This question has two parts: A calculation that is giving me a lot of troubles, and a theoretical one on weighted projective spaces.

1) I want to find the genus of the curve $C_7 \subset \mathbb{P}(1,2,3)$. Naive adjunction does not work, because this curve is not well-formed in the sense of [1]. To calculate the genus of the curve, it is possible to use the map $$ \mathbb{P}^2 \to \mathbb{P}(1,2,3) $$ defined by $$ [y_0:y_1:y_2] \to [y_0:y_1^2:y_2^3] $$ This map is a 6 to 1 ramified Galois covering with Galois group $\mathbb{Z}_2 \oplus \mathbb{Z}_3$. If the variables of $\mathbb{P}(1,2,3)$ are $[x:y:z]$. Then our map branches along the the hyperplanes $(z=0)$ and $(y=0)$. It induces a cover $D \to C_7$ where $D$ is a smooth plane curve of degree 7 and genus 15. It is possible to use Hurwitz theorem for finding that the genus of $C_7$ is one. However, I don't understand the ramification divisor! So, I cannot recover the calculation! (This is an example on [1, Note I.3.15.i])

2) A weighted projective space $\mathbb{P}(a_0, \ldots a_n)$ is well formed if $gcd(a_0 \ldots \hat{a}_i \ldots a_n)=1$ for every $i$. However, we find non well formed weighted projective spaces all the time. Even, those with $gcd(a_0, \ldots a_n) \neq 1$. What would be the adjunction formula in those cases?



[1] Working with weighted projective spaces. A.R. Fletcher

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    $\begingroup$ What curve is $C_7$? $\endgroup$ – Mariano Suárez-Álvarez Jul 16 '13 at 5:54
  • $\begingroup$ I meant the generic curve of degree seven. In this case, it is $x^7+xy^3+xz^2+x^3y^2...$ $\endgroup$ – mathSt Jul 16 '13 at 13:23
  • $\begingroup$ this now has another answer: mathoverflow.net/questions/215171/… $\endgroup$ – Tim Aug 27 '15 at 23:05

Question 1: Take $F$ to be the polynomial that defines $C_7$ as a hypersurface in your projective space. If you are looking for the cohomology to this (possibly singular) variety, then you should look at Dolgachev's Weighted Projective Varieties paper. In that paper, you can find a result (Theorem 4.3.2) where you can compute $h^{1,0}(C_7)$ by looking at elements of a certain degree of the graded ring $\mathbf{C}[x,y,z]/J_f$, where $J_f$ is the Jacobian ideal. In general, take the coordinate ring $S(Q) = \mathbb{C}[x_0, \ldots, x_n]$ with weights $q_0, \ldots, q_n$. We consider the quasihomogeneous polynomial $f$ of degree $d$ which defines a well-defined hypersurface $X = Z(F) \subset \mathbb{P}(q_0, \ldots, q_n)$. Then the primitive cohomology hodge numbers in the middle cohomology piece are: $$ h_0^{i, n-1-i}(X) = \dim (S(Q)/ J_f)_{(n-i)d - \sum_{i=0}^n q_i} $$

(I just noticed that the adjunction-like formula you might want is Theorem 3.3.4 in Dolgachev here.).

In the case that you're looking at the smooth resolution of a singular curve, then you can use Chen-Ruan cohomology of your orbifold to compute the cohomology of the resolution as you can view these hypersurfaces in weighted projective spaces as (smooth) DM stacks as long as they hypersurface is quasismooth (aka all the singularities the hypersurface has are inherited in some sense from the weighted projective space). Such computations are pretty okay for low dimension and some concrete examples are done in the paper by Chiodo and Ruan in section 3.2 of http://arxiv.org/pdf/0908.0908v2.pdf for threefolds.

