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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?

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  • $\begingroup$ 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 $F(M)$? $\endgroup$
    – Didier
    Dec 4, 2020 at 17:00
  • $\begingroup$ 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 $B$. $\endgroup$
    – Kamil
    Dec 4, 2020 at 17:08
  • $\begingroup$ If $i$ is the inclusion, then $i$ is an embedding $\endgroup$
    – Didier
    Dec 4, 2020 at 17:18

1 Answer 1

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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.

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  • $\begingroup$ 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 $M$? $\endgroup$
    – Kamil
    Dec 4, 2020 at 18:04
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    $\begingroup$ 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 $M$. $\endgroup$
    – Didier
    Dec 4, 2020 at 18:05

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