Zheng
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Unregistered User
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Jan 13 |
awarded | ● Student |
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Dec 30 |
comment |
Isometry of K3 surface. 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. |
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Dec 30 |
awarded | ● Editor |
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Dec 30 |
comment |
Isometry of K3 surface. @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. |
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Dec 30 |
comment |
Isometry of K3 surface. @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. |
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Dec 30 |
comment |
Isometry of K3 surface. @Robert Yes, it does. I added the definition. |
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Dec 30 |
revised |
Isometry of K3 surface. added 1 characters in body |
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Dec 30 |
asked | Isometry of K3 surface. |
