User alex - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T06:48:36Z http://mathoverflow.net/feeds/user/12336 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/94001/semisimplicity-of-automorphic-galois-representations/94176#94176 Answer by Alex for semisimplicity of automorphic Galois representations Alex 2012-04-16T03:36:52Z 2012-04-16T03:36:52Z <p>Do you mean the local Galois representations or the local Galois representations ?</p> <p>The global Galois representations they are constructing correspond to cuspidal automorphic representations of GL(n). They are expected to be always irreducible, though I'm not sure when this is known exactly. But it is known in the case that Harris and Taylor consider (when the automorphic representation is square integrable at a finite place), cf corollary 1.3 of the article "Compatibility of local and global Langlands correspondences" by Taylor and Yoshida.</p> <p>The local Galois representations are not expected to be semi-simple in general. They are expected to be Frobenius semi-simple (ie, the Frobenius elements are supposed to act semi-simply), but this is not known for $n\geq 3$. So, if you mean the local representations, then yes, very often people are just taking the Frobenius semi-simplifications of the representations that appear in the cohomology of Shimura varieties.</p> http://mathoverflow.net/questions/69035/the-category-of-l-adic-sheaves/93605#93605 Answer by Alex for The category of l-adic sheaves Alex 2012-04-09T21:07:56Z 2012-04-09T21:07:56Z <p>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 a second paper that is still in preparation, and I would not call their first paper easy to read. But it is certainly a modern presentation...) Reference : <a href="http://math.columbia.edu/~zheng/bc1.pdf" rel="nofollow">http://math.columbia.edu/~zheng/bc1.pdf</a></p> <p>By the way, they use Gabber's finiteness results, and there is now a nice reference for these too ! (This is really cool.) <a href="http://www.math.polytechnique.fr/~orgogozo/travaux_de_Gabber/" rel="nofollow">http://www.math.polytechnique.fr/~orgogozo/travaux_de_Gabber/</a></p> http://mathoverflow.net/questions/84632/equivalent-forms-of-the-proper-base-change-isomorphism/85748#85748 Answer by Alex for Equivalent forms of the proper base change isomorphism Alex 2012-01-15T16:36:48Z 2012-01-15T16:36:48Z <p>(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 $g_*q_!$ to the closed complement is $0$, you will want to use a base change isomorphism which is not true in general (it is true if $g$ is proper, but then (1) is trivially an isomorphism). </p> <p>Anyway, so (1) is not always an isomorphism. </p> <p>Neither is (2), and it doesn't matter whether $f$ is proper or not. Take $Z=\mathbb{A}_2$ (over some field, say), $X=$ one of the axes, $Z=$ the origin, with the obvious maps (so $P=Z$, and $f$ is a closed immersion, hence proper), and look what happens for the constant sheaf $\Lambda$ on $Z$ ($\Lambda$ is, for example, a finite ring of torsion prime to the characteristic of the base field). Then $f^!\Lambda=\Lambda(-1)[-2]$ by purity, so the rleft-hand side of (2) is $\Lambda(-1)[-2]$, while the right-hand side is $\Lambda$.</p> http://mathoverflow.net/questions/81040/do-etale-neighhbourhoods-of-a-subvariety-descend-along-base-field-extensions-doe/81089#81089 Answer by Alex for Do etale neighhbourhoods of a subvariety descend along base field extensions; does normalization commute with etale base change? Alex 2011-11-16T17:04:55Z 2011-11-16T17:04:55Z <ol> <li><p>I don't know. Do you have a reference for the affine case ?</p></li> <li><p>The variety obtained is called "integral closure" (or "normalization") of $V$ in $L$. ;-) A reference is EGA II 6.3, which seems pretty canonical. EGA II 6.3.4 tells you that for any scheme $X$, for any quasi-coherent $\mathcal{O}_X$-algebra $\mathcal{A}$, you can construct the integral closure of $X$ in $\mathcal{A}$, which is an affine scheme $X'$ over $X$.</p></li> <li><p>It seems that it should generalize trivially, maybe I am missing something ? If I understand your question correctly, you have a scheme $X$, a quasi-coherent <code>$\mathcal{O}_X$</code>-algebra $\mathcal{A}$, the integral closure $X'\rightarrow X$ of $X$ in $\mathcal{A}$ and an étale map $f:Y\rightarrow X$. You define $\mathcal{B}$ to be the pull-back of $\mathcal{A}$ (i.e. <code>$f^{-1}\mathcal{A}\otimes_{f^{-1}\mathcal{O}_X}\mathcal{O}_Y$</code>), and you want to show that $Y':=X'\times_X Y$ is the integral closure of $Y$ in $\mathcal{B}$. You can assume that $X$ is affine. Then, for every affine open $U$ of $Y$, the proposition you quote tells you that $Y'_U\rightarrow U$ is the integral closure of $U$ in $\mathcal{B}_U$, which, if I understand EGA II 6.3.4, is just saying that $Y'$ is the integral closure of $Y$ in $\mathcal{B}$.</p></li> </ol> http://mathoverflow.net/questions/80817/when-does-the-equivariant-homology-of-the-fixed-part-of-a-g-space-surject-onto/80823#80823 Answer by Alex for When does the equivariant homology of the fixed part of a $G$-space surject onto the equivariant homology of the whole space? Alex 2011-11-13T15:12:06Z 2011-11-13T15:12:06Z <p>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 $X^G$; they even calculate the image in some cases. The paper is "Equivariant cohomology, Koszul duality, and the localization theorem", it's available on Goresky's webpage.</p> http://mathoverflow.net/questions/80362/finiteness-of-normalization-of-noetherian-normal-domain/80374#80374 Answer by Alex for Finiteness of normalization of Noetherian normal domain Alex 2011-11-08T10:45:10Z 2011-11-08T10:45:10Z <p>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 is always finite over A, even if A is a dvr. There is a counterexample in theorem 100 of Kaplansky's "Commutative rings".</p> <p>By the way, an integral domain A such that the integral closure of A in any finite extension of its fields of fractions is finite over A is called a Japanese ring. The wikipedia article on Nagata rings gives examples of Japanese rings. Searching for "non Japanese discrete valuation rings" will give you other counterexamples to your question (or at least other references).</p> http://mathoverflow.net/questions/80036/when-do-adjunctions-preserve-equivalence/80043#80043 Answer by Alex for When do adjunctions preserve equivalence? Alex 2011-11-04T12:20:59Z 2011-11-04T12:20:59Z <p>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 $\mathcal{C}=\mathcal{C}'$ and $\mathcal{D}=0$. This seems to satisfy your hypotheses.</p> http://mathoverflow.net/questions/55790/bad-behaviour-of-perverse-sheaves-over-general-bases/79560#79560 Answer by Alex for Bad behaviour of perverse sheaves over 'general' bases? Alex 2011-10-30T23:37:49Z 2011-10-30T23:37:49Z <p>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 :</p> <p>(1) General definition of perverse t-structure, not using stratifications (using stratifications is probably a bad idea in general), not using finiteness conditions, no need for a base field either : see Gabber's article "Note on some t-structures". Section 8 is of particular interest, since this is where he defines perverse t-structures on étale n-torsion sheaves on a noetherian scheme X, provided that X admits a dualizing complex (a priori in the sense of SGA 5 I). He also gives a condition on the perversity that guarantees that the perverse truncation functors will respect constructibility.