Mostow rigidity for complex hyperbolic manifolds

A Riemannian manifold $$(X,g)$$ is hyperbolic if the sectional curvatures are constant and negative. A theorem of Mostow says that these manifolds are determined by their fundamental group.

Theorem (Real Mostow Rigidity) If $$X$$ and $$Y$$ are closed, hyperbolic $$n$$-manifolds with $$n \ge 3$$ and $$\pi_1 X \simeq \pi_1 Y$$ then there is a Riemannian homothety $$\varphi:X \simeq Y$$.

A Riemannian homothety is a diffeomorphism $$\varphi:X \simeq Y$$ with $$\varphi^*g_Y = \lambda \cdot g_X$$ for some $$\lambda > 0$$.

There is a clear analogue of the hypothesis of Mostow rigidity in the Kahler setting. Namely, a Kahler manifold $$(X,\omega,J)$$ is complex-hyperbolic if the holomorphic sectional curvatures are constant and negative. For a definition of the holomorphic sectional curvatures, see (for instance) Hyperbolic Complex Spaces by Kobayashi.

Main Question: My question is whether or not the following Kahler analogue is true.

Theorem (?) (Complex Mostow Rigidity) If $$X$$ and $$Y$$ are closed, complex-hyperbolic $$2n$$-manifolds with $$\pi_1 X \simeq \pi_1 Y$$ then there is a Kahler homothety $$\varphi:X \simeq Y$$.

A dimensional restriction may be necessary here, for instance $$n \ge 2$$ or $$n \ge 3$$. Again, I would like to know if this theorem is true and if so, where I can find a reference.

• Yes, Mostow rigidity covers all irreducible locally symmetric spaces of non-compact type and dimension $\ge 3$. – YCor Dec 10 '18 at 20:27
• (Note: you have to change "isometry" to "homothety" in your statement of Real Mostow rigidity, and idem in the complex case) – YCor Dec 10 '18 at 20:27
• ah yeah, it should be isometry up to scale. I'm not sure why ordinary Mostow rigidity should cover the case I'm considering here, since complex-hyperbolic spaces are not necessarily hyperbolic in the ordinary sense. The sectional curvatures are allowed to not be constant on 2-planes that are not complex (roughly speaking). Anyway, if you can clarify in an answer that would be great! – Julian Chaidez Dec 10 '18 at 20:39
• Who told you that Mostow rigidity is only in the "hyperbolic" setting? Oh I see, the Wikipedia article... don't rely too much on Wikipedia. No idea why the author of the Wikipedia article focusses on the constant curvature case. Mostow proved his result for irreducible locally symmetric spaces of non-compact type and dimension $\ge 3$. – YCor Dec 10 '18 at 20:44
• I mean in my previous comment that I don't think that the Wikipedia version is the "ordinary" one. You need nothing about symmetric spaces, except to know that complex hyperbolic spaces (on which complex-hyperbolic manifolds are modeled) are symmetric spaces. – YCor Dec 10 '18 at 21:11

Let $$G$$ (resp. $$G^\prime$$) be a semi-simple analytic group and $$\Gamma$$ (resp. $$\Gamma^\prime$$) an irreducible lattice in $$G$$ (resp. $$G^\prime$$). Assume that $$G, G^\prime$$ have trivial centers and no compact factors and $$G$$ is not locally isomorphic to $$SL(2,{\mathbb R})$$. Then any isomorphism $$\theta\colon \Gamma\to\Gamma^\prime$$ extends to an analytic isomorphism of $$G$$ onto $$G^\prime$$.
The case of complex hyperbolic manifolds corresponds to $$G=G^\prime=SU(n-1,1)$$. An analytic isomorphism is an isometry and Kähler.
• If you just want to classify simply connected manifolds of constant sectional curvature, this is known by Élie Cartan, there are only $S^n$, $E^n$ and $H^n$. (And you don‘t need the Kähler condition.) If you are asking for constant holomorphic sectional curvature, you have complex projective space, $C^n$ and complex hyperbolic space. The proof that there are no others is in Hawley: „Constant holomorphic sectional curvature“, Can. Math. J. 5 (1953), 53-56. – ThiKu Dec 11 '18 at 21:25