MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $S$ be a K3 surface and $\iota$ be anti-symplectic involution of $S$. Suppose that $g$ is a Kahler-Einstein metric on $S$. My question is;

Why $\iota$ is an isometry of $S$ with respect to $g$? Is this true for any holomorphic action of $S$?

Edit $\iota$ is called anti-symplectic if it acts on $\Omega^{2,0}$ as $-id$.

share|cite|improve this question
Does 'anti-symplectic involution of $S$' mean a holomorphic involution of $S$ that carries a holomorphic volume form on $S$ to its negative? – Robert Bryant Dec 30 '12 at 15:07
I don't understand the question. Are you implicitly assuming that $g$ is invariant under $\iota$? Why would the choice of complex structure affect whether a map is an isometry? – Johannes Nordström Dec 30 '12 at 15:12
@Robert Yes, it does. I added the definition. – Zheng Dec 30 '12 at 15:26
@Johannes I am not assuming that $g$ is invariant under $\iota$. As to the second question, you are right. $g$ is Kahler-Einstein for any complex structure obtained by hyperKahler rotation but $\iota$ is not necessarily holomorphic in other complex structure. – Zheng Dec 30 '12 at 15:31
@Johannes So the question should be "Assume $\iota$ is anti-symplectic for some complex structure, then is it isometry?". But I simplified my question above. Thank you for pointing out this. – Zheng Dec 30 '12 at 15:33

There is a unique Ricci-flat Kähler metric in each Kähler class of $S$. Thus, for any holomorphic automorphism $\iota$ of $S$, a Ricci-flat Kähler metric $g$ is invariant under $\iota$ if and only if its Kähler class $[\omega_g] \in H^{1,1}(S)$ is.

share|cite|improve this answer
One can average the Kahler form and get invariant Kahler class. For such a Kahler metric, $g$ is invariant and thus $\iota$ is an isometry. This works for any holomorphic $G$-action. – Zheng Dec 30 '12 at 23:04
Sure, that tells you that for any automorphism of $S$ (of finite order at least) there exists some invariant Kähler class, and hence an invariant Ricci-flat Kähler metric. But it does not mean that a given Ricci-flat Kähler metric is invariant, which is what your question seems to ask. – Johannes Nordström Dec 30 '12 at 23:36

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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