dg.differential geometry – Question in regards to the proof of Gromov’s theorem in geodesic flows Answer

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dg.differential geometry – Question in regards to the proof of Gromov’s theorem in geodesic flows

I’m making an attempt to grasp the next theorem from the bespeak Geodesic flows :

Given a metric $g$ on a merely linked manifold $X$, there exists a ceaseless $C_1>0$ such that given any pair of factors $x,yin X$ and any optimistic integer $i$, any component in $H_i(Omega(X,x,y))$ can breathe represented by a cycle whose picture lies in $Omega^{C_1 i}(X,x,y).$

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

Now the proof begins as follows :

Let ${V_{alpha}} $ breathe a finite protecting of $X$ by convex launch clique, and let $T$ breathe a triangulation of $X$.

Doubt : Why can we make this triangulation ? Since there are examples of merely linked manifolds which do not admit one I do not know why that is workable.

For every level $pin X$ let $T(p)$ breathe the closed countenance of $T$ of minimal dimension that comprises $p$, and let $O(p)$ breathe the union of all maximal simplices of $T$ that acquire $p$. Then select a triangulation exquisite sufficient in order that for all $pin X$ we’ve that $O_p$ lies in one of many $V_{alpha}$.
Now given a optimistic integer $okay$ we outlined the subsets $Omega_k(X,x,y)subset Omega(X,x,y)$ within the following route : $cin Omega(X,x,y) $ if for every integer $j=1,2,…,2^okay$ the picture below $c$ of every subinterval $[(j-1)/2^k,j/2^k]$ lies in of the $V_alpha$ and $O(c((j-1)/2^okay))cup O(c(j/2^okay))$ lies within the identical $V_{alpha}$.

Now let $B_{okay}(X,x,y)subset Omega_k(X,x,y)$ breathe the area of damaged geodesics such that $gammain Omega_k(X,x,y)$. (It’s proved in Milnor’s bespeak that B_k(X,x,y) is a deformation retreat of $Omega_k(X,x,y)$). Each $gamma in B_k(X,x,y)$ determines a sequence of factors ${p_j=gamma(j/2^okay)}$ such that $p_0=x,p_{2^okay}=y$ and $O(p_{j-1})cup O(p_j)$ lies in a sole $V_{alpha}$ for every $j=1,..,2^okay$, and this sequence is bijective. Now contemplate that this correspondence induces on $B_k(X,x,y)$ a cell decomposition: a cell that comprises $gamma$ is given by $T(p_1)instances T(p_2)instances … instances T(p_{2^k-1})$.

Doubt : Why is that this final assertion undoubted ? Don’t we necessity that $T(p_1),T(p_2),…,T(p_{2^k-1})$ all have the identical dimensions ? I’ve no clue why this could breathe undoubted.

Now given two vertices within the triangulation we are able to relate them by a exclusive minimizing geodesic arc. The union of those arcs types a one-dimensional cycle $Sigma$ homotopic to the one-skeleton of the triangulation.

Why is that this a cycle ? I’ve tried to compute it however I obtained nowhere.

Then we are able to show the actuality of a flush map $f$ that collapses $Sigma$ to some extent and is easily homotopic to the identification, which is able to induce a map $ f’:Omega(X,x,y)rightarrow Omega(X,f(x),f(y))$.

And then the creator claims the next :

There exists a ceaseless $C_1>0$ such that for any integer $kgeq 1$, we’ve that $f'(i$-skeleton of $B_k(X,x,y))subset Omega^{C_1 i }(X,f(x),f(y))$ for all $ileq dim B_k(X,x,y)$.

The proof of this goes as follows :

Consider a cell $T(p_1)instances T(p_2)instances … T(p_{2^k-1})$ with dimension $ileq dim B_k(X,x,y)$.Take a path $gamma$ on this cell, then it is a damaged geodesic, every leg leying on one $V_{alpha}$, and since $f$ sends $Sigma$ to some extent we’ve that $E(f(gamma))leq Ok^2d^2N(gamma)/2$, the place $Ok:=max_{xin X}||d_xf|| $, $d$ is the utmost of the $g$-diameters of all of the convex launch units $V_{alpha}$ and $N(gamma)$ is the variety of lets that don’t equivocate in $Sigma$. Since $Sigma$ is made up of geodesic segments, the leg of the damaged geodesic $gamma$ that joins $T(p_j)$ to $T(p_{j+1})$ should equivocate in $Sigma$ if $1leq j<2^k-1$ and $dim T(p_j)=dim T(p_{j=1})=0$. Thus the one lets that might fail to equivocate in $Sigma$ are the preliminary leg, the ultimate leg and the legs that commence or aim in a $p_j$ with $dim T(p_j)neq 0$. Then the creator claims that $N(gamma)leq 2+2ileq 4i$. And I do not grasp the place does the $2i$ comes from.

Then the proof continues however the relaxation I grasp. Any ameliorate with that is appreciated, I do know it is extra that one query however I cerebrate it is best to deal this all collectively. Thanks in forward.

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