Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

[Edit: I had a mistake in the numerology (took d=6,5 instead of d=5,4). Edit: I mistakenly identified my mistake, it is 6,5 but I got the indices shifted by one.]

Background: Polygon spaces

Given a polygon P in space we can consider the space of all its embeddings when the lengths of the edges are fixed and mod out by rigid motion of space. Such spaces are called polygon spaces: see, for example, the paper Polygon spaces and Grassmannians by Jean-Claude Hausmann and Allen Knutson. (See also the MO question Space of simple polygons on $n$-vertices as a set of points in $\mathbb{R}^{2n}$.)

The dimension of these spaces can be recovered by a simple numerology: The number of degrees of freedom is $3n$; from this you have to subtract $n$ constraints given by the length of edges, and the dimension 6 of rigid motions of space, and you are left with $2n-6$.

If you consider embeddings into the plane rather then into space then the dimension of your space is $2n-n-3 = n-3$. The idea that such spaces of polygons have symplectic structure is attributed in the Hausmann and Knutson's paper to Cappell. [AK adds: It seems to have first been observed by Klyachko.]

The fact that $n-3$ is half of $2n-6$ is no coincidence and indeed the space of planar embeddings correspond to Lagrangian submanifolds of the polygon spaces.

In between the simple numerology and the complicated spaces we can identify some intermediate objects: the tangent vector spaces. Those are the vector spaces of infinitesimal motions of an embedded polygon in $R^3$ or $R^2$. After modding out with infinitesimal motions arising from rigid motions of the whole space (or plane) we are left with vector spaces of dimensions $2n-6$ and $n-3$ respectively.

Summary: Talking about polygon spaces we can identify three levels:

Numerology: $2n-6$ degrees of freedom to embeddings of polygons into $R^3$, $n-3$ degrees of freedom to embeddings of polygons into $R^2$.

Linear algebra: The vector spaces of infinitesimal motions of polygons embedded in the plane and space.

Symplectic manifolds: The polygon spaces and their Lagrangian submanifolds.

The polygon spaces have various structures on them and are quite exciting.

The idea:

Polygons are triangulations of $S^1$. Extend the notion of a "polygon space" to certain embeddings of triangulated $(2k+1)$-dimensional spheres. I will mainly talk about the "next case" which are triangulations of $S^3$.

Problem: Describe an appropriate analog of "polygon spaces" for triangulations of 3-dimensional spheres.

Details: From numerology to spaces

Numerology --->>> Linear algebra (vector spaces) --->>> Varieties/spaces.

1) The Numerology:

The numerology refers to the dimension of our hypothetical analogs for polygon spaces.

Given a sequence of non-negative integers $(f_{-1}=1) ,f_0, f_1, f_2 \dots$ and an integer $d \ge 0$ we define a new sequence $g_0[d]$, $g_1[d]$ , $g_2[d], \dots$ as follows: $$g_0[d] = 1,$$ $$g_1[d] = f_0 - d,$$ $$g_2[d] = f_1 - (d-1) f_0 + {{d} \choose {2}},$$ $$g_3[d] = f_2 - (d-2) f_1 + {{d-1} \choose {2}} f_0 - {{d} \choose {3}},$$ etc. (But below we worry only about g_2 and g_3)

The sequence $f_0, f_1,\dots $ usually comes as the $f$-vector of some simplicial complex and the new sequences $g_i[d]$ are important in studying the combinatorics of these complexes.


Now, if we have a polygon with n vertices we have: $$f_0 = f_1 = n ,$$ $$g_2[3] = -(n-3)$$ and $$g_2[4] = -(2n-6),$$ which are the dimensions of the polygon spaces for embeddings in $R^2$ and $R^3$ respectively.

The next case: triangulations of $S^3$.

The next analogous case is to start with a triangulation $K$ of $S^3$ with $f_0$ vertices $f_1$ edges $f_2$ triangles and $f_3$ 3-simplices.

