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How to check that whether or not a surface in $\mathbb{C}^3$ is a K3 surface? Are there some method or math software to transform the equation of a surface to a normal form? For example, the following is a surface in $\mathbb{C}^3$:

\begin{align} 135\, x^2\, y\, z^2 + 360\, x^2\, y\, z + 540\, x^2\, z^2 + 180\, x^2\, z + 90\, x\, y^2\, z^2 + 654\, x\, y^2\, z + 360\, x\, y^2 + 360\, x\, y\, z^2 - 720\, x\, y\, z + 180\, x\, y + 240\, y\, z + 720\, y=0. \end{align} How to transform it to a normal form? Is it a K3 surface? Thank you very much.

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    $\begingroup$ K3 should be compact？ $\endgroup$
    – Chen Jiang
    Commented Aug 26, 2018 at 7:12
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    $\begingroup$ Compactify and compute singularities. If all of them are rational double points (aka Du Val singularities), the resolution is a K3 surface. $\endgroup$
    – Sasha
    Commented Aug 26, 2018 at 7:44
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    $\begingroup$ If there were a "normal form" for surfaces in $\mathbb{P}^3$, the life of algebraic geometers specialized in moduli spaces would be a lot easier. $\endgroup$
    – Alex Gavrilov
    Commented Aug 26, 2018 at 13:41

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I assume that the equation you've given is of interest to you, so here is my interpretation of Sasha's comment. The equation under consideration has degree $2$ in all three variables, so a compactification of this surface in $\mathbb CP^1\times \mathbb CP^1\times \mathbb CP^1$ is a divisor of degree $(2,2,2)$. So if by any chance it is smooth, then indeed it is a K3 surface. It might happen, however that the surface has singularities in $\mathbb C^3$ or in one of $7=2^3-1$ other charts. So one needs to find the solution of $F_x=F_y=F_z=F=0$ in all these charts. Probably this can be done with a help of some program. Once these singularities are found one needs to check if these are Du Val or not (for this they need to be isolated of course). This should not be super hard since the degree in each variable is $\le 2$.

PS. As Jianrong is pointing out, the surface is singular at the point $(0,0,-3)$. In order to see whether it is Du Val or not at this point, we calculate its Taylor series at the point. It turns out the the second term is the following:

$$60(-27x^2+42xy+4yz).$$ It is easy to see that this quadratic form has rank three, so we are lucky and the point $(0,0,-3)$ is the simplest possible singularity - an ordinary double point (of course Du Val).

Note that, in order to conclude whether this surface is $K3$ or not, we still need to check what kind of singularities it has at infinity (at three planes $x=\infty$, $y=\infty$, $z=\infty$).

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  • $\begingroup$ I don't know what you mean by degree, but don't all the terms of the equation on the bottom line not have degree 2 in all the variables. $\endgroup$
    – meh
    Commented Aug 26, 2018 at 17:34
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    $\begingroup$ I don't mean anything unusual by degree, just that each of variables $x,\,y,\, z$ enters the equation with degree $\le 2$. Just to give you an example, suppose we have an equation $1+xy+y^2+x^2y^2=0$. It can be made bi-homogeneous of degree $(2,2)$ in the following way: $x_2^2y_2^2+x_1x_2y_1y_2+x_2^2y_1^2+x_1^2y_1^2=0$. This equation defines a curve of degree $(2,2)$ in $\mathbb CP^1\times \mathbb CP^1$. Does this make sense? $\endgroup$
    – Dmitri Panov
    Commented Aug 26, 2018 at 17:51
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    $\begingroup$ @agiensky, I don't write it because the formula will be twice longer than the one written by the author of the post. For example $135x^2yz^2$ becomes $135x_1^2y_1y_2z_1^2$. But anyway, the claim is that any polynomial in $\mathbb C^n$ of degree at most $2$ in each variable with $3^n$ non-zero generic coeffs defines a smooth Callabi-Yau in it. You can check it for $n=2$, since in this case one gets elliptic curves. This is just the fact that the anti-canonical bundle of $(\mathbb CP^1)^n$ is the product of $O(2)$. $\endgroup$
    – Dmitri Panov
    Commented Aug 26, 2018 at 22:15
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    $\begingroup$ @aginensky: this point with the quintic also confused me, but you can see from what Dmitri says that this quintic will always be singular if each variable has degree \leq 2: since the generic such polynomial gives a smooth K3, it must give a singular quintic, otherwise the affine chart would exhibit a birational equivalence between something with Kodaira dimension 0 and something with Kodaira dimension 2 (but Kodaira dimension is a birational invariant). Finally, a nongeneric polynomial can't give a smooth quintic if the generic one gives a singular quintic. $\endgroup$
    – Jonny Evans
    Commented Aug 27, 2018 at 7:44
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    $\begingroup$ At $x=\infty$ it becomes $y_0 z_1 \left(4 z_0\left(y_0+2 y_1\right) +3 z_1 \left(4 y_0+y_1\right)\right)=0$, at $y=\infty$ it is $x_0 x_1 \left(20 z_0+3 z_1\right) \left(3 z_0+5 z_1\right)=0$, and at $z=\infty$ it is $x_1 \left(4 y_0+y_1\right) \left(3 x_1 y_0+2 x_0 y_1\right)=0$ $\endgroup$
    – მამუკა ჯიბლაძე
    Commented Aug 28, 2018 at 10:18

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