# pullback or push forward of logarithmic differential sheaf by cyclic cover

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.

• 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. Oct 20, 2011 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. Oct 20, 2011 at 15:22
• 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. Oct 21, 2011 at 8:19

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 $m>1$.

• Right, I was confused by the notation. I've deleted my previous comment.
– rita
Oct 20, 2011 at 19:54
• Thank you for the answer and sorry for the misleading notations. Question 1 is wrong. I should have been more careful. Oct 20, 2011 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.. Oct 21, 2011 at 7:40