Here (using Euler's theorem) it is easy to verify that for every $K$ we have $$g_3[6] = 2g_3[5].$$ (Both quantities are negative.)

Thus, the suggestion is that $-g_3[6]$ should be the dimension of a "generalized polygon space" for some sort of embedding of the triangulation $K$ (possibly into $ R^5$), and $-g_3[5]$ the dimension for "embedding" of $K$ (possibly into $R^4$). Hopefully, the latter will correspond to a Lagrangian submanifold of the former.

(Remark: in the comments Misha proposed making the embedding into certain CAT (1) spaces.)

Let me just verify the identity I described: $g_3[6] = f_2-4f_1+10f_0-20$ and $g_3[5]=f_2-3f_1+6f_0-10.$ For triangulated 3-spheres, Euler's theorem asserts that $f_0-f_1+f_2-f_3=0$ and also $f_2=2f_3$ and therefore $f_2=2f_1-2f_0$. It follows that $$g_3[6]=-2f_1+8f_0-20,$$ while $$g_3[5]=-f_1+4f_0-10.$$

2) The linear algebra

The linear algebra refers to the tangent vector spaces of our hypothetical spaces.


For a graph $G$ consider the quantity
$X$ = (the number of edges - $m$ the number of vertices + ${{m+1} \choose{2}})$.

$X$ ($=g_2$) is a lower bound (interesting of course only when $X$ is non-negative) on the space of infinitesimal stresses when the vertices of the graph are embedded in $R^m$. $-X$ is a lower bound on the number of infinitesimal flexes for such embeddings (this is interesting when $X$ is negative).

For polygons embedded in $R^2$ and $R^3$. $X$ is non-positive and $-X$ is the dimension of the infinitesimal flexes (not just a lower bound). Those are the tangents of the polygon spaces.

Triangulations of $S^3$.

For a simplicial 2-dimensional complex $K$ we can let $$Y=f_2 - (d-2) f_1 + {{d-1} \choose {2}}f_0 - {{d} \choose {3}}.$$ (We called it $g_3[d]$ before.) We care about $d=6$ so $$Y=f_2-4f_1+10f_0-20.$$

There are some known notions of high dimensional infinitesimal flexes and stresses which are bounded by $Y$ (or $-Y$, respectively).

There are conjectures for simplicial spheres, which are proven for simplicial polytopes, that these bounds are tight.

These spaces are perhaps tangent to some hypothetical polygon-like spaces (but only at very special points).

3) How the polygon-like space may look?

This particular suggestion probably does not work, but it allows to describe what I am up to.

We embed the vertices of $k$, the triangulated $S^3$ in some fixed location. We need 4 degrees of freedom for every edge and one constraint for every triangle.

For example, we can let the edges be arcs between the corresponding vertices which are parabolas; the "triangles" will be quadratic (describing a surfaces of minimum area) which extend the arcs corresponding to edges.

The condition is that the areas of the "triangles" are prescribed.

share|improve this question
At least for generic choice of a metric on a triangulated 3-sphere, maps to 4-space are locally rigid. This follow from Gluck's rigidity theorem for links of vertices in the 3-sphere, which are 2-spheres. Gluck proved his generic rigidity for maps of 2-spheres to $R^3$ but the same proof works for maps to $S^3$. I think negativity which you observed reflects lack of deformations. –  Misha Jun 8 '12 at 10:42
Dear Misha, Right! but note that I talk about a different notion of rigidity where the degrees of freedom correspond to the edges and the constraints correspond to the triangles. –  Gil Kalai Jun 8 '12 at 12:27
@Gil: Sorry, I misread your question. Here is a suggestion. Instead of realizing polyhedral spheres in flat space, you consider their realizations in a spherical building (here a spherical building for me is a certain CAT(1) metric space, not just a simplicial complex). Vertices are mapped to a fixed collection of pairwise antipodal points, say, vertices of the building. Edges are mapped to geodesics. Geodesics connecting pairs of antipodal points are no longer unique, dimension of the space of such geodesics can be read off from the building data. For instance, if the building is ... –  Misha Jun 8 '12 at 17:27
...associated with the incidence geometry of $RP^2$, then you get 1 degree of freedom for every edge, for $CP^2$ you get 2. To get one constraint for every 2-face you can, for instance, prescribe the area of the triangle spanned by the given edges. In the complex case you have another natural invariant, namely symplectic area of the triangle. Note that polygon spaces exist not only for Euclidean space but also for, say, 3-sphere (as well as hyperbolic space, etc). You get the same dimension count, also symplectic, even, Kahler, structure (Lie-Poisson for $H^3$). –  Misha Jun 8 '12 at 17:37
The configuration spaces of special triangulations of $S^2$ can be interesting. See flexible polyhedra and the Bellows Conjecture (Sabitov's Theorem, 1995). Variations give generically positive-dimensional spaces, such as polyhedra whose face areas are specified but not edge lengths, and perhaps these could indicate the structure you want within $S^{2k+1}$. –  Douglas Zare Jun 8 '12 at 21:45
show 5 more comments

