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](https://mathoverflow.net/questions/277560/the-structure-of-banach-manifolds-in-symplectic-geometry). I'll appreciate it if you know any reference on this issue. Thanks.



Find a metric so that a given submanifold is totally geodesic
Suppose $N$ is closed. Let $p:U\to N$ be a tubular neighborhood of $N$ in $M$. So $p:U\to N$ is a vector bundle. Now let $g\_N$ be a Riemannian metric on $N$ and $g\_U$ be a fiber metric on $U$, and let $\nabla$ be a linear connection on the vector bundle $p:U\to N$. Split $TU= V\oplus H$ into the vertical and horizontal bundle over $U$. Put $T(p)^\*g\_N$ on $H$, Put $g\_U$ on $V$ via the affine structure, and declare $V$ an $H$ to be orthogonal everywhere. This gives you metric $\tilde g$ on $U$ so that $p$ is a Riemannian submersion by construction.
Thus horizontal geodesics in $U$ project to geodesics of the same length in $N$, so $N$ is totally geodesic in $U$. Now use a partition of unity argument to get a metric on $M$ which equals $\tilde g$ near $N$. 


