Timeline for How to check that whether or not a surface is a K3 surface?
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
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| Aug 29, 2018 at 9:05 | comment | added | Jianrong Li | @მამუკაჯიბლაძე, thank you very much. According to your computations, what are the kind of singularities of the surface at infinity? | |
| Aug 28, 2018 at 10:18 | comment | added | მამუკა ჯიბლაძე | 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 $x_1 \left(4 y_0+y_1\right) \left(3 x_1 y_0+2 x_0 y_1\right)=0$ | |
| Aug 28, 2018 at 8:36 | history | edited | Dmitri Panov | CC BY-SA 4.0 |
added 654 characters in body
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| Aug 28, 2018 at 6:52 | comment | added | Jianrong Li | @Dmitri, thank you very much. I solved the equations $F_x=F_y=F_z=F=0$ and obtain a unique solution $(x,y,z)=(0,0,-3)$. How to check that whether or not it is Du Val singular? | |
| Aug 27, 2018 at 7:44 | comment | added | Jonny Evans | @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. | |
| Aug 26, 2018 at 22:15 | comment | added | Dmitri Panov | @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 $O(2)$. | |
| Aug 26, 2018 at 19:20 | comment | added | meh | We agree on degree, but I think degree $\leq 2$ and degree 2 are different things. I don't think things homogenize as nicely as you say. Why don't you write don't the multi-homogeneous equations ? Then it will be clear. One reason I'm suspicious is that one of the terms has degree 5, which means that the ordinary completion in ${P}^3$ will a quintic and hence only a K-3 surface if it has singularities. | |
| Aug 26, 2018 at 17:51 | comment | added | Dmitri Panov | 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? | |
| Aug 26, 2018 at 17:34 | comment | added | meh | 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. | |
| Aug 26, 2018 at 17:13 | history | answered | Dmitri Panov | CC BY-SA 4.0 |