Let $S$ be a K3 surface over $\mathbb{C}$, that is, $S$ is a simply connected compact smooth complex surface whose canonical bundle is trivial. I recall reading somewhere that any two such surfaces are diffeomorphic, however I can't for the life of me remember where, or how the proof goes.

Does anybody know a good reference to a proof, or can provide a proof?

Thanks, Dan

  • 5
    $\begingroup$ Just to add to the two references below: I believe that the moduli space of complex analytic K3s is connected of dimension 20. (The algebraic K3s are parameterized by a countable collection of 19-dimensional closed subspaces in this moduli space.) The connectedness is what ensures that all the K3s are diffeomorphic. $\endgroup$
    – Emerton
    Apr 21, 2010 at 13:00

2 Answers 2


I think this was first proved by Kodaira. See On the structure of complex analytic surfaces, 1. There Kodaira proves that any K3 surface is a deformation of a non-singular quartic surface in $\mathbb{CP}^3$. In particular, they are all diffeomorphic.


I find this reference quite readable:

Le Potier: "Simple connexité des surfaces K3", in Asterisque 126, 1985.

I haven't read Kodaira's paper (in Joel's answer), so I don't know whether it is the same argument, but Le Potier also deforms to quartic surfaces.


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