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



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> 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 $\Omega^{C\_1 i}(X,x,y).$
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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 $X$.



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> 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.
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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 $T(p\_1)\times T(p\_2)\times ... \times T(p\_{2^k-1})$.



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> 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.
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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.



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



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> 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 $i\leq \dim B\_k(X,x,y)$.
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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.



Question about the proof of Gromov's theorem in geodesic flows
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 $2+2i$.


