Verify if a geodesic stays on a submanifold

Question

Let $(M,g)$ be a Riemannian manifold and $N\subset M$ a submanifold.

On $N$ we can put the induce riemannian metric.

Is there a "smart way" to verify if $N$ with the induced metric is totally geodesic?

Answer 1

As Alex M. said, you can compute the second fundamental form to see if it is zero. I think it's also relevant to mention that the fixed point set of an isometry is totally geodesic. This is not so useful if you are handed a submanifold $N$ and want to know if $N$ is totally geodesic (as it is only sufficient but not necessary). But can be useful sometimes if you want to find some totally geodesic submanifolds.

Answer 2

First of all, there is no such thing as $c''(0)$ - second-order differentials are not defined in differential geometry.

Second, what you might try to do is to compute the second fundamental form of $N$ in $M$: it will be $0$ if and only if $N$ is totally geodesic in $M$. This is easy to say in theory, but significantly more complicated to do in practice.