User andrew mcintyre - MathOverflow

Is every finite group a group of "symmetries"?

2009-10-18T03:39:56Z

I was trying to explain finite groups to a non-mathematician, and was falling back on the "they're like symmetries of polyhedra" line. Which made me realize that I didn't know if this was actually true:

Does there exist, for every finite group G, a positive integer n and a convex subset S of R^n such that G is isomorphic to the group of isometries of R^n preserving S?

If the answer is yes (or for those groups for which the answer is yes), is there a simple construction for S?

I feel like this should have an obvious answer, that my sketchy knowledge of representations is not allowing me to see.

Answer by Andrew McIntyre for Atiyah-Singer index theorem

2009-10-19T16:42:11Z

If you have access to Atiyah's collected works at your library, try taking a look at those. There are a few transcribed lectures and short expository papers where he explains the context and motivation of the theorems. (If I remember correctly, they are classified as miscellanea and appear in the first volume, but there might also be some in the second volume. I don't have it at hand to check.)

He writes beautifully, and for myself, I didn't feel like I "got" the index theorem until I read these.

EDIT: Oops, I didn't read your question carefully enough: you are mostly looking for the analysis part. In that case I would just second the recommendation of the relevant chapter in Warner or Wells.

Comment by Andrew McIntyre

2010-02-22T15:58:00Z

Thanks! It seems like it would be difficult, in the equivariant embedding, to guarantee that the image has no accidental symmetries (i.e. its isometry group is no larger than G). If one knew this, it would give a sort of perverse answer to the original question: find a hyperbolic surface X (with relatively very large genus) whose isometry group is G, and embed X into some R^n such that the isometries of its image are exactly G.

Comment by Andrew McIntyre

2010-02-22T15:50:28Z

This construction shows that a finite group G of order n can be realized as exactly the isometries of a convex polyhedron in R^n. Now, suppose m is the least positive integer such that G has a faithful representation in GL(m,R). Typically m is much smaller than n. I am curious whether it is possible to use your strategy to produce a polyhedron in R^m, whose symmetries are exactly the elements of G, considered as elements in GL(m,R). If that could be done, it would be a solution to the original problem with minimal dimension.

Comment by Andrew McIntyre

2010-02-22T15:39:07Z

This is a really nice argument. I like that it is effective: given an explicit description of a group, one could easily find the vertices of the polyhedron P. It's even almost visualizable.

Comment by Andrew McIntyre

2010-02-19T15:08:40Z

Hee hee, I'm glad it troubled your sleep! I have an intuitive feeling that the inner product has too few degrees of freedom to kill symmetries as flexibly as needed. But I can't back up that feeling with an actual proof. It would be lovely if one could find a clean, representation theoretical answer to this question!

Comment by Andrew McIntyre

2010-02-19T15:03:41Z

Neat! Do you know if there are analogous results in higher dimensions? A refinement of the original question, (assuming that previous commenters and I are all correct, and the answer is yes), would be to find lower and upper bounds for the integer n, as a function of the order of the group (or some other data from the group). This would be sort of analogous to embedding theorems for manifolds. If the permutohedron argument suggested by Mariano S-A works, we would have the order of the group as an upper bound.

Comment by Andrew McIntyre

2010-02-16T20:12:02Z

@RB: Do you have a reference for every finite group being the automorphism group of a Riemann surface? I seem to recall hearing this as folklore, but it occurs to me that I have no idea how it is proved. That suggests another question: is there an "equivariant" embedding theorem for manifolds with symmetries? I.e. if a hyperbolic surface X has nontrivial automorphisms, is there an embedding of X in R^n for some n, such that every automorphism of X is induced by an isometry of R^n preserving the image of X?

Comment by Andrew McIntyre

2010-02-16T20:04:20Z

@RB and KO'B: I was aware of the variants you mention, and there are certainly more one can cook up. What I wanted to say to the non-mathematician is the most literal generalization of symmetry in the colloquial sense: that every finite group is the isometry group of some polytope in R^n.

Comment by Andrew McIntyre

2010-02-16T19:49:14Z

@kakaz: the conditions 1 and 2 you say should be added to the original question are, if I understand them correctly, contained in the word "isomorphic". Reid B and Anton G's answers above show that the answer is yes, with 1 and 2 as requirements.

Comment by Andrew McIntyre

2010-02-16T19:46:41Z

@Mariano S-A: Thanks, I didn't know about the permutohedron! I'll have to check details, but I think your suggestion answers the second part of my original question, giving a simple construction for the required convex set. Just give the group G as a subset of S_n, and then truncate the permutohedron appropriately to restrict its symmetries to the subgroup.

Comment by Andrew McIntyre

2010-02-15T15:48:40Z

You did not read the whole question. As you mention, any finite group is a subgroup of a symmetric group; answering the question for the symmetric group would answer it for all finite groups. However, I was not looking to visualize the group as symmetries of "some kind of graph" (which is straightforward), but as isometries of a convex set in R^n. Note that the previous commenters have given good answers.

Comment by Andrew McIntyre

2009-10-19T02:45:18Z

You know, I use that trick at least once a week in the context of Poincaré series. I don't know why I didn't see it here. Thanks!

Comment by Andrew McIntyre

2009-10-18T16:05:23Z

Thanks, this is ugly but also neat! I am convinced now. I am still wondering whether there is a "nice" construction. I don't know how to define "nice", but I am getting the feeling that, for any reasonable candidate for "nice", one could construct counterexamples (I suspect that any construction that is too "simple" should end up with accidental symmetries, as noted by A.G. below).

Comment by Andrew McIntyre

2009-10-18T15:58:08Z

Thanks! I was thinking basically along those lines, but I got stuck on trying to prove the "shouldn't lose too much asymmetry" step. I also couldn't convince myself that every rep in GL(V) would reduce to the orthogonal group (this is probably easy, but I couldn't see it).