MathOverflow will be down for maintenance for approximately 3 hours, starting Monday evening (06/24/2013) at approximately 9:00 PM Eastern time (UTC-4).

2 clarified wording

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 for 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.

1

# Is there a unified reason that there are an infinite number of geodesics between nonconjugate points on a compact manifold?

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 for $H_m(\Omega M)=0$ for $m\geq N$. 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.