Suppose that $N$ is a totally geodesic submanifold of a complete Riemannian manifold $(M,g)$. Is it the case that a geodesic segment that minimizes length in the submanifold $N$ also minimizes length in the ambient manifold $M$?
riemannian-geometry – Totally Geodesic Submanifolds in Riemannian Geometry
riemannian-geometry
Related Solutions
Contrary to another comment/answer, totally-geodesic $1$-manifolds in $S^3$ are not trivial because they can be disconnected. Great circles do not have to intersect, as can be seen by taking generic planes through the origin in $\mathbb{R}^4$.
Geodesics sometimes can't be isotoped to fibers of the same Hopf fibration. Great circle links were the subject of G. Walsh's thesis. One chapter classified great circle links up to $5$ components. There are $1,1,2,3,7$, respectively.
In higher dimensions, there are disconnected geodesic submanifolds of $S^n$ of dimension up to $(n-1)/2.$ The only place where you can have interesting linking is with totally geodesic $d$-spheres in $S^{2d+1}$, and I think the link theory is always nontrivial in that dimension.
You should be aware that, for $n\ge3$, the generic Riemannian metric $(M^n,g)$ has no totally geodesic hypersurfaces at all, even locally. Typically, when they do exist, it is for some geometric reason that makes them obvious. For example, if $(M^n,g)$ admits an isometry $\iota: M\to M$ of order $2$ (i.e., an involution) whose fixed point set is a hypersurface $H^{n-1}\subset M$, then $H$ will be totally geodesic (and it will be complete if $g$ is complete).
As for topological constraints, there really aren't any, because, for any closed, embedded hypersurface $H^{n-1}\subset M^n$, you can always construct a metric $g$ such that $H$ is totally geodesic in $(M,g)$; just take the metric to be a product metric on some tubular neighborhood of $H$.
As a practical matter, if someone hands you a Riemannian manifold $(M^n,g)$ that is described sufficiently explicitly that you can compute the various curvature tensors and functions to arbitrary orders and determine their vanishing loci, there is an algorithm to follow that will determine all of the totally geodesic hypersurfaces in $(M^n,g)$, and that is probably the best you can do in terms of giving explicit 'constraints' on $(M^n,g)$ for it to admit totally geodesic hypersurfaces (complete or not). The algorithm basically, goes like this: Let $\pi: \Sigma^{2n-1}\to M^n$ be the unit sphere bundle of $M$ with respect to the metric $g$. Then, using the Levi-Civita connection, construct an $(n{-}1)$-plane field $D\subset T\Sigma$ such that a curve $\nu:\mathbb{R}\to \Sigma$ of the form $\nu(t) = \bigl (x(t), e(t)\bigr)$, where $e(t)\in T_{x(t)}M$ is a unit vector, is tangent to $D$ if and only if $x'(t)\cdot e(t)=0$ and $e(t)$ is parallel along $x(t)$, i.e., $\nabla_{x'(t)}e(t) = 0$. Then a hypersurface $H\subset M$ is totally geodesic if and only if its 'unit normal graph' in $\Sigma$ is everywhere tangent to $D$. (It is easy to show that this plane field $D$ is Frobenius if and only if $(M^n,g)$ has constant sectional curvature, which explains why only those manifolds have totally geodesic hypersurfaces through every point perpendicular to every direction.) Now the problem is reduced to finding the $(n{-}1)$-manifolds in $\Sigma$ that are tangent to $D$, which is a standard problem.
It is probably worth summarizing what this process says about Riemannian $3$-manifolds $(M^3,g)$. When one goes through the above analysis, the first thing it tells one is that, if $H^2\subset M$ is a totally geodesic hypersurface with normal vector field $\nu:H\to \Sigma$, then, for all $x\in H^2$, the normal vector $\nu(x)\in T_xM$ must be an eigenvector of the Ricci curvature $\mathrm{Ric}(g)$. In particular, if $\mathrm{Ric}(g)$ has distinct eigenvalues everywhere (or, what would be generic, on a dense open set $U\subset M$), then one can locally write $$ g = {\omega_1}^2+{\omega_2}^2+{\omega_3}^2 \quad\text{and}\quad \mathrm{Ric}(g) = \lambda_1\,{\omega_1}^2+ \lambda_2\,{\omega_2}^2+ \lambda_3\,{\omega_3}^2 $$ for some functions $\lambda_1 < \lambda_2 < \lambda_3$ and some coframing $\omega_i$, and one of the $\omega_i$ will have to vanish on $H$. However, if $\omega_i$ vanishes on $H$, then $\mathrm{d}\omega_i$ must also vanish on $H$. Defining the functions $f_{ij}$ that satisfy $\omega_i\wedge\mathrm{d}\omega_j = f_{ij}\,\omega_1\wedge\omega_2\wedge\omega_3$, one sees that $f_{ii}$ must also vanish on $H$.
In the generic situation, each of the three equations $f_{ii} = 0$ will define a surface $H_i$ in $M$, and these $H_i$ are the only possible totally geodesic surfaces in $M$. In fact, $H_i$ will be totally geodesic if and only if $\omega_i$ vanishes on $H_i$ along with the three functions $f_{jk}$, $f_{kj}$ and $f_{jj}{-}f_{kk}$, where $(i,j,k)$ is a permutation of $(1,2,3)$. Thus, this provides the effective test in the case that the Ricci tensor has distinct eigenvalues.
In the degenerate case of multiple eigenvalues (or, say, that $f_{ii}=0$ does not define a surface in $M$), more differentiation is required to formulate an explicit test of this nature, but I'll leave that to you.
Best Answer
Let $M$ be the flat cylinder $\mathbb{R} \times S^1 \subset \mathbb{R} \times \mathbb{C}$ and $N = \{(t,e^{it}) | t \in \mathbb{R}\}$, which is a geodesic (hence a complete totally geodesic submanifold of $M$) minimizing between any two points of $N$ (among the geodesics of $N$). But the minimizing geodesic in $M$ between the points $(0,1)$ and $(2\pi,1)$ is the segment $\{(s,1) | s\in[0,2\pi]\}$.