</p> <p>(2) So you want dualizing complexes. Deligne has proved that they exist if X is of finite type over a regular basis S of dimension $\leq 1$ (SGA 4 1/2, [Th finitude], by the way the dualizing complex is what you might expect i.e. the exceptional inverse image of the constant sheaf on S). More generally, Gabber's recent (mostly unpublished) results include the existence of a dualizing complex over any noetherian excellent scheme that admits a dimension function (where "dualizing complex" has a slightly weaker definition that in SGA 5 I, but I don't think it matters for the application to (1)). A remark : for the first case (X of finite type of S regular of dimension $\leq 1$, the dimension function on X would simply be Artin's rectified dimension, if I remember well. In general, I think that the existence of a dimension function on X is equivalent to the fact that X is universally catenary (which is part of the definition of "excellent", so I am confused).</p> <p>(3) In any case, suppose you're okay with (1) and (2), so you have your dualizing complex, your perverse t-structure, you chose the perversity function so the truncation functors are compatible with constructibility, and let's say you even chose the self-dual perversity, that is, the perversity is related to the dimension function in the usual way. Then your category of constructible perverse sheaves should be stable by duality, and artinian (use theorem 8.3 of the article of Gabber I cited in (1)).</p> <p>(4) Now you want to study the exactness properties of the 4 operations, so probably you'll be happy to use Gabber's finiteness theorems so constructible things stay constructible, so you'll restrict to morphisms of finite type (and the schemes already have to be noetherian excellent). The exactness properties of direct images by affine maps also follow from results of Gabber (he generalized SGA 4 XIV; if you're of finite type over a trait, the result was proved by Gabber too, but much earlier). I think these are the central results, the t-exactness of shifted $f^*$ for $f$ smooth should just come from duality.</p> <p>(5) And finally, you want to go $\ell$-adic. Then I don't know anything better than Ekedahl's results (maybe they could be generalized to other schemes using Gabber's finiteness theorem ? I haven't tried so I can't be sure). So you'll finally need to restrict to the case where X is of finite type over S regular of dimension $\leq 1$. Note that in this case the existence of the dualizing complex and the finiteness results are already in SGA 4 1/2, the construction of the t-structures is in a <em>published</em> paper of Gabber, so the only possible problem is affine Lefschetz (the exactness properties of affine direct images), which as far as I understand has been known for more than 15 years in this case, but I'm unable to cite a reference right now.</p> <p>(6) Well, I guess you'd probably be interested in the nearby and vanishing cycles functors next. Over a dvr or a general base ? ;) Good luck anyway, I don't think I know any reference (published or not).</p> http://mathoverflow.net/questions/45948/is-there-a-k-structure-for-hodge-modules-over-a-k-variety/79553#79553 Answer by Alex for Is there a $k$-structure for Hodge modules over a $k$-variety? Alex 2011-10-30T22:42:57Z 2011-10-30T22:42:57Z <p>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 finite filtration $W$, then there are obvious functors $MFW(X)\rightarrow MFW(X_\mathbb{C})$ and $MHM(X_\mathbb{C})\rightarrow MFW(X_\mathbb{C})$, where $MHM(X_\mathbb{C})$ is the category of mixed Hodge modules on $X_\mathbb{C}$. So you can consider the fiber product of $MFW(X)$ and $MHM(X_\mathbb{C})$ over $MFW(X_\mathbb{C})$, and indeed things should work nicely for this category, at least if you believe Saito (the construction is taken from his paper "On the formalism of mixed sheaves", <a href="http://arxiv.org/abs/math/0611597" rel="nofollow">here</a>; it's example 1.8(ii)).</p> <p>(I don't think that this paper of Saito has appeared in any peer-reviewed journal, so you should check whatever you want to take from it. It's a pity because it's a nice reference.)</p> http://mathoverflow.net/questions/79376/reference-for-numerical-vs-homological-equivalence/79382#79382 Answer by Alex for Reference for Numerical vs Homological equivalence Alex 2011-10-28T12:32:33Z 2011-10-28T12:52:57Z <p>There's a quick proof in Yves André's book "Une introduction aux motifs" (proposition 3.4.6.1). </p> <p>Note that a stronger result is true : actually, algebraic equivalence coincides with numerical equivalence for divisors on $X$ (*)("Matsusaka's theorem"). The only reference I know for this result is Matsusaka's original article ("The criteria for algebraic equivalence and the torsion group", Amer. J. Math. 79 (1957), 53–66), and I don't know if you would consider it a good reference.</p> <p>(*) If your coefficients contain $\mathbb{Q}$, that is. If you take coefficients in $\mathbb{Z}$, then the result is that the group of divisors algebraically equivalent to zero is of finite index in the group of divisors numerically equivalent to zero.</p> http://mathoverflow.net/questions/76186/functoriality-properties-of-the-perverse-t-structure-for-torsion-constructible/76213#76213 Answer by Alex for Functoriality properties of the perverse $t$-structure for torsion (constructible complexes of) sheaves Alex 2011-09-23T15:55:07Z 2011-09-23T15:55:07Z <p>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 properties of the 4 operations is proved in BBD is to reduce to $\mathbb{Z}/\ell/\mathbb{Z}$ coefficients (see 4.0), so the answer to 1 is "obviously yes" and the answer to 2 is "BBD 4.1".</p> <p>As for the fact that the shifted constant sheaf on a lci variety is perverse, it only uses the results of BBD 4, and again, these are all true for $\mathbb{Z}/\ell\mathbb{Z}$ coefficients as explained in BBD 4.0. (The problems start appearing when the coeffients are not a field.) So in that case too the answer to 1 is "yes" and the answer to 2 is "BBD 4".</p> <p>NB : Here "BBD" refers to the book "Analyse et topologie sur les espaces singuliers I", by Beilinson-Bernstein-Deligne (aka Asterisque 100).</p> http://mathoverflow.net/questions/76069/finiteness-of-etale-cohomology-groups/76070#76070 Answer by Alex for Finiteness of étale Cohomology Groups Alex 2011-09-21T18:40:08Z 2011-09-21T18:40:08Z <p>(This was going to be a comment, but it's too long.)</p> <p>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 of the field but using small étale sites all time. (Though I'm not that familiar with Milne, as I use SGA 4 and 4 1/2 as references.)</p> <p>Anyway, I think the problem with the proof of corollary IV.2.8 is when he says it just follows from the Hochschild-Serre spectral sequence. I would say the Hochschild-Serre spectral sequence gives you a spectral sequence</p> <p><code>$E_2^{pq}=H^p(Gal(k_s/k),H^q(X\otimes k_s,F))\Longrightarrow H^{p+q}(X,F)$</code></p> <p>($k_s$ is the separable closure of $k$ as in Milne)</p> <p>Then Milne IV.2.1 tells you that the groups $H^q(X\otimes k_s,F)$ are finite, but it doesn't follow that their $Gal(k_s/k)$-cohomology is finite too. In fact Milne himself gives a counterexample in the part of his notes that you quote.</p> <p>(Does that make sense ?)