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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.

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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 $V$. $\quad\square$

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