How to glue perverse sheaves of abelian groups? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T15:39:24Z http://mathoverflow.net/feeds/question/99588 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/99588/how-to-glue-perverse-sheaves-of-abelian-groups How to glue perverse sheaves of abelian groups? Jan Weidner 2012-06-14T10:31:00Z 2012-10-07T00:13:30Z <p>Let $X$ be a complex algebraic variety and consider the category $P(X)$ of perverse sheaves of complex vector spaces.</p> <p>Let $f:X\rightarrow \mathbb C$ be a regular function, $Z$ its zero set and $U$ its complement.</p> <p>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. </p> <p>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.</p> <p>I suspect the answer is yes, because if I checked correctly the glueing construction in "<a href="http://arxiv.org/abs/math/0301098" rel="nofollow">Tilting exercises</a>" 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 "<a href="http://arxiv.org/abs/math/0301098" rel="nofollow">Tilting exercises</a>".</p> http://mathoverflow.net/questions/99588/how-to-glue-perverse-sheaves-of-abelian-groups/109030#109030 Answer by Ryan Reich for How to glue perverse sheaves of abelian groups? Ryan Reich 2012-10-07T00:13:30Z 2012-10-07T00:13:30Z <p>Everything I know about gluing perverse sheaves is written in my paper <a href="http://arxiv.org/abs/1002.1686" rel="nofollow">Notes on Beilinson's "How to glue perverse sheaves"</a>, which deconstructs the construction sufficiently that it is possible to identify the following minimal axioms for making gluing work:</p> <ol> <li><p>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 <em>is</em> 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.</p></li> <li><p>The triangle (4) must hold, but (and this is not stated in the paper) only necessarily for the full nearby cycles, not the unipotent parts. You can check that this modification is valid by looking at the proofs of Propositions 4.5&ndash;6.</p></li> <li><p>The functors $j_!$, $j_*$, $i^*$, and $i_*$ must be defined along open and closed immersions and have the expected properties.</p></li> </ol> <p>The paper uses unipotent nearby cycles exclusively (except for Theorem 4.4 and its prerequisites) but once axiom 1 is known there is no need to restrict to them. <em>However</em>, if you want to know anything about gluing with unipotent nearby cycles you first have to know that they <em>exist</em>; the definition is reasonably obvious but it is also useful that they are a direct summand of the full nearby cycles. And this fact seems to be a consequence of the theory of modules over a PID, which requires vector spaces.</p> <p>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 :)</p>