3
$\begingroup$

I was hesitant about posting this question here, but since it deals with a partially unanswered question already on this site I figured that this would be the best place for it. I apologise in advance if not.

The question deals with explicitly calculating the ramification divisor on a curve in weighted projective space. The example given in the question was that of $C_7\subset\mathbb{P}(1,2,3)$.

There is a (generally) degree-6 covering map $\pi\colon\mathbb{P}^2\to\mathbb{P}(1,2,3)$ given by $[x_0:x_1:x_2]\mapsto[x_0:x_1^2:x_2^3]_{(1,2,3)}$. Write $[y_0:y_1:y_2]_{(1,2,3)}$ for the coordinates on $\mathbb{P}=\mathbb{P}(1,2,3)$. Then $\pi$ branches along the hyperplanes $\{y_1=0\}$ and $\{y_2=0\}$, where it is 2-1 and 3-1, respectively.

However, by, say, the Griffiths-Dolgachev-Steenbrink formula (see the linked question), we can calculate that $g_{C_7}=1$, since the usual degree-genus formula gives us that $g_C=15$, where $C=\pi^{-1}(C_7)\subset\mathbb{P}^2$. So naively plugging things into Riemann-Hurwitz we see that $$2g_C-2=6(2g_{C_7}-2)+\deg(D)$$ where $D$ is the ramification divisor. Using the facts that we already know then, we see that $\deg(D)=28$. But how exactly is $D$ defined?

I've searched through bits of Dolgachev's Weighted Projective Varieties and Reid's Young Person's Guide to Canonical Singularities but can't seem to find an explicit answer.

Naively, it seems like there would be 14 points in total on $C_7$ that intersect the two hyperplanes, 7 with index 2 and 7 with index 3. Plugging this into the usual definition without any thought then gives us that $\deg D=7*(2-1)+7*(3-1)=21$.

Edit: I've now answered this question in slightly more generality, but only using pretty basic (i.e. accessible to an undergraduate) ideas, in Theorem 4.3.7 in this link.

$\endgroup$
4
+50
$\begingroup$

The curve $C_7$ passes through the points $P = (0 : 0 : 1)$ and $Q = (0: 1: 0)$. You can see this by looking at what monomials can be used to define $C$. The group of order $6$ acting on $\mathbf{P}^2$ fixes these two points, hence the map $C_7 \to C$ has ramification order $6$ at $P$ and $Q$. There are $6$ points on $C_7$ besides $P$ on the line $x_1 = 0$. There are $6$ points on $C_7$ besides $Q$ on the line $x_2 = 0$. Plugging this into the usual definition then gives us that $\deg D = (6 - 1) + (6 - 1) + 6 * (2 - 1) + 6 * (3 - 1) = 28$.

$\endgroup$
  • $\begingroup$ When you say that there are $6$ points on $C_7$ besides $P$ on the line $x_1=0$, do you mean counting multiplicities, or are you saying that there are six distinct points? $\endgroup$ – Tim Aug 26 '15 at 14:52
  • 2
    $\begingroup$ My suggestion: choose a sufficiently general equation for $C$ and compute for yourself. $\endgroup$ – Count Dracula Aug 26 '15 at 15:50
  • $\begingroup$ (Also, I think in your answer you're using $C$ and $C_7$ in the opposite way that I used them in my question? I use $C\subset\mathbb{P}^2$ and $C_7\subset\mathbb{P}(1,2,3)$. Just checking that I'm not getting confused here!) $\endgroup$ – Tim Aug 31 '15 at 23:36
  • $\begingroup$ Everything seems to work out nicely, but all I have left to prove is that $C_7$ (and thus $C$ (I think)) being smooth gives us that all intersections with $\{x_i=0\}$ hyperplanes are of multiplicity one (i.e. all distinct points). $\endgroup$ – Tim Sep 1 '15 at 19:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.