry, you don't need to hunt around for it.
I had exactly the same thing. I could see things like Z_12, it's just a sort of 'enhanced' symmetry in 6 dimensions but 7 doesn't have any common factors with 6 so it's not something you'd expect at just a glance. That's why I think those Dynkin diagrams are useful, they give a natural way to construct lattice roots and it turns out that one of them has a natural Z_7 symmetry, you don't need to hunt around for it.
QUOTE (mr_homm+Nov 26 2007, 07:25 AM)
My graduate work was mainly in algebraic topology
I keep meaning to learn a little about that, since it's tools are regularly used in things like cohomologies and in crazy geometries but every time I open an "An Intro to Algebraic Topology" book it scares me.
QUOTE (mr_homm+Nov 26 2007, 07:25 AM)
(loved homology groups, hated homotopy groups, never used holonomy groups),
Homology I pretty much get, on a meaning/definition level. Homotopy groups I just don't get. Every now and again I come across something which says "And the 3rd homotopy group of X is Z, so ...." as if that explains something...
QUOTE (mr_homm+Nov 26 2007, 07:25 AM)
You choose a Lie group, which represents a basic physical symmetry (right?), and its roots form a lattice, which you use to create a torus of appropriate dimension. The torus then in a sense encodes some information about the roots and therefore the Lie group. At the same time, the lattice places restrictions on the values of k in Z/k that will preserve the lattice and therefore can be used to create orbifolds. The Lie group determines the dimension of the torus and the menu of Z/k choices, and the particular choice made then finally determines the exact orbifold. So the orbifold encodes information about the underlying Lie group and the choice of modding group. Does that sound right?
Yep, sounds about right. In this case, I don't think the Lie group represents anything physical, it is just a convenient way of expressing a lattice vector setup with a particular symmetry. In usual QFT the Lie group represents an underlying symmetry of the system (phase invariance in the U(1) QED or colour symmetry in the SU(3) QCD) but that doesn't relate to the use of Lie groups here. The construction discussed in that paper and in ones like it has nothing to do with the Standard Model use of Lie groups. That comes later when you start constructing intersecting branes in order to break down a Caton Paton construction into the SM group.
QUOTE (mr_homm+Nov 26 2007, 07:25 AM)
Do you have a reference where they discuss how to get the holonomy group and mention this straight line transport? I'd like to take a look and try to see what's going on there.
http://gesalerico.ft.uam.es/paginaspersonales/angeluranga/lect14.pdf
Pages 5 and 6, particularly the picture on page 6. Note the curved path and then a straight line connection.
That straight line makes sense on a torus since it's constructed from a flat space, the straight line is uniquely defined, but if the manifold which is being orbifolded isn't flat then no such line exists.
http://gesalerico.ft.uam.es/paginaspersona.../firstpage.html is the set of lecture notes I got that from.
Pages 5 and 6, particularly the picture on page 6. Note the curved path and then a straight line connection.
That straight line makes sense on a torus since it's constructed from a flat space, the straight line is uniquely defined, but if the manifold which is being orbifolded isn't flat then no such line exists.
http://gesalerico.ft.uam.es/paginaspersona.../firstpage.html is the set of lecture notes I got that from.
QUOTE (mr_homm+Nov 26 2007, 07:25 AM)
...discussion about looped paths....
I agree with that, but I have issue with the holonomy becoming discrete. If I consider looped paths I can curve my path so that the contribution from the singularity is removed. The holonomy group remains it's usual subgroup of O(n).
QUOTE (mr_homm+Nov 26 2007, 07:25 AM)
In my experience, books have to make a choice between being hard-core technical or "friendly." In the former case, they assume you know nearly everything already, and in the latter case they reduce everything to the level of "Mr. Sun shines because of thing called fusion, which is like a really hot fire..." In neither case do they provide much wisdom about how or why the theory came to be constructed as it presently exists. Interactive would be great, but there's such a limited availability of "face time" with real experts, alas for the rest of us.[/i]
Yeah, I have two books on opposite sides of the spectrum, Nakahara - Geometry, Topology and Physics and Joyce - Riemannian Holonomy Groups and Calibrated Geometry. Nakahara is a graduate text on various geometry concepts in physics and the most things is excellent. For string things thought, it doesn't go deep enough, Joyce is unspeakably complex for most things. It spends about 6 chapters proving the Calabi conjecture.
The rest of your post I'll reply to later today. I have something to do now though...
The rest of your post I'll reply to later today. I have something to do now though...
what is the nature of strings and spacetimes? each one is generator of the other.
the particles are singularities of the infinity? and spacetime continuos can be reverted by T until the initial singularity?
An abelian differential S1$- non trivial holomorph form in a riemannian surface that define a translation structures,with ramifications in the zeros.A translation
structure permit to define an unitary parallel vector field oriented to upper.that induce a slow dynamic in the surface,the vertical flow.
can it deform a translation structure,contrayng the vertical direction( with factor of
constant contraction) and expanding correspondent to the horizontal direction.That
deformation define the flow of teitcmuller in space of moduli in the abelian differentiations.
the flow of teitcmuller can to be seen as a dynamics of renormalization of vertical flows is strongest mixture.is this,whether comport as is wereindependents.This permit
deduce many of the propeties of the vertical flow of surfaces of smooth translations through of analysis of the orbits of flows of teithmuller
excuse me bad english
congratulations for alpha and beta....
the particles are singularities of the infinity? and spacetime continuos can be reverted by T until the initial singularity?
An abelian differential S1$- non trivial holomorph form in a riemannian surface that define a translation structures,with ramifications in the zeros.A translation
structure permit to define an unitary parallel vector field oriented to upper.that induce a slow dynamic in the surface,the vertical flow.
can it deform a translation structure,contrayng the vertical direction( with factor of
constant contraction) and expanding correspondent to the horizontal direction.That
deformation define the flow of teitcmuller in space of moduli in the abelian differentiations.
the flow of teitcmuller can to be seen as a dynamics of renormalization of vertical flows is strongest mixture.is this,whether comport as is wereindependents.This permit
deduce many of the propeties of the vertical flow of surfaces of smooth translations through of analysis of the orbits of flows of teithmuller
excuse me bad english
congratulations for alpha and beta....
QUOTE (mr_homm+Nov 26 2007, 07:25 AM)
But what is the action of Z/2 on the torus? Z/2 can in general act by reversing the sign of k coordinates while preserving the sign of the remaining n-k coordinates. But if k is odd, then Z/2 is not orientation preserving, and you get the weird behavior that I mentioned before where the orbifold has a reflective boundary instead of a wraparound. It seems puzzling to me, not what you expect from an orbifold, but perhaps there was a flaw in my reasoning and it won't really happen that way.
By pure chance I came across this paper earlier today while browsing ArXiv for something interesting to read :
http://xxx.lanl.gov/pdf/0711.1582
Discusses a Z2 orbifold
But you're right, it acts as a reflection, not a rotational twist.
And as for Mott.Carl, for those who are unaware of his methods, he just spews gibberish and hopes people think his incoherence is due to a language barrier. It's not, he's just that deranged. Notice nothing he says follows on from what Mr_Homm and I have been discussing, it's because he doesn't understand but wants people to think he does.
By pure chance I came across this paper earlier today while browsing ArXiv for something interesting to read :
http://xxx.lanl.gov/pdf/0711.1582
Discusses a Z2 orbifold
But you're right, it acts as a reflection, not a rotational twist.
And as for Mott.Carl, for those who are unaware of his methods, he just spews gibberish and hopes people think his incoherence is due to a language barrier. It's not, he's just that deranged. Notice nothing he says follows on from what Mr_Homm and I have been discussing, it's because he doesn't understand but wants people to think he does.
hello mr_homm and AlphaNumeric 
i do not even know how to compose a question about what you guys are discussing without feeling somewhat embarrassed... but the best i can come up with is
why do you need to know what is being discussed ?
i do not even know how to compose a question about what you guys are discussing without feeling somewhat embarrassed... but the best i can come up with is
why do you need to know what is being discussed ?
QUOTE (kjw+Nov 26 2007, 09:07 PM)
why do you need to know what is being discussed ?
I would say that Mr_Homm is doing it for the intellectual pleasure of it, which is an excellent reason in itself (and one which the cranks here do not appreciate). I, however, have to do it because my PhD is based upon it.
String theory has extra dimensions which are curled up. String theory also includes supersymmetry, but just the right kind to be viable. The kind of supersymmetry you have depends on the curled up extra dimensions. To get the right kind you can have curled up dimensions which are either Calabi Yau or an orbifolded/orientifolded torus. Calabi Yaus are, in general, unspeakably complicated. Orbifolded/orientifolded tori are considerably easier to describe. The particular ramifications of the orbifolding process is what we're discussing.
Papers which use such notions and which I've been doing work related to include the following :
http://lanl.arxiv.org/abs/hep-th/0607015
http://lanl.arxiv.org/abs/hep-th/0508133
http://lanl.arxiv.org/abs/0705.3410
They all start with a general 6 dimensional torus, T^6, and then create an orbifold or an orientifold by quotienting out a discrete group, a particular choice of Z_p or Z_p x Z_q. Orientifolding is the next step in such things, involving doing something extra to an orbifold.
I can explain a bit more if you want, giving a bit more physical motivation behind why this is important to string theorists.
let me get comfy ...
I would say that Mr_Homm is doing it for the intellectual pleasure of it, which is an excellent reason in itself (and one which the cranks here do not appreciate). I, however, have to do it because my PhD is based upon it.
String theory has extra dimensions which are curled up. String theory also includes supersymmetry, but just the right kind to be viable. The kind of supersymmetry you have depends on the curled up extra dimensions. To get the right kind you can have curled up dimensions which are either Calabi Yau or an orbifolded/orientifolded torus. Calabi Yaus are, in general, unspeakably complicated. Orbifolded/orientifolded tori are considerably easier to describe. The particular ramifications of the orbifolding process is what we're discussing.
