User geordie williamson - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T20:38:05Z http://mathoverflow.net/feeds/user/919 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/126532/vanishing-of-restriction-of-constructible-sheaves/129244#129244 Answer by Geordie Williamson for Vanishing of !-restriction of constructible sheaves Geordie Williamson 2013-04-30T19:07:05Z 2013-04-30T19:07:05Z <p>I think the vanishing you want holds. For simplicity consider the case where $i$ is the inclusion of <code>$Z = \{ z \}$</code> a point. (One should be able to reduce to this case by taking a normal slice.)</p> <p>The question is local so we can replace $X$ by a small neighbourhood $U$ of $z$. Let $j$ denote the inclusion of $U - { z }$. Consider the distinguished triangle <code>$i_!i^!\mathcal{F} \to \mathcal{F} \to j_*j^*\mathcal{F} \to$</code>. Now let $S^{n-1}_\epsilon$ be a sphere of radius $\epsilon$ around $z$. By the constructibility assumption the stalk of <code>$j_*j^*\mathcal{F}$</code> at $z$ is equal, for small enough $\epsilon$, to the cohomology of <code>$S^{n-1}_\epsilon$</code> with values in the restriction of $\mathcal{F}$. By standard vanishing theorems this vanishes in degrees $\ge n$. Hence the cohomology of <code>$i_!i^!\mathcal{F}$</code> vanishes in degrees $> n$ as required.</p> http://mathoverflow.net/questions/128822/divisibility-of-all-entries-in-an-intersection-form Divisibility of all entries in an intersection form Geordie Williamson 2013-04-26T12:27:56Z 2013-04-26T12:27:56Z <blockquote> What are situations where one can conclude that all entries of an intersection form are divisible by an integer? </blockquote> <p>More precisely: $F \subset S$ is a proper connected (usually reducible) half-dimensional subvariety of an smooth variety $S$. In this situation one define the "refined intersection form" on the top degree homology of $F$. The top homology has a basis over $\mathbb{Z}$ given by the fundamental classes of components of $F$ of maximal dimension, amd the intersection form gives a pairing on this space.</p> <p>(To define this intersection form one uses the isomorphism of $H^{top}(F) = H^{n}(S, S \setminus F)$ and the product on relative cohomology. Informally one takes two components of $F$, moves them inside $S$ until they are transverse, and counts intersection points. For example, if $F$ is smooth one can take $S = T^*F$ and the self-intersection of $[F]$ is the Euler characteristic of $F$, up to $\pm 1$.)</p> <p>For reasons coming from representation theory we have many examples where all entries in the intersection form are divisible by a integer, which happens to be the order of a finite group $W(e)$.</p> <blockquote> Is this a familiar phenomenon somewhere else? What geometric tools are used to prove such divisibility? </blockquote> <p>(In our situation the finite group $W(e)$ acts on the whole set-up, and my naive guess was that we should consider the quotient by $W(e)$ and proceed as in <a href="http://mathoverflow.net/questions/104253/intersection-form-on-quotient-manifold" rel="nofollow">this question</a>. However there are many finite groups which act, and whose orders have nothing to do with $|W(e)|$. So if this is the explanation, then $W(e)$ should be special in some way.)</p> <p>For those with some background in geometric representation theory here is a more precise description of my situation:</p> <ul> <li> $\mathcal{N}$ is the nilpotent cone in a complex semi-simple Lie algebra; <li> $e \in \mathcal{N}$ and $S_e$ the intersection of $\mathcal{N}$ with a Slodowy slice through $e$; <li> $\pi : T^*(G/B) \to \mathcal{N}$ is the Springer resolution; <li> $\pi : S \to S_e$ is the corresponding resolution of $S_e$; <li> $F = \pi^{-1}(e)$ is the Springer fibre inside $S$. </ul> <p>We can show in type $A$ that all entries of the intersection form are divisible by $W(e)$, the Weyl group of the reductive part of the centraliser of $e$. However the proof is entirely non-geometric. We do not know if this is true in other types.