Assume we are on a smooth, complete Riemannian manifold $(M,g), dim(M) \geq 3$. What are the specific geometric/topological constraints for such a manifold to admit complete, totally geodesic hypersurfaces? Please, I admit that I'm a rookie so any simple illustrations and sources are more than welcome, especially for the case of lowest dimension $(dim(M) = 3)$.
Existence of Totally Geodesic Hypersurfaces in Differential Geometry
dg.differential-geometrydifferential-topologyriemannian-geometry
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Edited 2 January 2011 to clarify:
I don't know if there is an established meaning for general position in this context: it would be nice to have a clarification of terminology (as we've seen from the comments discussion). One possible definition is that a set of subspaces is in topological general position if for any small perturbation, there is a homeomorphism taking one configuration to the other. For 2-planes in $\mathbb R^4$, this just means that any two intersect in a point. If you intersect the subspaces with a small sphere about the origin, this gives a great circle link in $S^3$; for any $n$, there is a finite set of possible link types for these links.
Example for topological general position
For topological general position, there is a relatively simple example satisfying the question: take $n$ complex 1-dimensional subspaces of $\mathbb C^2$. The intersection with $S^3$ gives $n$ circles in a Hopf fibration. Now modify the Euclidean metric of $\mathbb C^2$ by changing the metrics of concentric spheres to homogeneous metrics on $S^3$ obtained by keeping the Hopf fibers rigid, but uniformly expanding their orthogonal planes by an increasing function of radius $r$. These metrics are invariant by $U(2)$, with nonpositive curvature. (This behaviour is closely related to the geometry of $\mathbb{CP}^2$ and of complex hyperbolic space $\mathbb {CH}^2$: the family of concentric spheres in either case has this same family of metrics on $S^3$ up to a constant that depends on $r$).
Possible stronger requirements for general position
At first it may be tempting to strengthen the condition of topological general position to require that for any small enough perturbation there is a linear map carrying one pattern to another, but this condition is unreasonably strong. The linear equivalence condition is satisfied for almost every triple of 2-dimensional subspaces of a 4-dimensional vector space, since two of the subspaces can be transformed into orthogonal coordinate planes, whereupon the third subspace is the graph of a map from one to the other; by a linear transformation, this can be arranged to map basis vectors to basis vectors. When there is a fourth subspace, the map associated with the third followed by the inverse of the map associated with the fourth is a linear map from a plane to itself, and its characteristic polynomial is an invariant that varies in any open neighborhood of configurations. Note, however, that there are open sets of quadruples of 2-dimensional subspaces for which this self-map has a pair of complex conjugate eigenvalues; when this happens, the quadruple of subspaces is linearly equivalent to a quadruple of complex 1-dimensional subspaces of $\mathbb C^2$.
There are many possible intermediate conditions that could be imposed, depending on what set of special situations you focus on. If you consider all linear maps defined by triples of planes, as above, you might want to require that the rank of any composition is constant in a neighborhood. However, for many (most) configurations of $n$ planes, when $n$ is high enough, these generate a dense set of linear maps among the planes, and small perturbations can change the rank of something. A more plausible requirement is that the dimension of the Zariski closure of the set of all compositions of these maps is constant, or perhaps that the dimension of the Zariski closure of the set of all compositions associated with triples in a subset remains constant. But this definition is special to the setting of half-dimensional subspaces of an even-dimensional space so it's not ideal. It's not hard to modify this for a more general setting, but there are too many possible choices: it's not self-evident that there is one `right' definiton.
One way around the issue is to rephrase the question: is there an open set, or a set of positive measure, or a generic set (intersection of open dense sets) of $n$ 2-dimensional subspaces of $\mathbb R^4$, such that ___ [a given property is true].
Examples for an open set of subspaces
Now make a random $C^2$-small perturbation of the Hopf fibration to be a foliation of $S^3$ by geodesics, so that the perpendicular plane field $\tau$ is still a contact plane field. We can construct new metrics in a neighborhood of the origin in $\mathbb R^4$ by modifying only in the 2-planes orthogonal to the family of 2-planes that are cones on these geodesics, scaling these planes by multiplication by a function $f(r)$ that is concave upward (to aim for negative curvature). If $f$ has $C^\infty$ contact to the constant function 1 at $r = 0$, the resulting metric is smooth. It is flat in each plane that is a cone on leaf of the foliation.
I think that for appropriately chosen $f$ these metrics have nonpositive sectional curvature, but I'll need to come back to this later to back it up (or tear it down), unless someone else will do it.
The first question is false as stated. By Artin's encoding, geodesics on $SL_{2}(\mathbb{R})/SL_{2}(\mathbb{Z})$ corresponding to continued fractions, and the geodesic flow corresponds to the shift. It's easy to find one fraction where you'll see any given prefix (hence dense), but you won't be equidistributed (say think about larger and larger blocks composed out of $1$'s).
The situation is the same even for cocompact (hyperbolic) homogeneous spaces, and relays on the fact that the corresponding dynamical system is a Bernoulli system, see for example the survey by Katok in the Clay Pisa proceedings for more information about the encoding.
In the case where the manifold is a Nilmanifold, the answer is indeed true, which follows from say Furstenberg's theorem about skew-products (when you use both the topological version and the ergodic version). Finer (quantitative) results are probably attained by Green-Tao (see Tao's post about the Nilmanifold version of Ratner's theorem). In the toral case, this boils done to merely Fourier series computations and Weyl's equidistribution criterion or so.
In the higher rank (semisimple) case, things get more complicated, as one might think about multi-parameter actions, and then the measure-classification theorem by Lindenstrauss kicks in, but it was observed by Furstenberg in the $60$'s (and maybe before that) that even for multi-parameter actions, there might be dense but not equidistributed orbits. Maybe the easiest toy model to think of is to think about the multiplicative action of $<2,3>$ as a semi-group on the torus $\mathbb{R}/\mathbb{Z}$, and start at say a Liouville number for base $6$. This action is some $S$-adic analouge for a higher-rank multi-parameter diagonalizable action.
Edit - to address the revised question, here the geometrical settings are being addressed more intimately. In the case of homogeneous spaces ($G/\Gamma$, or you can take the appropriate locally symmetric space as well), where $G$ is semi-simple say, then the geodesic flow is ergodic (it follows for example from the Howe-Moore theorem, or from the Bernoullicity theorem I've mentioned above). As a result, a simple application of the pointwise ergodic theorem will tell you that for almost every point and every direction (the approperiate measures here will be the Liouville measure on the unit tangent bundle, which is really where the geodesic flow "lives"), the orbit is equidistributed. For the variable curvature case, as long as some natural conditions are met (say an upper bound on the sectional curvature making it negative everywhere), the dynamical picture is pretty much the same (but the proofs are significantly more involved, as you don't have rep. theory at hand). Again in the Nilmanifold case, the situation is much more simple, the toy model for that is tori, where the question of rationality implies both density and equidistribution.
I will address the Andre-Oort question in the comments, as I'm not an expert on this subject.
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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.