8
$\begingroup$

I have been reading BBD and Geordie Williamson's “An illustrated guide to perverse sheaves”. The latter has been tremendously helpful to get some intuition for the former.

Let $K$ be some field, and for future use we fix a prime number $\ell$ that is invertible in $K$. (I am particularly interested in finitely generated fields. But I think that this does not matter for what follows.)

Let $f \colon X \to D$ be a Lefschetz pencil of relative dimension $n$. With this I mean: $D = \mathbb{P}^1_K$; $f$ is smooth of relative dimension $n$ over a non-empty open $j \colon U \hookrightarrow D$, and over $i \colon S = (D\setminus U) \to D$ the fibres of $f$ have exactly one ordinary double point and no other singularities, and have dimension $n$. $\def\QQl{\mathbb{Q}_\ell}\def\pR{{}^p\!R}$

Such Lefschetz pencils have been studied in detail, and Picard–Lefschetz theory gives a detailed description of the sheaves $R^qf_*\QQl$. However, I now want to understand the perverse sheaves $\pR^qf_*(\QQl[n+1])$.

Q1. Is there a description of these perverse sheaves in the literature?

The perverse sheaves $F^q = \pR^qf_*(\QQl[n+1])$ are complexes $F^q_{-1} \to F^q_0$, since they are perverse sheaves over a curve. Also, $F^q_0$ is supported on $S$.

For $q \ne n, n+1$ the classical sheaf $R^qf_*\QQl$ is constant. I have the strong feeling that this means that for $q \ne -1,0,1$ the perverse sheaf $F^q$ is isomorphic to $(R^{n+q}f_*\QQl)[1]$. In other words $F^q_{-1}$ is the constant sheaf $(R^{n+q}f_*\QQl)$ and $F^q_0 = 0$.

For $q = -1,0,1$ my guesses are:

  • $F^q$ is also constant for $q = \pm1$. We have $F^{\pm1}_0 = 0$. The sheaf $F^{-1}_{-1}$ is $R^{n-1}f_*\QQl$, whereas $F^1_{-1}$ is $j_*j^*R^{n+1}f_*\QQl$.
  • For $q = 0$ the description depends on whether the vanishing cycles are trivial or not.
    • If the vanishing cycles are non-trivial, then $F^0_0 = 0$ and $F^0_{-1}$ is $R^nf_*\QQl$. In this case $L = j^*R^nf_*\QQl$ is a local system with non-trivial monodromy, and $F^0 = R^nf_*\QQl = j_{!*}L[1]$.
    • If the vanishing cycles are trivial, then $n$ is odd and $F^0_0 = \bigoplus_{s \in S} \QQl(\tfrac{n+1}{2})_s$. The sheaf $F^0_{-1}$ is isomorphic to $R^{n}f_*\QQl$ which in this case is constant.

So far this picture is only based on my (probably misleading) intuition and on what I know about the classical picture.

Q2. How much of my guess is correct? If it is wrong, how should I fix it? And what is the proper way to prove the correct picture?

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.