Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

As far as I remember, there 'should exist' an exact etale realization functor from the category of mixed motivic sheaves (over a base scheme $S$) to the category of perverse $l$-adic sheaves over $S$. However, I was not able to find this conjecture in the literature. I would be deeply grateful for any references or details and comments here!

Upd. I would like to understand here the interplay between the residue fields of a scheme and their separable closures.

It seems that I know most of the main reference on the subject; yet I cannot find a very precise answer to my question (possibly it's still there but is difficult to see).

share|improve this question
    
Is this asserted in Huber's book, or does she restrict $S$ to be the spectrum of a field? –  S. Carnahan Dec 18 '10 at 13:14
    
Most of the book is dedicated to motives over a field; possibly, in some small part of it 'relative' motives are considered. –  Mikhail Bondarko Dec 18 '10 at 18:18
    
A (slightly informal) version of such a conjecture is discussed in Jannsen's article in Motives 1 (section 4.8), and he refers to Beilinson's height pairings paper for a reference. –  Bhargav Dec 18 '10 at 18:35
    
I looked at the Beilinson's height pairing paper; yet I was not able to find a precise formulation. –  Mikhail Bondarko Dec 18 '10 at 18:52

1 Answer 1

up vote 0 down vote accepted

For triangulated category of geometric motives over a regular scheme $S$, the $\ell$-adic realisation has been constructed by Florian Ivorra in his thesis. I think the functor is expected to be t-exact for the motivic and perverse t-structure but don't know if it has been explictly written as a conjecture. There is also a chapter about the perverse t-structures in the motivic setting in Ayoub's thesis.

share|improve this answer
    
I am affraid that there could be some complicated details here.:) –  Mikhail Bondarko Dec 18 '10 at 18:17

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.