Is there a unified reason that there are an infinite number of geodesics between nonconjugate points on a compact manifold? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T14:18:59Z http://mathoverflow.net/feeds/question/57082 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/57082/is-there-a-unified-reason-that-there-are-an-infinite-number-of-geodesics-between Is there a unified reason that there are an infinite number of geodesics between nonconjugate points on a compact manifold? Aaron Mazel-Gee 2011-03-02T07:20:08Z 2012-01-22T12:18:29Z <p>The proof of this statement seems to break into two really different arguments. So, I'm wondering if there is a better argument that can explain them both, or whether it's really just two theorems that happen to be easy to say at the same time. Both rely on a bit of Morse theory, namely that (assuming $p$ and $q$ are nonconjugate) we get a CW-complex for the space of paths $\Omega (M;p,q)$ from $p$ to $q$ which has one cell for each geodesic from $p$ to $q$, whose dimension is the index of the geodesic (i.e. the number of points along it that are conjugate to the starting point, with multiplicity).</p> <p><strong>Case 1 ($|\pi_1(M)|&lt;\infty$):</strong> When $\pi_1(M)=0$, applying the Serre spectral sequence to the path fibration $\Omega M \rightarrow \mathcal{P}M \rightarrow M$ shows that it would be a contradiction if we ever had $H_m(\Omega M)=0$ for all $m\geq N$. So the statement follows from cellular homology. If $0&lt;|\pi_1(M)|&lt;\infty$, pull back the metric on $M$ to its universal cover $\tilde{M}$, which is also compact. Choose $\tilde{p}\in \pi^{-1}(p)$ and $\tilde{q}\in \pi^{-1}(q)$. Then from what we have just said, there are an infinite number of geodesics on $\tilde{M}$ from $\tilde{p}$ to $\tilde{q}$, and these project to geodesics from $p$ to $q$. (I don't want to use the Serre spectral sequence when the base isn't simply connected, if I can help it!)</p> <p><strong>Case 2 ($|\pi_1(M)|=\infty$):</strong> Note that $\pi_1(M)=\pi_0(\Omega(M))$, so any CW-decomposition of $\Omega (M)$ must have an infinite number of cells.</p> <p>I'm pretty sure that since my manifold is complete, in Case 2 I could also have just said "lift a representative of each element of $\pi_1$ (concatenated with some fixed path from $p$ to $q$), homotope it to a geodesic, and project back down" but I'm not positive I can do that. In any case, that still feels kind of different from the argument in Case 1, but maybe there's something here I'm just not seeing.</p> http://mathoverflow.net/questions/57082/is-there-a-unified-reason-that-there-are-an-infinite-number-of-geodesics-between/86345#86345 Answer by Renato G Bettiol for Is there a unified reason that there are an infinite number of geodesics between nonconjugate points on a compact manifold? Renato G Bettiol 2012-01-22T01:28:07Z 2012-01-22T01:28:07Z <p>It seems to me that the OP's last remarks about the difference of the cases when $\pi_1(M)$ is finite or infinite already give the answer to the question. Namely, the two cases are not that different, in that you either use that the universal covering $\tilde M$ satisfies the same hypotheses as $M$; or that the two points have infinitely many pre-images and by completeness of $\tilde M$ there are infinitely many geodesics joining them (that get projected down to the desired infinitely many geodesics on $M$). Maybe it can be more "didactic" to slightly reorganize proof as follows:</p> <h2>Proof:</h2> <p>The crucial fact is that, from the Morse relations,</p> <blockquote> <p>the number of geodesics joining two non-conjugate points $p$ and $q$ of index $m$ is greater or equal to the $m^{th}$ Betti number of the loop space of $M$ with coefficients in (any field) $\mathbf K$.