Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Theorem 2 [1, p.46] Let $X$ be a non-singular projective algebraic variety of dimension $n$, and $D$ a numerically effective $\mathbb{Q}$-divisor such that $(D^n)>0$. We assume that the support of the non-integral part $D-[D]$ is a divisor with normal crossing on $X$. Then $H^i(X,O_X([-D]))=0$ for $i<n$.

Proof. We write $-D=[-D]+\sum d_jE_j$, where $0<d_j<1$ and $\sum E_j$ is a divisor with normal crossings. Let $p_j,q_j$ be positive integers such that $d_j=p_j/q_j$. Applying Lemma 5 to the $q_j$ attached to $E_j$, we obtain non-singular covering $f:X'\to X$ such that $f^*E_j=q_jE_j'$. Let $f^*(-D)$ be the divisor on $X'$ defined to be $f^*([-D])+\sum p_jE_j'$. Then $f^*(-D)$ on $X'$ satisfies the condition of Theorem 1. Hence $H^i(X,f_*(O_{X'}(f^*(-D))))=H^i(X',O_{X'}(f^*(-D)))=0$ for $i<n$.

All is Ok for the above.

On the other hand, the trace map: $f_*(O_{X'})\to O_X$ extends to a surjective homomorphism $f_*O_{X'}(\sum p_jE_j)\to O_X$ since $d_j<1$. Hence $O_X$ is a direct summand of $f_*O_{X'}(\sum p_jE_j)$, and $O_X([-D])$ is that of $f_*O_{X'}(f^*(-D))$. Therefore, $H^i(X,O_X([-D]))=0$. Q.E.D.

I am not clear about how the trace map is $f_*(O_{X'})\to O_X$ is defined and how it extends? And how is this related to $d_j<1$? I had thought about the surjective morphism $O_{\sum q_jE_j}\to O_{\sum p_jE_j}\to 0$

Reference

[1] Yujiro Kawamata, A Generalization of Kodaira-Ramanujam's Vanishing theorem, Math. Ann. 262 (1982) pp 43-46 (link)

share|improve this question
2  
Around a smooth point of $D$, with equation $f=0$, $X'$ is defined by (copies of) the cyclic covering given by equation $y^q=f(x)$. For simplicity, neglect this story of copies. Then the ring of functions on $X'$ is locally free over the ring of function on $X$, with basis $1,y,...,y^{q-1}$. The trace maps $1$ to the constant $q$, and the other elements of the basis to $0$. Since $d<1$, one has $p\leq q-1$ and (at the level of function fields) it maps elements of the form $y^{-a} u$, with $0\leq a\leq p$ and $u\in \mathscr O_{X'}$ to a regular function on $X$. This is the desired exension. –  ACL May 27 at 8:55

1 Answer 1

up vote 2 down vote accepted

ACL already basically answered this in the comments (including what to do in local coordinates), but maybe this is worth explaining in a bit more detail and in more generality.

Whenever one has a finite map between varieties $f : X' \to X$ we have an induced map of fraction fields $K(X) \subseteq K(X')$. Then $K(X')$ is a finite dimensional $K(X)$ vector space and each element $z \in K(X')$ gives a $K(X)$-linear map $\phi_z : K(X') \xrightarrow{\cdot z} K(X')$. The trace of $z$ is defined to be the trace of the linear map $\phi_z$. This map is non-zero if and only if $f$ is separable (so in your context, it is definitely nonzero).

This gives us a $K(X)$-linear map $T : K(X') \to K(X)$. Now, it is an easy exercise (in Atiyah-Macdonald) that $T$ sends $f_* O_{X'}$ to $O_X$ as long as $X$ is normal. From now on, let's assume that both $X$ and $X'$ are normal (which is your situation as well). It is also easy to see that $T(f_* O_{X'}) = O_X$ since $T(1)$ is just the degree of the map mod the characteristic (which is not zero in characteristic zero).

More generally though, if one chooses $K_X$ to be a canonical divisor on $X$ and sets $K_{X'} = f^* K_X + \text{Ram}$ where $\text{Ram}$ is the ramification divisor, then $T$ also sends $f_* O_{X'}(K_{X'})$ to $O_X(K_X)$ and it's easy to see that this is equivalent to sending $f_* O_{X'}(\text{Ram})$ to $O_X$ (if $K_X$ is Cartier, as in your case, then it's trivial, if $K_X$ is not Cartier, it is Cartier on the smooth locus and then you can reflexify elsewhere).

Ok, so how does this help you?

Note $X' \to X$ is a cyclic cover, and its ramified over $D$ and the ramification divisor is probably exactly $\sum (q_j-1) E_j'$ since the ramification is tame (we are in characteristic zero) -- I didn't look up the appropriate lemmas in Kawamata's paper, but this is how it should work. Regardless, the point is that $\sum p_j E_j$ is less than the ramification divisor and so we have

$$ f_* O_{X'}( \sum p_j E_j) \hookrightarrow f_* O_{X'}(\text{Ram}) \xrightarrow{T} O_X $$

which I think is exactly the map you want.

For references for this, I think it is nearly all in Serre's Local Fields among other places. But if you need more references, I can track down some information.

Final comment

It is probably worth observing that in fact $f_* O_{X'}(\text{Ram})$ is the largest $O_X$-module subsheaf of $f_* K(X')$ that you can restrict trace to such that the image is still contained $O_X$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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