</p> http://mathoverflow.net/questions/121996/applications-for-intersection-cohomology-and-for-the-decomposition-theorem-for/122082#122082 Answer by Geordie Williamson for Applications for intersection (co)homology and for the Decomposition Theorem for students? Geordie Williamson 2013-02-17T17:07:32Z 2013-02-17T17:07:32Z <p>I gave a course on the decomposition theorem (and its failure with positive char coefficients) in Bochum. Notes and exercises are here: <a href="http://people.mpim-bonn.mpg.de/geordie/bochum/" rel="nofollow">http://people.mpim-bonn.mpg.de/geordie/bochum/</a></p> <p>I found the example of Weierstraß family of elliptic curves quite instructive. One considers the family $Y^2 = (X-\lambda Z)(X-Z)X$ of elliptic curves over $\mathbb{C}$ (parametrised by $\lambda$).</p> <p>In a series of exercises (available at the above link) the topology of this example can be pursued in several steps:</p> <p>1) one tries to understand the local system of $H^1$ away from the singular points 0 and 1. Using ramified covers of $\mathbb{P}^1$ students can calculate the monodromy around the singular points.</p> <p>(Here one sees the first remarkable fact predicted by the decomposition theorem: even though the monodromy is unipotent around each singularity, the global representation of the free group on two letters is simple. So the decomposition theorem fails in the complex analytic category.)</p> <p>2) one can calculate the the cohomology of the singular points and sees that it agrees with the invariants of the monodromy.</p> <p>(Here one sees why the definition of the IC sheaf is "correct" for local systems on $\mathbb{C}$ minus points. This is also an example of the invariant cycle theorem, as Donu points out.)</p> <p>(More advanced: This also gives an example of a non pointwise pure IC sheaf.)</p> <p>3) The local systems of $H^0$ and $H^2$ are constant. One can use the hard Lefschetz theorem along the fibres of the map to deduce the splitting of the direct image.</p> <p>(Here one sees the relative hard Lefschetz theorem in play.)</p> <p>Other examples that I find really instructive:</p> <p>1) resolutions of Kleinian surface singularities. (These are also discussed in the notes above.) Here one can connect the decomposition theorem to the non-degeneracy of intersection forms, as in the beautiful work of de Cataldo and Migliorini.</p> <p>There are notes from a beautiful course Migliorini gave here:</p> <p><a href="http://people.mpim-bonn.mpg.de/geordie/Migliorini.pdf" rel="nofollow">http://people.mpim-bonn.mpg.de/geordie/Migliorini.pdf</a></p> <p>2) I find the proof of Deligne's theorem on the decomposition theorem for smooth maps quite instructive. Here one really sees "relative Hodge theory" in action: the hard Lefschetz theorem implies the degeneration of the spectral sequence, and to get the semi-simplicity of the local systems of cohomology one needs to use the fact that they are all polarised by an ample line bundle on the fibres, hence admit invariant forms, hence are semi-simple.</p> <p>3) Another example where the situation is much simpler is the Hilbert scheme of points on a curve, viewed as a resolution of the symmetric power (the Hilbert-Chow morphism).</p> http://mathoverflow.net/questions/109885/second-homotopy-groups-of-3-complexes-and-fenns-spiders Second homotopy groups of 3-complexes and Fenn's spiders. Geordie Williamson 2012-10-17T08:46:16Z 2012-10-17T20:08:39Z <p>Let $X$ be a finite CW complex then with one zero cell. Then (up to homotopy) the two skeleton of X is the same as a group presentation, via the Cayley complex construction. For a while I had been searching for some planar description of the second homotopy group, which would allow a concrete combinatorial description of the fundamental 2-groupoid of X (up to equivalence).</p> <p>I found many discussions close to what I needed, before stumbling on the (IMHO beautiful) book "Techniques of geometric topology" by Roger Fenn. In Chapter 2 he gives a description of $\pi_2(X)$ of a 3-complex in terms of certain diagrams modulo local relations. Each relation in the 2-complex gives a "relation spider" and the second homotopy group of $X$ is the group of isotopy classes of planar diagrams generated by these spider diagrams modulo certain "universal" local relations (analogous to $gg^{-1} = 1$ in the $\pi_1$ case) and relations given by the 3-cells of $X$. (The spider diagrams are roughly dual to Van Kampen diagrams.)</p> <p>My questions are:</p> <p>1) are spider diagrams Fenn's invention? Perhaps this way of thinking about $\pi_2$ was folklore?</p> <p>2) what are other sources describing $\pi_2$ (or even better the fundamental 2-groupoid) concretely (ideally diagrammatically) for small dimensional complexes?