Consider the weighted projective space $\mathbb P(5,6,8,19)$ with weighted homogeneous coordinates $x,y,z,w$, in this order. I want to construct an explicit quasi-smooth embedding of the weighted K3 surface $X_{38} \subset \mathbb P(5,6,8,19)$, in order to find and classify its singularities.
(The strategy that I am using is defined in Iano-Fletcher's “Working with weighted complete intersections”, and it relies on the following fact. Let $p$ be an arbitrary point on an hypersurface $X$ of a well-formed weighted projective space $\mathbb P$. If $\mathbb P$ near $p$ is locally isomorphic to $\mathbb C^{n+1} / G$ near the origin, then $X$ near $p$ is locally isomorphic to $\mathbb C^n / G$ near the origin. In particular, the singular points of $X$ are precisely the singular points of $\mathbb P$ that happen to be in $X$.)
I came up with the following polynomial:
$$f(x,y,z,w) = zx^6 – zy^5 + yz^4 + w^2 + \lambda x^4 y^3$$
where $\lambda \in \mathbb C^\star$ is a nonzero constant to be determined. Notice that, if we set $\lambda = 0$, then $(1,1,0,0) \in \mathbb C^4$ is a nonzero critical point of $f$, hence the embedding is not quasi-smooth. However, if we could ignore this issue, then the singularities of $X_{38}$ would be easy to find and classify:
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$[0:1:1:0]$ is a singularity of type $\frac 12 (1,1)$, hence an $A_1$ singularity.
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$[1:0:0:0]$ is a singularity of type $\frac 15 (1,-1)$, hence an $A_4$ singularity.
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$[0:1:0:0]$ is a singularity of type $\frac 16 (-1,1)$, hence an $A_5$ singularity.
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$[0:0:1:0]$ is a singularity of type $\frac 18 (-3,3)$, hence an $A_7$ singularity.
From now onwards, we will call these four points the special points.
Notice that $x^4 y^3$ and its first partial derivatives vanish at the special points. Therefore, varying $\lambda$ will change neither the identified singularities nor their classification. Moreover, it is easy to check that, for $\lambda \in \mathbb C^\star$, we obtain a hypersurface that is quasi-smooth at the special points. However, at least in principle, we might be introducing other critical points elsewhere.
Therefore, my question is: How can I prove that, for some $\lambda \in \mathbb C^\star$, the weighted projective hypersurface defined by $f$ is quasi-smooth? I am aware of Bertini's theorem for linear systems in ordinary projective space, but I do not know of any analogue that works in weighted projective space.
EDIT 1: I just thought of a possible solution. Consider the ordinary (not weighted!) projective space $\mathbb P^4$ with homogeneous coordinates $x,y,z,w,t$. Then $\mathbb C^4$ is just the open affine $t = 1$ of $\mathbb P^4$. Homogenizing $f$, we obtain
$$f^\#(x,y,z,w,t) = zx^6 – tzy^5 + t^2 yz^4 + t^5 w^2 + \lambda x^4 y^3$$
This defines a linear system of divisors of $\mathbb P^4$ parametrized by $\lambda \in \mathbb P^1$. By Bertini's theorem, for almost all $\lambda \in \mathbb C^\star$, we obtain a hypersurface of $\mathbb P^4$ that is smooth away from the system's base locus, which, if I am not mistaken, is the closure in $\mathbb P^4$ of the curves in $\mathbb C^4$ that project to the special points.
Thus, by adding a suitable multiple of the correction term $x^4 y^3$, we have obtained a surface $X_{38} \subset \mathbb P(5,6,8,19)$ that is quasi-smooth away from the special points. Since we already knew that $X_{38}$ would be quasi-smooth at the special points as well, $X_{38}$ is globally quasi-smooth.
QUESTION: Does this work?
EDIT 2: Well, I was mistaken. A line is determined by two points. Similarly, the base locus of this linear system is determined by the hypersurfaces obtained for two distinct values of $\lambda$. But two hypersurfaces in $\mathbb P^4$ cannot possibly intersect in just a union of curves.
To make this strategy workable, I need a larger linear system, determined by linear combinations of three surfaces. But at this point I am running out of imagination…
EDIT 3: Actually, I do not need a larger linear system. What I need is a larger brain to come up with the right ideas faster.
The base locus of the linear system given above is the intersection of $X_{38}$ with the union of the planes $x = 0$ and $y = 0$.
On the base locus, the partial derivatives of $x^4 y^3$ are identically zero, hence the the partial derivatives of $f$ do not depend on $\lambda$. Thus, for any point $P$ of the base locus, the following two propositions are equivalent:
- $P$ is a quasi-smooth point of $X_{38}$ when $\lambda = 0$.
- $P$ is a quasi-smooth point of $X_{38}$ for any $\lambda \in \mathbb C$.
Since the problematic point $[1:1:0:0]$ is not in the base locus, we are done.
QUESTION: Does this work?
Best Answer
Your last strategy (Edit 3) basically works.
By Bertini, for some $\lambda$, this hypersurface is smooth away from the locus where either $x=0$ or $y=0$.
On the locus where $y=0$, the derivative with respect to $z$ is $x^6$, so the hypersurface is smooth away from the locus where $x=0$. So it suffices to consider the case $x=0$.
Regardless, the derivative with respect to $w$ is $2w$, so there are no singularities unless $w=0$.
Setting $w=x=0$, the equation becomes
$$y z^4 - zy^5$$ which factors into distinct linear factors and thus has no singularities (unless $y=z=0=x=w$), and you are done.