Let $X$ be a compact Kahler manifold with $c_1(X) = 0$. Any Kahler metric $\omega$ on $X$ gives a Laplacian $\Delta_\omega$ and the $(1,1)$-form $\omega$ is harmonic with respect to this Laplacian.

  1. Take $\omega$ to be Ricci-flat. Is any other Ricci-flat metric $\alpha$ harmonic with respect to $\omega$?

  2. If this is not true in general, and $X$ is a manifold on which this holds, then is $X$ a quotient of a torus?

The first question has a positive response in the case of tori, which is where it comes from in the first place. There the Ricci-flat metrics are exactly those Kahler metrics that have constant coefficients, and the $(1,1)$-forms which are harmonic with respect to a such a metric have constant coefficients. So a Ricci-flat metric on a torus is harmonic with respect to any other Ricci-flat metric.

  • $\begingroup$ Hey little brother, nice question! I don't have much time, but as soon as possible I'll think about it! Good night! $\endgroup$ – diverietti Dec 5 '10 at 23:06
  • $\begingroup$ I'm starting to think the second question is true. If you reverse the question, then you're looking at a manifold $X$ which admits a Kahler metric $\omega$ which is harmonic with respect to every Ricci-flat $\alpha$. I have a feeling this imposes so many conditions on the metric $\omega$ that it must in fact be flat, thus the manifold would be a (quotient of) a torus. $\endgroup$ – Gunnar Þór Magnússon Dec 9 '10 at 15:12

First let me give some cheap examples of CY manifolds that satisfy condition of 1) while they are not tori quotients.

Example. Let $X$ be any CY manifold with $h^{1,1}(X)=1$, for example $X$ can be a quintic in $\mathbb CP^4$. Then the conclusion of 1) holds, because all Kahler Ricci flat metric on $X$ are proportional. Now, to get an example with $h^{1,1}(X)=2$ one can just take a direct product $X_1\times X_2$ of CY manifolds satisfying $h^{1,1}(X_i)=1$.

So the list of manifolds with $c_1=0$ satisfying 2) is larger than tori quotients. But maybe all these examples can be understood. For example, it should be easy to show that a $K3$ surface does not satisfy 1). I'll give a brief proof in the case of a generic K3 that does not have $-2$ curves.

EDDITED. I put more details here so that it becomes clear that was said previously is correct.

The good thing about harmonic forms is that the sum of two harmonic forms is harmonic. By Nakai-Moishenzon (for Kahler surfaces), the Kahler cone of a K3 without $-2$ curves coincides with a connected component of $(1,1)$ classes with positive square in $H^{1,1}$. Now, by Yau in any such class there is a unique Ricci-flat metric. And also obviously there exists a unique harmonic metric. Suppose that assumption 1) holds for our K3. Then it is obvious that this means that each harmonic form $w$ with $\int_{K3}w^2>0$ in the correct component of the cone is Ricci flat. Let me get a contradiction from this

So let $w_1$ and $w_2$ be two Kahler Ricci-flat forms harmonic with respect to the metric defined by $w$ on a $K3$ surface, and moreover that $aw_1+bw_2$ is Ricci flat again. Then we know that $(aw_1+aw_2)^2$ is proportional to $\Omega\wedge \bar\Omega$, where $\Omega$ is the complex volume form on $K3$. Consider now the family of forms $w_1-tw_2$, $t>0$. Take the first moment $t_0$ when we have $\int_{K3}(w_1-t_0w_2)^2=0$. But since $(w_1-tw_2)^2=c_t\Omega \wedge \bar\Omega$ for some $c_t>0$ for all $t$ less than $t_0$ (since this form is Ricci flat) we conclude that $(w_1-t_0w_2)^2$ is equal to zero point-wise. So the kernel of $w_1-tw_2$ should define a holomorphic folitation on $K3$. Since for $K3$ it holds $h^{1,1}=20$, we should get a tremendous amount of holomorphic foliations on it, but this is clearly impossible.

This is still not a 100% complete reasoning, but I am sure it can be completed. So it seems to me that the complete list of manifolds satisfying 1) are all manifolds that are finite quorients of a torus times a collection of CY manfiolds with $h^{1,1}=1$. The idea is simple: whenever you have two Ricci-flat forms such that $w_1+tw_2$ is Ricci flat for small $t$, this should lead to a local metric splitting of the manifold into direct product. Then we should just use De-Rahm decomposition theorem.

  • $\begingroup$ I don't understand the 4th paragraph. You have $w_1$ and $w_2$ are Ricci-flat Kahler forms, harmonic with respect to $w$. Then it is true that for any $t$, $w_1 - t w_2$ is still harmonic with respect to $w$. However, how do you know that the value of $t$ which makes $\int_{K3} (w_1 - t w_2)^2 = 0$ is one for which $w_1 - t w_2$ is not only positive definite, but also Ricci-flat? To conclude that $(w_1 - t w_2)^2 = c \Omega \wedge \bar \Omega$ for a constant $c$, one needs to know that the metric associated to $w_1 - t w_2$ is still Ricci-flat. $\endgroup$ – Spiro Karigiannis Dec 12 '10 at 0:05
  • $\begingroup$ But your explanation in the first two paragraphs about why (2) is not true if very nice and correct. $\endgroup$ – Spiro Karigiannis Dec 12 '10 at 0:05
  • $\begingroup$ Spiro, I edited the answer. I assumed, that the K3 is generic, but this is not necessary. The point is just that since for small $t$ $w_1-tw_2$ is Ricci flat, then $(w_1-tw_2)^2$ is proportional to the $\Omega\wedge \bar \Omega$. But if this is true for small $t$, this is clearly always true. (I guess it is enough to have this proportionality just for a finite number of values of t) So, for $t$ such that $\int_{K3}(w_1-tw_2)^2=0$ we have $(w_1-tw_2)^2=0$ pointwise. $\endgroup$ – Dmitri Panov Dec 12 '10 at 1:48
  • $\begingroup$ Okay, now I understand. Very nice. +1 $\endgroup$ – Spiro Karigiannis Dec 12 '10 at 3:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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