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Let $N \subset K \subset M$ be submanifolds with $N$ totally geodesic in $K$ and $K$ totally geodesic in $M$. Show that $N$ is totally geodesic in $M$.

This is trivial if you use the formulation that a submanifold is totally geodesic $\iff$ every geodesic starting from $p$ in a submanifold is also a geodesic in the ambient manifold. But I'm curious how to show this rigorously using fundamental forms. I've been working on this for awhile and here is my attempt.

Use the inclusions $N \hookrightarrow K \hookrightarrow M$ to identify $T_pN \subset T_pK \subset T_pM$ for $p \in N$ and make the following orthogonal decompositions:

(i) $T_pM=T_pN \oplus (T_pN)^{\perp_{M}}$

(ii) $T_pM=T_pK \oplus (T_pK)^{\perp_{M}}$

(iii) $T_pK=T_pN \oplus (T_pN)^{\perp_{K}}$

where by $\perp_{M}$ I mean the orthogonal complement inside $M$.

I've observed the following relations which I think are correct:

$(T_pK)^{\perp_{M}} \subset (T_pN)^{\perp_{K}} \subset (T_pN)^{\perp_{M}}$.

Okay, to show $N$ is totally geodesic in $M$ I need to show that for $\eta \in (T_p N)^{\perp_{M}}$ and $x \in T_pM$ that the fundamental form $H_\eta(x,x)=0$. I thought this would fall out at this point from the inclusions of the orthogonal complements, but the inclusions seem to be going in the wrong direction for what I had in mind.

Am I on the right track here?

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  • $\begingroup$ Your first inclusion is wrong: if $v$ is perpendicular to $K$, then it is NOT tangent to $K$. $\endgroup$ – Jason DeVito Apr 9 '18 at 2:10

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