# Image of a totally geodesic surface under immersion equals image of exponential map?

I was wondering the following. Suppose we have some ambient manifold $$B$$, and we have a totally geodesic surface (meaning second fundamental form vanishes) $$M \subset B$$.

I know from a proposition that due to $$M$$ being totally geodesic, the geodesics of $$M$$ are also geodesics in $$B$$.

Now also suppose that $$B$$ is a complete Riemannian manifold. I want to know the following: Suppose we have an immersion $$F: M \rightarrow B$$. If we let $$p \in M$$ be a point, is it true in general that $$F(M) = \exp_p (T_p M)$$? If so, how can we prove this? Or is it trivial?

• I don't get your question. Do you mean that $M$ is a totally geodesic submanifold, and then you look at another immersion $F(M) \subset B$? If so, why would even $p$ be a point of $F(M)$? Dec 4 '20 at 17:00
• I'm not sure. I am wondering whether there is a way to characterize the image $i(M)$ where $i$ is the inclusion, $i$ is an immersion, and $M \subset B$ is a totally geodesic submanifold of dimension $2$. Then I want to characterize the image $i(M)$ in terms of the Riemannian exponential, which is defined on every tangent space of the ambient manifold $B$. Dec 4 '20 at 17:08
• If $i$ is the inclusion, then $i$ is an embedding Dec 4 '20 at 17:18

Say $$(M',g)$$ is a complete riemannian manifold.
Suppose $$M \subset M'$$ is a totally geodesic submanifold. This means that if $$p \in M$$ and $$v \in T_pM$$, then the geodesic $$\gamma_{p,v} : t \mapsto \exp_p(tv)$$ lies in $$M$$, where $$\exp_p$$ denotes the exponential map of $$M'$$ at $$p$$.
This shows that if $$p \in M$$, $$\exp_p(T_pM) \subset M$$. If $$M$$ is connected, then every point in $$M$$ can be connected to $$p$$ by a geodesic in $$M$$, that is, $$M \subset \exp_p(T_pM)$$, and the the equality follows.
• But I don't understand for example why the geodesic $\gamma_{p,v}$ cannot leave $M$ eventually? Is it not possible that initially it lies in $M$, but then leaves $M$? Dec 4 '20 at 18:04
• Because it is the definition of totally geodesic submanifold : the geodesic in $M$ with initial conditions $p,v$ is by definition a geodesic in $M$. As it is a totally geodesic submanifold, it is also a geodesic in $M'$. By unicity of geodesics, any geodesic in $M'$ locally contained in $M$ stays in $M$. Dec 4 '20 at 18:05