A question about kawamata's proof of vanishing for big and nef $\mathbb{Q}$ divisors

Question

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)

Answer 1

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 $O_X$.