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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).

Case 1 ($|\pi_1(M)|<\infty$): 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 $0<|\pi_1(M)|<\infty$, 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!)

Case 2 ($|\pi_1(M)|=\infty$): Note that $\pi_1(M)=\pi_0(\Omega(M))$, so any CW-decomposition of $\Omega (M)$ must have an infinite number of cells.

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.

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Nitpick: what is $N$ in $m \geq N$? –  K.J. Moi Mar 2 '11 at 8:42
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I like the "two theorems that happen to be easy to say at the same time". The point in both proofs is that $\Omega M$ is not homotopy equivalent to a finite CW complex. In the simply connected case, you show that $\Omega M$ is infinite dimensional, in the other case, you show that it has infinitely many components. I do not believe you can unify both arguments: can you think of a uniform reason why $\Omega T^2 \simeq Z^2$ and $\Omega S^2$ are both infinite CW complexes? –  Johannes Ebert Mar 2 '11 at 9:19
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Well, you can prove case 2 for any complete manifold (not even compact), while case 1 is false for non-compact manifolds (take $\mathbb{R}^n$), so I guess any proof of the complete statement will have to make a distinction between the two cases. Case 2 can be done by proving every continuous path in a complete manifold between two points is homotopic to a geodesic. –  Jan Jitse Venselaar Mar 2 '11 at 9:26
    
@ K.J.: Edited for clarity. @ Johannes: I cannot! :o) @ Jan: That's true, I hadn't thought of that. This seems to be about the observation I was looking for. –  Aaron Mazel-Gee Mar 2 '11 at 17:36
    
I agree with Jan's comment. In one sense, the original question is already a "unified" perspective, which is to look at geodesics as critical points on a path space and then prove that the path space has a lot of topology. But as Jan points out, the second Case is way easier and doesn't use any Morse theory at all. You can just find a minimizing geodesic in each homotopy class. The first Case is a far more sophisticated theorem since you are finding geodesics of high index rather than just minimizers, whose existence is straightforward. –  Dan Lee Mar 2 '11 at 18:26
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2 Answers

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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:

Proof:

The crucial fact is that, from the Morse relations,

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) $\mathbf K$.

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 $0< j< n$ such that $H_{i+j}(\Omega_{x,x}(M);\mathbf K)\neq0$. Thus $\Omega_{x,x}(M)$ has infinitely many non zero Betti numbers, hence there are infinitely many geodesics joining $p$ to $q$.

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 $M$.


An interesting remark:

In a very similar way, one can prove that the number of geodesics between two distinct non-conjugate points in a non-necessarily compact, but contractible, manifold is either odd or infinite. (e.g., in $\mathbf R^n$ it is always odd).

Proof. 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.

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I like this answer; the Morse relations certainly seem to play a fundamental role here. As for your remark at the end though, if $\Omega_{p,p}(M) \simeq *$ then shouldn't its $0^{th}$ Betti number be 1 and all others be 0? Although this seems to contradict your observation for $\mathbb{R}^n$, since it says that $\sum k_m = 2Q(1)$, so I must be missing something... –  Aaron Mazel-Gee May 16 '12 at 23:27
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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.

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.

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.

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I have thoroughly thought about this argument before, as well discussed it with @Jan Jitse. The problem seems to me that the geodesic flow is defined on the noncompact domain $\mathbb{R} \times SM$. As such, the derivative of the flow (generated by the Jacobi field, I assume, or at least closely related) need not have a bounded derivative. Thus it may happen that the newly parametrized geodesic is actually the same as a previous one (after arc-length reparametrization). –  Jaap Eldering Jan 22 '12 at 15:30
    
@Benoît: How do you choose a Jacobi field $Y$ that vanishes at $x$ and "points towards $y$ at $\xi^{t_1}(x,v)$"? It seems to me that "pointing towards $y$" means that $Y$ is the tangent field to the geodesic $t\mapsto\xi^{t}(x,v)$ (that doesn't vanish at $x$). Also, since the Jacobi equation is a 2nd order linear ODE, why should it be possible to solve a "boundary value problem" like this? Finally, what do you mean by "shifting $v$ in the direction of $Y'(0)$? –  Renato G Bettiol Jan 22 '12 at 17:24
    
@Benoît: Another remark is that, e.g. in the case of the round sphere, the infinitely many geodesics joining two non-conjugate points are obtained by iterating the periodic geodesic through them. Your construction would seem to give geometrically distinct geodesics (i.e., with different images), which cannot happen in this case. –  Renato G Bettiol Jan 22 '12 at 17:26
    
I agree that the new geodesic may very well end up just being the old geodesic, which should be clear from Renato's example of $S^2$. @Renato [in response to your first comment]: Between any two non-conjugate points, there's a unique Jacobi field with either (a) specified initial value and initial derivative, or (b) specified initial and final values. And $v,Y'(0)\in T_xM$ so it makes sense to add them. –  Aaron Mazel-Gee Jan 23 '12 at 10:09
    
I do not get the objection about the new geodesic possibly being the same than the preceding one, since the proposed process produces arbitrarily long geodesics. The worst that can happen is that the geodesic is in fact periodic, but then we are just in the same case than the spherical one. Either we consider that making $n$ periods gives a different (length-parametrized) geodesic, and we are done in this case, or we consider only the geometric locus of the geodesic, and the result is false as is seen in the case of the round sphere. –  Benoît Kloeckner Jan 24 '12 at 12:46
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