</p> </blockquote> <p>If $\pi_1(M)=0$ and $\dim M=n$, then $H_n(M;\mathbf K)=\mathbf K$ since $M$ is orientable and $H_i(M;\mathbf K)=0$, $i>n$. By Serre's result mentioned above, this implies that the singular homology of the loop space $\Omega_{x,x}(M)$ of $M$ with arbitrary base point $x\in M$ satisfies the following property: for any integer $i\geq 0$, there exists an integer $0&lt; j&lt; n$ such that $H_{i+j}(\Omega_{x,x}(M);\mathbf K)\neq0$. Thus $\Omega_{x,x}(M)$ has <em>infinitely many non zero Betti numbers</em>, hence there are <em>infinitely many geodesics</em> joining $p$ to $q$.</p> <p>If $\pi_1(M)\neq0$, consider the universal covering $\tilde M$ of $M$ with the pull-back metric and notice it is complete and since $\tilde M\to M$ is a local isometry, points in the pre images of $p$ and $q$ are not conjugate to each other. Then if $\pi_1(M)$ is finite, $\tilde M$ is compact and hence satisfies the same hypotheses as $M$, so there are infinitely many geodesics joining any two pre images of $p$ and $q$ by $\tilde M\to M$, which clearly project to infinitely many geodesics joining $p$ and $q$. If $\pi_1(M)$ is infinite, there are infinitely many pre images of $q$ by $\tilde M\to M$, hence infinitely many geodesics joining them to a pre image of $p$, because $\tilde M$ is complete. These clearly project to infinitely many geodesics joining $p$ to $q$ on $M$.</p> <hr> <h2>An interesting remark:</h2> <p>In a very similar way, one can prove that the number of geodesics between two distinct non-conjugate points in a non-necessarily compact, <em>but contractible</em>, manifold is either <strong>odd or infinite</strong>. (e.g., in $\mathbf R^n$ it is always odd).</p> <p><em>Proof.</em> If $M$ is contractible, also $\Omega_{p,p}(M)$ is contractible hence all Betti numbers of $\Omega_{p,p}(M)$ are 1. The Morse relations state that there exists a formal power series $Q$ so that $$\sum k_m\lambda^m=b_\lambda (\Omega_{p,p}(M),\mathbf K)+(1+\lambda)Q(\lambda), \quad \lambda\in\mathbf R$$ where $k_m$ is the number of geodesics between $p$ and $q$ of index $m$. Setting $\lambda=1$, we conclude that the total number of geodesics joining $p$ to $q$ is equal to $2Q(1)+1$, which is either odd or infinite.</p> http://mathoverflow.net/questions/57082/is-there-a-unified-reason-that-there-are-an-infinite-number-of-geodesics-between/86368#86368 Answer by Benoît Kloeckner for Is there a unified reason that there are an infinite number of geodesics between nonconjugate points on a compact manifold? Benoît Kloeckner 2012-01-22T12:18:29Z 2012-01-22T12:18:29Z <p>It seems to me that one could design a dynamical unified argument. The following might be turned into a proof, but there are some points to check.</p> <p>Let $x,y$ be your two non-conjugate points, and let $v$ be the unit tangent vector at $x$ that defines the shortest geodesic from $x$ to $y$. Using compactness, and denoting by $(\xi^t)$ the geodesic flow on $SM$ (the unit tangent bundle), one should be able to find a time $t_1$ such that $\xi^{t_1}(x,v)$ comes very close to $y$. Now, take the Jacobi vector field $Y$ along the geodesic under study that vanishes at $t=0$ and points toward $y$ at $t=t_1$ (it should exist since $x$ and $y$ are non-conjugate, an open condition if I am not mistaken). Shift $v$ in the direction of $Y'(0)$: this gives a new geodesic parametrized on $[0,t_1]$ that comes closer to $y$. What you have to check is that you can push it so as it touches $y$. This is equivalent to ask that $y$ is in the image of a domain around $t_1 v$ where $\exp_x$ is a local diffeomorphism.</p> <p>Then you can, from each geodesic obtained that way, construct a new one that has greater length, so you get an infinite number of geodesics.</p>