</p> http://mathoverflow.net/questions/75632/which-shimura-varieties-are-known-to-be-automorphic/75684#75684 Answer by Alex for Which Shimura varieties are known to be automorphic? Alex 2011-09-17T16:57:29Z 2011-09-17T16:57:29Z <p>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 - I can develop later if you want) :</p> <p>(1) Do some kind of point-counting over finite fields. For PEL Shimura varieties of type A and C, Kottwitz has done it. Actually it's a bit more, you calculate the trace of a Hecke operator times a power of the Frobenius (at a place of good reduction) on the cohomology with compact support.</p> <p>(2) "Stabilize" the resulting formula so you can compare it with Arthur's stable trace formula. Here there is a choice. As the reason we expect the trace of a Hecke operator to compare well to the trace formula is Matsushima's formula, as Matsushima's formula for noncompact Shimura varieties (due to Borel and Casselman in that case) is a formula for the L^2 cohomology of the Shimura variety, and as the algebraic avatar of this L^2 cohomology is the intersection cohomology of the minimal compactification, you can choose to first extend the result of (1) to this intersection cohomology, and then "stabilize" whatever you get. Or you can ignore this and choose to work with compact support cohomology, at the cost of greater complication on the trace formula side. Laumon followed that approach for Siegel modular threefolds, but as far as I know, in other cases people generally choose the first approach. Anyway, this is not a problem for compact Shimura varieties. For compact PEL Shimura varieties of type A and C, this "stabilization" part is also due to Kottwitz (in the first part of his beautiful Ann Arbor article "Shimura varieties and $\lambda$-adic representations"), we also know some noncompact PEL cases of type A and C by Morel's work, and basically all other type A and C PEL cases should reduce to Kottwitz's calculations, but I don't think this is written anywhere.</p> <p>Oh, and you need the fundamental lemma for this part.</p> <p>(3) Now you still need to compare the result of (2) with the stable trace formula, so obviously you need to know the stable trace formula (here you need the weighted fundamental lemma), and to make sense of the results and get a nice formula for your zeta functions you also want to know Arthur's conjectures on the classification of discrete automorphic representations. How you get the zeta function formula if you assume Arthur's conjectures is explained in the second part of Kottwitz's previously-cited beautiful Ann Arbor article. </p> <p>So the question is, what do we know about Arthur's conjectures ? Well, they are accessible. Arthur is supposed to be writing a proof in the case of symplectic groups, and he is actually making progress on it. Note however that for Siegel modular varieties, you'll need general symplectic groups, so there will be a further reduction step even after Arthur finishes writing his book. (But we are nearing a proof of the automorphy of the zeta function. Yay !) There are a few young and brave ones who are planning to tackle the case of unitary groups (if I remember well, Sug Woo Shin, Tasho Kaletha, Paul-James White and Alberto Minguez). A last word of caution, all this (Arthur's work and the others' future work) depends on the stabilization of the twisted trace formula, which is at the moment not totally written down, I'm afraid, but there's a group of Serious People in Paris (like Clozel, Waldspurger etc) who have vowed to take care of it.</p> <p>So, to sum up, It's Complicated, but we seem to be close for PEL Shimura varieties of type A and C, especially Siegel modular varieties. Also, if you just want the zeta function to be a product of automorphic L-functions with complex exponents (and not integral exponents), then I think that's known for PEL case A (maybe not written in all cases, though); whether you can make these exponents rational without too much additional work, I am not sure (it seems that this should be an easier thing than proving they're integers).</p> http://mathoverflow.net/questions/54516/is-there-an-easy-proof-of-the-fact-that-the-intermediate-image-functor-respects-w/54542#54542 Answer by Alex for Is there an easy proof of the fact that the intermediate image functor respects weights? Alex 2011-02-06T18:19:45Z 2011-02-08T16:46:47Z <p>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 :</p> <ul> <li><p>If $f$ is a morphism of schemes, then $f_*$ sends a complex of weight $\geq a$ to a complex of weight $\geq a$, and $f_!$ sends a complex of weight $\leq a$ to a complex of weight $\leq a$ (a very natural property of weights; of course that's not so easy to prove for weights of $\ell$-adic complexes, and it is the main result of Deligne's Weil II). Cf BBD 5.1.14.</p></li> <li><p>If $K$ is an $\ell$-adic complex, then $K$ is of weight $\leq a$ (resp. $\geq a$) if and only if, for every $k\in\mathbb{Z}$, the $k$th perverse cohomology sheaf of $K$ (call it <code>${}^pH^k K$</code>) is of weight $\leq a+k$ (resp. $\geq a+k$). Cf BBD 5.4.1. Again, hard to prove, but a natural enough property of weights, and a reason in my opinion why perverse sheaves are so much more natural than constructible sheaves (one out of many).</p></li> <li><p>If $j$ is a locally closed immersion (more generally, a quasi-finite map), then <code>$j_{!*}$</code> is the image of <code>${}^pH^0j_!$</code> in <code>${}^pH^0j_*$</code>. This is the definition of the intermediate extension.</p></li> </ul> <p>Now the result you want is obvious : Take <code>$j:X\rightarrow Y$</code> a quasi-finite morphism. If the perverse sheaf $K$ on $X$ is of weight $\leq a$, then <code>$j_!K$</code> is of weight $\leq a$ (as a complex), so the perverse sheaf <code>${}^pH^0j_!K$</code> is of weight $\leq a$, and so is its quotient <code>$j_{!*}K$</code>. Likewise for weights $\geq a$, using this time $j_*$. </p> <p>Note that you could also define $j_{!*}K$ (for $K$ pure of weight $a$) as the weight $\leq a$ part of $j_*K$, or as the weight $\geq a$ part of $j_!K$. I think it's not too hard to recover the usual properties of $j_{!*}K$ from that definition, but I would have to think more to see how to make it work for mixed (but not pure) perverse sheaves.</p> <p>Edited to add two remarks :</p> <p>(1) I don't think that it is so hard to go from the affine case to the general case. Consider an open embedding $j:U\rightarrow X$, let $i:Y\rightarrow X$ be the complement. Let $\pi:X'\rightarrow X$ be the blowup of $Y$ in $X$, and $j':U\rightarrow X'$ be the inclusion. Then $j'$ is affine, and, for every perverse sheaf $K$ on $U$, <code>$j_{!*}K$</code> is a direct factor of <code>${}^pH^0\pi_*j'_{!*}K$</code>, so the result for <code>$j_{!*}K$</code> follows if you know it for <code>$j'_{!*}K$</code>, without any need of BBD 5.3.1. (You don't need the decomposition theorem to prove my claim. It is an exercise in perverse sheaves to prove that the map <code>${}^pH^0\pi_*j'_!K={}^p H^0j_!K\rightarrow j_{!*}K$</code> factors through a map <code>${}^pH^0\pi_*j'_{!*}K\rightarrow j_{!*}K$</code>. Likewise, or by duality, there is a natural map <code>$j_{!*}K\rightarrow{}^pH^0\pi_*j_{!*}K$</code>. The composition <code>$j_{!*}K\rightarrow{}^pH^0\pi_*j'_{!*}K\rightarrow j_{!*}K$</code> is the identity when restricted to $U$, so it is the identity.)