TITLE: Find a metric so that a given submanifold is totally geodesic
QUESTION [8 upvotes]: Let $M$ be a smooth manifold, and $N$ is a submanifold. My question is simply that, does there exist a metric $g$ on $M$ so that $N$ is totally geodesic?
In general the answer might be 'no'. But it might be possible that, when we assumed additional conditions on $M$ or $N$, the answer became 'yes'. This question is kind of related to a previous question here. I'll appreciate it if you know any reference on this issue. Thanks.

REPLY [3 votes]: I happened to think of an alternative proof long after this question was answered, but I decided to post it anyway now.
As mentioned in the comments, if $N$ is properly embedded and has trivial normal bundle then taking any product metric on a tubular neighborhood and extending with a partition of unity does the trick.
If the normal bundle $E$ of $N$ is nontrivial, let $E'$ be an inverse bundle so that $E \oplus E' \cong N \times \mathbb{R}^n$. Again giving $N\times \mathbb{R}^n$ any product metric renders $N\times \{0\}$ totally geodesic in $N \times \mathbb{R}^n$. A glance at the second fundamental form reveals that $N\times \{0\}$ is also totally geodesic in $E$ (identifying $E$ with its image in $N \times \mathbb{R}^n$). Now pull back the induced metric on $E$ to a tubular neighborhood $U$ of $N$, and extend to $M$ with a partition of unity.