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Suppose I have an affine equation $f(x, y) = 0$ which after homogenizing becomes $f(X, Y, Z) = 0$ in $\mathbb{P}^{3}$. Are there ways to check that $f$ represents a K3 surface?

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Btw, if you're interested in K3 surfaces then there are good introductions available in Beauville's "Complex algebraic surfaces" (and his Astérisque K3 surface seminar notes, but they are hardcore) and in Huybrecht's notes (available for free on his website). Avoid Barth, Peters, Van de Ven on your first pass; it's a reference book and isn't actually meant to be read. –  Gunnar Þór Magnússon Jun 12 '11 at 17:37

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up vote 9 down vote accepted

There sure are. Let's denote by $X$ the surface defined by $f$, and let's suppose that $X$ is smooth. For any compact complex surface, being K3 is equivalent to being simply connected and having trivial canonical bundle.

A hypersurface in $\mathbb P^n$ is connected and simply connected by the Lefschetz theorem, so we only need to find a condition ensuring that the canonical bundle is trivial.

This condition is given by the adjunction formula, which says that if the polynomial $f$ is of degree $d$, then

$$ K_X = ( K_{\mathbb P^3} \otimes \mathcal O(d) )_{|X} = \mathcal O_X(d-4). $$

This bundle is trivial if and only if $d = 4$, or in other words, if $f$ is a quartic.

A fun exercise involving the adjuction formula is to see that there are very few K3 surfaces given as complete intersections in $\mathbb P^n$. In fact, they only exist in dimension 4, 5 and 6 if I remember correctly.

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K3-surfaces X being complete intersections in m-dimensional projective space exist only for m = 3,4,5; they have multidegree respectively 4, (2,3), (2,2,2): If the complete intersection X has multidegree (d_1,...,d_n) in n+2-dimensional projective space, than d_1 +...+ d_n = n+3 is required for the canonical bundle of X to be trivial; avoiding hyperplane sections one can assume d_j >= 2 for each j. One easily checks that such complete intersections have vanishing irregularity because the cohomology of projective space is well-known. –  jo wehler Sep 18 at 8:35

I am convinced that $f(X,Y,Z)$ represents a K3 surface iff it is a quartic (fourth-degree) polynomial, see page 16 of this review of K3 surfaces:


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