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.



Is there a unified reason that there are an infinite number of geodesics between nonconjugate points on a compact manifold?
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:
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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 $M$.




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An interesting remark:
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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.