Papers which use such notions and which I've been doing work related to include the following :
http://lanl.arxiv.org/abs/hep-th/0607015
http://lanl.arxiv.org/abs/hep-th/0508133
http://lanl.arxiv.org/abs/0705.3410
They all start with a general 6 dimensional torus, T^6, and then create an orbifold or an orientifold by quotienting out a discrete group, a particular choice of Z_p or Z_p x Z_q. Orientifolding is the next step in such things, involving doing something extra to an orbifold.
I can explain a bit more if you want, giving a bit more physical motivation behind why this is important to string theorists.
QUOTE (kjw+Nov 26 2007, 09:07 PM)
i do not even know how to compose a question about what you guys are discussing without feeling somewhat embarrassed... but the best i can come up with is
Nothing to feel embarrassed about. A year ago I knew nothing about this particular area. Hell, 3 months ago I didn't know very much about the particulars we've been talking about. Besides, I don't know much about a lot of physics. Listening to someone talk about their PhD area might give the impression they know shedloads about everything but it's not true Learning a lot about a tiny tiny area of physics isn't quite the same.
Learning a lot about a tiny tiny area of physics isn't quite the same.
QUOTE
AlphaNumeric Posted: Today at 7:03 AM I can explain a bit more if you want, giving a bit more physical motivation behind why this is important to string theorists.
i would greatly appreciate it. let me get comfy ...
QUOTE (AlphaNumeric+Nov 26 2007, 12:19 AM)
I had exactly the same thing. I could see things like Z_12, it's just a sort of 'enhanced' symmetry in 6 dimensions but 7 doesn't have any common factors with 6 so it's not something you'd expect at just a glance. That's why I think those Dynkin diagrams are useful, they give a natural way to construct lattice roots and it turns out that one of them has a natural Z_7 symmetry, you don't need to hunt around for it.
You're right, the Dynkin diagrams do give a nice way to see what lattice symmetries are out there. The "reason" (in the sense of a plausible geometric visualization) for the Z/7 symmetry just occured to me. It's a relative of the Penrose tiling, which you may recall can be constructed by slicing a 5-d cubic lattice with a 2-d plane. Of course, that's nonperiodic, so perhaps that's not a good example. Closer to the mark is slicing a 3-d cubic lattice with a 2-d plane normal to the line (1,1,1) (i.e. the space diagonal of the lattice). Cubes of course have a 3 fold symmetry around their space diagonal, which is inherited by the lattice. In a similar way, you can take a 7-d cubic lattice, and slice through it with a 6-d plane normal to the line (1,1,1,1,1,1,1) (again, the space diagonal). This will produce a 7-fold symmetry in the 6-d lattice. So my intuition is happy again.
Try Munkries Topology for topological preliminaries at the undergrad level and Munkries Algebraic Topology. The latter is the book I used in grad school, and it is very "approachable" for a topology book. Munkries' undergrad book is also very good. These are rather old now, but not much has changed at that level I suspect.
Try Munkries Topology for topological preliminaries at the undergrad level and Munkries Algebraic Topology. The latter is the book I used in grad school, and it is very "approachable" for a topology book. Munkries' undergrad book is also very good. These are rather old now, but not much has changed at that level I suspect.
Lie group represents anything physical, it is just a convenient way of expressing a lattice vector setup with a particular symmetry. In usual QFT the Lie group represents an underlying symmetry of the system (phase invariance in the U(1) QED or colour symmetry in the SU(3) QCD) but that doesn't relate to the use of Lie groups here. The construction discussed in that paper and in ones like it has nothing to do with the Standard Model use of Lie groups. That comes later when you start constructing intersecting branes in order to break down a Caton Paton construction into the SM group.
Thanks for clearing that up. I did have the QFT use of Lie groups in mind, and was unaware of this other role. I'm afraid you've lost me with the Caton Paton constuctions and the SM group, as I'm unfamiliar with the terminology here.
http://gesalerico.ft.uam.es/paginaspersona...anga/lect14.pdf
Pages 5 and 6, particularly the picture on page 6. Note the curved path and then a straight line connection.
That straight line makes sense on a torus since it's constructed from a flat space, the straight line is uniquely defined, but if the manifold which is being orbifolded isn't flat then no such line exists.
Thanks, I'll take a look at it soon. But I just got busy again after the Thanksgiving holiday break, so suddenly my posts will get slower and sparser.
Could you not use a geodesic of miminal length instead of a straight line? I don't know what value that might have, since to get a geodesic, you would have to know enough already about the curvature of the manifold that you could probably compute the holonomy directly from that.
Thanks, I'll take a look at it soon. But I just got busy again after the Thanksgiving holiday break, so suddenly my posts will get slower and sparser.
Could you not use a geodesic of miminal length instead of a straight line? I don't know what value that might have, since to get a geodesic, you would have to know enough already about the curvature of the manifold that you could probably compute the holonomy directly from that.
http://gesalerico.ft.uam.es/paginaspersona.../firstpage.html is the set of lecture notes I got that from.
I agree with that, but I have issue with the holonomy becoming discrete. If I consider looped paths I can curve my path so that the contribution from the singularity is removed. The holonomy group remains it's usual subgroup of O(n).
The straight line path bothers me less than the notion of a loop path AROUND the singular point, since this concept does not generalize to dimensions > 2. As you say, the path can be deformed to avoid the contribution from the singular point, or wrap it to catch the contribution several times. I don't think you can associate the holonomy group with a single path, but only with the ALL paths near the singular point. That lets the holonomy group be discrete, since each path (depending on where it travels) gets a different DISCRETE increment of rotation on its parallel transported vector. The collective set of these discrete increments gives the holonomy group, right? I'll read the notes you linked to here and see if that clarifies things for me.
That's interesting! I'll take a look at it, time permitting.
That's interesting! I'll take a look at it, time permitting.
I would say that Mr_Homm is doing it for the intellectual pleasure of it, which is an excellent reason in itself (and one which the cranks here do not appreciate). I, however, have to do it because my PhD is based upon it.
Exactly right. It would be nice if fame and eternal glory were mine through some great discovery, but that isn't too likely now it it? It is enough for me to try to understand things, simply because they are fascinating and beautiful.
AlphaNumeric Posted: Today at 7:03 AM I can explain a bit more if you want, giving a bit more physical motivation behind why this is important to string theorists.
i would greatly appreciate it.
let me get comfy ...
Me too! I can hold up my end of the conversation on the purely mathematical aspects (mostly), but the physical interpretation of a lot of it is still rather vague to me. I would be happy to hear a little more about it.
More later.
--Stuart Anderson
You're right, the Dynkin diagrams do give a nice way to see what lattice symmetries are out there. The "reason" (in the sense of a plausible geometric visualization) for the Z/7 symmetry just occured to me. It's a relative of the Penrose tiling, which you may recall can be constructed by slicing a 5-d cubic lattice with a 2-d plane. Of course, that's nonperiodic, so perhaps that's not a good example. Closer to the mark is slicing a 3-d cubic lattice with a 2-d plane normal to the line (1,1,1) (i.e. the space diagonal of the lattice). Cubes of course have a 3 fold symmetry around their space diagonal, which is inherited by the lattice. In a similar way, you can take a 7-d cubic lattice, and slice through it with a 6-d plane normal to the line (1,1,1,1,1,1,1) (again, the space diagonal). This will produce a 7-fold symmetry in the 6-d lattice. So my intuition is happy again.
QUOTE
I keep meaning to learn a little about that, since it's tools are regularly used in things like cohomologies and in crazy geometries but every time I open an "An Intro to Algebraic Topology" book it scares me.
Try Munkries Topology for topological preliminaries at the undergrad level and Munkries Algebraic Topology. The latter is the book I used in grad school, and it is very "approachable" for a topology book. Munkries' undergrad book is also very good. These are rather old now, but not much has changed at that level I suspect.
QUOTE (->
| QUOTE |
| I keep meaning to learn a little about that, since it's tools are regularly used in things like cohomologies and in crazy geometries but every time I open an "An Intro to Algebraic Topology" book it scares me. |
Try Munkries Topology for topological preliminaries at the undergrad level and Munkries Algebraic Topology. The latter is the book I used in grad school, and it is very "approachable" for a topology book. Munkries' undergrad book is also very good. These are rather old now, but not much has changed at that level I suspect.
Lie group represents anything physical, it is just a convenient way of expressing a lattice vector setup with a particular symmetry. In usual QFT the Lie group represents an underlying symmetry of the system (phase invariance in the U(1) QED or colour symmetry in the SU(3) QCD) but that doesn't relate to the use of Lie groups here. The construction discussed in that paper and in ones like it has nothing to do with the Standard Model use of Lie groups. That comes later when you start constructing intersecting branes in order to break down a Caton Paton construction into the SM group.
Thanks for clearing that up. I did have the QFT use of Lie groups in mind, and was unaware of this other role. I'm afraid you've lost me with the Caton Paton constuctions and the SM group, as I'm unfamiliar with the terminology here.
QUOTE
http://gesalerico.ft.uam.es/paginaspersona...anga/lect14.pdf
Pages 5 and 6, particularly the picture on page 6. Note the curved path and then a straight line connection.
That straight line makes sense on a torus since it's constructed from a flat space, the straight line is uniquely defined, but if the manifold which is being orbifolded isn't flat then no such line exists.
Thanks, I'll take a look at it soon. But I just got busy again after the Thanksgiving holiday break, so suddenly my posts will get slower and sparser.
Could you not use a geodesic of miminal length instead of a straight line? I don't know what value that might have, since to get a geodesic, you would have to know enough already about the curvature of the manifold that you could probably compute the holonomy directly from that.
