I'm curious about the following question and have not been able to find any literature on the topic: Suppose that $M$ is a closed negatively curved Riemannian manifold with a "large" quantity of totally geodesic submanifolds. Is there anything one can say about such a manifold?


To be more precise, consider such an $M$ (diffeomorphic to a complex hyperbolic manifold) carrying an almost complex structure $J$ which looks like a complex hyperbolic manifold in the following sense: for each $x \in M$ and each $v \in T\_{x}M$ there is a totally geodesic surface $S$ tangent to the plane spanned by $v$ and $J(v)$ at $x$. It seems to be a natural question to ask if $M$ is itself complex hyperbolic (which would follow for example if $M$ was a Kahler manifold). I have no intuition as to whether this should be an easy or difficult problem so I'm curious (and thankful in advance) if anyone has some guidance/references.


Edit: I have made my question in the second paragraph more explicit below, since I think the original version may be ambiguous, 


Let $X$ be a compact quotient of complex hyperbolic space $\mathbb{C} H^{n}$ ($n \geq 2$). Let $M$ be a Riemannian manifold obtained by taking a small smooth perturbation of the symmetric metric on $X$ (same underlying space, different metrics). Now suppose we are given an almost complex structure $J$ - not necessarily the standard one - such that for each $x \in M$ and each $v \in T\_{x}M$ there is a totally geodesic subsurface $S$ of $M$ which is tangent to the plane spanned by $v$ and $J(v)$ inside of $T\_{x}M$. Is $M$ isometric (up to rescaling) to $X$?


As far as I can tell from searching, Riemannian manifolds typically have very few higher dimensional totally geodesic submanifolds so the condition being imposed above is quite strong. 


Edit 2: As pointed out in the comments, I actually want to assume these subsurfaces are also $J$-holomorphic, i.e., $TS \subset TM$ is preserved by $J$ for each surface $S$. 



Negatively curved manifolds with many totally geodesic submanifolds
Here is a partial answer: Let $(M^{2n},g,J)$ be a compact Riemannian manifold endowed with a *$g$-orthogonal* almost complex structure $J$ with the property that, for every nonzero $v\in T\_pM$ there exists a $J$-holomorphic curve $C\subset M$ passing through $p$ with tangent space spanned by $v$ and $Jv$ that is totally geodesic. Moreover, suppose that the scalar curvature of $g$ is negative. Then, up to a constant scale factor, $(M,g,J)$ is isomorphic to a compact quotient of $\mathbb{CH}^n$.


Note that the only additional hypotheses that I have added to the OP's problem are that the almost complex structure $J$ be $g$-orthogonal *and* that the totally geodesic surfaces should be $J$-holomorphic.


The argument goes as follows: 


First, one shows by a local calculation that, if $(M^{2n},g,J)$ has the desired property of having sufficiently many totally $g$-geodesic surfaces that are $J$-holomorphic curves that there is one tangent to each $J$-complex line in $TM$, then $(M^{2n},g,J)$ must be nearly Kähler.


Second, a complete, simply-connected nearly Kähler manifold is known to be the product of a Kahler manifold and a *strict* nearly Kähler manifold. (See Proposition 2.1 of (*Nearly Kähler geometry and Riemannian foliations*, by Paul-Andi Nagy, arXiv:math/0203038v1). Meanwhile, by a theorem of Paul-Andi Nagy (*On nearly Kähler geometry*, arXiv:math/0110065), if $(N^{2n},g,J)$ is a complete strict nearly Kähler, then the scalar curvature of $g$ is constant and strictly positive. Since we are assuming that our given compact $(M^{2n},g,J)$ has negative curvature, it follows that its simply connected cover cannot have any strict nearly Kähler factor. Hence, $(M^{2n},g,J)$ must be Kähler.


Third, once one is in the Kähler category, it is easy to show that the existence of totally geodesic $J$-holomorphic curves tangent to every complex line implies that the metric has constant holomorphic sectional curvature, i.e., it is a complex 'space form'. Since the curvature is negative, it follows that, up to scale, it must be isometric to the standard Kähler structure on $\mathbb{CH}^n$.


Note: Without the negative curvature assumption, there do exist strictly nearly Kähler examples. E.g., the $6$-sphere with its standard $\mathrm{G}\_2$-invariant metric and (non-integrable) almost complex structure has a totally geodesic $2$-sphere tangent to any given complex line in its tangent space.


