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I have a few questions relating to the BBD decomposition theorem.

I have come across the following two versions of the decomposition theorem.

Version 1. Let $f : X \to Y$ be a proper map of complex algebraic varieties and suppose that $X$ is smooth. Let $IC_X = \mathbb{C}_X[d_X]$ be the "constant perverse sheaf" on $X$. Then the derived direct image $f_*IC_X$ is a (finite) direct sum of simple perverse sheaves on $Y$, i.e., a direct sum of shifted intersection cohomology complexes $IC(Z,\chi)[i]$, where $Z$ ranges over irreducible closed subvarieties of $X$ and $\chi$ over irreducible local systems on a Zariski open subset of the regular part of $Z$.

There is also a different version.

Version 2. Let $f: X \to Y$ be a proper map of complex algebraic varieties. Let $K$ be a constructible complex which is semisimple of geometric origin. Then $f_*K$ is also semisimple of geometric origin.

I understand that the second version is more general than the first. What puzzles me a little is that the first version of the theorem does not mention perverse sheaves of geometric origin at all. I know that if $X$ is irreducible or smooth then $IC_X$ is in fact semisimple of geometric origin. Now version 1 seems to suggest that, in the setting of that version of the theorem, every simple perverse sheaf on $Y$ is of geometric origin. Is this true? Or are some simple perverse sheaves not of geometric origin, so they cannot occur in the decomposition, but version 1 of the theorem fails to mention this fact? Or does one perhaps need to impose some stronger assumptions on the variety $Y$, like projectivity? I would also appreciate any further comments/insight about how these two versions of the theorem relate to each other and precisely what assumptions one needs to make for these theorems to hold.

There is also a refinement of the first version of the theorem.

Version 1.a. In the setting of Version 1, suppose that there exists an algebraic stratification of $Y = \sqcup_\lambda S_\lambda$ such that $f : f^{-1}(S_\lambda) \to S_\lambda$ is a locally trivial topological fibration. Then the direct summands in the decomposition of $f_*IC_X$ are of the form $IC(\overline{S_\lambda},\chi)$, where $\chi$ is an irreducible local system on the stratum $S_\lambda$.

Question: is there an easy way to see how one can deduce Version 1.a from Version 1?

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  • $\begingroup$ I'm confused by your confusion. Version 1 doesn't mention that the summands are of geometric origin, but I believe they are obviously are by the definition of the geometric origin, so there's no need to mention it. Even if it is a stronger statement, that would just mean that Version doesn't state the theorem in its strongest form, which we've already established. $\endgroup$
    – Ben Webster ♦
    Commented Jul 30, 2015 at 1:29
  • $\begingroup$ @BenWebster That's right, $IC_X$ is of geometric origin in Version 1. I was rather wondering about the $IC(Z,\chi)$'s occuring in the decomposition. Version 2 states they should be of geometric origin, Version 1 does not. So my question was: is there some (hidden) assumption in Version 1 which guarantees that any simple perverse sheaf is necessarily of geometric origin? It seems to me now that this is not the case - Version 1 is probably indeed weaker, and fails to mention that only $IC(Z,\chi)$'s of geometric origin (i.e., where the monodromy of $\chi$ is finite) can occur. $\endgroup$
    – Balerion_the_black
    Commented Jul 30, 2015 at 2:06
  • $\begingroup$ I don't think $\chi$ must have finite monodromy. Do you mean that the local monodromy is quasi-finite? $\endgroup$
    – Will Sawin
    Commented Jul 30, 2015 at 4:33
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    $\begingroup$ I agree with Ben Webster - what's your definition of geometric origin? I would say that $K$ is of geometric origin if it can be obtained from the trivial perverse sheaf over a point by applying the standard six functors and forming subquotients; in particular, every summand of $f_\ast K$ in Version 2 ought to be of geometric origin by definition, and the only nontrivial part of the statement is semisimplicity. $\endgroup$
    – Dan Petersen
    Commented Jul 30, 2015 at 10:54
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    $\begingroup$ @Toan at a glance I don’t think I understand what you’re asking. Maybe you should ask this as a new question. $\endgroup$
    – Dan Petersen
    Commented Oct 19 at 12:25

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