</p> <p>I am aware that all of this can be viewed as a concrete example (for $n = 2$) of the dictionary between n-groupoids and n-types. However because of the applications I have in mind I am only looking for "concrete" sources!</p> http://mathoverflow.net/questions/108614/intermediate-extension-functor-exact/109124#109124 Answer by Geordie Williamson for Intermediate extension functor exact? Geordie Williamson 2012-10-08T09:00:31Z 2012-10-08T18:54:36Z <p>I think that the answer for 2) is no.</p> <p>Consider the affine Grassmannian for $SL_2$ and let $X$ denote the Schubert variety corresponding to the three dimensional representation of $PSL_2$. Then $X$ is stratified by Iwahori orbits into three affine strata of dimensions 2, 1 and 0. Let $U$ denote the open stratum.</p> <p>Consider the exact sequence of local systems</p> <p>$0 \to \mathbb{Z}_2 \to \mathbb{Z}_2 \to \mathbb{F_2} \to 0$</p> <p>on $U$. Shifting by $[2]$ we get an exact sequence of perverse sheaves on $U$ and if we apply $j_{!*}$ we get a sequence</p> <p>$0 \to IC(X,\mathbb{Z}_2) \stackrel{2}{\to} IC(X,\mathbb{Z}_2) \to IC(X,\mathbb{F}_2) \to 0$</p> <p>(Here one needs to be careful: because $\mathbb{Z}_2$ is not a field there are two possible $t$-structures (denoted $p$ and $p^+$) which are interchanged by duality. See Daniel Juteau's paper "Decomposition numbers for perverse sheaves" arXiv:0803.2326.) Here I am considering the perversity $p$.</p> <p>I claim that this sequence cannot be exact. Indeed, the stalks of $IC(X,\mathbb{Z}_2)$ are $\mathbb{Z}_2$ in degree -2, and 0 elsewhere. Whereas the stalks of $IC(X,\mathbb{F}_2)$ are $\mathbb{F}_2$ in degrees $-2$, $-1$ and zero elsewhere.</p> <p>(These stalks calculations are explained in detail in arXiv:0901.3322.)</p> http://mathoverflow.net/questions/92221/intrinsic-characterization-of-soergel-bimodules/99880#99880 Answer by Geordie Williamson for Intrinsic characterization of Soergel bimodules? Geordie Williamson 2012-06-18T08:20:14Z 2012-06-18T08:20:14Z <p>There is an intrinsic characterisation which is probably more complicated than what you are looking for. As Ben says, Soergel bimodules are pretty subtle things ...</p> <p>Because Soergel bimodules are (finitely generated) $R$-bimodules one can think about them as coherent sheaves on $V \times V$ (where $V = Spec R$). Inside $V \times V$ one has for any $w \in S_n$ its reversed graph:</p> <p><code>$Gr_w = \{ (wv, v) \;| \;v \in V \}$</code></p> <p>Hence, given any subset of $U \subset S_n$ one can talk about "sections of an $R$-bimodule $M$ with support in $U$": those sections of $M$ which have support in the union of the graphs of all elements of $U$. In this way, for any subset $I$ of $W$ one can consider $\Gamma_I M \subset M$.</p> <p>Your point (2) means that Soergel any Soergel bimodule satisfies $\Gamma_{S_n} M = M$ (that is, every element is supported on the union of all the graphs of elements of $S_n$). It follows that any Soergel bimodule has a canonical filtration indexed by the ideals of the poset $S_n$. A basic fact is that if one considers the quotient</p> <p>$\Gamma_{\le w / &lt; w} (M) := \Gamma_{\le w} M / \Gamma_{&lt; w} M$</p> <p>this is a free left $R$-module, isomorphic as a bimodule to a direct sum of copies of $R_w$ (the standard'' bimodule with normal left action and right action twisted by $w$). This is proved in Soergel's "Kazhdan-Lusztig-Polynome und unzerlegbare Bimoduln über Polynomringen" and is also discussed and generalised in my "Singular Soergel bimodules".</p> <p>So now one can consider all bimodules which satisfy the above property. One this category one can put an exact structure: a sequence is exact if whenever one applies the functor $\Gamma_{\le w / &lt; w}$ one obtains a split exact sequence of $R$-bimodules (necessarily isomorphic to direct sums of shifts of $R_w$'s).</p> <p>Then the claim is that Soergel bimodules are the injective objects in this exact structure. I don't think this is written down anywhere. In the very similar language of moment graphs it is proved by Peter Fiebig in "Sheaves on moment graphs and a localization of Verma flags" here:</p> <p>arxiv.org/abs/math.RT/0505108</p> <p>(I might be mixing things up a bit. I think Peter considers the opposite filtration, which is why he gets projective objects. Anyway, if this is really what you're looking for then I can try to provide some more detail.)