QUOTE (->
| QUOTE |
http://gesalerico.ft.uam.es/paginaspersona...anga/lect14.pdf Pages 5 and 6, particularly the picture on page 6. Note the curved path and then a straight line connection. That straight line makes sense on a torus since it's constructed from a flat space, the straight line is uniquely defined, but if the manifold which is being orbifolded isn't flat then no such line exists. |
Thanks, I'll take a look at it soon. But I just got busy again after the Thanksgiving holiday break, so suddenly my posts will get slower and sparser.
Could you not use a geodesic of miminal length instead of a straight line? I don't know what value that might have, since to get a geodesic, you would have to know enough already about the curvature of the manifold that you could probably compute the holonomy directly from that.
http://gesalerico.ft.uam.es/paginaspersona.../firstpage.html is the set of lecture notes I got that from.
I agree with that, but I have issue with the holonomy becoming discrete. If I consider looped paths I can curve my path so that the contribution from the singularity is removed. The holonomy group remains it's usual subgroup of O(n).
The straight line path bothers me less than the notion of a loop path AROUND the singular point, since this concept does not generalize to dimensions > 2. As you say, the path can be deformed to avoid the contribution from the singular point, or wrap it to catch the contribution several times. I don't think you can associate the holonomy group with a single path, but only with the ALL paths near the singular point. That lets the holonomy group be discrete, since each path (depending on where it travels) gets a different DISCRETE increment of rotation on its parallel transported vector. The collective set of these discrete increments gives the holonomy group, right? I'll read the notes you linked to here and see if that clarifies things for me.
QUOTE
By pure chance I came across this paper earlier today while browsing ArXiv for something interesting to read :
http://xxx.lanl.gov/pdf/0711.1582
Discusses a Z2 orbifold
But you're right, it acts as a reflection, not a rotational twist.
http://xxx.lanl.gov/pdf/0711.1582
Discusses a Z2 orbifold
But you're right, it acts as a reflection, not a rotational twist.
That's interesting! I'll take a look at it, time permitting.
QUOTE (->
| QUOTE |
| By pure chance I came across this paper earlier today while browsing ArXiv for something interesting to read : http://xxx.lanl.gov/pdf/0711.1582 Discusses a Z2 orbifold But you're right, it acts as a reflection, not a rotational twist. |
That's interesting! I'll take a look at it, time permitting.
I would say that Mr_Homm is doing it for the intellectual pleasure of it, which is an excellent reason in itself (and one which the cranks here do not appreciate). I, however, have to do it because my PhD is based upon it.
Exactly right. It would be nice if fame and eternal glory were mine through some great discovery, but that isn't too likely now it it? It is enough for me to try to understand things, simply because they are fascinating and beautiful.
QUOTE (kjw+ Nov 26 2007, 07:04 PM)
QUOTE
AlphaNumeric Posted: Today at 7:03 AM I can explain a bit more if you want, giving a bit more physical motivation behind why this is important to string theorists.
i would greatly appreciate it.
let me get comfy ...
Me too! I can hold up my end of the conversation on the purely mathematical aspects (mostly), but the physical interpretation of a lot of it is still rather vague to me. I would be happy to hear a little more about it.
More later.
--Stuart Anderson
QUOTE (kjw+Nov 27 2007, 04:04 AM)
i would greatly appreciate it.
In the late 70s and very early 80s bosonic string theory had been constructed, initially with the hope it might explain the interaction between quarks. This was because the force between two quarks seemed to act like a string or tube between them, if you pulled them further apart the force attracting them increased, as if connected by a bit of elastic string.
Eventually QCD won out, using the notion of colour confinement and gluons to construct 'flux tubes'. String theory then entered a bit of a lull because bosonic string theory had problems with tachyons, as well as being totally unphysical because it lacked fermions. Someone then realised that if you put in fermions, a number of amazing things happened. The tachyons disappeared and the theory was supersymmetric. Previously, supersymmetry (which had been developed for usual QFT around the same time as bosonic string theory) had to be tagged on by hand, you could have unequal numbers of fermions and bosons. In string theory, if you had both kinds of particles, you had to have equal numbers!
However, there are varying degrees of supersymmetry. N=0 means 'no supersymmetry'. N=1 means 'a little bit' (in very non-technical terms) and it goes all the way up to N=8, "that's ****loads of supersymmetry!". The bigger the value of N, the more extra particles there were predicted, as if the particles we can currently see are a smaller and smaller tip of an iceberg. N=0, we can see all the iceberg, N=1 we see about half of it, N=8 we only see the very edge of it. The value of N also alters the particle content. N>1 is excluded by experiments because it doesn't account for the left/right handed nature of the Weak force (known as 'chirality'). Usual constructions of superstring theory had N=2. A bit of a problem.
Another problem was that superstring theory had lots of dimensions, 10 of them (9 space, 1 time). But we only see 4 (3+1) so the only way that makes sense is if 6 of them are too small for us to see with experiments. In QFT (and relativity in general), energy and length are inversely proportional (if you set c=G=1, your units are such that units of mass = 1/(units of length)), so that in order to access smaller and smaller length interactions, you need more and more energy. This makes sense if you think about how we need bigger and bigger accelerators to probe smaller and smaller objects. So if these dimensions hadn't appeared yet, they must be very small. Experiments say smaller than 10^(-15) metres, theory says as small as 10^(-35) metres! We'd need an accelerator larger than the universe to get that kind of energy!
So how to get past these two problems? Well one problem fixes the other. We have to make 6 of the dimensions small, 'compactified'. But how we do that affects the symmetry of our theory and so alters the amount of supersymmetry. If you compactify your 6 extra dimensions in the simplest way, onto 6 circles (ie a 6d torus, T^6), then you quadruple the amount of supersymmetry your theory has. So a 10d string theory with N=2 goes to a 4+6d string theory with N=8. Bugger, that didn't help.
Well perhaps our ideas about compactification need to be refined a little? Perhaps we need to make the small dimensions a little more complicated. This is where holonomy comes in. We want a theory with N=1 supersymmetry. This is akin to saying that we want a unique spinor in our theory under parallel transport around a closed loop in our space-time. In n dimensions, parallel transport around a closed loop acts like an operator on the spinor, specifically like a membler of a subgroup, G, of SO(n). In flat space-time G=SO(n). However, in complicated spaces you get G < SO(n). It turns out that in order to get a working theory, you need an even number of dimensions in your compact space, so n=2m, and to get the right kind of supersymmetry you need an holonomy group G = SU(m).
Through some rather unpleasant differential geometry (see Joyce), you can prove that a manifold with SU(m) holonomy is Kahler and Ricci flat and has vanishing 1st Chern Class and that's known as a Calabi-Yau space (since Calabi predicted and Yau proved those properties are equivalent). Thus 3 dimensional (3 complex dimensions) Calabi Yau spaces form the core of string theory interest in compact dimensions. Orbifolded and orientifolded tori are a kind of step in the right direction. A general Calabi Yau doesn't have anywhere near the nice description that tori do and you'd have to work through a huge number of cases. At present the orbi/orientifold constructions allow us to see the kind of problems which will arise in full Calabi Yau models and learn how to solve some of them in a nicer environment.
Mr Homm, the difference between an orbifold and an orientifold is that an orientifold has had an extra quotienting process applied to it. Rather than just an orbifold group G, an orientifold also have the group (-1)^(F_L)Ωσ quotiented out. (-1)^(F_L) is the 'left fermionic oscillation counter'. It's +1 if there's even numbers of fermionic oscillators going 'to the left' and -1 if odd. Ω is the worldsheet parity operator and σ is an involution (ie a Z_2 map). Sometimes orbifolding isn't enough or you want to have additional structure.
In the late 70s and very early 80s bosonic string theory had been constructed, initially with the hope it might explain the interaction between quarks. This was because the force between two quarks seemed to act like a string or tube between them, if you pulled them further apart the force attracting them increased, as if connected by a bit of elastic string.
Eventually QCD won out, using the notion of colour confinement and gluons to construct 'flux tubes'. String theory then entered a bit of a lull because bosonic string theory had problems with tachyons, as well as being totally unphysical because it lacked fermions. Someone then realised that if you put in fermions, a number of amazing things happened. The tachyons disappeared and the theory was supersymmetric. Previously, supersymmetry (which had been developed for usual QFT around the same time as bosonic string theory) had to be tagged on by hand, you could have unequal numbers of fermions and bosons. In string theory, if you had both kinds of particles, you had to have equal numbers!
However, there are varying degrees of supersymmetry. N=0 means 'no supersymmetry'. N=1 means 'a little bit' (in very non-technical terms) and it goes all the way up to N=8, "that's ****loads of supersymmetry!". The bigger the value of N, the more extra particles there were predicted, as if the particles we can currently see are a smaller and smaller tip of an iceberg. N=0, we can see all the iceberg, N=1 we see about half of it, N=8 we only see the very edge of it. The value of N also alters the particle content. N>1 is excluded by experiments because it doesn't account for the left/right handed nature of the Weak force (known as 'chirality'). Usual constructions of superstring theory had N=2. A bit of a problem.
Another problem was that superstring theory had lots of dimensions, 10 of them (9 space, 1 time). But we only see 4 (3+1) so the only way that makes sense is if 6 of them are too small for us to see with experiments. In QFT (and relativity in general), energy and length are inversely proportional (if you set c=G=1, your units are such that units of mass = 1/(units of length)), so that in order to access smaller and smaller length interactions, you need more and more energy. This makes sense if you think about how we need bigger and bigger accelerators to probe smaller and smaller objects. So if these dimensions hadn't appeared yet, they must be very small. Experiments say smaller than 10^(-15) metres, theory says as small as 10^(-35) metres! We'd need an accelerator larger than the universe to get that kind of energy!