</p> <p>(2) If $K$ is pure, there is a slightly different way to prove what you want (you might be able to do something if $K$ is mixed too, but I didn't try to work it out). Notation : $j$ is an open immersion from $U$ to $X$. First, the problem is local in $X$, so you can assume that $X$ is affine. Then $Y:=X-U$ is defined by a finite number of functions on $X$. By induction over the number of functions necessary to define $Y$, you can reduce to the case where there exists a function <code>$f:X\rightarrow\mathbb{A}^1$</code> such that <code>$Y=f^{-1}(0)$</code>. Now you can use the result of Beilinson-Bernstein (cf "A proof of Jantzen conjectures") that the Jantzen filtration on <code>$j_!K$</code> coincides with (a shift of) the weight filtration if $K$ is pure. The Jantzen filtration on <code>$j_!K$</code> is induced by the monodromy filtration on the maximal extension <code>$\Xi_f K$</code>, and it is an exercise to identify the quotient <code>$j_{!*}K$</code> of <code>$j_!K$</code> with one of the graded pieces of this filtration and to conclude that it has the expected weight. This proof avoids BBD 5.2, but it relies on the article of Beilinson-Bernstein instead; as fat as I can tell, the methods Beilinson-Bernstein use to prove the result that you need are natural extensions of the methods of Weil II, and you have to assume Weil II anyway, so maybe this is slightly more natural.</p> http://mathoverflow.net/questions/54623/is-it-easy-to-define-weights-for-q-l-sheaves-over-finite-type-z1-l-schemes/54780#54780 Answer by Alex for Is it easy to define weights for $Q_l$-sheaves over finite type $Z[1/l]$-schemes? Alex 2011-02-08T15:52:15Z 2011-02-08T15:52:15Z <p>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 about which morphisms of schemes you allow".</p> <p>Let me be more precise. Huber wants to consider only complexes of sheaves on $S_0$ that are constructible with respect to a stratification all of whose strata are smooth over $U_0$ (I will explain why later). But direct images by morphisms of $U_0$-schemes do not respect this kind of constructibility in general, they only respect it after you restrict to an open subscheme of $U_0$, so she has to do that and take the direct limit over all open subschemes of $U_0$. See the end of section 2, where she explains that you could stay on $U_0$ if you put supplementary conditions on the morphisms of schemes (for example, proper and smooth morphisms between smooth $U_0$-schemes will be fine).</p> <p>Now here are the reasons why she chooses this restrictive notion of constructibility :</p> <p>(1) She needs it to show that her definition of the perverse $t$-structure works. See the remark after theorem 2.5. I am not sure that this is necessary, because now we have more general ways to construct $t$-structures on categories of complexes of sheaves (see Gabber's "Note on some $t$-structures"), although there is still the problem of checking that your category of perverse sheaves satisfies the usual conditions (finiteness, description of simple objects etc) or of showing that you don't really need these things.</p> <p>(2) She defines weights by restricting her complexes over the fibers over closed points of $U_0$, and she wants to prove all the properties of weights by citing BBD (Beilinson-Bernstein-Deligne, Astérisque 100). To do that, she needs the fact that restriction to fibers over closed points of $U_0$ will commute with the 4 (or 6, or 7) operations on her categories of complexes, and this would not be true for more general complexes. What she does, actually, is just to cite Deligne's generic base change theorem, which will immediately give you the compatibility you want, but at the cost of shrinking the base (i.e., of shrinking $U_0$).</p> <p>(3) Related to (2) : Given the way she defines weights, she wants her weights to be the same over each closed point of $U_0$. That is of course not going to be true if she allows stratifications with strata that are not necessarily smooth over $U_0$ : you could have a sheaf that is concentrated over one closed point of $U_0$, for example.</p> <p>I am not saying that it is necessary to do what she does, just that it makes your life much simpler. It might be possible to develop a theory of weights for general constructible sheaves (i.e., constructible with respect to a general stratification). But you will not be able to reduce everything to BBD anymore because you won't be able to use generic base change, and you will likely run into very hard problems like the weight-monodromy conjecture when you try to prove that the 4 operations preserve mixed sheaves and that they have the expected effect on weights.</p> <p>Now if you're looking at motives instead of $\ell$-adic complexes, the situation is very different because your approach is different : all motives are already mixed, the fact that, say, $f_*$ increases weights is somehow built in the very definition of weights, etc. Then the problem is of course to related weights for motives and weights for $\ell$-adic complexes. Note that the weight-monodromy conjectures follows from the "standard" conjectures of motives, so if we had the full formalism of motives all our problems would probably disappear (it doesn't much help us at this point but it might at least make us feel better to know that the world is conjectured to be coherent).</p> http://mathoverflow.net/questions/54330/if-one-wants-to-work-with-q-l-adic-sheaves-should-the-scheme-be-of-finite-type/54339#54339 Answer by Alex for If one wants to work with $Q_l$-adic sheaves, should the scheme be of finite type over a 1-dimensional one? Alex 2011-02-04T17:45:37Z 2011-02-04T17:45:37Z <p>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 not written it down. There was a seminar in the Ecole Polytechnique four years ago (I think) about his proof, and the goal was to write a full proof (probably a book), but it doesn't look as if they are done yet. You can find statements and outlines of proofs in notes on Illusie's webpage (e.g., <a href="http://www.math.u-psud.fr/~illusie/Illusie_Tokyo.pdf" rel="nofollow">http://www.math.u-psud.fr/~illusie/Illusie_Tokyo.pdf</a>), there are some notes about at least parts of the proofs and useful statements on Joël Riou's webpage (http://www.math.u-psud.fr/~riou/), I think some of the ideas are showing up in fabrice Orgogozo's work (for example, Orgogozo announced a result about uniform constructibility for torsion sheaves and I'm pretty sure it uses Gabber's ideas, but I don't find a preprint on his webpage). Actually, I think Gabber's uniformization theorem has been used in other works; maybe Frederic Deglise's work ?</p> <p>(Oh, and all the references are in French. Enjoy. ;-))</p> http://mathoverflow.net/questions/53378/are-two-conjectural-descriptions-of-the-motivic-t-structure-via-cohomology-and-v/53509#53509 Answer by Alex for Are two conjectural descriptions of the motivic t-structure (via cohomology and via affine varieties) known to be equivalent? Alex 2011-01-27T17:08:20Z 2011-01-27T17:08:20Z <p>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 any of the equivalent definitions that are available); I am taking $\mathbb{Q}$-coefficients and looking only at compact objects, although I'm not sure the last is necessary. Also, I am not good with homological notation, so I will use cohomological notation all along, beware ! For example, for me $X[1]$ will mean "$X$ put in degree $-1$" and the motive of $\mathbb{G}_m$ will be $\mathbb{Q}\oplus\mathbb{Q}(-1)[-1]$ (where $\mathbb{Q}$ is the unit for the tensor product, i.e., the motive of $Spec(k)$). Sorry, but I'm too afraid to make a mistake if I try to translate.