Since there has been only one response to this (via direct e-mail to me), let me try to rephrase it so it doesn't sound so technical.* Because it really isn't. Let's define a "regular map" on an orientable topological surface M to be a tiling of the surface by n-gons for some fixed n, such that for any two such polygons P and Q, and for any of the n rotational positions that P can be mapped to Q (without flipping it over), there is a homeomorphism of the surface carrying P to Q that way. Note that the tiling here means only some graph G (edges and nodes) on the surface M such that the complement M \ G is a disjoint union of topological (open) n-gons. As long as the open polygons satisfy the symmetry condition above. So that, e.g., representing a torus as a single square with its opposite edges identified is fine. PUZZLE: Let M be homeomorphic to the torus T. What is the smallest positive integer K such that there does not exist a regular map on T having K polygons? --Dan ___________________________________________________ * Okay, maybe it doesn't sound less technical, but I hope all the details are clear. On Mar 21, 2014, at 10:23 AM, Dan Asimov <dasimov@earthlink.net> wrote:

For our purposes, a “regular map” on an metric orientable surface M (assumed compact and without boundary) is a tiling of the surface by congruent regular n-gons, such that for any two n-gons P and Q and any orientation-preserving isometry P -> Q (there are n of these), there is an automorphism f: M -> M of the tiling such that f restricted to P is this isometry P -> Q.

(There must be an easier way to say this. But a regular map defined as just a tiling of M by congruent polygons need not have the isometry property.)

PUZZLE: If M = the torus T, what is the least positive integer L such that there exists no regular map on T having exactly L polygons?

(Note: We allow the boundary of any tile to overlap any part of itself. So even one lone square can be a regular map of T.)

—Dan

If the matrices form a group, then their eigenvalues must lie on the unit circle, as otherwise the inverse would have an eigenvalue outside.  A sufficient condition is that each of the generating matrices is unitary (with respect to some choice of basis), since then the group they generate is a subgroup of the unitary group.  But this is not a necessary condition, as the  matrix [[1,1],[0,1]] satisfies the requirement, but is not unitary in any basis.   --  Gene

________________________________ From: Dan Asimov <dasimov@earthlink.net> To: math-fun <math-fun@mailman.xmission.com> Sent: Friday, March 21, 2014 8:58 AM Subject: Re: [math-fun] matrix group with bounded eigenvalues

If the eigenvalues are *strictly* inside the unit circle, then I would expect the matrices to each be a contraction mapping on their vector space (with some contraction constant L < 1 equal to the maximum |eigenvalue|).

So then wouldn’t a composite of finitely many of these matrices also be a contraction mapping, with contraction constant equal to the maximum of the finitely many contraction constants pertaining to each matrix in the composite?

—Dan

On Mar 21, 2014, at 3:23 AM, Fred Lunnon <fred.lunnon@gmail.com> wrote:

...

  Given a finite set of complex matrices generating an infinite group, I need to establish that every element of the group has all eigenvalues within the unit circle.

  Such problems must have been well investigated --- perhaps (shudder) by math physicists --- but I have no idea where to start looking.

  [It's actually a semigroup of integer matrices with constraints on allowed products, and circle radius exponential in the number of product factors --- but those details are almost certainly unimportant.]

WFL

I think I’ve verified what Jim says 32 ways from Sunday. I get (81/200)*(3+sqrt(5)) or about 2.12 for the largest eigenvalue. This is certainly counterintuitive from my naïve viewpoint. —Dan On Mar 21, 2014, at 9:36 AM, James Propp <jamespropp@gmail.com> wrote:

I don't think this Dan'a guess is true. Consider the non-commuting matrices .9 .9 .0 .9 and .9 .0 .9 .9 Each of them has eigenvalues strictly inside the unit disk, but their product doesn't.

(Unless I made a mistake, in which case, please let me know!)

Jim

On Friday, March 21, 2014, Dan Asimov <dasimov@earthlink.net> wrote:

...

If the eigenvalues are *strictly* inside the unit circle, then I would expect the matrices to each be a contraction mapping on their vector space (with some contraction constant L < 1 equal to the maximum |eigenvalue|).

So then wouldn't a composite of finitely many of these matrices also be a contraction mapping, with contraction constant equal to the maximum of the finitely many contraction constants pertaining to each matrix in the composite?