So how to get past these two problems? Well one problem fixes the other. We have to make 6 of the dimensions small, 'compactified'. But how we do that affects the symmetry of our theory and so alters the amount of supersymmetry. If you compactify your 6 extra dimensions in the simplest way, onto 6 circles (ie a 6d torus, T^6), then you quadruple the amount of supersymmetry your theory has. So a 10d string theory with N=2 goes to a 4+6d string theory with N=8. Bugger, that didn't help.
Well perhaps our ideas about compactification need to be refined a little? Perhaps we need to make the small dimensions a little more complicated. This is where holonomy comes in. We want a theory with N=1 supersymmetry. This is akin to saying that we want a unique spinor in our theory under parallel transport around a closed loop in our space-time. In n dimensions, parallel transport around a closed loop acts like an operator on the spinor, specifically like a membler of a subgroup, G, of SO(n). In flat space-time G=SO(n). However, in complicated spaces you get G < SO(n). It turns out that in order to get a working theory, you need an even number of dimensions in your compact space, so n=2m, and to get the right kind of supersymmetry you need an holonomy group G = SU(m).
Through some rather unpleasant differential geometry (see Joyce), you can prove that a manifold with SU(m) holonomy is Kahler and Ricci flat and has vanishing 1st Chern Class and that's known as a Calabi-Yau space (since Calabi predicted and Yau proved those properties are equivalent). Thus 3 dimensional (3 complex dimensions) Calabi Yau spaces form the core of string theory interest in compact dimensions. Orbifolded and orientifolded tori are a kind of step in the right direction. A general Calabi Yau doesn't have anywhere near the nice description that tori do and you'd have to work through a huge number of cases. At present the orbi/orientifold constructions allow us to see the kind of problems which will arise in full Calabi Yau models and learn how to solve some of them in a nicer environment.
Mr Homm, the difference between an orbifold and an orientifold is that an orientifold has had an extra quotienting process applied to it. Rather than just an orbifold group G, an orientifold also have the group (-1)^(F_L)Ωσ quotiented out. (-1)^(F_L) is the 'left fermionic oscillation counter'. It's +1 if there's even numbers of fermionic oscillators going 'to the left' and -1 if odd. Ω is the worldsheet parity operator and σ is an involution (ie a Z_2 map). Sometimes orbifolding isn't enough or you want to have additional structure.
QUOTE
AlphaNumeric Posted on Today at 7:23 AM However, there are varying degrees of supersymmetry. N=0 means 'no supersymmetry'. N=1 means 'a little bit' (in very non-technical terms) and it goes all the way up to N=8, "that's ****loads of supersymmetry!". The bigger the value of N, the more extra particles there were predicted, as if the particles we can currently see are a smaller and smaller tip of an iceberg. N=0, we can see all the iceberg, N=1 we see about half of it, N=8 we only see the very edge of it. The value of N also alters the particle content. N>1 is excluded by experiments because it doesn't account for the left/right handed nature of the Weak force (known as 'chirality').
is this saying that at N=0 there is no symmetry because instead of there being 2 types of elementary particles (bosons and fermions) there is just 1 type of elementary particle ie a string ? QUOTE (->
| QUOTE |
| AlphaNumeric Posted on Today at 7:23 AM However, there are varying degrees of supersymmetry. N=0 means 'no supersymmetry'. N=1 means 'a little bit' (in very non-technical terms) and it goes all the way up to N=8, "that's ****loads of supersymmetry!". The bigger the value of N, the more extra particles there were predicted, as if the particles we can currently see are a smaller and smaller tip of an iceberg. N=0, we can see all the iceberg, N=1 we see about half of it, N=8 we only see the very edge of it. The value of N also alters the particle content. N>1 is excluded by experiments because it doesn't account for the left/right handed nature of the Weak force (known as 'chirality'). |
is this saying that at N=0 there is no symmetry because instead of there being 2 types of elementary particles (bosons and fermions) there is just 1 type of elementary particle ie a string ?
Another problem was that superstring theory had lots of dimensions, 10 of them (9 space, 1 time).
Another problem was that superstring theory had lots of dimensions, 10 of them (9 space, 1 time).
this suggests that the dimension of time limited to our 3 spatial dimensions and excluded from the compacted dimensions. is it sensible to ask is there time, or similar dimension, in the compacted dimensions?
QUOTE (kjw+Nov 28 2007, 10:33 AM)
is this saying that at N=0 there is no symmetry because instead of there being 2 types of elementary particles (bosons and fermions) there is just 1 type of elementary particle ie a string ?
No, there's both fermions and bosons but not an equal amount of them. Usual field theory like QED or QCD is N=0. The Minimally Symmetric Standard Model (MSSM) is N=1. Type IIB string theory in 10 dimensions in N=2. If you want conformal Yang Mills you need N=4.
No, there's both fermions and bosons but not an equal amount of them. Usual field theory like QED or QCD is N=0. The Minimally Symmetric Standard Model (MSSM) is N=1. Type IIB string theory in 10 dimensions in N=2. If you want conformal Yang Mills you need N=4.
QUOTE (kjw+Nov 28 2007, 10:33 AM)
this suggests that the dimension of time limited to our 3 spatial dimensions and excluded from the compacted dimensions. is it sensible to ask is there time, or similar dimension, in the compacted dimensions?
Time still works for the compact dimensions as it does for our dimensions. It's lumped as (3+1)+6 because the time dimension isn't compact, it doesn't loop around, like a compact dimension does.
Ok, now that I have some time again, I'm resurrecting this orbifold thread.
QUOTE
(AlphaNumeric @ Nov 26 2007, 12:19 AM)
Now that I have time to look at this, the explanation is so simple it's almost funny. You're misreading the diagram! The straight line is (if you look closely) double headed. It indicates correspondence of points that are identified under the quotient group, not a new segment of the path. That is, the curved section of path is all there is, and its final point IS its initial point modulo Z/3. The maker of the diagram should perhaps have given some less subtle visual cues to tell you that the straight line is not indicating the same kind of thing as the curved path.
So the answer is that there IS NO straight path segment. The curved part of the path is an example of the kind of path mentioned elsewhere in that set of notes, which is closed on the orbifold but not closed on the torus. The picture is an "exploded view" with the cone unwrapped, so to speak, so the path doesn't LOOK closed without the straight segment. A most misleading diagram.
Perhaps because I'm looking at this from a different perspective, it seems natural for the holonomy group to be discrete, and I had to think for a bit to see how it could be continuous. Here are my thoughts on the question:
On a general manifold, holonomy is measured by traversing a loop while parallel transporting a vector. Upon completing the loop, the vector may not be in its original orientation. Therefore, there is an element of a rotation group which would act to rotate the original orientation into the new one. Considering all possible loops through the manifold, the holonomy group is the set of all rotations which occur.
Now the amount of rotation observed on a given loop is related to the curvature of the manifold within the loop. This same anomaly in parallel transport is responsible for gyrosopic precession (Lense-Thirring effect) in GR, I believe. Please correct me if I am wrong. Because of this, if the manifold has any nonzero area with nonvanishing curvature, then a loop around such an area will show a nonzero rotation. Shrinking the loop to a point (always possible, because you can choose a patch sufficiently small that it has trivial topology, so curves on it are contractible) reduces the curved area within the loop, which reduces the angle of rotation.
(A simple case is the unit sphere S2, in which the excess of a triangle's angle sum above pi is precisely equal to the area enclosed. This excess is exactly equal to the angular rotation of a vector parallel transported around the triangle. Since for a sphere, curvature is constant, this is proportional to the integrated curvature within the triangle. Hence in this simple case, enclosed total curvature governs the magnitude of the rotation.)
Therefore, for any rotation theta, all smaller rotations automatically occur, simply by contracting the loop to a point. Simply traversing the same loop multiple times will of course multiply the rotations, so for a given angle, all multiples of it are automatically also possible. Since you can both multiply by integers and shrink by arbitrary factors, you will get all possible angles from 0 to 2pi. So if a given infinitesimal rotation occurs at all, it gives a full one-parameter subgroup of SO(n). If the manifold is non-orientable, you can get stuff in the other component of O(n) as well. Rotations around some axes may never occur at all, so you do not always get all of O(n).
Now that looks like a proof that the holonomy group must be continuous, but that merely means that the assumptions should be examined. Multiple loop traversal is ALWAYS possible, because if you can traverse a closed loop once, you are positioned to do it again. So this assumption is always good, but it is also compatible with a discrete holonomy group, since powers of an element inherit group membership from the closure property of the group.
It is also always possible to find a small patch around any nonsingular point in a manifold, with the patch diffeomorphic to euclidean space, and hence all sufficiently small curves are contractible. However, the contraction of the curve will only reduce the angle of rotation of the parallel transported vector if the curvature of the manifold is smooth, i.e. is distributed across the patch. That way, shrinking the loop must reduce the quantity of curvature enclosed.
At this point, an obvious counter-analogy comes to mind. Where have you seen a function whose integral value is independent of shrinking the region of integration? Obviously, the Dirac delta. If the manifold has zero curvature everywhere, but delta curvature at the singular points, then contracting a loop will do no good. The angle of rotation will not reduce. In fact, if the loop does not go around the singular point, then the rotation will be zero (because the manifold is locally flat), but if the loop does go around the singular point, it will pick up the entire curvature concentrated at that point. It's just like poles and residues in contour integral theory. Therefore, the rotation is all-or-nothing. You still get multiples of whatever rotation occurs, but you cannot shrink the rotation. Hence you get a discrete holonomy group.
It is interesting to look at the process of taking a Calabi-Yau to an orbifold. To start with, the curvature is distributed across the space. Now pick a set of points to fix (the exact set determined by the structure of the manifold) and deform the manifold to straighten out its curvature at locations far from the selected points. The curvature of the manifold will now be concentrated near these points. It should be possible to make the manifold precisely flat outside small caps around these points. (This just reverses the process of reverting an orbifold to a Calabi-Yau by replacing the singular points with small smooth caps.) Let the caps be exceedingly small.
