# How to glue perverse sheaves of abelian groups?

Let $X$ be a complex algebraic variety and consider the category $P(X)$ of perverse sheaves of complex vector spaces.

Let $f:X\rightarrow \mathbb C$ be a regular function, $Z$ its zero set and $U$ its complement.

A glueing data consists of a tuple $({\cal F}_U,{\cal F}_Z, u,v)$, where ${\cal F}_U$ and ${\cal F}_Z$ are perverse sheaves on $U$ and $Z$ and $u,v$ are maps going between the nearby cycles $\psi({\cal F}_U)$ and ${\cal F}_Z$ such that $u\circ v=1-T$, where $T$ is the monodromy operator on $\psi({\cal F}_U)$. These glueing data form a category $Glue(U,Z)$ in an obvious way.

Now it is known, that $Glue(U,Z)$ is equivalent to $P(X)$. My question is, does this still hold if we take instead of perverse sheaves of complex vectorspaces perverse sheaves of say abelian groups? I am also interested in the variant where one takes unipotent nearby cycles.

I suspect the answer is yes, because if I checked correctly the glueing construction in "Tilting exercises" works over the integers. On the other hand I must admit that I neither understand the details of the above glueing construction very well nor its precise relation to the one in "Tilting exercises".

1. The nearby cycles functor $\psi[-1]$ must be t-exact for the perverse t-structure. I prove this using its unipotent part $R\psi^\text{un}$ and the triangle $$i^* j_* \to R\psi^\text{un} \xrightarrow{1 - T} R\psi^\text{un} \to \qquad (4)$$ (numbering as in the paper), which is valid without the unipotent superscript but is not useful for this particular theorem. However, this is the only place that unipotence is required in this triangle.
3. The functors $j_!$, $j_\*$, $i^\*$, and $i_\*$ must be defined along open and closed immersions and have the expected properties.
To see how axioms 1, 2, and 3 imply gluing, consult the last paragraph of my paper. I would say (not knowing anything about non-field coefficients) that they all probably remain true, particularly 2 and 3; most likely 1 does as well, though the proof may be more difficult without being able to use the Jordan decomposition of Lemma 4.2 together with the perversity of $\psi^\text{un}[-1]$ that is shown in Lemma 1.2, which relies on (of course) properties of $\psi^\text{un}$ that seem to rely on a more limited form of Jordan decomposition as well. However, this is not the only proof of axiom 1 that exists, merely the easiest :)