</p> <p>By the way, the condition that the subsequent quotients in this filtration be split has other applications. In this paper</p> <p><a href="http://arxiv.org/abs/1205.4206" rel="nofollow">http://arxiv.org/abs/1205.4206</a></p> <p>we examine when Rouquier complexes satisfy this property. It turns out that this is the case if (probably: and only if) the braid is a positive lift of an element of the Weyl group.</p> http://mathoverflow.net/questions/83877/are-kazhdan-lusztig-r-polynomials-the-poincare-polynomials-of-the-corresponding/83897#83897 Answer by Geordie Williamson for Are Kazhdan-Lusztig $R$-polynomials the Poincare polynomials of the corresponding affine varieties Geordie Williamson 2011-12-19T20:39:09Z 2011-12-19T21:18:14Z <p>I think that this is one of these things that looks plausible in small examples but is false. For example, this would imply that the coefficients of R polynomials are alternating in $q$. This is implied by another conjecture called the Gabber-Joseph conjecture (roughly: coefficients of R-poynomials give dimensions of Ext groups between Verma modules), which is false. See "A counterexample to the Gabber-Joseph conjecture" by Brian Boe.</p> http://mathoverflow.net/questions/77348/subexpressions-of-reduced-words-in-coxeter-groups Subexpressions of reduced words in Coxeter groups Geordie Williamson 2011-10-06T10:20:59Z 2011-10-13T18:15:49Z <p>Let $\underline{w} = [s_1, s_2, \dots ,s_n]$ be a reduced expression in a Coxeter group $W$. Given $x$ in $W$ one can consider the set $\Pi(\underline{w},x)$ consisting of all subexpressions of $\underline{w}$ with product $x$. (A subexpression of $\underline{w}$ is a sequence $[t_1, t_2, \dots, t_n]$ such that $t_i$ is either the identity or $s_i$ for all $i$. Given a subexpression $[t_1, \dots, t_n]$ its product is (obviously) $t_1 \dots t_n$.</p> <p>There are lots ways of thinking about subepxpressions. Probably one of the nicest is as a path in the Coxeter complex, where at time $= i$ one either chooses to either stay put or to cross a wall coloured by $s_i$. In this interpretation the product is the end-point of the path.</p> <p>My question is the following:</p> <p>How should one go about calculating the set $\Pi(\underline{w},x)$?</p> <p>Of course there is the obvious answer: enumerate all $2^n$-subsequences and calculate all products (or perhaps take some obvious subset given, for example by length restrictions). However, this obviously gets difficult quickly.</p> <p>I am wondering if anybody has studied this problem from a computational point of view. In small examples it seems like there are a lot of tricks that give an answer much faster than the brute-force method above.</p> <p>(My motivation is that I have recently been writing software to do calculations in the Hecke algebra, where this problem is the bottleneck. Often I have a long reduced expression (length 40 or so) and think that one should be able to calculate $\Pi(\underline{w},x)$ for elements which are not too much smaller than w (say length 30). However either it's hard or I'm missing something!)</p> <p>Myself and my computer would like to thank you in advance for any labour saving tips!</p> http://mathoverflow.net/questions/77348/subexpressions-of-reduced-words-in-coxeter-groups/77970#77970 Answer by Geordie Williamson for Subexpressions of reduced words in Coxeter groups Geordie Williamson 2011-10-12T22:34:06Z 2011-10-13T18:15:49Z <p>I have found a much more efficient way of solving this problem on computer. Having asked the question I guess I should provide a brief account. However I feel like the algorithm is technical and not very enlightening, and so I'll be brief. Please leave comments if you would like more detail.</p> <p>First we consider an important special case: Is there an efficient algorithm to find all subexpressions of a fixed (reduced) expression $\underline{w} = [s_1, \dots, s_m]$ whose product is the identity? In the language of the question, how does one calculate $\Pi(\underline{w}, id)$?</p> <p>Of course, there is one canonical such subexpression: $(id,id,…,id)$. Moreover, given any subexpression for the identity we can apply "cancellation moves" to get to this expression.