Preliminaries: Let $(X,\omega,J)$ be a closed Kahler manifold. That is, $X$ is a closed $2n$-manifold, $\omega$ is a symplectic form and $J$ is a compatible (integrable) complex structure. Suppose that $[\omega] = \lambda \cdot c_1(X,J)$ with $\lambda \in \mathbb{R}$, where $[\omega] \in H^2(X)$ is the de Rham class of $\omega$ and $c_1(X,J)$ is the 1st Chern class of $X$.

When $\lambda \le 0$, famous theorems of Yau and Aubin (in the $\lambda < 0$ case) state that there exists a unique Kahler (symplectic) form $\omega_{KE}$ on $X$ that is compatible with $J$ and that satisfies the Kahler-Einstein condition $\omega_{KE} = \lambda \cdot \text{Ric}[\omega_{KE}]$. Here $\text{Ric}[\omega_{KE}]$ is the Ricci form.

Main Questions: My question is essentially: how unique is a Kahler-Einstein complex structure if it is compatible with a fixed symplectic form? More precisely, let $(X,\omega)$ be a closed symplectic manifold and let $J$ be a compatible complex structure with respect to which $\omega$ is Kahler-Einstein.

Question 1: Is $J$ unique in the sense that, for any other compatible Kahler-Einstein $J'$ on $(X,\omega)$, there is a biholomorphic symplectomorphism $\varphi:X \to X$, i.e. a diffeomorphism $\varphi$ such that $\varphi^*\omega = \omega$ and $T\varphi \circ J = J' \circ T\varphi$?

I would also be interested in the following restricted cases of Question 1.

Question 2: Is Question 1 true if we restrict to either the case where $[\omega] = -c_1(X,J)$? What if we assume $\text{dim}_{\mathbb{C}}(X) = 2$? What if we assume both?

Finally, I would be interested in other curvature conditions guaranteeing a uniqueness property in the spirit of Question 1. For instance, we could ask the following.

Question 3: Let $(X,\omega,J)$ be a Kahler manifold with constant holomorphic sectional curvature. Then is $J$ unique in the sense of Question 1?

  • $\begingroup$ What about Riemann surfaces? Any two area forms that integrate one are symplectomorphic, however there is a moduli of hyperbolic unit volume metrics $\endgroup$ Dec 10, 2018 at 23:26
  • $\begingroup$ that's a good point, without a dimensional restriction Question 1 is definitely false even when $\lambda < 0$. maybe I'll add that as an edit. $\endgroup$ Dec 11, 2018 at 0:16

2 Answers 2


Concerning 2) one can, of course, take the product of two curves of higher genus to get a counter-example. In general, to have a statement as you want, one should look for rigid complex surfaces of general type (with ample $K$), because as soon as such a surface has a deformation, we get a counterexample (by Yau Aubin).

Concerning 3), the situation depends on whether the curvature is zero or not. The universal cover of a manifold of constant holomorphic sectional curvature is either i) $\mathbb CP^n$ with Fubini-Study metric, or ii) $\mathbb C^n$ with a flat metric or iii) the open unit ball $\mathbb B^n$ with complex hyperbolic metric. See the discussion here (don't be mislead by the title :) ):

Kahler manifolds with constant bisectional curvature

In the case i) it is a well known theorem, that a Kaehler manifold diffeomorphic to $\mathbb CP^n$ is in fact biholomoprhic to $\mathbb CP^n$. So the answer to your question is positive.

In the case ii) we can take two elliptic curves with a flat metric of area $0$, that are not biholomorphic. So the answer to the question is negative.

Finally, I believe that the answer is also negative in case iii), provided the dimension is greater than $1$. This is certainly true for complex dimension $2$. Indeed, by Yau's theorem any complex surface diffeomorphic to a ball quotient is in fact biholomorphic to it, and also the complex structure on a ball quotient is rigid.

  • $\begingroup$ Excellent, this is all very helpful. I probably should have thought about Questions 1 and 2 more before asking, Riemann surfaces seem to be a good source of counter-examples there. $\endgroup$ Dec 12, 2018 at 1:47
  • $\begingroup$ For the case (iii), do you mean that the answer is positive for negative sectional curvature? If so, I think it might follow from Mostow- rigidity applied to the isometry group of the complex hyperbolic ball $\mathbb{B}^n$. This was actually my motivation for this question: mathoverflow.net/questions/317362/mostow-rigidity-for-complex-hyperbolic-manifolds. $\endgroup$ Dec 12, 2018 at 1:47
  • $\begingroup$ Julian, concerning your second comment, what I wrote is different from Mostov rigidity. The claim is the following: if a complex 2-surface is homeomorphic to a ball quotient, then it is byholomorphic to it. Mostov rigidity is a statement about two ball quotients - if they have the same fundamental group they are isometric. But a priori one could imaging that there is a complex surface that is diffeomorphic to a ball quotient, but is not byholomorphic to a ball quotient. Yau's theorem rules this out. $\endgroup$ Dec 12, 2018 at 9:14

Let $X$ be a smooth complex del Pezzo surface which forms non-trivial family (for example the cubic surface). Then by a Theorem of Tian every variety in the family will have a KE-metric ( Tian, G. On Calabi’s conjecture for complex surfaces with positive first Chern class, Invent. Math. 101 (1990), no. 1, 101–172).

Note that Tian's theorem states that all del Pezzo surfaces which are not a blow-up of $\mathbb{P}^{2}$ in 1 or 2 points have a KE-metric, but the 1 and 2 point blow-ups come in trivial families.

It is proven in Example 3.9 here (https://people.math.ethz.ch/~salamon/PREPRINTS/unique.pdf), that for any two symplectic forms $\omega_{1},\omega_{2}$ on a del Pezzo surface $X$ such that $[\omega_{1}] = [\omega_{2}] \in H^{2}(X,\mathbb{R})$ there exists a diffeomorphism $f : X \rightarrow X$ such that $f^{*}(\omega_{2}) = \omega_{1}$ (see the conclusion at the bottom of page 16). Hence the underlying symplectic form of each of these KE manifolds are the same, but they cannot be identified biholomorphically, whilst simultaneously preserving the symplectic structure. This provides a negative answer to question 1.

  • $\begingroup$ That's a nice way of generating counter-examples for the case where $\lambda > 0$. Maybe there's still some hope for Questions 2 and 3. $\endgroup$ Dec 10, 2018 at 19:30

Your Answer

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