1 Answer 1

up vote 14 down vote accepted

Before describing my construction, I will explain why it is natural: In Gil's setup, we are supposed to fix a certain set of vertices (points in some space $X$) and vary edges connecting them. If edges are geodesics, we then need $X$ where geodesics with fixed end-points are far from being unique. Round $n$-spheres $S^n$ provide natural examples, where antipodal points are connected by $n-1$ dimensional families of geodesics. However, in the sphere, every point is antipodal to a unique point, so $S^n$ is not good enough. Spherical buildings are a natural generalizations of round spheres where antipodes are not unique.

Here is a general recipe for getting spaces of $d$-dimensional space of geodesics connecting "antipodal" points.

Let $G$ be a real Lie group which is the set of real points of an algebraic semi-simple group over reals (I assume that $G$ has no compact factors). I will assume that $r$, the (real) rank of $G$, is $>1$. Let $R$ be the root system of $G$ and $R_+$ be the set of positive roots with respect to a choice of Borel subgroup $B\subset G$. Let $W$ be the Weyl group of $G$.

One then associates with $G$ a spherical (Tits) building $X$. One can think of $X$ combinatorially as a cell complex encoding inclusions between parabolic subgroups of $G$. I will think of $X$ as a metrized cell complex, where each cell is isometric to a cell in the Coxeter complex $(S,W)$, here $S$ is the sphere of dimension $r-1$. Given such metric, one can talk about geodesic segments in $X$, which are (globally) length-minimizing paths. The group $G$ acts on $X$ by isometries, preserving the combinatorial structure of $X$. Then every geodesic in $X$ has length $\le \pi$ and is contained in an apartment in $X$, which is an isometrically embedded copy of $S$ in $X$. Chambers in $X$ are the facets, they are stabilized by the Borel subgroups of $G$.

More generally, stabilizers $G_\sigma$ of cells $\sigma$ in $X$ are parabolic subgroups in $G$ (by the definition of $X$). I will use the notation $G_x$ for the $G$-stabilizer of a point $x$ in $X$. Then $G_x=G_\sigma$, where $\sigma\subset X$ is the smallest cel containing $x$. Two oriented geodesics are called congruent if there exists an orientation-preserving isometry between them induced by an element of $G$. Then, given an oriented geodesic $\gamma\subset X$, the space of $\Gamma_\gamma$ of oriented geodesics in $X$ congruent to $\gamma$ identifies naturally with $G/G_\Gamma$, where $G_\gamma$ is the stabilizer of $\gamma$ in $G$. It is easy to see that $G_\gamma$ is again an algebraic subgroup of $G$, so $\Gamma_\gamma$ is a homogeneous manifold. This explains why sets of congruent geodesics are useful for defining "edges'' between points in $X$.

