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Assume $\Gamma$ be a Bieberbach group which acts on $\mathbb R^n$ (i.e. a discrete subgroup of isometries of $n$-dimensional Euclidean space with a compact fundamental domain). Denote by $M(\Gamma)$ the number of maximal finite subgroups (up to conjugation) in $\Gamma$. Is it true that $M(\Gamma)\le 2^n$?

Things I can do: There is a simple geometric observation (due to Perelman) which shows that if $N(\Gamma)$ is the number of orbits of isolated fixed point of some subgroups of $\Gamma$ then $N(\Gamma)\le2^n$. Clearly, each such point corresponds to a maximal finite subgroup. Thus, $N(\Gamma)\le M(\Gamma)$, but in all examples I know I still have $M(\Gamma)\le 2^n$ (and I believe it is allways true).

The formulation is completely algebraic so maybe it has a completely algebraic solution...

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It will be very nice if you could include in your question a hint on how to prove this geometric observation due to Perelman! And also, what is the minimal dimension for which you don't know the answer to the question? –  Dmitri Feb 1 '10 at 22:51
    
Sorry for a naive/elementary question: are all such groups finite extensions of some sublattice of ${\mathbb R}^n$? –  Yemon Choi Feb 2 '10 at 7:14
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yes, they are semidirect products of a finite group (stabilizer of 0) by a lattice of rank n preserved by this group. –  Pavel Etingof Feb 2 '10 at 12:39
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Bieberbach groups are extensions of $Z^n$ by a finite group, but I don't think that this extension splits in general (contradicting Etingof's comment), e.g. 22* in Conway's notation. One remark is that I think one ought to be able to reduce this question to Bieberbach orbifolds which don't fiber (i.e., they are conformally rigid). For example, if a Euclidean orbifold fibers with 1-dimensional fiber, then the maximal stabilizers should sit over the maximal stabilizers of one lower dimension, with multiplicity at most 2 (corresponding to the two fixed points of a dihedral fiber). –  Ian Agol Apr 13 '10 at 3:15

2 Answers 2

up vote 10 down vote accepted

Dima, I can not write a comment (yet) so I will start an answer to my own question.

You may assume that $\Gamma$ acts by isometries, so $A=\mathbb R^n/\Gamma$ is an Alexandrov space. For each maximal subgroup $F$ one can take its fixed point set $S_F$ in $\mathbb R^n$. $S_F$ is an (affine) subspace and image (say $E_F$) in $\mathbb R^n/\Gamma$ is a singular set (so called extremal subset of Alexandrov space). The maximality of $F$ implies that $E_F$ contains no proper extremal sets (a smaller subset is fixed by bigger group). (In fact $E_F$ is a flat manifold and its has a neighborhood which isometric for a product $E_F\times Cone$.)

So the question boils down to finding maximal number of such extremal sets in $A$. A particular case of such sets are isolated singular points. Perelman's theorem states that number of "one-point extremal sets" in an Alexandrov space with curvature $\ge 0$ is at most $2^n$. The proof repeats a proof of Erdős problem: if you have $m$ points in $\mathbb R^n$ such that all angles in all triangles $\le \pi/2$ then $m\le 2^n$. We take homothety with coefficient 1/2 for each point, then images of convex hull don't have common internal points (otherwise it would occur obtuse angle), then compairing volume of convex hull and its images gives estimate for number of points.

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Nina, thanks a lot for the answer! I never heard about Erdos problem, it is really cute! –  Dmitri Feb 2 '10 at 22:14

That is not an answer. I want to give an example where the argument of Erdős does not work directly.

Consider an action of group $\Gamma$ on $\mathbb R^3$ generated by the reflections $r_1, r_2$ and $r_3$ correspondingly in the lines $x=z=0$ and $x+1=z=0$ and $x-y=z-1=0$.

Each of the reflections $r_i$ generate a maxiamal $\mathbb Z_2$-subgroups, all of them are nonconjugate. These groups corespond to three singular circles, say $\Sigma_i$ in the factor $X=\mathbb R^3/\Gamma$. ($X$ is homeomorphic to $S^3$ and $\Sigma_1$, $\Sigma_2$, $\Sigma_3$ form Borromean rings, but all this is not important.)

Let us try to mimic argument of Erdős. Take subsets $X_i$ of $X$ of midpoints $m$ between $x\in X$ and a closest $x_0\in\Sigma_i$ to $x$. As in the argument of Erdős we have* $\mathrm{vol}\\, X_i>\tfrac{1}{2^3}\cdot\mathrm{vol}\\, X$. BUT $X_1\cap X_3$ has interior points and here argument brakes into parts.

Comments

  • Since fixed point sets are 1-dimensional, it would be enough to take $m\in [xx_0]$ such that $\tfrac{|mx_0|}{|xx_0|}=\tfrac1{2\sqrt[3]{2}}$. But even in this case one has interior points in $X_1\cap X_2$ (the borderline in this example seems to be $\tfrac13$).

  • There is a natural bisecting hyperplane for any two affine subspaces. We may use it to cut a cylinder domain around each fixed point set of a maximal subgroup. The projection of these cylinders in $X$ gives Voronoi-like domains, but they do not cover whole space in general --- that is OK as far as we have lower bound on their volumes...

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Since there seems to be no way to comment on deletion, let me explain where an error in my answer was: when lemma about acute angles was applied, things that should be orthogonal were not. Anton noted this in a comment, but that comment has gone along with the answer. –  Sergei Ivanov Apr 15 '10 at 19:04
    
Anton, btw, in your example there are 1/2-homotheties with non-overlapping images. You just divide the space not into Voronoi regions of lines (they are weird) but into regions bounded by the most natural planes separating pairs of lines. I wonder if something like this makes sense in the general case... –  Sergei Ivanov Apr 15 '10 at 19:10
    
@Sergei, I also think that "something like this might make sense in the general case"; if yes then it might be done via linear algebra (but not geometry). –  Anton Petrunin Apr 15 '10 at 20:11
    
@Anton: I don't quite understand your example. The glide-rotation and reflection through x=z=0 generate a dihedral group acting on the z-axis, so there is another rotation (through x=y, z=1/2). Thus, I'm not sure your circle count is correct. This is an example of a fibered group (since all three generators preserve the vertical direction), and it fibers over the 244 triangle orbifold. –  Ian Agol Apr 16 '10 at 4:32
    
@Agol, you are right, I corrected the answer --- hopefully it is OK now. –  Anton Petrunin Apr 18 '10 at 16:49

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