</p> <p>So, first here is Hanamura's definition of the $t$-structure. He assumes that all the Grothendieck standard conjectures, Murre's conjecture and the vanishing conjectures are true, and this implies in particular that any realization functor $H:DM(k)\longrightarrow D^b(F)$ (where $F$ is an appropriate field of coefficients) is faithful. He defines a $t$-structure, say <code>$({}^H D^{\leq 0},{}^H D^{\geq 0})$</code> on $DM(k)$ by taking the inverse image by $H$ of the usual $t$-structure on $D^b(F)$. Of course, you have to prove that it is indeed a $t$-structure (and independent of the realization functor), and he does this. He calls the heart the category of mixed motives over $k$, say $MM(k)$. If $X$ is a variety over $k$, we can associate to it a (cohomological) motive $M(X)$ in $DM(k)$, and I will denote by $H^k_M(X)\in MM(X)$ the cohomology objects of $M(X)$ for Hanamura's $t$-structure. Hanamura also proves that every mixed motive has a weight filtration, that is a filtration whose graded parts are pure motives, and he proves that pure motives are semi-simple and that all irreducible pure motives are direct factors of motives of the form $H_M^k(X)(a)$, where $X$ is a smooth projective variety.</p> <p>Now to Voedvodsky's definition. I have tried to understand it, then rewrite it in cohomological notation, so directions of maps and shifts may have changed, but I think I still got the spirit of it. What he does is something like this : Define a full subcategory <code>${}^VD^{\geq 0}$</code> of $DM(X)$ by the condition that an object $M$ is in it if and only if, for every affine scheme $f:U\rightarrow Spec(k)$ that is purely of dimension $n$, for every $m>n$ and every $a\geq 0$, <code>$Hom_{DM(k)}(f_*\mathbb{Q}_U(a)[m],M)=0$</code>. I would like to make a few remarks. First, what I denoted by $f_*\mathbb{Q}_U$ is just my $M(U)$ of the preceding paragraph, but I wrote it like this because it will make the generalization to a general base scheme $S$ more clear; my notation makes sense if I allow myself to remember that we now have categories of motives over a very general base and the 6 operations on them (and if I say that $\mathbb{Q}_U$ is the unit motive in the category of motives over $U$). Second remark, I added twists whereas Voedvodsky's definition doesn't have any. The reason I did this is because Voedvodsky makes a definition only for effective motives, and I didn't see how to make ${}^VD^{\geq 0}$ stable by $(1)$ unless I added it in the definition (but maybe it is not necessary). Third remark, remember, cohomological notation, and for me passing from effective to general motives means inverting $\mathbb{Q}(-1)$, not $\mathbb{Q}(1)$ (in my world, $\mathbb{Q}(-1)$ is effective).</p> <p>Ah, yes, and then Voedvodsky defines ${}^VD^{\leq 0}$ as the left orthogonal of <code>${}^V D^{\geq 1}:={}^VD^{\geq 0}[-1]$</code>.</p> <p>Anyway, what is the motivation for Voedvodsky's definition ? Here are a few principles. First, motivic $t$-structure on motives over a base $S$ should correspond (by the realization functors) to the (selfdual) perverse $t$-structure on complexes of sheaves over $S$. Second, if $f$ is an affine map of schemes, then <code>${f_*$}</code> is right $t$-exact for the perverse $t$-structures. Third, for any scheme $U$, the constant sheaves over $U$ are concentrated in perverse cohomology degree $\leq dim(U)$. So, if I come back to my situation above : $f:U\longrightarrow Spec(k)$ is an affine variety over $k$, purely of dimension $n$, $a\geq 0$, $m>n$, then <code>$\mathbb{Q}_U(a)[m]$</code> should be concentrated in degree $&lt;0$ for the motivic $t$-structure on the category of motives over $U$, and so $f_*\mathbb{Q}_U(a)[m]$ should be concentrated in degree $&lt;0$ for the motivic $t$-structure on $DM(k)$, and it should be left orthogonal to elements that are concentrated in degree $\geq 0$. What Voedvodsky says is that this is enough to characterize the elements concentrated in degree $\geq 0$.</p> <p>From this, the natural generalization of Voedvodsky's definition to a general base scheme $S$ is obvious : replace affine schemes $U\longrightarrow Spec(k)$ by affine maps $U\longrightarrow S$ (or maps $U\longrightarrow S$ such that $U$ is affine, I don't think it will make a difference).</p> <p>So, are the two $t$-structures the same ? I think so. A first obvious observation is that <code>${}^HD^{\geq 0}\subset{}^VD^{\geq 0}$</code>, that is, every object of <code>${}^HD^{\geq 0}$</code> is right orthogonal to motives <code>${{f_*}\mathbb{Q}(a)[m]}$</code> as above. This follows from the faithfulness of the realization functor and the fact that this would be true in the usual categories of sheaves (see the remarks above). We also know that <code>${}^HD^{\geq 0}$</code> is the right orthogonal of <code>${}^HD^{\leq -1}$</code>, by the definition of a $t$-structure. So, what we have to see is that ${}^VD^{\geq 0}$ is right orthogonal to ${}^HD^{\leq -1}$, that is, that a motive that is right orthogonal to every $f_*\mathbb{Q}_U(a)[m]$ as above is right orthogonal to the whole ${}^HD^{\leq -1}$.</p> <p>Here is one way to do this : Let $C$ be the smallest full additive subcategory of ${}^HD^{\leq -1}$ that is stable by isomorphism, direct summand, extension and contains all the objects of the form $f_*\mathbb{Q}(a)[m]$ as above. It is enough to show that $C={}^HD^{\leq -1}$. Noting that ${}^HD^{\leq -1}$ is generated (in the same way : direct sumands, isomorphisms, extensions) by objects of the form $H^k_M(X)(b)[l]$, for $X$ a smooth projective variety, $l>k$ and $b\in\mathbb{Z}$, I think that this is an easy exercise, playing with hyperplane sections of smooth projective varieties. (I had a bit a trouble with the fact that $C$ is stable by Tate twists. We know that $M(U)(-1)$ is a direct factor of $M(U\times\mathbb{G}_m)[1]$, so stability by $(-1)$ is not a problem. But I couldn't show stability by $(1)$ unless I put it in the definition.)</p> http://mathoverflow.net/questions/53220/could-the-kunneth-decomposition-of-a-motif-depend-on-the-choice-of-l/53435#53435 Answer by Alex for Could the Kunneth decomposition of a motif depend on the choice of $l$? Alex 2011-01-27T02:00:52Z 2011-01-27T02:00:52Z <p>Let me develop YBL's answer a bit. (I wanted to make this a comment but it was too long...)</p> <p>Consider a smooth variety $U$ over $\mathbb{F}_p$ with function field $K$ such that your motive and its two Künneth decompositions extend over U. Take a $\mathbb{F}_q$-rational point $x$ of $U$, and look at the specialization of everybody at $x$ (see André-Kahn, <em>Construction inconditionnelle de groupes de Galois motiviques</em>, section 3 for the definition of good reduction and specialization of motives). Then you get a motive over $\mathbb{F}_q$ with two Künneth decompositions, but these have to be the same for the reason YBL gave : weights (more precisely, the projectors on the components of the Künneth decomposition are given in this case by rational polynomials in the graph of Frobenius that are independent of the Weil cohomology, see part III of Katz-Messing, <em>Some consequences of the Riemann hypothesis for varieties over finite fields</em>). So your two decompositions are the same on the specialization, but the specialization functor is faithful, so the two decompositions are the same on the original motive.