Now consider the curves that stay outside the caps, in the flat part of the manifold. These curves must either enclose all or none of the curvature associated with each cap, and so considering these curves only, you would again get a discrete holonomy group. The only curves which can produce very small rotations of their parallel transported vectors are the curves which dip into a cap and "scoop up" a little of its curvature. As the cap radius shrinks to zero, these curves become ever more unlikely, a smaller subset of all curves. In the limit of infinitesimal radius, the only thing a curve could do is to dive straight into the cap, make a sharp turn at the singular point, and come back out of the cap along a different direction than it entered. This would carve a "pie wedge" out of the cap and produce a small rotation angle.
Therefore, in the limit of zero radius, the only curves that produce angles other than those of the discrete holonomy group are those curves that have a sharp corner exactly at a singular point. This is a set of measure zero among all curves (in some appropriate measure). But the singular points are not normal points of the orbifold, and so curves are not supposed to pass through them. Therefore, the only class of curves that produce arbitrary rotations is not only a set of measure zero, but is actually excluded from consideration altogether. The continuous part of the holonomy group has been squeezed entirely out of existence!
I think that clears up holonomy.
Now that I have some free time again, I've looked at this paper. Some of the terminology is unfamiliar to me, and some of the physical background as well, but where it talks about manifolds and quotient groups it is easy to follow. I do have a couple of comments about it though:
First, on page 4 where they discuss the S1/Z2 "toy model" orbifold as an introductory example, they mention that Z2 acts by reflection, but do not discuss why. On page 3, they point out that the modding group should be a discreet non-freely acting symmetry, by which I assume that the symmetry must produce fixed points. That explains their choice in this case, since the 180 degree rotation symmetry is also an action of Z2 on S1, but it turns S1 into P1, the one dimensional projective space with S1 as a double cover, and has no fixed points. (This is essentially the same process as putting a half twist in a rubber band and folding the figure 8 onto itself to make a smaller circle that the rubber band "covers" twice, hence the term "double cover.")
On the other hand, the situation for higher dimensions is quite different. For S1XS1 = T2, Z2 acting as rotation produces a fixed point, while acting as reflection it produces a fixed line. So their example is slightly misleading, because to get fixed points on S1 you choose reflection, but on T2 you choose rotation.
Later in this paper there is stuff that is just plain WRONG. At the bottom of page 8, they say, "Z2 then acts as a reflection (or, equivalently, a 180 degree rotation represented by the purple arc) about a lattice node which one should call the 'origin'." Reflection and 180 degree rotation are INEQUIVALENT representations of Z2. Rotation produces the set of fixed points in their illustration on the top of page 9, but lattice reflection around the e2 line would produce the same shaded region with the two edges parallel to e2 as fixed LINES, rather than simply 4 fixed points.
The "pillow" or "ravioli" description at the top of page 9 is correct for rotational Z/2. For reflection Z/2, when you "fold the top over" the sides do not close because their orientations do not match. You get a tube with reflective boundaries instead of a pillow.
Later, on page 12, they correctly note that the orbifold will have fixed planes when the modding group is not prime, which was something I mentioned in an earlier post.
Now that I have some free time again, I've looked at this paper. Some of the terminology is unfamiliar to me, and some of the physical background as well, but where it talks about manifolds and quotient groups it is easy to follow. I do have a couple of comments about it though:
First, on page 4 where they discuss the S1/Z2 "toy model" orbifold as an introductory example, they mention that Z2 acts by reflection, but do not discuss why. On page 3, they point out that the modding group should be a discreet non-freely acting symmetry, by which I assume that the symmetry must produce fixed points. That explains their choice in this case, since the 180 degree rotation symmetry is also an action of Z2 on S1, but it turns S1 into P1, the one dimensional projective space with S1 as a double cover, and has no fixed points. (This is essentially the same process as putting a half twist in a rubber band and folding the figure 8 onto itself to make a smaller circle that the rubber band "covers" twice, hence the term "double cover.")
On the other hand, the situation for higher dimensions is quite different. For S1XS1 = T2, Z2 acting as rotation produces a fixed point, while acting as reflection it produces a fixed line. So their example is slightly misleading, because to get fixed points on S1 you choose reflection, but on T2 you choose rotation.
Later in this paper there is stuff that is just plain WRONG. At the bottom of page 8, they say, "Z2 then acts as a reflection (or, equivalently, a 180 degree rotation represented by the purple arc) about a lattice node which one should call the 'origin'." Reflection and 180 degree rotation are INEQUIVALENT representations of Z2. Rotation produces the set of fixed points in their illustration on the top of page 9, but lattice reflection around the e2 line would produce the same shaded region with the two edges parallel to e2 as fixed LINES, rather than simply 4 fixed points.
The "pillow" or "ravioli" description at the top of page 9 is correct for rotational Z/2. For reflection Z/2, when you "fold the top over" the sides do not close because their orientations do not match. You get a tube with reflective boundaries instead of a pillow.
Later, on page 12, they correctly note that the orbifold will have fixed planes when the modding group is not prime, which was something I mentioned in an earlier post.
(AlphaNumeric @ Nov 27 2007, 01:23 PM)
snipped a very nice outline of string history
Well perhaps our ideas about compactification need to be refined a little? Perhaps we need to make the small dimensions a little more complicated. This is where holonomy comes in. We want a theory with N=1 supersymmetry. This is akin to saying that we want a unique spinor in our theory under parallel transport around a closed loop in our space-time. In n dimensions, parallel transport around a closed loop acts like an operator on the spinor, specifically like a membler of a subgroup, G, of SO(n). In flat space-time G=SO(n). However, in complicated spaces you get G < SO(n). It turns out that in order to get a working theory, you need an even number of dimensions in your compact space, so n=2m, and to get the right kind of supersymmetry you need an holonomy group G = SU(m).
This is where I start to perceive the problem you were having with holonomy. I think I see that the number of dimensions in the compact space ought to be even, because odd dimensions force at least one fixed line (rather than point) or else at least one copy of Z/2 acting as a reflection in order to avoid this line. Either one seems undesirable. Also, unless n=2m, I intuitively wouldn't expect to find an action of SU(m) that doesn't have a fixed line (but perhaps my intuition would be surprised -- it has happened before!).
However, the fact that the holonomy group was continuous surprised me at first, because my starting point in looking at holonomy was with the orbifolds, where the space was flat except at the singular points. Once I made the connection (pun intended) with the curvature, the status of the holonomy groups became clear.
Mr Homm, the difference between an orbifold and an orientifold is that an orientifold has had an extra quotienting process applied to it. Rather than just an orbifold group G, an orientifold also have the group (-1)^(F_L)Ωσ quotiented out. (-1)^(F_L) is the 'left fermionic oscillation counter'. It's +1 if there's even numbers of fermionic oscillators going 'to the left' and -1 if odd. Ω is the worldsheet parity operator and σ is an involution (ie a Z_2 map). Sometimes orbifolding isn't enough or you want to have additional structure.
I'm not familiar with the notion of left and right oscillators, so this raises a question in my mind: The orbifold is just a pointed manifold with a certain topological structure. It would seem that the oscillators are an additional structure added to the underlying orbifold, so the orientifold quotient group seems to be acting on the orbifold + additional structure, rather than on just the base orbifold structure itself. Or am I missing something here? Are the oscillators somehow an outgrowth of the structure of the orbifold itself? If that is true, then could not the extra quotient process in the orientifold be defined without reference to the oscillators, but purely in terms of the properties of the orbifold as a pointed manifold? Just wondering.
I'm not familiar with the notion of left and right oscillators, so this raises a question in my mind: The orbifold is just a pointed manifold with a certain topological structure. It would seem that the oscillators are an additional structure added to the underlying orbifold, so the orientifold quotient group seems to be acting on the orbifold + additional structure, rather than on just the base orbifold structure itself. Or am I missing something here? Are the oscillators somehow an outgrowth of the structure of the orbifold itself? If that is true, then could not the extra quotient process in the orientifold be defined without reference to the oscillators, but purely in terms of the properties of the orbifold as a pointed manifold? Just wondering.
(AlphaNumeric @ Nov 28 2007, 01:58 AM)
Time still works for the compact dimensions as it does for our dimensions. It's lumped as (3+1)+6 because the time dimension isn't compact, it doesn't loop around, like a compact dimension does.
Let me add to your explanation that each dimension interacts with all the others, and the separation here is purely a matter of how we choose to think about the manifold. For instance, let's think about a "toy universe" which has dimensions length, angle, and time. Length is along a line, angle is around a circle, and time is also along a line. You can visualize the space as being a long tube, with a coordinate x along its length, theta along the circle, and time flowing as usual. You can also (as an alternative point of view) think of the space as basically a line, with a circle attached to the line at each point, and all the circles lining up to form a tube. You're now thinking of the circles as an extra bit attached to the line, and if you now "zoom out far away," the circular dimension would be too small to see and you could ignore them. That would leave length and time. You could also take the opposite point of view that the space is basically a circle, with a line attached to it at every point, and all the lines lining up to form a tube. In this case, zooming out doesn't help, but suppose that motion along the line is physically irrelevant (for some reason). Then you would be left with angle and time.
@AlphaNumeric:
By the way, did you ever have a chance to look at the second half of my earlier (very long) post? I would be interested to know whether you found my scheme of representing the modding action of G on the torus by a path on a grid to be helpful or illuminating at all. Now that questions about the discrete nature of the holonomy group are out of the way (I assume), it might be worth a look. I have though about it some more, and it is clear to me that you could use it for lot's of other modding groups besides Z/3. I could easily produce grids for any of the many groups mentioned in your link
http://lanl.arxiv.org/PS_cache/hep-th/pdf/0609/0609013v2.pdf
and will do so if you have any interest. It it is not helpful, I'll not bother, but go on to other interesting questions about orbifolds or string theory in general, of which I have many to keep me happily occupied!
