I am trying to get through the proof of the index theorem.

The background: I have been stuck for quite a while on the following point which Milnor says is evident:
Let $\gamma: [0,1]\rightarrow M$ be a geodesic in a Riemannian manifold. Let $(t_0=0, t_1,…,t_k=1)$ be a partition of $[0,1]$ so that $\gamma$ sends $[t_i, t_{i+1}]$ into an open set U with the property that any two points in U can be connected by a distance minimizing geodesic which depends smoothly on the two endpoints. If $\tau$ is between $t_j$ and $t_{j+1}$ then the space of "broken" Jacobi fields along $\gamma|_{[0,\tau]}$ (i.e. those piecewise smooth $V$ which are Jacobi fields along each piece of the partition of $[0, \tau]$) which vanish at $t=0$ and $t=\tau$ is isomorphic as a real vector space to the direct sum $T_{\gamma (t_1)}M\oplus...\oplus T_{\gamma (t_j)}M$ . Call this latter sum $\Sigma$. Then the Hessian of the energy function associated with $\gamma|_{[0,\tau]}$ (call it $E_\tau$) can be viewed as a bilinear form on $\Sigma$.

My question: I want to know why this bilinear form should vary continuously with $(\tau, V, W) \in (t_j,t_{j+1})\times \Sigma \times \Sigma$. I.e. if $V_\tau$ and $W_\tau$ are the broken Jacobi fields along $\gamma|_{[0,\tau]}$ associated with $V, W\in \Sigma$, why is $(t_j,t_{j+1})\times \Sigma \times \Sigma \rightarrow \mathbb{R}, (\tau, V, W) \mapsto E_\tau (V_\tau , W_\tau )$ continuous?
Where I am struck: From the second variation formula it seems as though I should start by proving that $D_t(V_\tau|_{[t_j,\tau]} )|_{t_j}$ varies continuously with $(\tau, V)$. I'm having trouble showing this though.


After Sleeping for 12 Hours: This seems like a possible solution. I'd be grateful for any corroboration because I wouldnt expect Milnor to dismiss something as "evident" if this were the easiest way to prove it:

Using the second variation formula it seems as though the problem reduces to the following: Let $\gamma: [0,1]\rightarrow U$ be a geodesic s.t. any two points in $U$ are connected by a minimizing geodesic which depends smoothly on the endpoints. Show that $(0,1)\times T_{\gamma(0)}\rightarrow T_{\gamma(0)}, (\tau, V)\mapsto D_t(V_\tau)|_{t=0}$ is continuous (where $V_\tau$ is the unique Jacobi field along $\gamma$ with $V_\tau(0)= V, V_\tau(\tau)=0$).

$proof$: After choosing a parallel orthonormal frame along $\gamma$, Jacobi fields are just projections of integral curves to a vector field on $[0,1]\times \mathbb{R}^{2n}$. If $0<\tau_0<1$ then by the ODE theorem and the linearity of the Jacobi equation, there exists $\epsilon>0$ such that

$\Theta: (\tau_0-\epsilon,\tau_0+\epsilon)\times \mathbb{R}^{2n}\rightarrow (\tau_0-\epsilon,\tau_0+\epsilon)\times\mathbb{R}^{2n}$

$(\tau, V,W) \mapsto (\tau, X_{t=\tau}, D_t(X)|_{t=\tau})$

is a smooth map. Where $X$ is the unique Jacobi field along $\gamma$ with $X_{t=0}=V, D_t(X)|_{t=0}=W$ (All of these vectors are coordinates w.r.t. the parallel frame). Now postcompose $\Theta$ with projection onto $X_{t=\tau}$ and call this composition $\theta$. Then I claim that the implicit function theorem applies for the zero vector in the image of $\theta$. Specifically for any $V_0\in \mathbb{R}^n$ let $W_0$ such that $\theta(\tau_0,V_0,W_0)=0$ (that such a vector $W_0$ exists is a fact about Jacobi fields along geodesics in open sets like $U$). The hypotheses of the implicit function theorem are met because $W\mapsto \theta(\tau_0, V_0, W)$ is a linear isomorphism. So we get a smooth map $(\tau_0-\delta, \tau_0+\delta)\times B(V_0, \delta)\rightarrow \mathbb{R}^n$ which gives the (coordinates of the) unique vector in $T_{\gamma(0)}M$ which is the covariant derivative of the Jacobi field vanishing at $\tau\in (\tau_0-\delta, \tau_0+\delta)$ and whose value at $t=0$ is $V\in B(V_0, \delta)$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.