</p> http://mathoverflow.net/questions/53124/snake-lemma-in-category-of-groups/53139#53139 Answer by Alex for snake lemma in category of groups Alex 2011-01-24T22:22:40Z 2011-01-24T22:22:40Z <p>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 cokernels, that is, that their images are normal subgroups of their targets, then the snake lemma is true with the same proof as in the abelian category case (replacing the $0$'s by $1$'s and the minus signs by inversions, as J. Polak pointed out). </p> http://mathoverflow.net/questions/53028/are-irreducible-representations-with-large-fixed-subspaces-trivial/53071#53071 Answer by Alex for are irreducible representations with large fixed subspaces trivial? Alex 2011-01-24T15:22:15Z 2011-01-24T15:30:19Z <p>I don't know the answer, but here are a few remarks :</p> <p>(0) As has been pointed out before, "irreducible" and "indecomposable" are not the same for representations in positive characteristic. "Irreducible" is a stronger property. (For example, irreducible representations of commutative groups are always $1$-dimensional, whereas indecomposable representations don't have to be. No commutative group is going to give a counterexample.)</p> <p>(1) The answer is "yes" if we look at representations over a field of characteristic prime to the order of $G$ (because representations theory of $G$ will be the same as in characteristic zero).</p> <p>(2) If we want to construct a counterexample, then we should not take $G$ to be a $p$-group (as in the two attempts above). Because then $k$ will have to be of characteristic $p$, but then every finite-dimensional representation of $G$ over $k$ has a nonzero fixed vector.</p> <p>(3) More generally, it is not possible to construct a counterexample in characteristic $p$ if $G$ has a normal $p$-Sylow $H$ (let $W$ be the subspace of vectors fixed by $H$, it is nonzero because $H$ is a $p$-group, it is stable by $G$ because $H$ is normal, hence it has to be the whole space because the representation is irreducible, but then we are reduced to the case of $G/H$, which has order prime to $p$, and see (1)).</p> <p>That means, I'm afraid, no easy counterexamples with very small groups.</p> http://mathoverflow.net/questions/52798/conjugacy-class-of-a-full-jordan-block-over-integers/53024#53024 Answer by Alex for Conjugacy class of a full Jordan block over integers Alex 2011-01-24T03:02:58Z 2011-01-24T03:02:58Z <p>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 if $Im(A^{n-1})=Ker(A)$. (Here and afterwards, I am seeing $A$ as an endomorphism of $\mathbb{Z}^n$ and taking kernels and images there.)</p> <p>It is easy to see that this condition is necessary. To see that it is sufficient, you can go as follows :</p> <p>(1) Let $A$ be as before, but assume only that $A^n=0$. Then, for every <code>$k\in\{1,\dots,n-1\}$</code>, the map <code>$Ker(A^k)/(Im(A^{n-k})+Ker(A))\longrightarrow Ker(A^{k-1})/Im(A^{n-k+1})$</code> induced by $A$ is injective. This is very easy.</p> <p>(2) Now assume that $Ker(A)=Im(A^{n-1})$. From (1), we deduce that, for every <code>$k\in\{1,\dots,n-1\}$</code>, <code>$Ker(A^k)=Im(A^{n-k})$</code>.</p> <p>(3) Assume again that $Ker(A)=Im(A^{n-1})$. Note that this implies that $A^n=0$ and $A^{n-1}\not =0$, so that $Ker(A)$ is a $\mathbb{Z}$-submodule of $\mathbb{Z}^n$ such that <code>$Ker(A)\otimes_{\mathbb{Z}}\mathbb{Q}$</code> is $1$-dimensional; from this we get that $Ker(A)$ is a free $\mathbb{Z}$-module of rank $1$. Pick an element $e_n\in\mathbb{Z}^n$ such that $A^{n-1}e_n$ is a generator of $Ker(A)$. For every <code>$k\in\{1,\dots,n-1\}$</code>, let $e_k=A^{n-k}e_n$. Then I say that $(e_1,\dots,e_n)$ is a basis of $\mathbb{Z}^n$. (And $A$ has the form we want in this basis, so we're done.)</p> <p>You can prove this last fact by induction on $n$. It is obvious for $n=1$. If $n>1$ and we know the fact for $n-1$, then, by the induction hypothesis applied to the restriction of $A$ to $Im A=Ker(A^{n-1})$ (seen as an endomorphism of this space), we get that $(e_1,\dots,e_{n-1})$ is a basis of $Im(A)$. As $A(e_i)=e_{i-1}$, this implies that $\mathbb{Z}^n=\mathbb{Z}e_2+\dots+\mathbb{Z}e_{n-1}+Ker(f)$, but $Ker(f)=\mathbb{Z}e_1$, so we're done.</p> <p>Note that the following conditions are equivalent :</p> <p>(a) $Ker(A)=Im(A^{n-1})$</p> <p>(b) $Im(A)=Ker(A^{n-1})$</p> <p>(c) $A^n=0$ and $Coker(A)$ is isomorphic to $\mathbb{Z}$.</p> <p>So, in this particular case, the invariant given by Torsten Ekedahl is enough.</p> <p>In the general case, there are other invariants, and I don't know the general answer. For example, consider a $n\times n$ matrix $A$ with coefficients in $\mathbb{Z}$, and assume that it is similar (over $\mathbb{Z}$) to a matrix with $0$'s on the diagonal and under the diagonal, nonzero integers $b_1,\dots,b_{n-1}$ on the upper diagonal (no divisibility condition) and some integers (zero or nonzero) elsewhere. Then the integers $b_1,\dots,b_{n-1}$ are uniquely determined up to sign, because $|b_1\dots b_k|$ is the order of the quotient <code>$Ker(A)/A^k(Ker(A^{k+1}))$</code>. But these are still not the only invariants. For example, if $n=3$, we can see by direct calculation that the $GL_3(\mathbb{Z})$-conjugacy classes of the matrices <code>$$\left(\begin{matrix}0 &amp; b_1 &amp; c \\ 0 &amp; 0 &amp; b_2 \\ 0 &amp; 0 &amp; 0\end{matrix}\right)$$</code> and <code>$$\left(\begin{matrix}0 &amp; b_1 &amp; d \\ 0 &amp; 0 &amp; b_2 \\ 0 &amp; 0 &amp; 0\end{matrix}\right)$$</code> (with $b_1,b_2$ positive integers) are the same if and only if $c=d$ modulo $gcd(b_1,b_2)$. For bigger $n$, it becomes more and more complicated, and I'm not sure what to do. (Perhaps some more experiments would indicate an answer ? A quick way to obtain necessary conditions is also to look at what happens for conjugacy over $\mathbb{Z}_p$; the problem becomes much simpler there, especially with your divisibility hypothesis, because there's only one prime in $\mathbb{Z}_p$. But I don't know if you can get sufficient conditions this way.)</p> http://mathoverflow.net/questions/30311/whats-an-example-of-an-intersection-cohomology-sheaf-whose-stalks-are-pure-but-n/52712#52712 Answer by Alex for What's an example of an intersection cohomology sheaf whose stalks are pure but not pointwise pure? Alex 2011-01-21T02:36:14Z 2011-01-21T02:36:14Z <p>Here is an example. Sorry it's so complicated. (There's probably a simpler one, but my mind works in complicated ways, it seems.)</p> <p>Consider a Siegel modular threefold $U$, i.e., a Shimura variety for the general symplectic group $GSp(4)$ with some level $n\geq 3$. (So that $U$ is smooth and quasi-projective over $\mathbb{Q}$.) Let $X$ be the Baily-Borel-Satake (aka minimal) compactification of $U$. Then $X$ has a stratification $X=U\cup X_1\cup X_2$, where $X_1$ is a finite disjoint union of modular curves and $X_2$ is a finite disjoint union of points (spectra of abelian extensions of $\mathbb{Q}$). Let's denote the inclusion of $U$ in $X$ by $j$, and define the intersection complex on $X$ by $IC=(j_{!*}\mathbb{Q}_\ell[2])[-2]$ (I like my intersection complexes to be concentrated in nonnegative degree). Then I say that the cohomology of the stalk of $IC$ at a point of $X_2$ is not pure (or pointwise pure, as you put it).