4:08 AM
Garbage, what I mean is, $dE$ has critical points on the geodesics joining $p$ and $q$. $dE_\gamma = 0$ iff $\gamma$ is a geodesic. Now if $p$ and $q$ are conjugate points, then the subspace of $\Omega_{p, q} M$ consisting of all geodesics joining $p$ and $q$ is a finite-dimensional submanifold, whose tangent space is the space of Jacobi fields
Let's work out the relationship with Morse theory actually. What is the Hessian of $E$?
The obvious thing is to choose a two-parameter variation. $hE_\gamma(X, Y) = \partial_v \partial_u E(\gamma_{u, v})$ where $\gamma_{u, v}$ is a 2-parameter variation of $\gamma$ with $\partial_u \gamma_{u, v}(t) |_{u = 0} = X(\gamma_{u, v}(t))$ and $\partial_v \gamma_{u, v}(t) |_{v = 0} = Y(\gamma_{u, v}(t))$
I bet the submanifolds of geodesics are the stable and unstable submanifolds of the Morse function $E$
Let's try. $\partial_u \partial_v \int \|\gamma'_{u, v}\|^2$. I will shorted $\gamma'_{u, v}$ to $\gamma'$ from now on. Differentiate under integral sign: $2 \partial_u \langle \nabla_v \gamma', \gamma' \rangle = 2\langle \nabla_u \nabla_v \gamma', \gamma' \rangle + 2\langle \nabla_u \gamma', \nabla_v \gamma' \rangle$
I can do many switcharoonies, but the question is where should I do it
The second term should obviously be $\langle \nabla_{\gamma'} u, \nabla_{\gamma'} v \rangle$
$\nabla_u \nabla_v \gamma' = \nabla_u \nabla_{\gamma'} v = R(u, \gamma') v + \nabla_{\gamma'} \nabla_u v$
$\langle \nabla_{\gamma'} \nabla_u v, \gamma' \rangle = \partial_t \langle \nabla_u v, \gamma' \rangle - \langle \nabla_u v, \nabla_{\gamma'} \gamma' \rangle$ and the second term dies because $\gamma$ is geodesic, first term dies in the integral by fundamental theorem of calculus and good behavior of $X$ and $Y$ at the endpoints
So the final thing is $hE_\gamma(X, Y) = \int \langle R(X, \gamma') Y, \gamma' \rangle + \langle \nabla_{\gamma'} X, \nabla_{\gamma'} Y \rangle$
Oh shoot okay. If we switcharoonie around a little we get $hE_\gamma(X, Y) = 0$ for all $Y$ iff $X$ is a Jacobi field
It was the nullspace of the Hessian!
Of course that makes sense, I am an idiot.
But then $E$ is not a Morse function. Hessian of $E$ is not nondegenerate at many critical points
But maybe that's the point; it's Morse upto finite rank kernel
Can't expect to have discrete critical points when the domain is infinite dimensional; finite dimensional critical submanifolds is the next best case
I knew all this but forgot lmao. That's got to be what an infinite dimensional Morse function is; the Hessian has finite kernel
Welp time to go back to Milnor