Question 2: For when $\text{gcd}(a_0, \ldots, a_n) \neq 1$, I suggest you first think about what the $\mathbb{C}^*$ action looks like and try to simplify the weighted projective space. :-)

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    $\begingroup$ It's worth noting that you should expect a formula like the one Dolgachev wrote, due to the formula of Griffiths that for a hypersurface of dimension $n$ and degree $d$ has $H^{p,n-p}\cong R_{(n-p+1)d-(n+2)}$, where $R$ is the Jacobian ring, essentially the exact formula above, corrected for the weights. $\endgroup$ – Charles Siegel Jul 17 '13 at 2:42
  • $\begingroup$ Oops! Yeah the result being discussed here is usually called the Griffiths-Dolgachev-Steenbrink formula. $\endgroup$ – Tyler Kelly Jul 17 '13 at 2:44
  • $\begingroup$ I posted a reply as an "answer". It was too long for fitting here. $\endgroup$ – mathSt Jul 18 '13 at 0:18

This is not an answer, but rather a reply to Tyler's answer. It was too long for posting it as a comment, and it is not an edition of the question. So I posted here.

First, thanks Tyler for answering. On question 1, you are right the Griffiths-Dolgachev-Steenbrink formula calculates that $g(C_7)=1$ (see [1]). However, I was actually trying to understand the technique of using the Hurwitz theorem together with the covering $\mathbb{P}^2 \to \mathbb{P}(1,2,3)$. The map is generically 6 to 1. It is 2 to 1 over the line (z=0) and it is 3 to 1 over the line (y=0). However, the ramification divisor $R$ of the map $D \to C_7$ is not clear for me. Particularly $$ 2g_D-2= 28 =6(2g_{C_7}-2)+deg(R)=deg(R) $$ Why is $deg(R)=28$ and what is the description of $R$? On other hand, the Theorem 3.3.4 in Dolgachecv does not work if the curve is not well formed. Indeed, our $C_7$ is quasismooth, but we obtain $$ \omega_{C_7} = \mathcal{O}_X(7-1-2-3)=\mathcal{O}_X(1) $$ This is incompatible with $C_7$ being an elliptic curve!
On simplifying the $\mathbb{C}^*$ action, consider this case from Reid ([2, Example 3.7-3.8]): $$ D:=(x^a+y^b+z^c =0) \subset \mathbb{P}(bc,ac,ab) \; \; \text{ such that } \;\; gcd(bc,ac,ab)=1 $$ We can simplify the weighted projective space as in Section 1.3.1 of Dolgachev: $$ \mathbb{P}(bc,ac,ab) \cong \mathbb{P}(b,a,ab) \cong \mathbb{P}(b,1,b) \cong \mathbb{P}(1,1,1) $$ The curve $D$ transform as $$ (x^a+y^b+z^c=0) \to (x^a+y^b+z=0) \to (x+y^b+z=0) \to (x+y+z=0) $$ However, we lost something along the way: Three orbifolds points of order $a$ , $b$, and $c$! Therefore, I developed some suspicion on simplifying the weighted projective spaces :P and it will be nice to have a version of Theorem 3.3.4 that works for ANY set of positive weights and ANY hypersurface.


[1] Professor Carlson wrote an nice Sage code for calculating those hodge numbers: https://github.com/jxxcarlson/math_research/blob/master/hodge.sage}.

[2] http://homepages.warwick.ac.uk/~masda/surf/more/grad.pdf

  • $\begingroup$ Just to say that I am also very interested in an answer to this, especially the first part about the ramification divisor. It seems like some kind of weighted Bézout would come into play, but I'm not too sure on the details. Would very much like to hear how to calculate the ramification divisor! $\endgroup$ – Tim Aug 18 '15 at 13:36
  • $\begingroup$ this is now a question: mathoverflow.net/questions/215171/… $\endgroup$ – Tim Aug 21 '15 at 14:10

As a quick note: your first question inspired some (certainly not original) research on my behalf. The question is answered in slightly more generality, but only using pretty basic (i.e. accessible to an undergraduate) ideas. See Theorem 4.3.7 in this link.


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