Two points in $X$ are called antipodal if they are within distance $\pi$. The key property of $X$ is that it has abundance of antipodal points and geodesics between such points are far from being unique. First of all, if $x, y$ are antipodal and belong to an apartment $S\subset X$ then we have $r-2$-dimensional space of geodesics between $x$ and $y$ contained in $S$. However, most of these will not be congruent to each other. More interesting construction of geodesics connecting $x$ to $y$ is as follows: Let $\xi$ be a germ of a geodesic in $X$ emanating from $x$. Then $\xi$ can be extended (uniquely) to a geodesic $\gamma_\xi$ connecting $x$ to $y$. One obtains congruent geodesics $\gamma_\xi$ by using germs $\xi$ which belong to a single $G_x$-orbit, where $G_x=G_\sigma$ is a parabolic subgroup of $G$ ($\sigma$ is the minimal cell in $X$ containing $x$).

Thus, the space of congruent geodesics connecting $x$ to $y$ (with germs at $x$ of type $\xi$) is naturally identified with the quotient $$ G_x/G_\xi=G_\sigma/G_\tau, $$ where $\tau$ is the smallest cell in $X$ containing $\xi$.

Our goal is then to find buildings $X$ and pairs of cells $(\sigma, \tau)$, with $\sigma\subset \tau$, so that:

  1. $dim(G_\sigma/G_\tau)=4$.

  2. The Levi subgroup of $G_\sigma$ admits an epimorphism to a group locally isomorphic to $SL(2, {\mathbb R})$ so that $G_\tau$ is contained in the kernel.

I will do so when $G$ is split, since computations are easier in this case. Then each parabolic subgroup $P$ of $G$ corresponds (up to conjugation) to a root subsystem $R'$ in $R$, where ${R}'$ is obtained by omitting some nodes in the Dynkin diagram of $R$. Then $P$ has dimension $$|R_+'| + |R_+| + r$$ Note that $|R_+| + r$ is the dimension of the Borel subgroup $B$ of $G$, the smallest possible dimension for a parabolic. In particular, $$ d(R')=dim(P)-dim(B)= |{R'}_+|. $$

Thus, if we take, say, $G_\tau=B$, then I can get any value for $$ d(R')=dim(P)-dim(B)=dim(G_\sigma/G_\tau) $$ (not exceeding, say, $r/2$) by taking $R'$ to be a direct sum of rank 1 root subsystems in $R$ (so that Dynkin diagram for $R'$ contains no edges). This will satisfy condition 1, of course.

Here is a bit more interesting example. Let $G=SL(5, {\mathbb R})$, $r=4$. Take $R'$ obtained by removing the one of the middle nodes from Dynkin diagram of $R$ so that $R'\cong A_1\oplus A_2$ and the Levi factor of $P$ is locally isomorphic to $SL(2, {\mathbb R})\times SL(3, {\mathbb R})$. Then $d(R')=1+3=4$. In particular, $P$ is a maximal parabolic and, hence, $\sigma=x$ is a vertex of $X$. Alternatively, one can use $G=SL(3, {\mathbb R})\times SL(3, {\mathbb R})$, where it does not matter which node you remove, you still get $R'\cong A_1\oplus A_2$. Below, I will consider $G= SL(5, {\mathbb R})$.

Now, the space of geodesics $\Sigma_{x,y,\xi}$ connecting $x$ to an antipodal point $y$ and having a fixed regular congruence type, is a smooth 4-dimensional manifold. (Regular congruence type corresponds to the assumption that the germ $\xi$ is not contained in any wall of $X$, so $\tau$ is a chamber in $X$ and $G_\tau$ is Borel.) This is the degree of freedom for edges in $X$ that Gil asked for.

Now, I can explain why the condition 1 above is useful. Suppose that $H$ is a group which admits an epimorphism $\rho: H\to SL(2, {\mathbb R})$. Then $H$ contains a codimension 1 subgroup $H'$ obtained as the preimage of a Borel in $SL(2, {\mathbb R})$. In the context of the pair $P=G_\sigma, G_\tau$, I get another parabolic $P'=H'$ of codimension 1 in $P=H$ and containing $G_\tau$. Now, using $P'$-orbit of the above germ $\xi$ (instead of the $P$-orbit), I will get a 3-dimension submanifold in the manifold of geodesics $\Sigma_{x,y,\xi}$.

