2
$\begingroup$

Problem. I want to prove the following theorem from these notes on Riemannian geometry:

Theorem. Let $(N, h)$ be a Riemannian manifold, let $p\in N$ and let $V$ be an $m$-dimensional linear subspace of $T_pN$. Then there exists (locally) at most one totally geodesic submanifold $M$ of $(N, h)$, such that $T_pM = V$.

Attempt. It turns out this might work better than I first thought, so instead of posting it here, I will try posting my attempted solution as an answer.

$\endgroup$
0
$\begingroup$

Attempted solution. Suppose that $M$ is such a totally geodesic submanifold of $N$, with $T_pM=V$, and suppose that $g$ is the induced metric on $M$. We want to show that, locally, $M$ is completely determined by $(N,h)$ and $V$.

We first introduce a bit of notation:

Notation. For each $v\in T_pM=V$, let $\alpha_v\colon (-a_v,b_v)\to M$ denote the unique maximal geodesic in $M$ that satisfies $\alpha_{v}(0)=p$ and $\alpha_{v}'(0)=v$. Similarly, for each $v\in T_pN$, let $\gamma_v\colon (-\epsilon_v,\eta_v)\to N$ denote the unique maximal geodesic in $N$ that satisfies $\gamma_{v}(0)=p$ and $\gamma_{v}'(0)=v$.

It's a well-known fact [see for instance Wikipedia or Ch. 7 here] that there exists some $r>0$ such that for $B_r(0)=\{v\in T_pM:g_p(v,v)<r\}$, the following restriction of the exponential map, $$\left.\exp_p\right|_{B_r(0)}\colon B_r(0)\to \exp_p(U)\subseteq M\,,$$ defined by $\exp_p(v)=\alpha_v(1)$, is a diffeomorphism. In particular, this means that for the open neighbourhood $\exp_p(B_r(0))$ of $p$, $M$ is determined by the values of $\alpha_v(1)$ for $v\in B_r(0)\subseteq T_pM=V$.

However, it is clear from the fact that $M$ is totally geodesic, that $\alpha_{p,v}$ is a geodesic also in the big manifold $N$, and hence, by the uniqueness of a maximal geodesic with a given derivative in a given point, we must have $\alpha_v=\left.{\gamma_v}\right|_{(a_v,b_v)}$. This means that for $v\in B_r(0)$, $$\exp_p(v)=\alpha_{p,v}(1)=\gamma_{p,v}(1).$$ So in the neighbourhood $\exp_p(B_r(0))$, $M$ is determined by the values of $\gamma_v(1)$ for $v\in B_r(0)\subseteq V$, i.e. completely determined by $(N,h)$ and $V$. $\quad\square$

$\endgroup$

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.