The statement is: Show that if $M^k$ is a closed, totally geodesic submanifold of $H^n$ (hyperbolic $n$ space), then $M^k$ is isometric to $H^k$.
My friend and I thought to use the "Cartan Theorem". This requires setting notation.
Let $M$ and $\tilde{M}$ be two $n$-dimensional Riemannian manifolds, say $p \in M$ and $\tilde{p} \in \tilde{M}$ are arbitrary points. Choose a linear isometry $i:T_pM \rightarrow T_{\tilde{p}}\tilde{M}$ Let $V \subset M$ be a normal neighborhood of $p$ small enough so that $\exp_{\tilde{p}}$ is defined on $i \circ \exp_p^{-1}(V)$.
Define a map $f:V \rightarrow \tilde{M}$ by $f(q) = \exp_{\tilde{p}} \circ i \circ \exp_p^{-1}(q)$ for $q \in V$.
Now, for all $q \in V$, there exists a unique unit-speed geodesic $\gamma:[0,t] \rightarrow M$ such that $\gamma(0) = p$ and $\gamma(t) = q$. Denote by $P_t$ the parallel transport along $\gamma$ from $\gamma(0)$ to $\gamma(t)$. Define another map $\phi_t:T_qM \rightarrow T_{f(q)}\tilde{M}$ by:
$\phi_t(v) = \tilde{P}_t \circ i \circ P_t^{-1}(v)$ for all $v \in T_qM$
where $\tilde{P}_t$ is the parallel transport along the normalized geodesic $\tilde{\gamma}:[0,t] \rightarrow \tilde{M}$ given by $\tilde{\gamma}(0) = \tilde{p}$, $\tilde{\gamma}'(0) = i(\gamma'(0))$. Finally, denote by $R$ and $\tilde{R}$ the curvature tensors for $M$ and $\tilde{M}$ respectively.
We can now state Cartan's Theorem:
With notation as above, if for all $q \in V$ and all $x,y,u,v \in T_qM$ we have:
$\langle R(x,y)u,v \rangle = \langle \tilde{R}(\phi_t(x), \phi_t(y))\phi_t(u), \phi_t(v) \rangle$
then $f:V \rightarrow f(V) \subset \tilde{M}$ is a local isometry, and $df_p = i$.
Our work:
I've heard that in negative curvature, things like conjugate points aren't a problem, and the injectivity radius is large or infinity, so I figured $\exp$ would be a global diffeomorphism for $H^n$ (which I didn't confirm). The hope was that the $f$ in Cartan's Theorem could give us an isometry onto $H^n$.
We looked at choosing the linear isometry $i$ to be the inclusion of the tangent space of the submanifold $T_pM^k$ restricted to its image in $T_pH^n$. In that case, since the second fundamental form vanishes because $M^k$ is totally geodesic, the curvature tensors are equal via the Gauss formula. $\phi_t$ ends up being the identity (again, vanishing second fundamental form -> same covariant derivatives -> same parallel transport) with this choice of $i$, and so the hypotheses of the theorem are met, and we get that $f$ is an isometry.
Because of the way $f$ is defined, and since we chose the tangent space inclusion as $i$, I believe this only shows that $M^k$ is isometric to its image in $H^n$, which is not enough. I thought that maybe we could choose $i$ to be a different isometry whose image is all of $T_pH^n$, but it seems obvious that this would fail to preserve length, or even be a diffeomorphism if $n \neq k$.
Maybe another approach would be to show that the image of $f$ is clopen in a connected manifold? We aren't sure where to go next. Everything we've done so far has only relied on "totally geodesic", so we need to bring in properties of hyperbolic space somehow. Thoughts?
Edit: I've also realized that they aren't asking for an isometry onto $H^n$, they're asking for an isometry onto $H^k$. I think our proof works when $n = k$ because the tangent space inclusion is a linear isomorphism (if $\exp$ really is a global diffeo). I wonder what you get when exponentiating subspaces of a tangent space in $H^n$? Things that are isometric to lower-dimensional hyperbolic spaces perhaps?