--Stuart Anderson
QUOTE (->
| QUOTE |
| (AlphaNumeric @ Nov 26 2007, 12:19 AM) (mr_homm @ Nov 26 2007, 07:25 AM) Do you have a reference where they discuss how to get the holonomy group and mention this straight line transport? I'd like to take a look and try to see what's going on there. http://gesalerico.ft.uam.es/paginaspersona...anga/lect14.pdf Pages 5 and 6, particularly the picture on page 6. Note the curved path and then a straight line connection. That straight line makes sense on a torus since it's constructed from a flat space, the straight line is uniquely defined, but if the manifold which is being orbifolded isn't flat then no such line exists. |
Now that I have time to look at this, the explanation is so simple it's almost funny. You're misreading the diagram! The straight line is (if you look closely) double headed. It indicates correspondence of points that are identified under the quotient group, not a new segment of the path. That is, the curved section of path is all there is, and its final point IS its initial point modulo Z/3. The maker of the diagram should perhaps have given some less subtle visual cues to tell you that the straight line is not indicating the same kind of thing as the curved path.
So the answer is that there IS NO straight path segment. The curved part of the path is an example of the kind of path mentioned elsewhere in that set of notes, which is closed on the orbifold but not closed on the torus. The picture is an "exploded view" with the cone unwrapped, so to speak, so the path doesn't LOOK closed without the straight segment. A most misleading diagram.
QUOTE
QUOTE (->
| QUOTE |
(mr_homm @ Nov 26 2007, 07:25 AM) ...discussion about looped paths.... I agree with that, but I have issue with the holonomy becoming discrete. If I consider looped paths I can curve my path so that the contribution from the singularity is removed. The holonomy group remains it's usual subgroup of O(n). |
Perhaps because I'm looking at this from a different perspective, it seems natural for the holonomy group to be discrete, and I had to think for a bit to see how it could be continuous. Here are my thoughts on the question:
On a general manifold, holonomy is measured by traversing a loop while parallel transporting a vector. Upon completing the loop, the vector may not be in its original orientation. Therefore, there is an element of a rotation group which would act to rotate the original orientation into the new one. Considering all possible loops through the manifold, the holonomy group is the set of all rotations which occur.
Now the amount of rotation observed on a given loop is related to the curvature of the manifold within the loop. This same anomaly in parallel transport is responsible for gyrosopic precession (Lense-Thirring effect) in GR, I believe. Please correct me if I am wrong. Because of this, if the manifold has any nonzero area with nonvanishing curvature, then a loop around such an area will show a nonzero rotation. Shrinking the loop to a point (always possible, because you can choose a patch sufficiently small that it has trivial topology, so curves on it are contractible) reduces the curved area within the loop, which reduces the angle of rotation.
(A simple case is the unit sphere S2, in which the excess of a triangle's angle sum above pi is precisely equal to the area enclosed. This excess is exactly equal to the angular rotation of a vector parallel transported around the triangle. Since for a sphere, curvature is constant, this is proportional to the integrated curvature within the triangle. Hence in this simple case, enclosed total curvature governs the magnitude of the rotation.)
Therefore, for any rotation theta, all smaller rotations automatically occur, simply by contracting the loop to a point. Simply traversing the same loop multiple times will of course multiply the rotations, so for a given angle, all multiples of it are automatically also possible. Since you can both multiply by integers and shrink by arbitrary factors, you will get all possible angles from 0 to 2pi. So if a given infinitesimal rotation occurs at all, it gives a full one-parameter subgroup of SO(n). If the manifold is non-orientable, you can get stuff in the other component of O(n) as well. Rotations around some axes may never occur at all, so you do not always get all of O(n).
Now that looks like a proof that the holonomy group must be continuous, but that merely means that the assumptions should be examined. Multiple loop traversal is ALWAYS possible, because if you can traverse a closed loop once, you are positioned to do it again. So this assumption is always good, but it is also compatible with a discrete holonomy group, since powers of an element inherit group membership from the closure property of the group.
It is also always possible to find a small patch around any nonsingular point in a manifold, with the patch diffeomorphic to euclidean space, and hence all sufficiently small curves are contractible. However, the contraction of the curve will only reduce the angle of rotation of the parallel transported vector if the curvature of the manifold is smooth, i.e. is distributed across the patch. That way, shrinking the loop must reduce the quantity of curvature enclosed.
At this point, an obvious counter-analogy comes to mind. Where have you seen a function whose integral value is independent of shrinking the region of integration? Obviously, the Dirac delta. If the manifold has zero curvature everywhere, but delta curvature at the singular points, then contracting a loop will do no good. The angle of rotation will not reduce. In fact, if the loop does not go around the singular point, then the rotation will be zero (because the manifold is locally flat), but if the loop does go around the singular point, it will pick up the entire curvature concentrated at that point. It's just like poles and residues in contour integral theory. Therefore, the rotation is all-or-nothing. You still get multiples of whatever rotation occurs, but you cannot shrink the rotation. Hence you get a discrete holonomy group.
It is interesting to look at the process of taking a Calabi-Yau to an orbifold. To start with, the curvature is distributed across the space. Now pick a set of points to fix (the exact set determined by the structure of the manifold) and deform the manifold to straighten out its curvature at locations far from the selected points. The curvature of the manifold will now be concentrated near these points. It should be possible to make the manifold precisely flat outside small caps around these points. (This just reverses the process of reverting an orbifold to a Calabi-Yau by replacing the singular points with small smooth caps.) Let the caps be exceedingly small.
Now consider the curves that stay outside the caps, in the flat part of the manifold. These curves must either enclose all or none of the curvature associated with each cap, and so considering these curves only, you would again get a discrete holonomy group. The only curves which can produce very small rotations of their parallel transported vectors are the curves which dip into a cap and "scoop up" a little of its curvature. As the cap radius shrinks to zero, these curves become ever more unlikely, a smaller subset of all curves. In the limit of infinitesimal radius, the only thing a curve could do is to dive straight into the cap, make a sharp turn at the singular point, and come back out of the cap along a different direction than it entered. This would carve a "pie wedge" out of the cap and produce a small rotation angle.
Therefore, in the limit of zero radius, the only curves that produce angles other than those of the discrete holonomy group are those curves that have a sharp corner exactly at a singular point. This is a set of measure zero among all curves (in some appropriate measure). But the singular points are not normal points of the orbifold, and so curves are not supposed to pass through them. Therefore, the only class of curves that produce arbitrary rotations is not only a set of measure zero, but is actually excluded from consideration altogether. The continuous part of the holonomy group has been squeezed entirely out of existence!
I think that clears up holonomy.
QUOTE
(AlphaNumeric@Nov 26 2007, 08:54 AM)
By pure chance I came across this paper earlier today while browsing ArXiv for something interesting to read :
http://xxx.lanl.gov/pdf/0711.1582
Discusses a Z2 orbifold
But you're right, it acts as a reflection, not a rotational twist.
By pure chance I came across this paper earlier today while browsing ArXiv for something interesting to read :
http://xxx.lanl.gov/pdf/0711.1582
Discusses a Z2 orbifold
But you're right, it acts as a reflection, not a rotational twist.
Now that I have some free time again, I've looked at this paper. Some of the terminology is unfamiliar to me, and some of the physical background as well, but where it talks about manifolds and quotient groups it is easy to follow. I do have a couple of comments about it though:
First, on page 4 where they discuss the S1/Z2 "toy model" orbifold as an introductory example, they mention that Z2 acts by reflection, but do not discuss why. On page 3, they point out that the modding group should be a discreet non-freely acting symmetry, by which I assume that the symmetry must produce fixed points. That explains their choice in this case, since the 180 degree rotation symmetry is also an action of Z2 on S1, but it turns S1 into P1, the one dimensional projective space with S1 as a double cover, and has no fixed points. (This is essentially the same process as putting a half twist in a rubber band and folding the figure 8 onto itself to make a smaller circle that the rubber band "covers" twice, hence the term "double cover.")
On the other hand, the situation for higher dimensions is quite different. For S1XS1 = T2, Z2 acting as rotation produces a fixed point, while acting as reflection it produces a fixed line. So their example is slightly misleading, because to get fixed points on S1 you choose reflection, but on T2 you choose rotation.
Later in this paper there is stuff that is just plain WRONG. At the bottom of page 8, they say, "Z2 then acts as a reflection (or, equivalently, a 180 degree rotation represented by the purple arc) about a lattice node which one should call the 'origin'." Reflection and 180 degree rotation are INEQUIVALENT representations of Z2. Rotation produces the set of fixed points in their illustration on the top of page 9, but lattice reflection around the e2 line would produce the same shaded region with the two edges parallel to e2 as fixed LINES, rather than simply 4 fixed points.
The "pillow" or "ravioli" description at the top of page 9 is correct for rotational Z/2. For reflection Z/2, when you "fold the top over" the sides do not close because their orientations do not match. You get a tube with reflective boundaries instead of a pillow.
Later, on page 12, they correctly note that the orbifold will have fixed planes when the modding group is not prime, which was something I mentioned in an earlier post.