Thus, we got spaces of edges connecting $x$ to $y$ which have dimensions $4$ and $3$ respectively, which is what Gil asked for.

Our next goal is to impose restrictions on 2-faces of the 3-sphere, i.e., triangles $T_i$ whose vertices are antipodal vertices $x_1, x_2, x_3$ in $X$ and whose edges are described above. Let $\xi_i, i=1,2,3$ denote the germs (at $x_i$) of the oriented edges $[x_i,x_{i+1}]$, each germ $\xi_i$ is contained in a unique chamber $\sigma_i$ in $X$. Generically, the chambers $\sigma_i$ are mutually antipodal. Let $C_3(X)$ denote the space of ordered triples of pairwise antipodal chambers in $X$. This space is birational to $U/T$, where $U$ is unipotent radical of $B$ and $T\subset G$ is maximal torus normalizing $U$ and acting on $U$ via conjugation. Hence, $C_3(X)$ has dimension $|R_+|-r$, in our example $G=SL(5, {\mathbb R})$, it is $10-4=6$.

Thus, we have plenty of functions on $C_3(X)$ to choose from in order to impose one restriction for each triangle. Below is a somewhat random choice, based on my reading of the paper by Fock and Goncharov "Moduli spaces of local systems and higher Teichmuller theory". A chamber in $X$ (in the case of $G=GL(n, {\mathbb R})$ is given by a complete flag $F=(V_0\subset V_1\subset ... \subset V_n)$ in $V={\mathbb R}^n$. If $n=3$, then $C_3(X)$ is 1-dimensional with a single (most natural) invariant of a a triple of flags $(F_1,F_2,F_3)$ given by Goncharov's triple ratio,

$$ \rho(F_1,F_2,F_3)=\frac{\langle f_1, v_2 \rangle \langle f_2, v_3 \rangle \langle f_3, v_1 \rangle} {\langle f_1, v_3 \rangle \langle f_2, v_1 \rangle \langle f_3, v_2 \rangle} $$

Here $f_k$'s are linear functions on ${\mathbb R}^3$ whose kernels are 2-planes in the flags $F_k$ and $v_k$'s are basis vectors in the lines in the flags $F_k$. For $n\ge 4$, given a triple of flags $(F_1,F_2,F_3)$, taking quotient of $V$ by the subspace $V_{n-3}$ appearing in the $i$-th flag, I reduce the dimension to $3$ and, hence, can get the triple ratio $\rho_i$ of the image of my flag in the resulting 3-dimensional space. Now, I will take function $h: C_3(X)\to {\mathbb R}$ given by

$$ h(F_1,F_2,F_3)=\rho^2_1+\rho^2_2+\rho^2_3. $$

Then, I impose one restriction $h(T_i)=t_i\in {\mathbb R}_+$ for each triangular face $T_i$ in the 3-sphere. This works for general $n$, but I will restrict to the case $n=5$.

At the moment, I have no idea where the symplectic structure on the resulting space of maps of $S^3$ to $X$ would come from (actually, I do, I just do not know how to make it work). Instead of $G=SL(5, {\mathbb R})$ one may have to use another Lie group. The easiest way to get a symplectic structure is to use some form of symplectic reduction, which is how it was done first by Klyachko and then in my paper with Millson "The symplectic geometry of polygons in Euclidean space", Journal of Diff. Geometry, Vol. 44 (1996) p. 479-513, or in my paper with Millson and Treloar "The symplectic geometry of polygons in hyperbolic 3-space", Asian Journal of Math., Vol. 4 (2000), N1, p. 123-164. The latter used Poisson Lie theory, which, somehow, looks more promising.

share|improve this answer
Dear Misha, wow! remarkable! I guess the first thing to check is if the numerology works right even for the lower order terms... –  Gil Kalai Jun 18 '12 at 11:38
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.