</p> <p>Before I do the messy calculation, let me give a heuristic reason. If we allowed ourselves to restrict everything to $Spec(\mathbb{C})$ and use $\mathbb{C}$-coefficients (which we can't because we want to look at weights of Frobenius), then we could use the calculation of the stalks of the intersection cohomology complex by Goresky, Harder, MacPherson and Nair in their article <em>Local intersection cohomology of Baily-Borel compactifications</em> (available on Goresky's webpage). In section 7 of this article, they work out the case of Siegel modular threefolds in some detail. In particular, they show (formula 7.7.2) that, if $x\in X_2$, then $IC_x$ has as a direct factor $H^*(X_\ell,\mathbb{C})$, where $X_\ell$ is a locally symmetric space associated to the linear part of one of the maximal Levi subgroups of $GSp(4)$; the linear part is in this case $GL(2)$. Morally, the cohomology of such "linear" locally symmetric spaces is made up of Artin motives, so the factor $H^*(X_\ell,\mathbb{C})$ should be of weight $0$ in every degree. But $H^1(X_\ell,\mathbb{C})$ contributes to $H^1(IC_x)$, so this vector space is not pure of weight $1$.</p> <p>Okay, now for the formal proof. Unfortunately, there is no result as precise as the calculation of Goresky-Harder-MacPherson-Nair for the $\ell$-adic intersection complex on the variety $X$ over $\mathbb{Q}$, but, if we reduce modulo a prime of good reduction, then we can use the results of Morel's article <em>Complexes pondérés sur les compactifications de Baily-Borel : le cas des variétés modulaires de Siegel</em>. So now I will only consider the reductions of the various varieties modulo $p$, where $p$ is a big enough prime (here, it just means that $p$ does not divide the level).</p> <p>I will need to introduce some notation. To make my life easier, I define the symplectic group with a symplectic form such that the intersection of $GSp(4)$ with the group of upper triangular matrices in $GL(4)$ is a Borel subgroup of $GSp(4)$ (say, an antidiagonal form). Let $N$ be the unipotent radical of this Borel subgroup. Let $N_2$ be the center of $N$, so that $N_2$ is the intersection of $GSp(4)$ and of the group <code>$\left(\begin{matrix}1 &amp; 0 &amp; * &amp; * \\ 0 &amp; 1 &amp; * &amp; * \\ 0 &amp; 0 &amp; 1 &amp; 0 \\ 0 &amp; 0 &amp; 0 &amp; 1\end{matrix}\right)$</code></p> <p>Let me also consider the following two cocharacters of $GSp(4)$ : <code>$w_1:\lambda\longmapsto \left(\begin{array}{cccc}\lambda^2 &amp; 0 &amp; 0 &amp; 0 \\ 0 &amp; \lambda &amp; 0 &amp; 0 \\ 0 &amp; 0 &amp; \lambda &amp; 0 \\ 0 &amp; 0 &amp; 0 &amp; 1\end{array}\right)$</code> and <code>$w_2:\lambda\longmapsto \left(\begin{array}{cccc}\lambda &amp; 0 &amp; 0 &amp; 0 \\ 0 &amp; \lambda &amp; 0 &amp; 0 \\ 0 &amp; 0 &amp; 1 &amp; 0 \\ 0 &amp; 0 &amp; 0 &amp; 1\end{array}\right)$</code></p> <p>Note that the images of $w_1$ and $w_2$ normalize $N$ and $N_2$, so we get two actions of $\mathbb{G}_m$ on $H^*(Lie(N),\mathbb{Q}_\ell)$ and $H^*(Lie(N_2),\mathbb{Q}_\ell)$. </p> <p>If $a$ and $b$ are integers, I will write <code>$H^*(Lie(N),\mathbb{Q}_\ell)_{\geq a,&lt;b}$</code> for the part of <code>$H^*(Lie(N), \mathbb{Q}_\ell)$</code> where $w_2(\mathbb{G}_m)$ acts by characters $\lambda\longmapsto \lambda^k$ with $k\geq a$ and $w_1(\mathbb{G}_m)$ acts by characters <code>$\lambda\longmapsto\lambda^k$</code> with <code>$k&lt;b$</code>. </p> <p>Likewise, I will write <code>$H^*(Lie(N_2),\mathbb{Q}_\ell)_{\geq a}$</code> for the part of <code>$H^*(Lie(N_2), \mathbb{Q}_\ell)$</code> where $w_2(\mathbb{G}_m)$ acts by characters $\lambda\longmapsto \lambda^k$ with $k\geq a$. (I don't care about the other action.)</p> <p>Let as before $x$ be a point of $X_2$. From theorem 4.2.1 of Morel's article, I get a distinguished triangle : <code>$H^*(IC_x)\longrightarrow K_1\longrightarrow K_2,$</code> where $K_1$ is finite direct sum of complexes of the form <code>$H^*(\Gamma_\ell,H^*(Lie(N_2),\mathbb{Q}_\ell)_{\geq -3})$</code> and $K_2$ is a finite direct sum of complexes <code>$H^*(Lie(N),\mathbb{Q}_\ell)_{\geq -3,&lt;-1}$</code>. Here, <code>$\Gamma_\ell$</code> is an arithmetic subgroup of <code>$GL(2,\mathbb{Q})$</code>, where we see $GL(2)$ as a subgroup of $GSp(4)$ via the map <code>$g\longmapsto\left(\begin{array}{cc}g &amp; 0 \\0 &amp; {}^tg^{-1}\end{array}\right)$</code>; then $GL(2)$ acts by conjugation on $N_2$, hence it acts on <code>$H^*(Lie(N_2),\mathbb{Q}_\ell)$</code>, and it commutes with <code>$w_2(\mathbb{G}_m)$</code>, so it stabilizes <code>$H^*(Lie(N_2),\mathbb{Q}_\ell)_{\geq -3}$</code>. Note that the arithmetic subgroup <code>$\Gamma_\ell$</code> could vary in the finite direct sum, but it won't matter.</p> <p>(Note also that the theorem I cited only gives you an equality in a Grothendieck group. However, if instead of Morel's 3.3.5 (that it relies on), you instead invoke the stronger proposition 3.3.4 (ii), then you can see that this equality comes from a distinguished triangle.)</p> <p>Another thing we need to know is that the (Frobenius) weights on $K_1$ and $K_2$ are given by minus the weights of $w_2(\mathbb{G}_m)$; that is, if $w_2(\mathbb{G}_m)$ acts on some part of $K_1$ or $K_2$ by the character $\lambda\longmapsto\lambda^k$, then the (Frobenius) weight of this part is $-k$. (This is proved for example in an article of Pink, and the reference is given in proposition 2.1.4 of the article of Morel I'm citing.)</p> <p>Now I need to look a little more closely at the action of my two tori on the first cohomology groups of $Lie(N)$ and $Lie(N_2)$. They both act trivially on the $H^0$ groups, so in particular $H^0(K_1)$ has (Frobenius) weight $0$ and $H^0(K_2)=0$ (because <code>$H^0(Lie(N_2),\mathbb {Q}_\ell)_{\geq -3}=H^0(Lie(N_2),\mathbb {Q}_\ell)$</code> and <code>$H^0(Lie(N),\mathbb {Q}_\ell)_{\geq -3,&lt;-1}=0$</code>).</p> <p>The torus $w_2(\mathbb{G}_m)$ acts by $\lambda\longmapsto\lambda$ on $Lie(N_2)$, so it acts by $\lambda\longmapsto \lambda^{-1}$ on <code>$H^1(Lie(N_2),\mathbb{Q}_\ell)$</code>, and <code>$H^1(Lie(N_2),\mathbb{Q}_\ell)_{\geq -3}=H^1(Lie(N_2),\mathbb{Q}_\ell)$</code>.</p> <p>Let $V=N/N_2$; this is a $1$-dimensional unipotent group. Using the spectral sequence <code>$E_2^{pq}=H^p(Lie(V),H^q(Lie(N_2),\mathbb{Q}_\ell))\Longrightarrow H^{p+q}(Lie(N),\mathbb{Q}_\ell)$</code> and the fact that $H^p(Lie(V),\ )$ vanishes for $p>1$, we get an exact sequence <code>$0\longrightarrow H^1(Lie(V),H^0(Lie(N_2),\mathbb{Q}_\ell))\longrightarrow H^1(Lie(N),\mathbb{Q}_\ell)\longrightarrow H^0(Lie(V),H^1(Lie(N_2),\mathbb{Q} _\ell))$</code>. On the third term, $w_1(\mathbb{G}_m)$ acts by $\lambda\longmapsto\lambda^k$ for <code>$0\leq k\leq 2$</code> and $w_2(\mathbb{G}_m)$ acts by $\lambda\longmapsto \lambda^{-1}$. On the first term, $w_1(\mathbb{G}_m)$ acts by $\lambda \longmapsto\lambda^{-1}$ and $w_2(\mathbb{G}_m)$ acts trivially. (This follows by looking at the weights of the tori in $Lie(N_2)$ and $Lie(V)$.) In particular, we find that <code>$H^1(Lie(N),\mathbb{Q}_\ell)_{\geq -3,&lt;-1}$</code> injects into <code>$H^0(Lie(V),H^1(Lie(N_2),\mathbb{Q}_\ell))$</code>, and this implies that $H^1(K_2)$ has (Frobenius) weight $1$.</p> <p>The distinguished triangle above gives us an exact sequence <code>$0=H^0(K_2)\longrightarrow H^1(IC_x)\longrightarrow H^1(K_1)\longrightarrow H^1(K_2)$</code>. Remembering the formula for $K_1$ above, we see that $H^1(K_1)$ is a finite direct sum of groups of the form <code>$H^0(\Gamma_\ell,H^1(Lie(N_2),\mathbb{Q}_\ell))$</code> and <code>$H^1(\Gamma_\ell,H^0(Lie(N_2),\mathbb{Q}_\ell))$</code> (with $\Gamma_\ell$ as above). The first summand has (Frobenius) weight $1$, and the second summand has (Frobenius) weight $0$.