Alex
• Member for 11 years

As far as I understand, what you need to do to prove that the zeta function of a Shimura variety is automorphic is (I'm ignoring the bad primes here, but I think that we now know enough about them too ...

The snake lemma doesn't make sense in the category of groups in general, because maps don't always have cokernels. If you assume that the three vertical maps in your snake lemma diagram do have ...

I don't know the answer, but here are a few remarks : (0) As has been pointed out before, "irreducible" and "indecomposable" are not the same for representations in positive characteristic. "...

Here is an example. Sorry it's so complicated. (There's probably a simpler one, but my mind works in complicated ways, it seems.) Consider a Siegel modular threefold $U$, i.e., a Shimura variety for ...

No, it's not true. You can say some things about B even if L/K is not separable : B is a dvr if A is, B is a Dedekind ring if A is (that is called the Krull-Akizuki theorem). But it's not true that B ...

The proof in BBD is not that complicated, and it doesn't matter much whether $j$ is affine or not. It uses the three following facts : If $f$ is a morphism of schemes, then $f_*$ sends a complex of ...

Okay, I'm not familiar with Beilinson's paper, but here's my take on this. First let's recall the two definitions. I will denote the triangulated category of motives over a field $k$ by $DM(k)$ (for ...

Gabber has announced a proof of the finiteness theorem for (direct images of constructible sheaves by) morphisms of finite type between general noetherian schemes years ago, but, being Gabber, he has ...

The answer is very likely "yes", but you will need to put together some technical articles (and unpublished results) that may not have yet been put together. Here are the key ingredients, as I see it :...

I think that in this generality the answer is "no". For example, take $\mathcal{C}'=\mathcal{D}'$ equal to some additive category, take the identity as the equivalence, take $\mathcal{C}=\mathcal{C}'$ ...

If you want only $\mathbb{Z}/\ell\mathbb{Z}$ coefficients (not general $\mathbb{Z}/\ell^m\mathbb{Z}$), then there is only one middle perverse t-structure, which is good. The way the exactness ...

I think that all the statements are true, except for 5.3.9 (ii). Remark 5.3.10 says that all the statements in 5 up to and including 5.3.8 are true for $\mathbb{Q}_\ell$-coefficients with the same ...

Do you mean the local Galois representations or the local Galois representations ? The global Galois representations they are constructing correspond to cuspidal automorphic representations of GL(n). ...

Zheng and Liu are using $\infty$-categories to study constructible sheaves on stacks, and they have a $\ell$-adic version too. (Though most of the details for the $\ell$-Adic version should appear in ...

The answer to the question in your title is, I think : "in general, no". The answer to your last question is : "well, it depends how you have defined the objects, and you will have to be very careful ...

I don't know. Do you have a reference for the affine case ? The variety obtained is called "integral closure" (or "normalization") of $V$ in $L$. ;-) A reference is EGA II 6.3, which seems pretty ...

There's a quick proof in Yves André's book "Une introduction aux motifs" (proposition 3.4.6.1). Note that a stronger result is true : actually, algebraic equivalence coincides with numerical ...

(This was going to be a comment, but it's too long.) I don't see how the big étale site appears, even in the proof of cor IV.2.8 of Milne. Seems like he's just base changing to the integral closure ...

I think that the answer is "yes". If you denote by $MFW(X)$ (resp. $MFW(X_\mathbb{C})$) the category of regular holonomic $D$-modules on $X$ (resp. $X_\mathbb{C}$) with a good filtration $F$ and a ...

Let me develop YBL's answer a bit. (I wanted to make this a comment but it was too long...) Consider a smooth variety $U$ over $\mathbb{F}_p$ with function field $K$ such that your motive and its two ...
(1) is not always an isomorphism when $f$ is an open immersion. (Take $X=Y$ equal to an open subscheme of $Z$, with the obvious maps.) Here is why : when you try to show that the restriction of $g_*q_!... View answer 1 votes The answer to your first (less general) question is this : Let$A$be a$n\times n$matrix with coefficients in$\mathbb{Z}$. Then$A$is similar over$\mathbb{Z}$to a full Jordan block if and only ... View answer 0 votes Is equivariant homology dual to equivariant cohomology ? Because there is a paper of Goresky-Kottwitz-MacPherson that gives condition for the equivariant cohomology of$X$to inject into that of$X^G\$;...