Let $R$ be a henselian DVR with fraction field $K$ and residue field $k$ of characteristic $p>0$. Let $\overline K$ be an algebraic closure of $K$, $\overline R$ the normalization of $R$ in $\overline K$ and $\overline k$ the residue field of $\overline R$. Let $G=\mathrm{Gal}(\overline K/K)$ be the absolute Galois group of $K$. Write $S, \eta, \overline{\eta}, s, \overline s$ for the $\mathrm{Spec}$ respectively of $R, K, \overline K, k, \overline k$. Let $\Lambda =\mathbb Z_{\ell}$ with $\ell \not = p$ (or $\mathbb Q_{\ell}$ or any torsion sheaf where $p$ is invertible). Let $X$ be a proper scheme over $S$. We consider the nearby cycle sheaf $\mathrm R\Psi\Lambda$ on the geometric special fiber $X_{\overline s}$.

Assume now that $X$ has semi-stable reduction, implying that the special fiber $X_s = \bigcup_{1\leq i \leq r} Y_i$ is a normal crossing divisor. For $E\subset \{1,\ldots,r\}$ let $Y_E := \bigcap_{i\in E} Y_i$ and for $1\leq m \leq r$ define $Y^{(m)} = \bigsqcup_{\#E=m} Y_E$.

In his paper "Exposé 1: autour du théorème de monodromie locale", Illusie explains how to compute the nearby cycle sheaf $\mathrm R\Psi\Lambda$. Théorème 3.2.(c) states that there are isomorphisms of $G-\Lambda-$sheaves on $X_{\overline s}$

  • $\mathrm R^0\Psi\Lambda \simeq \Lambda,$
  • $\mathrm R^1\Psi\Lambda \simeq \left(\bigoplus_{i} \Lambda_{Y_i}/\Lambda\right)(-1)$ where $\Lambda$ injects diagonally in the sum,
  • $\mathrm R^q\Psi\Lambda \simeq \bigwedge^q \mathrm R^1\Psi\Lambda.$

I have a doubt regarding this statement that I'd like to clear up. It is my geometric intuition that $\mathrm R^q\Psi\Lambda$ should be supported on $Y^{(q)}_{\overline s}$, but according to the statement is looks like it is a constant sheaf on the whole of $X_{\overline s}$. May I ask which is true?

(Edit) In regards to Will Sawin's reply, I understood the origin of my confusion. I thought that the notation $\Lambda_{Y_i}$ merely meant a copy of the constant sheaf $\Lambda$ (on the whole special fiber) for each $Y_i$, but it actually denotes the constant sheaf $\Lambda$ supported on $Y_i$.


1 Answer 1


According to this statement, $\mathrm R^q\Psi\Lambda$ is supported on points where $\mathrm R^1\Psi\Lambda $ has rank at least $q$ (by a property of wedge powers) and thus supported on points where $\bigoplus_{i} \Lambda_{Y_i}$ has rank at least $q+1$ (by a property of quotients) and thus on $Y^{(q+1)}$ (since each summand has rank $1$ at $x$ if $x\in Y_i$ and rank $0$ at $x$ otherwise).

  • $\begingroup$ Ah, got it! Thank you very much, it's all clear now. $\endgroup$
    – Suzet
    Oct 11, 2022 at 0:25

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