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Let $U$ be a smooth variety, $m >1$ be a positive integer and $D_{m,U} \in | {-}m K_U|$ be a smooth irreducible divisor. Let $\pi: V_m:= Spec \bigoplus_{i=0}^{m-1} \mathcal{O}_U(i K_U) \rightarrow U$ be a cyclic cover determined by $D_{m,U}$. Assume that $|-K_U|$ contains a smooth irreducible member $D_U$ and put $D_{V_m}:= \pi_m^{-1}(D_U)$. Consider the logarithmic differential sheaves $\Omega^1_{V_m}(\log D_{V_m})$ and $\Omega^1_U(\log D_U)$.

Question 1 $\Omega^1_{V_m} (\log D_{V_m}) \simeq \pi_m^* \Omega^1_U(\log D_U)$ ?

Question 2 Let $(\pi_m)_* \Omega^1_{V_m}(\log D_{V_m}) =: \bigoplus \mathcal{F}_i$ be the eigendecompposition with respect to the $\mathbb{Z}/ m\mathbb{Z}$-action where $\mathcal{F}_i$ is the sheaf of sections of eigenvalue $\zeta^i$ ($\zeta$ is the $m$-th primitive root of unity). Is $\mathcal{F}_1 \simeq \Omega^1_U (\log D_U)(-D_U)$? If it's not, can you describe $\mathcal{F}_i$ explicitly?

I saw the book by Esnault-Viehweg. Are there other references on that topic?

(add) I can see that Question 1 is wrong by the arguments in the Sándor Kovács' answer.

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Question 1 seems right. You can check this in local coordinates. I don't think Question 2 is quite right, it should probably be $\Omega^1(\log D)\otimes K$. You can use the projection formula, then reduce it analyzing $\pi_*\mathcal{O}$. This whole story is largely the creation of Esnault and Viehweg. So their book on Vanishing theorems and various papers are the best references in my opinion. –  Donu Arapura Oct 20 '11 at 14:17
    
Thank you for the reply. I think $\Omega^1_U(\log D_U) \otimes K_U \simeq \Omega^1(\log D_U) \otimes \mathcal{O}_U(- D_U)$. Note that $D_U$ is not the branch divisor. I saw the similar formula for the branch divisor, but I think it's not my case. Is Question 1 still correct? Anyway, I will try to check locally. –  tarosano Oct 20 '11 at 15:22
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My comment was somewhat hastily written. Sándor has gotten to the heart of the matter. I stand by my earlier claim regarding references. Specifically, see Viehweg, "Vanishing theorems", Crelles (1982), and Esnault-Viehweg, Revetement Cycliques I, II, in addition to their book. –  Donu Arapura Oct 21 '11 at 8:19
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1 Answer

up vote 4 down vote accepted

Question 1 indeed seems right until one reads your comment. It seems right, because we are thinking that $D_U$ is the branch divisor. It is indeed right if you take $D_U=D_{m,U}$ in place of your choice.

On the other hand, as stated, Question 1 could not be correct!! Observe that the right hand side is independent of $m$ while the left hand side can get bigger and bigger as $m$ grows.

The crux of the matter is that $D_{V_m}$ should be compared to $D_{m,U}$ and not $D_U$.

Here is some evidence: Observe that $B=D_{m,U}$ is the branch divisor of $\pi_m$. Let $R=\frac 1m\pi^*B$ (a $\mathbb Q$-divisor if you want) be the $m^\text{th}$ root of $B$. Then we have the (logarithmic) ramification formula: $$K_{V_m}+ R \sim \pi^*(K_U + B).\tag{$*$}$$

Next observe that $B\sim mD_U$ and hence $D_U$ cannot contain $B$ (which is assumed to be irreducible), so $D_{V_m}=\pi^*D_U$ and then $D_{V_m}\sim_{\mathbb Q}R$.

So, we obtain by ($*$) that $$\det\Omega_{V_m}(\log D_{V_m})\sim_{\mathbb Q}\det \Omega_{V_m}(\log R)\sim \pi^*(\det\Omega_U(\log B)),\tag{+}$$ which is strictly bigger than $\pi^*(\det\Omega_U(\log D_U))$.

Remarks

1) It is entirely unimportant that the linear system comes from $-K_U$. This was not used anywhere, but in the last sentence one may notice that with that definition $\det\Omega_U(\log D_U)\simeq \mathscr O_U$, which might make it easier to believe that the claim and proof here is correct.

2) If one defines $D_U=D_{m,U}$, then Question 1 is correct, this is an easy local calculation.

3) Under #2 above Question 1 implies Question 2. The proof is, as Donu suggests, simply using the projection formula and then simply applying the same formula for the push-forward of the structure sheaf for a cyclic cover.

Addition: The recent edit to your question seems to indicate that perhaps you still expect that Question 2 might be true even if Question 1 is not. I don't think that's going to happen either. (I thought this would be kind of clear from the above).

By the above computation and remarks one can see that $$\pi_*\Omega_{V_m}(\log R)\sim \Omega_U(\log B)\otimes \pi_*\mathscr O_{V_m}\sim \Omega_U(\log B)\otimes (\oplus_j\mathscr O_{U}(-jB)),$$ so the $\zeta$-eigensheaf for this pushforward is $\Omega_U(\log B)(-B)$.

Now since $\det\Omega_{V_m}(\log D_{V_m})\sim_{\mathbb Q}\det \Omega_{V_m}(\log R)$, counting degrees should show you that the answer to Question 2 is also negative, unless $\dim U=1$. On the other hand, if $\dim U=1$, you will be hard pressed to find a $D_{m,U}$ that is both smooth and irreducible for $m>1$.

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Right, I was confused by the notation. I've deleted my previous comment. –  rita Oct 20 '11 at 19:54
    
Thank you for the answer and sorry for the misleading notations. Question 1 is wrong. I should have been more careful. –  tarosano Oct 20 '11 at 20:46
    
Thank you for the addition. I'm still curious about what $\mathcal{F}_i$ is. However, I should consider the original problem again and look for the necessary statement.. –  tarosano Oct 21 '11 at 7:40
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