Suppose I am in the setting of the decomposition theorem, i.e., we have the decomposition of the direct image $f_*\mathbb Q_\ell$, where $f:X\to Y$ is proper. Then the direct image decomposes into a direct sum of shifted $IC$ extensions of semisimple local systems.

I would like to know a kind of converse to this question, that is, given a (semi)simple perverse sheaf, say some intermediate extension of a local system on $S\subset Y$, how can I determine whether it occurs in the decomposition above?

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    $\begingroup$ Well, if you sheaf is an intermediate extension from a certain $S$, it certainly makes sense to restrict $f_*Q_l$ to $S$. $\endgroup$ – Mikhail Bondarko Feb 28 '15 at 9:50
  • $\begingroup$ Moreover, you can restrict further to the generic point $Spec K$ of $S$. Then it remains to study the corresponding continuous $Q_l$-representations of the Galois group of $K$. $\endgroup$ – Mikhail Bondarko Feb 28 '15 at 21:46
  • $\begingroup$ Maybe I'm not understanding this, but how does the restriction tell me whether my local system on $S$ occurs in the decomposition of $f^*\mathbb Q_l$? $\endgroup$ – TA Wong Mar 1 '15 at 5:10
  • $\begingroup$ $j_{!*}L$ occurs in the decomposition of a semi-simple perverse $P$ if and only if $L$ is in the decomposition of $j^*P$. $\endgroup$ – Mikhail Bondarko Mar 1 '15 at 9:53
  • $\begingroup$ @TAWong ... into a direct sum of shifted IC extensions of semisimple local systems. $\endgroup$ – Daniel Juteau Mar 14 '15 at 20:34

In general it is a difficult problem. For example, the core of Ngô's proof of the fundamental lemma is his support theorem which implies that in the context of the Hitchin fibration, all simple constituents of the direct image have full support.

If the morphism is semismall, then one has a direct sum of IC's without shift and the situation is simpler. Only relevant strata (for which the semismall inequality is an equality) can have local systems whose IC extension can contribute to the direct image. This was discussed in the other question. The top cohomology of a stratum has an action of the fundamental group of the stratum, by permutation of the top dimensional irreducible components (hence the action factors through the finite group of permutation of the components, which implies easily the semisimplicity in this case). For a relevant stratum, this representation corresponds to the local system you have to take the IC of. So the irreducible constituents of the representation tell you which simple IC's appear. For example, for a relevant stratum, the trivial local system always appears. Example: the Springer resolution. The simple IC's that appear are the image of the Springer correspondence. I don't know a general argument to compute the image without actually computing the correspondence.

In the general case, I assume you know the stalks the IC's which don't have full support (assuming you are doing an induction and you don't already know the big IC's: otherwise, the explanations below might or might not be more relevant). Then you want to spot the factors which occur with a negative shift (so they're shifted to the right). Take the biggest stratum for which the perverse conditions are not satisfied on the right, take the highest nonvanishing cohomology sheaf of the restriction of the direct image to that stratum (it is a local system). Then you know the IC of this (maybe reducible) local sytem, with the appropriate shift, is a direct summand. You may split it off the direct image, and by duality also the IC of the dual local system with the opposite shift. Go on like this until you are left with a perverse sheaf, and do as in the preceding paragraph (the IC's of the local systems which are on the maximally allowed cohomological degrees).

You can try this in small examples, but if you are doing something general it does not seem very nice. Are you working in a particular situation? Then try to use the particularities of the situation. For example, see the proof that the KL basis of the Hecke algebra corresponds to IC sheaves on the flag variety, in Springer's Bourbaki seminar.

On the other hand, if you already know all the stalks of all IC's (including the ones which are fully supported), and if you can keep track of degrees in the Grothendieck group (maybe exploiting some purity properties, like in the KL setting), then all you need to do is to compute the stalks of the direct image (the cohomology of the fibers), and find the decomposition in the (enriched by degree) Grothendieck group. If you can't keep track of the degrees, you will only know the alternating sum of the contributions, some cancellations might occur.

It's hard to explain all this without a blackboard, but I hope it helps! Don't hesitate to ask for more explanations. I (and others) might be able to give you more specific answers, depending on the particular situation.

  • $\begingroup$ Thanks! This was quite helpful. I am in fact looking at a special case of the Hitchin Fibration where the Support Theorem is already available. I have a direct image from a different base, and want to check that it occurs in the decomposition of the constant sheaf. Do you know some references that may be relevant? $\endgroup$ – TA Wong Mar 16 '15 at 14:22
  • $\begingroup$ Ok. The first paragraph of my answer was very imprecise and only based on my recollection of Ngô's talks, it was just meant as an indicator of the difficulty in general. Well if you are in that case then Ngô's paper (Le lemme fondamental pour les algèbres de Lie, Publ. Math. IHES 111, 2010) is the definitive source. See in chapter 7 for the support theorem, and section 7.8 for the application to the Hitchin fibration. Indeed nontrivial $\kappa$ the support can be the (closed) image of the Hitchin base for some endoscopic group $H$. I had forgotten that bit! ... $\endgroup$ – Daniel Juteau Mar 16 '15 at 21:57
  • $\begingroup$ Moreover, the conclusions hold a priori only on the "good" part, where the $\delta$ condition is satisfied. The choice of the divisor $D$ in 8.6.4 ensures that the "bad" part is small enough. I hope that you also have the freedom to choose $D$, otherwise, it will be difficult to prove (from what I understood, Ngô conjectures that the "bad" part is always empty, and it is true in characteristic $0$, but he could not prove it in characteristic $p$). Disclaimer: I'm by no means an expert... $\endgroup$ – Daniel Juteau Mar 16 '15 at 22:04
  • $\begingroup$ The trouble I am having is the this: by the support theorem I know that if a perverse factor occurs in the decomposition, then its support is completely determined (by such and such $\kappa$..). Now I have a perverse sheaf coming from elsewhere, and I want to say that it is isomorphic to one such perverse factor. Thanks for your answers! They are helpful to think about. $\endgroup$ – TA Wong Mar 21 '15 at 5:23
  • $\begingroup$ usually, to show that they are isomorphic, you have to consider some nice enough open locus of your strata where you can compute the corresponding function via the function-sheaf dictionnary. Then, you have to see if they match. $\endgroup$ – prochet Apr 8 '15 at 16:31

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