</p> <p>But we have seen that $H^1(K_2)$ has weight $1$. So the second summand is sent to $0$ in $H^1(K_2)$, and $H^1(IC_x)$ contains a finite direct sum of groups of the form <code>$H^1(\Gamma_\ell,H^0(Lie(N_2),\mathbb{Q}_\ell))$</code>, in particular in contains a part of weight $0$, so it is not pure of weight $1$. (Note that all we did was prove an algebraic version of a very small part of the result of Goresky-Harder-MacPherson-Nair.)</p> http://mathoverflow.net/questions/31266/which-statements-in-section-5-of-bbd-will-fail-if-we-consider-mathbbq-l-adic/52571#52571 Answer by Alex for Which statements in section 5 of BBD will fail if we consider $\mathbb{Q}_l$-adic sheaves there? Alex 2011-01-20T02:02:38Z 2011-01-20T02:02:38Z <p>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 <code>$\mathbb{Q}_\ell$</code>-coefficients with the same proof, and that 5.3.9 (i) is still true but with a slightly more complicated proof. I am pretty sure that the proof of corollary 5.3.10 works too for $\mathbb{Q}_\ell$-coefficients. I cannot see any reason why the proofs in 5.4 would not also work, but I have read them quickly. Anyway, proposition 5.1.15 is certainly fine.</p> http://mathoverflow.net/questions/76186/functoriality-properties-of-the-perverse-t-structure-for-torsion-constructible/76213#76213 Comment by Alex Alex 2012-05-08T22:31:41Z 2012-05-08T22:31:41Z Yes, when I wrote my answer I had not realized that $\mathbb{Z}/\ell^m\mathbb{Z}$ coefficients work just like $\mathbb{Z}/\ell\mathbb{Z}$ for many things. There is a self-dual t-structure, BBD 4.1 is still fine I think, though you have to be careful if you want to take tensor products. http://mathoverflow.net/questions/81204/does-2x2y-rightarrow-xy-imply-the-axiom-of-choice/94263#94263 Comment by Alex Alex 2012-04-17T10:10:59Z 2012-04-17T10:10:59Z You're right. I'll get rid of my answer. http://mathoverflow.net/questions/87766/do-change-of-coefficients-functors-for-sheaves-commute-with-the-four-functors Comment by Alex Alex 2012-02-07T11:04:22Z 2012-02-07T11:04:22Z What do you call 'coefficient change' ? The forgetful functor or its adjoint ? And what is the category of 'all' etale sheaves ? Is it $\mathbb{Z}_\ell$-sheaves or torsion sheaves ? http://mathoverflow.net/questions/86622/the-arthur-trace-formula-for-function-fields Comment by Alex Alex 2012-01-25T19:43:07Z 2012-01-25T19:43:07Z Because Laumon just wanted a formula that would suffice for the applications he had in mind and was as easy to prove as possible ? I don't know any place where the trace formula for functions fields is worked out in full generality. (Not the non-invariant one, not the invariant one, and certainly not the stable one.) Laurent Lafforgue likely had to work out a more general case than Laumon (in his book &quot;Chtoucas de Drinfeld et conjecture de Ramanujan-Petersson&quot;), but that would only be for $GL(n)$, probably. http://mathoverflow.net/questions/83431/online-reference-for-bridge-between-mathbb-c-and-mathbb-f Comment by Alex Alex 2011-12-14T16:14:34Z 2011-12-14T16:14:34Z <a href="http://www.math.harvard.edu/~gaitsgde/grad_2009/BBD.djvu" rel="nofollow">math.harvard.edu/~gaitsgde/grad_2009/BBD.djvu</a> http://mathoverflow.net/questions/3139/how-do-i-compare-the-different-notions-of-fourier-transform-for-sheaves/3144#3144 Comment by Alex Alex 2011-11-21T00:55:43Z 2011-11-21T00:55:43Z Laumon has an article on Fourier transform for <code>$\mathbb{G}&#95;m$</code>-equivariant sheaves (&quot;La transformation de Fourier homog&#232;ne&quot;), but I don't remember if he says anything about Riemann-Hilbert (I think not, but it might still be a useful reference). http://mathoverflow.net/questions/80944/vanishing-constant-term-in-powers-of-a-laurent-polynomial Comment by Alex Alex 2011-11-15T19:27:25Z 2011-11-15T19:27:25Z If $K$ is a field of characteristic $0$ and $f\in K[x,x^{-1}]$, then $f$ is also in $L[x,x^{-1}]$, where $L$ is the subfield of $K$ generated by the coefficients of $f$. As $L$ is of finite type over $\mathbb{Q}$, it can be embedded in $\mathbb{C}$, so you can see $f$ as an element of $\mathbb{C}[x,x^{-1}]$ and apply the result that you are citing. So the result that you want seems to at least be true. But I'm guessing that you wanted another proof. http://mathoverflow.net/questions/62834/why-is-the-simple-trace-formula-a-weaker-tool-than-the-arthur-trace-formula Comment by Alex Alex 2011-11-10T01:25:17Z 2011-11-10T01:25:17Z Also the calculation of the zeta functions of Shimura varieties. http://mathoverflow.net/questions/80362/finiteness-of-normalization-of-noetherian-normal-domain/80374#80374 Comment by Alex Alex 2011-11-09T19:25:36Z 2011-11-09T19:25:36Z Depends on your definition of &quot;geometry&quot;. Discrete valuation rings of equal characteristic $p$ don't have to be excellent, and if you do analytic geometry in positive characteristic you can encounter them. http://mathoverflow.net/questions/80489/real-orbits-of-highest-weight-vectors/80504#80504 Comment by Alex Alex 2011-11-09T19:22:49Z 2011-11-09T19:22:49Z Sorry for the math. I meant &quot;you can see $\lambda$ as a map from $B/U$ to <code>$\mathbb{C}^\times$</code>&quot;. http://mathoverflow.net/questions/80489/real-orbits-of-highest-weight-vectors/80504#80504 Comment by Alex Alex 2011-11-09T19:21:11Z 2011-11-09T19:21:11Z $U$ is the unipotent radical of $B$, $G_m$ is $\mathbb{C}^\times$ (in this case), $H=B/U$ is isomorphic to a maximal torus of $G$, so you can see $\lambda$ as a map from $B/U$ to $&#249;mathbb{C}^\times$. If $v$ is a highest weight vector for $B$, then $U$ acts trivially on $v$, and $B/U$ acts by multiplication by this $\lambda:B/U\rightarrow\C^\times$, because that is the definition of a highest weight vector. Sasha's message means : 1/ yes, $G$ acts transitively on the set of highest weight vectors 2/ the stabilizer of a highest weight vector $v$ for $B$ is the $K_\lambda$ he or she defined. http://mathoverflow.net/questions/80489/real-orbits-of-highest-weight-vectors Comment by Alex Alex 2011-11-09T19:14:47Z 2011-11-09T19:14:47Z If $v'$ is a highest weight vector for the Borel $B'$, then the elements of $B'$ will act as scalings on $v'$. That's the scalings you need. http://mathoverflow.net/questions/80344/infinity-groupoid-on-the-etale-site-of-a-scheme Comment by Alex Alex 2011-11-08T10:23:45Z 2011-11-08T10:23:45Z Artin and Mazur, &quot;Etale homotopy type&quot;, Lecture Notes in Mathematics 100. See also Friedlander, &quot;&#201;tale homotopy of simplicial schemes&quot;, Annals of Mathematics Studies 104. http://mathoverflow.net/questions/79667/when-is-mathbbg-mr-enough-to-determine-r Comment by Alex Alex 2011-11-01T02:34:07Z 2011-11-01T02:34:07Z Actually, no, I don't think it works for finite fields. Take $R=\mathbb{F}_p$, where $p$ is an prime number $\geq 5$. Then $R^\times$ is a cyclic group of order $p-1$, so it has nontrivial subgroups. But $R$ does not have any nontrivial subrings. http://mathoverflow.net/questions/79667/when-is-mathbbg-mr-enough-to-determine-r/79668#79668 Comment by Alex Alex 2011-11-01T02:31:45Z 2011-11-01T02:31:45Z If S is required to be a subfield, then of course you get uniqueness, but I still don't see a natural condition that would guarantee existence. Besides, you wrote that your motivating example if the absolute Galois group of $\mathbb{Q}$, but you would need your subgroup to be commutative if it has to be the group of units in a field. Maybe I totally misunderstood, but I don't really see what you're trying to do.