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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 $S$.

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  • $\begingroup$ Are you assuming that $J$ is orthogonal with respect to the underlying metric? $\endgroup$ – Robert Bryant Apr 27 '17 at 19:04
  • $\begingroup$ Do you have any actual examples of such? (other than $\mathbb{C}P^n?$ $\endgroup$ – Igor Rivin Apr 27 '17 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 '17 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 '17 at 19:19
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    $\begingroup$ @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. $\endgroup$ – Robert Bryant Apr 27 '17 at 22:02
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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.

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  • $\begingroup$ 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 '17 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. $\endgroup$ – Robert Bryant Apr 30 '17 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 '17 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 '17 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. $\endgroup$ – Robert Bryant May 7 '17 at 15:35

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