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

  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 $2+2i$.

  • $\begingroup$ 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. $\endgroup$ Feb 21, 2021 at 17:21
  • $\begingroup$ Thanks for the answer I think it's clear now ! $\endgroup$
    – user174565
    Feb 22, 2021 at 11:00

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