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This is a question about chapter 3 of Esnault-Viehweg "Lectures on vanishing theorems".

Let $k$ be an algebraically closed field. Let $X$ be a smooth, projective variety over $k$ and $$ D=\sum a_j D_j $$ an effective normal crossing divisor. Assume that there exists an integer $d \geq 1$ and an invertible sheaf $L$ such that $\mathcal{O}_X(D)=L^{\otimes d}$. For $i=0, \ldots, d-1$ define $$ L^{(i)}=L^i \otimes \mathcal{O}_X(-[\frac{i}{d}\cdot D]) $$ where $[\Delta]$ of a divisor with rational coefficients is the divisor with integer coefficients obtained by taking the integral part of each coefficient.

Esnault-Viehweg proved the following theorem:

Theorem. The sheaf $L^{(i)^{-1}}$ has an integrable logarithmic connection $$ \nabla^{(i)}: L^{(i)^{-1}} \to \Omega^1_X(\log D^{(i)}) \otimes L^{(i)^{-1}} $$ with poles along $D^{(i)}$, the sum over the components $D_j$ such that $\frac{i a_j}{d} \notin \mathbb{Z}$.

Moreover, if char(k)=0, the spectral sequence $$ E_1^{a, b}:=H^b(X, \Omega^a_X(\log D^{(i)}) \otimes L^{(i)^{-1}}) \Longrightarrow \mathbb{H}^{a+b}(X, DR(\nabla^{(i)})) $$ degenerates at $E_1$.

Some comments:

(1) If $\pi: Y \to X$ is the associated cyclic cover branched over $D$, then the connection $\nabla^{(i)^{-1}}$ is the subconnection of $\pi_\ast(\mathcal{O}_Y, d)$ where $\mu_d$ acts by $\xi^i$.

(2) The second part of the theorem is also true for $char(k)=p\neq 0$ if $X$ and $D$ admit a lifting to $W_2(k)$ and $p \geq \dim X$, but I'm only interesting in the characteristic 0 case.

And finally my question:

Does the theorem hold if, instead of assuming $k$ algebraically closed of characteristic 0 I only assume that $k$ contains $\mu_d$. This is clearly necessary to get the eigenvalue decomposition. Is it enough to have the degenerescence of the spectral sequence or there is a subtle point somewhere?

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Yes. Lemma: If $E$ is a spectral sequence of finite dim vector spaces over a field $k$, such that $E_1\otimes L=E_\infty\otimes L$ for some field extension $L$, then $E_1= E_\infty$. Proof: compute dimensions. –  Donu Arapura Mar 31 '13 at 0:19

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