# 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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