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 $S$.

  • $\begingroup$ Are you assuming that $J$ is orthogonal with respect to the underlying metric? $\endgroup$ Apr 27, 2017 at 19:04
  • $\begingroup$ Do you have any actual examples of such? (other than $\mathbb{C}P^n?$ $\endgroup$
    – Igor Rivin
    Apr 27, 2017 at 19:05
  • $\begingroup$ @IgorRivin the example I have in mind is a cocompact quotient of the noncompact dual $\mathbb{C}H^{n}$ of $\mathbb{C}P^{n}$. $\endgroup$
    – Clark
    Apr 27, 2017 at 19:16
  • $\begingroup$ A product of complex hyperbolic manifolds is not complex hyperbolic, if by complex hyperbolic you mean locally isometric to $\mathbb{CH}^n$. $\endgroup$
    – Ben McKay
    Apr 27, 2017 at 19:19
  • 1
    $\begingroup$ @Clark: Since you aren't assuming that the almost complex structure $J$ is related to the metric in any way, one can construct trivial examples that are not complex hyperbolic space forms: Take a real compact hyperbolic manifold $M^{2n}$ whose tangent bundle admits some almost complex structure $J$. Then, since every tangent $2$-plane is tangent to a totally geodesic surface, there is such a surface tangent to every $J$-invariant $2$-plane. Since the real hyperbolic $2n$-ball is not isometric to $\mathbb{CH}^n$ (even locally), this is a new example. $\endgroup$ Apr 27, 2017 at 22:02

1 Answer 1


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.

  • $\begingroup$ Thanks! I checked the situation I was considering and my structure $J$ is indeed $g$-orthogonal; as edited in my post, I am also assuming that these totally geodesic surfaces are $J$-holomorphic. So this does give a complete answer to the question I originally had in mind. $\endgroup$
    – Clark
    Apr 29, 2017 at 17:49
  • $\begingroup$ @Clark: If you are interested in seeing the details of any of the steps in the above argument, let me know. When I have time, I can put them in; they aren't that hard. $\endgroup$ Apr 30, 2017 at 0:25
  • $\begingroup$ I believe I can fill in the details, although I want to clarify a few things since I don't really know much complex geometry. It looks like the proof could also be arranged to say "nearly Kahler + negative sectional curvature => complex space form" since compact negatively curved Kahler manifolds are all biholomorphic to $\mathbb{C}H^{n}$ and in this case the $J$-holomorphic surfaces are only used at the beginning to get the nearly Kahler condition. The first and third steps appear to be local Riemannian geometry calculations which I can do. $\endgroup$
    – Clark
    May 7, 2017 at 14:36
  • $\begingroup$ I also am curious to know if the theorems of Paul-Andi Nagy as applied here necessarily need to use global properties of the metric as opposed to local calculations. $\endgroup$
    – Clark
    May 7, 2017 at 14:36
  • $\begingroup$ @Clark: If I remember correctly, the result is that a nearly Kähler manifold is locally a product of a Kähler manifold and a strict nearly Kähler manifold, i.e., you don't need compactness or completeness for that. You do need the totally geodesic $J$-holomorphic curves in part three to get constant holomorphic sectional curvature, though. So, yes, I think that the whole thing can be done locally without assuming negative sectional curvature. The change is that now you'll have to add in the $G_2$-invariant nearly Kähler $S^6$ as a (local) solution. $\endgroup$ May 7, 2017 at 15:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.