# Find a metric so that a given submanifold is totally geodesic

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.

• I would guess that the answer is yes. For example, if $N$ has a trivial normal bundle, you could take any Riemannian metric on $N$ cross with the standart metric on a disc. To get a metric on the whole of $M$ you can use convex combinations near the boundary of the normal bundle and use partitions of unity and so on. This does not change the metric near $N$, so that it stays a totally geodesic submanifold. – HenrikRüping Jul 30 '17 at 16:45

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 $N$.
• or simply take an involution of $i: U\to U$ where point $q=exp_p(v)$ goes to $exp_p(-v)$ ($p\in N$ is nearest to $q$); and take a metric which is sum $g+i^*g$, then continue outside $U$ arbitrarily. Then, easy exercise, $N$ totally geodesic. – valeri Jul 30 '17 at 18:38
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.