QUOTE (->
| QUOTE |
| (AlphaNumeric@Nov 26 2007, 08:54 AM) By pure chance I came across this paper earlier today while browsing ArXiv for something interesting to read : http://xxx.lanl.gov/pdf/0711.1582 Discusses a Z2 orbifold But you're right, it acts as a reflection, not a rotational twist. |
Now that I have some free time again, I've looked at this paper. Some of the terminology is unfamiliar to me, and some of the physical background as well, but where it talks about manifolds and quotient groups it is easy to follow. I do have a couple of comments about it though:
First, on page 4 where they discuss the S1/Z2 "toy model" orbifold as an introductory example, they mention that Z2 acts by reflection, but do not discuss why. On page 3, they point out that the modding group should be a discreet non-freely acting symmetry, by which I assume that the symmetry must produce fixed points. That explains their choice in this case, since the 180 degree rotation symmetry is also an action of Z2 on S1, but it turns S1 into P1, the one dimensional projective space with S1 as a double cover, and has no fixed points. (This is essentially the same process as putting a half twist in a rubber band and folding the figure 8 onto itself to make a smaller circle that the rubber band "covers" twice, hence the term "double cover.")
On the other hand, the situation for higher dimensions is quite different. For S1XS1 = T2, Z2 acting as rotation produces a fixed point, while acting as reflection it produces a fixed line. So their example is slightly misleading, because to get fixed points on S1 you choose reflection, but on T2 you choose rotation.
Later in this paper there is stuff that is just plain WRONG. At the bottom of page 8, they say, "Z2 then acts as a reflection (or, equivalently, a 180 degree rotation represented by the purple arc) about a lattice node which one should call the 'origin'." Reflection and 180 degree rotation are INEQUIVALENT representations of Z2. Rotation produces the set of fixed points in their illustration on the top of page 9, but lattice reflection around the e2 line would produce the same shaded region with the two edges parallel to e2 as fixed LINES, rather than simply 4 fixed points.
The "pillow" or "ravioli" description at the top of page 9 is correct for rotational Z/2. For reflection Z/2, when you "fold the top over" the sides do not close because their orientations do not match. You get a tube with reflective boundaries instead of a pillow.
Later, on page 12, they correctly note that the orbifold will have fixed planes when the modding group is not prime, which was something I mentioned in an earlier post.
(AlphaNumeric @ Nov 27 2007, 01:23 PM)
snipped a very nice outline of string history
Well perhaps our ideas about compactification need to be refined a little? Perhaps we need to make the small dimensions a little more complicated. This is where holonomy comes in. We want a theory with N=1 supersymmetry. This is akin to saying that we want a unique spinor in our theory under parallel transport around a closed loop in our space-time. In n dimensions, parallel transport around a closed loop acts like an operator on the spinor, specifically like a membler of a subgroup, G, of SO(n). In flat space-time G=SO(n). However, in complicated spaces you get G < SO(n). It turns out that in order to get a working theory, you need an even number of dimensions in your compact space, so n=2m, and to get the right kind of supersymmetry you need an holonomy group G = SU(m).
This is where I start to perceive the problem you were having with holonomy. I think I see that the number of dimensions in the compact space ought to be even, because odd dimensions force at least one fixed line (rather than point) or else at least one copy of Z/2 acting as a reflection in order to avoid this line. Either one seems undesirable. Also, unless n=2m, I intuitively wouldn't expect to find an action of SU(m) that doesn't have a fixed line (but perhaps my intuition would be surprised -- it has happened before!).
However, the fact that the holonomy group was continuous surprised me at first, because my starting point in looking at holonomy was with the orbifolds, where the space was flat except at the singular points. Once I made the connection (pun intended) with the curvature, the status of the holonomy groups became clear.
QUOTE
Mr Homm, the difference between an orbifold and an orientifold is that an orientifold has had an extra quotienting process applied to it. Rather than just an orbifold group G, an orientifold also have the group (-1)^(F_L)Ωσ quotiented out. (-1)^(F_L) is the 'left fermionic oscillation counter'. It's +1 if there's even numbers of fermionic oscillators going 'to the left' and -1 if odd. Ω is the worldsheet parity operator and σ is an involution (ie a Z_2 map). Sometimes orbifolding isn't enough or you want to have additional structure.
I'm not familiar with the notion of left and right oscillators, so this raises a question in my mind: The orbifold is just a pointed manifold with a certain topological structure. It would seem that the oscillators are an additional structure added to the underlying orbifold, so the orientifold quotient group seems to be acting on the orbifold + additional structure, rather than on just the base orbifold structure itself. Or am I missing something here? Are the oscillators somehow an outgrowth of the structure of the orbifold itself? If that is true, then could not the extra quotient process in the orientifold be defined without reference to the oscillators, but purely in terms of the properties of the orbifold as a pointed manifold? Just wondering.
QUOTE (->
| QUOTE |
Mr Homm, the difference between an orbifold and an orientifold is that an orientifold has had an extra quotienting process applied to it. Rather than just an orbifold group G, an orientifold also have the group (-1)^(F_L)Ωσ quotiented out. (-1)^(F_L) is the 'left fermionic oscillation counter'. It's +1 if there's even numbers of fermionic oscillators going 'to the left' and -1 if odd. Ω is the worldsheet parity operator and σ is an involution (ie a Z_2 map). Sometimes orbifolding isn't enough or you want to have additional structure. |
I'm not familiar with the notion of left and right oscillators, so this raises a question in my mind: The orbifold is just a pointed manifold with a certain topological structure. It would seem that the oscillators are an additional structure added to the underlying orbifold, so the orientifold quotient group seems to be acting on the orbifold + additional structure, rather than on just the base orbifold structure itself. Or am I missing something here? Are the oscillators somehow an outgrowth of the structure of the orbifold itself? If that is true, then could not the extra quotient process in the orientifold be defined without reference to the oscillators, but purely in terms of the properties of the orbifold as a pointed manifold? Just wondering.
(AlphaNumeric @ Nov 28 2007, 01:58 AM)
QUOTE
(kjw @ Nov 28 2007, 10:33 AM)
this suggests that the dimension of time limited to our 3 spatial dimensions and excluded from the compacted dimensions. is it sensible to ask is there time, or similar dimension, in the compacted dimensions?
this suggests that the dimension of time limited to our 3 spatial dimensions and excluded from the compacted dimensions. is it sensible to ask is there time, or similar dimension, in the compacted dimensions?
Time still works for the compact dimensions as it does for our dimensions. It's lumped as (3+1)+6 because the time dimension isn't compact, it doesn't loop around, like a compact dimension does.
Let me add to your explanation that each dimension interacts with all the others, and the separation here is purely a matter of how we choose to think about the manifold. For instance, let's think about a "toy universe" which has dimensions length, angle, and time. Length is along a line, angle is around a circle, and time is also along a line. You can visualize the space as being a long tube, with a coordinate x along its length, theta along the circle, and time flowing as usual. You can also (as an alternative point of view) think of the space as basically a line, with a circle attached to the line at each point, and all the circles lining up to form a tube. You're now thinking of the circles as an extra bit attached to the line, and if you now "zoom out far away," the circular dimension would be too small to see and you could ignore them. That would leave length and time. You could also take the opposite point of view that the space is basically a circle, with a line attached to it at every point, and all the lines lining up to form a tube. In this case, zooming out doesn't help, but suppose that motion along the line is physically irrelevant (for some reason). Then you would be left with angle and time.
@AlphaNumeric:
By the way, did you ever have a chance to look at the second half of my earlier (very long) post? I would be interested to know whether you found my scheme of representing the modding action of G on the torus by a path on a grid to be helpful or illuminating at all. Now that questions about the discrete nature of the holonomy group are out of the way (I assume), it might be worth a look. I have though about it some more, and it is clear to me that you could use it for lot's of other modding groups besides Z/3. I could easily produce grids for any of the many groups mentioned in your link
http://lanl.arxiv.org/PS_cache/hep-th/pdf/0609/0609013v2.pdf
and will do so if you have any interest. It it is not helpful, I'll not bother, but go on to other interesting questions about orbifolds or string theory in general, of which I have many to keep me happily occupied!
--Stuart Anderson
I was wondering if you could discuss the similarities/differences between the math (such as with the complex dimensions) and/or geometrical visualization of Penrose's twistor space and the Calibi-Yau manifold. I understand that Witten has considered bringing together twistor space with string/M theory. Thanks.
@math/physics continuum:
I'm afraid I don't know anything about Penrose twistor space, so I cannot make any useful statements comparing them with Calabi-Yau manifolds. Do you have a good reference where I could find out more about twistors?
--Stuart Anderson
I'm afraid I don't know anything about Penrose twistor space, so I cannot make any useful statements comparing them with Calabi-Yau manifolds. Do you have a good reference where I could find out more about twistors?
--Stuart Anderson
Huggett & Tod's book is good, but not great. Alternatively look into Penrose's original paper, but the theory has been developed a fair amount since then.
There are several ways to define twistor space, I think the most simple being via a spinor valued solution to the `twistor equation' on Minkowski space. Alternatively (perhaps more simply, depending on background) twistor space can be viewed as the representation space for the spinor rep of SU(2,2). The first of these is fairly direct, and soon it brings you to look at hyperplanes in complexified Minkowski space. Looking at certain equivalence classes of these hyperplanes leads leads to the connection with complex projective spaces... which is the kind of thing twistor theory is famous (or not so famous?) for.
Essentially, if you know what you're doing with spinors, then you can get a general idea of the basics of twistor theory without doing too much work.
There are several ways to define twistor space, I think the most simple being via a spinor valued solution to the `twistor equation' on Minkowski space. Alternatively (perhaps more simply, depending on background) twistor space can be viewed as the representation space for the spinor rep of SU(2,2). The first of these is fairly direct, and soon it brings you to look at hyperplanes in complexified Minkowski space. Looking at certain equivalence classes of these hyperplanes leads leads to the connection with complex projective spaces... which is the kind of thing twistor theory is famous (or not so famous?) for.
Essentially, if you know what you're doing with spinors, then you can get a general idea of the basics of twistor theory without doing too much work.
Thanks, Euler, I'll start from there, although I may not get back to it immediately, since I can't study EVERYTHING at once, much as I'd like to.
