# Question about the proof of Gromov's theorem in geodesic flows

I am trying to understand the following theorem from the book Geodesic flows :

Given a metric $$g$$ on a simply connected manifold $$X$$, there exists a constant $$C_1>0$$ such that given any pair of points $$x,y\in X$$ and any positive integer $$i$$, any element in $$H_i(\Omega(X,x,y))$$ can be represented by a cycle whose image lies in $$\Omega^{C_1 i}(X,x,y).$$

Here $$\Omega(X,x,y)$$ is the set of paths in $$X$$ that start in $$x$$ and end in $$y$$, and $$\Omega(X,x,y)=E^{-1}(]-\infty,C_1i])$$ where $$E: \Omega(X,x,y)\rightarrow \mathbb{R}$$ is $$E(\gamma)=\frac{1}{2}\int_{0}^{1}|\dot \gamma|dt$$.

Now the proof starts as follows :

Let $$\{V_{\alpha}\}$$ be a finite covering of $$X$$ by convex open set, and let $$T$$ be a triangulation of $$X$$.

Doubt : Why can we make this triangulation ? Since there are examples of simply connected manifolds which don't admit one I don't know why this is possible.

For each point $$p\in X$$ let $$T(p)$$ be the closed face of $$T$$ of minimum dimension that contains $$p$$, and let $$O(p)$$ be the union of all maximal simplices of $$T$$ that contain $$p$$. Then choose a triangulation fine enough so that for all $$p\in X$$ we have that $$O_p$$ lies in one of the $$V_{\alpha}$$. Now given a positive integer $$k$$ we defined the subsets $$\Omega_k(X,x,y)\subset \Omega(X,x,y)$$ in the following way : $$c\in \Omega(X,x,y)$$ if for each integer $$j=1,2,...,2^k$$ the image under $$c$$ of each subinterval $$[(j-1)/2^k,j/2^k]$$ lies in of the $$V_\alpha$$ and $$O(c((j-1)/2^k))\cup O(c(j/2^k))$$ lies in the same $$V_{\alpha}$$.

Now let $$B_{k}(X,x,y)\subset \Omega_k(X,x,y)$$ be the space of broken geodesics such that $$\gamma\in \Omega_k(X,x,y)$$. (It's proved in Milnor's book that B_k(X,x,y) is a deformation retract of $$\Omega_k(X,x,y)$$). Each $$\gamma \in B_k(X,x,y)$$ determines a sequence of points $$\{p_j=\gamma(j/2^k)\}$$ such that $$p_0=x,p_{2^k}=y$$ and $$O(p_{j-1})\cup O(p_j)$$ lies in a single $$V_{\alpha}$$ for each $$j=1,..,2^k$$, and this sequence is bijective. Now observe that this correspondence induces on $$B_k(X,x,y)$$ a cell decomposition: a cell that contains $$\gamma$$ is given by $$T(p_1)\times T(p_2)\times ... \times T(p_{2^k-1})$$.

Doubt : Why is this last statement true ? Don't we need that $$T(p_1),T(p_2),...,T(p_{2^k-1})$$ all have the same dimensions ? I have no clue why this would be true.

Now given two vertices in the triangulation we can connect them by a unique minimizing geodesic arc. The union of these arcs forms a one-dimensional cycle $$\Sigma$$ homotopic to the one-skeleton of the triangulation.

Why is this a cycle ? I have tried to compute it but I got nowhere.

Then we can prove the existence of a smooth map $$f$$ that collapses $$\Sigma$$ to a point and is smoothly homotopic to the identity, which will induce a map $$f':\Omega(X,x,y)\rightarrow \Omega(X,f(x),f(y))$$.

And then the author claims the following :

There exists a constant $$C_1>0$$ such that for any integer $$k\geq 1$$, we have that $$f'(i$$-skeleton of $$B_k(X,x,y))\subset \Omega^{C_1 i }(X,f(x),f(y))$$ for all $$i\leq \dim B_k(X,x,y)$$.

The proof of this goes as follows :

Consider a cell $$T(p_1)\times T(p_2)\times ... T(p_{2^k-1})$$ with dimension $$i\leq \dim B_k(X,x,y)$$.Take a path $$\gamma$$ in this cell, then it's a broken geodesic, each leg leying on one $$V_{\alpha}$$, and since $$f$$ sends $$\Sigma$$ to a point we have that $$E(f(\gamma))\leq K^2d^2N(\gamma)/2$$, where $$K:=\max_{x\in X}||d_xf||$$, $$d$$ is the maximum of the $$g$$-diameters of all the convex open sets $$V_{\alpha}$$ and $$N(\gamma)$$ is the number of lets that do not lie in $$\Sigma$$. Since $$\Sigma$$ is made up of geodesic segments, the leg of the broken geodesic $$\gamma$$ that joins $$T(p_j)$$ to $$T(p_{j+1})$$ must lie in $$\Sigma$$ if $$1\leq j<2^k-1$$ and $$\dim T(p_j)=\dim T(p_{j=1})=0$$. Thus the only lets that could fail to lie in $$\Sigma$$ are the initial leg, the final leg and the legs that begin or end in a $$p_j$$ with $$\dim T(p_j)\neq 0$$. Then the author claims that $$N(\gamma)\leq 2+2i\leq 4i$$. And I don't understand where does the $$2i$$ comes from.

Then the proof continues but the rest I understand. Any help with this is appreciated, I know it's more that one question but I think it's best to treat this all together. Thanks in advance.

## 1 Answer

1. Since the manifold is Riemannian, it is in particular smooth and has a $$C^1$$ triangulation: see Whitehead's "On $$C^1$$ Complexes" for example (the original triangulation theorem in this case seems to be due to Cairns). If a manifold can't be triangulated it has no smooth structure.

2. Recall the cell structure on a product of complexes: a cell is just a product of cells. Here we take only a subset of the cells in the $$(2^k-1)$$-fold product, consisting of those products $$\sigma_1\times\ldots\times\sigma_{2^k-1}$$ such that all maximal simplices containing $$\sigma_i,\sigma_{i+1}$$ are both contained in the same $$V_\alpha$$ for some $$\alpha$$. This is still a complex. The result follows from the fact that each $$p_i$$ is contained in the relative interior of the simplex $$T(p_i)$$, so that the interior of the product of these simplices contains the tuple $$(p_0,\ldots,p_{2^k-1})$$. The given product is the unique minimal cell which contains $$\gamma$$.

3. It seems $$\Sigma$$ can only be a cycle in the homology if each vertex participates in an even number of edges. However, I don't think this is essential to the rest of the proof. It is enough that it is a union of cycles.

4. $$i$$ is the sum of dimensions of the simplices in the product. A leg of the path is in $$\Sigma$$ unless it begins in $$x$$ or ends in $$y$$ (two such legs) or it begins or ends at a simplex of positive dimension (at most $$2$$ for each cell of positive dimension). There are at most $$i$$ simplices of positive dimension, so we get $$2+2i$$.

• I am sure there is room for more detail, and I can probably add some on request. There are too many questions here for me to do this pre-emptively. – Geva Yashfe Feb 21 at 17:21
• Thanks for the answer I think it's clear now ! – Lost Feb 22 at 11:00