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Let $f\colon X \to Y$ be a finite map of smooth surfaces. Let the divisor $D$ of $Y$ be the branch locus of $f$. We assume that $D$ is a union of nonsingular curves intersecting transversally with no three components meeting at one point.

Claim:

$2K_X = f^*(2K_Y + D),$

where $K_X$ is the canonical divisor of $X$ and $K_Y$ is the same for $Y$.

How do you prove this claim?

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3 Answers 3

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Felipe is right. $K_X = f^* K_Y + R$. Thus $$2K_X = 2(f^*K_Y + R).$$

However, your formula might still be right depending on the context. If $X \to Y$ is a 2-to-1 cover with tame ramification, then the ramification index at each component of $R$ is $2$. Thus $2R = f^* D$.

Plugging this in you get: $$2K_X = 2f^*K_Y + 2R = f^*(2K_Y + D).$$ That's exactly what you wanted, and might be what's going on in your particular situation.

However, if your cover is 3-1 (characteristic $\neq 3$) and for simplicity if we assume that the ramification index at each component of the ramification divisor is 3 , then $R = (3-1)\text{Supp}(R) = 2\text{Supp}(R)$. It follows that $f^* D = 3 \text{Supp}(R) = (3/2)R$. Therefore

$$2K_X = f^*2K_Y + 2R = f^*2K_Y +(4/3)f^*D = f^*2K_Y + 4 \text{Supp}(R) .$$

Which is not what you want.

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  • $\begingroup$ The maps I'm considering are 8-to-1 and 27-to-1 covers. $\endgroup$
    – user14211
    Apr 8, 2011 at 1:20
  • $\begingroup$ Do you know more about the ramification? What the ramification indices are? $\endgroup$ Apr 8, 2011 at 1:22
  • $\begingroup$ As far as I know, there is no information given about the ramification indices. I found the formula in Vakil's paper on Murphy's law. It is in the proof of Theorem 4.4. $\endgroup$
    – user14211
    Apr 8, 2011 at 1:26
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    $\begingroup$ Hm, I'm not sure what to tell you. I'd have to read the surrounding information carefully to see what's going on. I assume you already looked up the relevant information in Pardini (theorem 2.1)? $\endgroup$ Apr 8, 2011 at 4:08
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I am not sure your formula is quite correct, I'm worried about those $2$'s. A correct formula is $K_X = f^*(K_Y)+R$ where $R$ is the ramification divisor, which is a divisor on $X$ with same support as $f^*(D)$. The proof is similar to the proof of Hurwitz's formula for curves. Pull-back a differential two-form from $Y$ to $X$ and compute divisors on both surfaces. I think you can find the result in Iitaka's book but I don't have it with me at the moment, so I can't check.

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  • $\begingroup$ I got the formula from Vakil's Murphy's Law paper. I've reproduced it as written. The formula is in the proof of Theorem 4.4. $\endgroup$
    – user14211
    Apr 8, 2011 at 1:21
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As Felipe and Karl correctly pointed out, the formula you write is not true in general. However, if you read Vakil's paper, you see that he's considering a very particular situation, namely Galois coverings with Galois group $G=(\mathbb{Z}_p)^3$, where $p=2,3$.

Let me explain how the formula works in a simpler case, namely $G=(\mathbb{Z}_2)^2$, the so-called bidouble covers (see Catanese' paper [Ca1] in Vakil bibliography). Then you can try to prove it in Vakil's cases (maybe after reading Pardini's paper on abelian covers).

Let us call $\chi_i$, $i=1,2,3$ the three non-trivial characters of $G$. Then the branch locus $D$ of $f$ can be written as

$D=D_1 + D_2 +D_3$,

where $D_i$ correspond to $\chi_i$. The divisors $D_i$ are smooth and intersect transversally.

Now we can factor $f$ as

$X \stackrel{g}{\longrightarrow} Z \stackrel{h}{\longrightarrow} Y$,

where $h$ and $g$ are double covers branched over $D_1+D_2$ and $h^*D_3$, respectively. Note that in general the intermediate cover $Z$ is singular!

By using the formulae for double covers, we can write

$2K_Z=h^*(2K_Y+D_1+D_2), \quad 2K_X=g^*(2K_Z+h^*D_3)$

that is, putting things together,

$2K_X=g^*h^*(2K_Y+D_1+D_2+D_3)=f^*(2K_Y+D)$.

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