--Stuart Anderson
--Stuart Anderson
QUOTE (mr_homm+Dec 19 2007, 05:21 PM)
@math/physics continuum:
I'm afraid I don't know anything about Penrose twistor space,
Hi Mr Homm,
Excellent choice Mr Homm.
Cheers
darren
I'm afraid I don't know anything about Penrose twistor space,
Hi Mr Homm,
Excellent choice Mr Homm.
Cheers
darren
QUOTE (mr_homm+Dec 19 2007, 01:38 AM)
Now that I have time to look at this, the explanation is so simple it's almost funny. You're misreading the diagram! The straight line is (if you look closely) double headed. It indicates correspondence of points that are identified under the quotient group, not a new segment of the path. That is, the curved section of path is all there is, and its final point IS its initial point modulo Z/3. The maker of the diagram should perhaps have given some less subtle visual cues to tell you that the straight line is not indicating the same kind of thing as the curved path.
So the answer is that there IS NO straight path segment. The curved part of the path is an example of the kind of path mentioned elsewhere in that set of notes, which is closed on the orbifold but not closed on the torus. The picture is an "exploded view" with the cone unwrapped, so to speak, so the path doesn't LOOK closed without the straight segment. A most misleading diagram.
Unfortunately, that's precisely how I was reading it (though perhaps I didn't explain it well enough). The problem I have is that I can construct a curved path which is closed modulo Z_3 but not on the torus which results in the shift in the vector being anything I want, not just 120, 240 and 360 degrees. It's only when I'm restricted to straight lines that I seem to get the discrete transformation.
Give me a while and I'll knock up a few pictures to demonstrate my thinking. I know I'm getting it wrong somehow, but it's just when I use the notion of holonomy culled from Green, Schwarz and Witten on this issue, it doesn't seem to sync up.
So the answer is that there IS NO straight path segment. The curved part of the path is an example of the kind of path mentioned elsewhere in that set of notes, which is closed on the orbifold but not closed on the torus. The picture is an "exploded view" with the cone unwrapped, so to speak, so the path doesn't LOOK closed without the straight segment. A most misleading diagram.
Unfortunately, that's precisely how I was reading it (though perhaps I didn't explain it well enough). The problem I have is that I can construct a curved path which is closed modulo Z_3 but not on the torus which results in the shift in the vector being anything I want, not just 120, 240 and 360 degrees. It's only when I'm restricted to straight lines that I seem to get the discrete transformation.
Give me a while and I'll knock up a few pictures to demonstrate my thinking. I know I'm getting it wrong somehow, but it's just when I use the notion of holonomy culled from Green, Schwarz and Witten on this issue, it doesn't seem to sync up.
QUOTE (mr_homm+Dec 19 2007, 01:38 AM)
Because of this, if the manifold has any nonzero area with nonvanishing curvature, then a loop around such an area will show a nonzero rotation. Shrinking the loop to a point (always possible, because you can choose a patch sufficiently small that it has trivial topology, so curves on it are contractible) reduces the curved area within the loop, which reduces the angle of rotation.
Is this true in general? Again, I can think of curves which enclose a non-zero area but which bend in such a way that they don't give a rotation (a circular loop).
If you put constraints on how you build up your loop then certain interesting properties appear, as your example gives :
If you put constraints on how you build up your loop then certain interesting properties appear, as your example gives :
QUOTE (mr_homm+Dec 19 2007, 01:38 AM)
(A simple case is the unit sphere S2, in which the excess of a triangle's angle sum above pi is precisely equal to the area enclosed. This excess is exactly equal to the angular rotation of a vector parallel transported around the triangle. Since for a sphere, curvature is constant, this is proportional to the integrated curvature within the triangle. Hence in this simple case, enclosed total curvature governs the magnitude of the rotation.)
This was a question in our 3rd year GR course, as it happens. Euler and I were supervision partners for that course
In this case, you've constructed your loop from 3 straight lines, with a sharp turn between each one, not a single smoothly changing loop.
In this case, you've constructed your loop from 3 straight lines, with a sharp turn between each one, not a single smoothly changing loop.
QUOTE (mr_homm+Dec 19 2007, 01:38 AM)
At this point, an obvious counter-analogy comes to mind. Where have you seen a function whose integral value is independent of shrinking the region of integration? Obviously, the Dirac delta. If the manifold has zero curvature everywhere, but delta curvature at the singular points, then contracting a loop will do no good. The angle of rotation will not reduce. In fact, if the loop does not go around the singular point, then the rotation will be zero (because the manifold is locally flat), but if the loop does go around the singular point, it will pick up the entire curvature concentrated at that point. It's just like poles and residues in contour integral theory. Therefore, the rotation is all-or-nothing. You still get multiples of whatever rotation occurs, but you cannot shrink the rotation. Hence you get a discrete holonomy group.
This makes perfect sense to me and your analogy to residues and poles makes a great deal of sense, except that that isn't how I've been thinking of holonomy. Perhaps that's my problem, I've been considering holonomy as purely a matter of how the curve returns back to it's base point (which really then only needs local considerations), not how the curve moves through the manifold and 'measures' the holonomy all over the manifold.
Passing around the singularity is definitely a "Hey, something major happened!" case, but my notion of holonomy doesn't take that into account because once you're back near your base point you can twist the path to return to the base point at any orientation you like.
If you only work with straight line transportation (as books like Nakahara use to define the Riemann Curvature Tensor and torsion), then I see it. Parallel transport around a closed straight lined polygon will pick up the Z_3 'kick' of 120 degrees. This is because the path itself cannot reorientate the vector it's parallelly transporting. A curved path can. That's my issue.
Passing around the singularity is definitely a "Hey, something major happened!" case, but my notion of holonomy doesn't take that into account because once you're back near your base point you can twist the path to return to the base point at any orientation you like.
If you only work with straight line transportation (as books like Nakahara use to define the Riemann Curvature Tensor and torsion), then I see it. Parallel transport around a closed straight lined polygon will pick up the Z_3 'kick' of 120 degrees. This is because the path itself cannot reorientate the vector it's parallelly transporting. A curved path can. That's my issue.
QUOTE (mr_homm+Dec 19 2007, 01:38 AM)
Therefore, in the limit of zero radius, the only curves that produce angles other than those of the discrete holonomy group are those curves that have a sharp corner exactly at a singular point. This is a set of measure zero among all curves (in some appropriate measure). But the singular points are not normal points of the orbifold, and so curves are not supposed to pass through them. Therefore, the only class of curves that produce arbitrary rotations is not only a set of measure zero, but is actually excluded from consideration altogether. The continuous part of the holonomy group has been squeezed entirely out of existence!
I think I should be changing my view on holonomy. The way you describe it (very well) seems to imply that it's better to consider it a quantity almost like an integral. You pick a path to move along and the holonomy is not so much a measure of the local properties of the path near it's base point (as I was considering) but the total measure along the path of some curvature quantity (vaguely related to R_abcd etc). This then takes into consideration when the path loops around huge distortions (ie singularities in curvature).
QUOTE (mr_homm+Dec 19 2007, 01:38 AM)
Later in this paper there is stuff that is just plain WRONG. At the bottom of page 8, they say, "Z2 then acts as a reflection (or, equivalently, a 180 degree rotation represented by the purple arc) about a lattice node which one should call the 'origin'." Reflection and 180 degree rotation are INEQUIVALENT representations of Z2. Rotation produces the set of fixed points in their illustration on the top of page 9, but lattice reflection around the e2 line would produce the same shaded region with the two edges parallel to e2 as fixed LINES, rather than simply 4 fixed points.
The structure of the orbifold is difference but you get the same 'fundamental region' under both representations.
A friend had to learn this stuff for his PhD and he and I ended up drawing diagrams like Figure 5 on a napkin in the student bar one evening
Motion through the regions is different, depending on wether you make your Z_2 symmetry either the reflection down the middle of the parallelogram (as shown in Figure 5) or you use a Z_2 rotation to identity the dark grey region with it's rotation (through 180 degrees about the origin). Either way, you've equated one half of the lattice section with the other half and you get all the same fixed points.
I think that's what they were going for, because they don't specifically talk about dynamics dependent on which part of which line identifies where, other than the fixed points.
transporting. A curved path can. That's my issue.
A friend had to learn this stuff for his PhD and he and I ended up drawing diagrams like Figure 5 on a napkin in the student bar one evening
Motion through the regions is different, depending on wether you make your Z_2 symmetry either the reflection down the middle of the parallelogram (as shown in Figure 5) or you use a Z_2 rotation to identity the dark grey region with it's rotation (through 180 degrees about the origin). Either way, you've equated one half of the lattice section with the other half and you get all the same fixed points.
I think that's what they were going for, because they don't specifically talk about dynamics dependent on which part of which line identifies where, other than the fixed points.
transporting. A curved path can. That's my issue.
QUOTE (mr_homm+Dec 19 2007, 01:38 AM)
However, the fact that the holonomy group was continuous surprised me at first, because my starting point in looking at holonomy was with the orbifolds, where the space was flat except at the singular points. Once I made the connection (pun intended) with the curvature, the status of the holonomy groups became clear.
*groan* I doubt there's anyone other than you, me and Euler who got that...
Nakahara makes specific the link between local curvature and holonomy (well, the restricted holonomy group), showing how SU(m) implies Ricci flat.
Nakahara makes specific the link between local curvature and holonomy (well, the restricted holonomy group), showing how SU(m) implies Ricci flat.
QUOTE (mr_homm+Dec 19 2007, 01:38 AM)
'm not familiar with the notion of left and right oscillators, so this raises a question in my mind
If x is the position variable of a given direction your left and right oscillators are those which obey the wave equation with variables of the form x+vt and x-vt only. ie they are pure solutions to the wave equation, in a particular direction of motion.
QUOTE (mr_homm+Dec 19 2007, 01:38 AM)
It would seem that the oscillators are an additional structure added to the underlying orbifold, so the orientifold