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If $M$ is a complete manifold and $N\subset M$ is a closed, embedded submanifold with the induced Riemannian metric, show that $N$ is complete.

I really don't know where to start. This is not homework, please help! Thank you very much.

  • Hint: Start by showing that a closed subset of a complete metric space, endowed with the induced metric, is a complete metric space.

    Then, in this proof, replace every instance of "metric space" with "complete manifold".

    – Adam Azzam Jan 11 '14 at 03:45
  • Yes, geodesic completeness is equivalent to completeness in the sense of metric spaces, using the induced metric space structure coming from the Riemannian metric. This is (part of) the Hopf-Rinow Theorem. – Ted Shifrin Jan 11 '14 at 03:55
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    @A. Adam Azzam maybe I'm wrong but this is not that simple. The distance function on N induced by the metric of M is not the same as the Riemannian distance of M restricting to N. –  Jan 11 '14 at 04:02
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    @user120543 Yes, but a Cauchy sequences in the two metrics coincide (and you only need one inclusion). – Max Jan 11 '14 at 04:12
  • Oh okay! Thanks a lot, I'm really dumb not to see this. –  Jan 11 '14 at 04:15

1 Answers1

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First, the intrinsic topology of $N$ agrees with the subspace topology of $N$. Next, you need to observe that the Riemannian distance functions $d_N, d_M$ are related (on $N$) by the inequality $$ d_N\ge d_M. $$ Now, apply the Cauchy criterion for completeness as well as the fact that $N$ is closed in $M$.

Moishe Kohan
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  • Let $g, \hat g$ be inner product w.r.t. $M, N$ respectively. Every Cauchy sequence ${p_n}$ w.r.t. $\hat g$ is a Cauchy sequence w.r.t. $g$, so $p_n \rightarrow p_0$ for some $p_0 \in P$ w.r.t. $g$. However, this doesn't show that $p_n \rightarrow p_0$ w.r.t. $\hat g$. Any idea how to elaborate further? – James C May 16 '21 at 20:04
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    @PaulPogba: Use the first sentence of my answer and the definition of the subspace topology. Also, $P=N$ since $N$ is a closed subset of $M$. – Moishe Kohan May 16 '21 at 20:28