</p> <p>By a "cancellation move" I mean the following: write $\pi[i,j]$ for the product $t_it_{i+1}…t_j$. Suppose that</p> <p>$s_i \pi[i+1,j-1] = \pi[i+1,j-1] s_j$</p> <p>then a "cancellation move" is one of the following:</p> <p>$\pi = [\dots ,s_i, \dots, id, \dots] \mapsto \pi ' = [\dots, id,…,s_j, \dots]$</p> <p>$\pi = [\dots ,s_i, \dots, s_j, \dots ]\mapsto \pi' = [\dots ,id, \dots, id, \dots]$</p> <p>where $\pi$ remains unchanged except at the $i^{th}$ and $j^{th}$ place. Obviously, applying a cancellation move to a subexpression does not change the product. Also, cancellation moves either increase the number of $id$'s or increase the number of $id$'s to the left. It follows that by repeatedly applying cancellation moves to any subexpression for the identity we end up at the canonical subexpression.</p> <p>For example, consider the Coxeter group with simple reflections $s$, $t$ and $u$ and only braid relation $(st)^3 = id$ (so that $su$ and $tu$ have infinite order). Let $\underline{w} = [s,t,s,u,t,s,t]$. Then the following is a sequence of cancellation moves:</p> <p>$[s,t,s,id,t,s,t] \mapsto [id,t,s,id,id,s,t] \mapsto [id,t,id,id,id,id,t] \mapsto [id, id, id, id, id, id, id]$.</p> <p>Now one can reverse this, and define "reverse cancellation moves". Applying all possible reverse cancellation moves to the canonical subexpression yields all subexpressions for the identity. (This is easily programmed.)</p> <p>We now turn to the general case. The above "cancellation moves" make sense for any subexpression and one can show that again there is a canonical subexpression. (It is characterised by the fact that the subexpression is reduced, and the $t_i \ne id$ occur as far to the right as possible.) Again, once one has this canonical subexpression then it is a simple matter to reconstruct all subexpressions by reverse cancellation moves.</p> <p>Of course this begs the question: can one find this canonical subexpression efficiently? Yes. Firstly, consider $R(x)$ the right descent set of $x$, and choose $s \in R(x)$ such that s occurs as far as possible to the right in $\underline{w}$. Now, delete everything to the right and including the right-most occurrence of $s$ and repeat with $x$ replaced by $xs$. (This also gives a reasonably efficient algorithm to decide whether $x \le y$ in the Bruhat order.)</p> <p>(One can ask what this canonical expression "means". Here is one possible explanation: because Schubert varieties are normal the fibre of the Bott-Samelson resolution over point corresponding to $x \le w$ is connected and hence its homology is one-dimensional in degree zero. Now subexpressions also index BB-cells, and hence a basis for the cohomology of the fibre. This "canonical subexpression" corresponds to a generator of the one-dimensional degree zero part.)</p> <p>For example, take $W = S_7$ (with simple reflections = simple transpositions) and take the following reduced expression for the longest element: </p> <p>$w_0 = 121321432154321654321$</p> <p>then there are 6408 subexpressions with product the identity. On my laptop the above algorithm takes 3.33 seconds to find them. A brute force attack takes 93 seconds.</p> <p>Here is another example which I actually cared about in my calculations. Take the reduced expression</p> <p>$\underline{w} = 13572613574352461357$</p> <p>in $S_8$. There are 80 subexpressions with product the longest element in the standard parabolic subgroup generated by all simple transpositions except 4. The above algorithm takes 0.03 seconds to find all of them, whereas the naive approach takes roughly 15 seconds. (So here the above algorithm is 500 times as fast.)</p> <p>I imagine that the differences become (even) more pronounced with longer words.</p> http://mathoverflow.net/questions/74017/parabolic-convolution-of-perverse-sheaves-in-terms-of-the-hecke-algebra/76539#76539 Answer by Geordie Williamson for Parabolic convolution of perverse sheaves in terms of the Hecke algebra Geordie Williamson 2011-09-27T17:59:29Z 2011-09-27T17:59:29Z <p>Let $G$ be a connected reductive algebraic group (over $\mathbb{C}$) and fix a Borel subgroup $B \subset G$. One can consider the 2-category with objects parabolic subgroups $P \supset B$ and 1-morphisms $P \to Q$ given by $D^b_{P\times Q}(G)$ (the $P \times Q$-equivariant derived category of $G$ with respect to the action $(p,q) \cdot g = pgq^{-1}$ for $p \in P$, $q \in Q$ and $g \in G$). 2-morphisms are the morphisms of $D^b_{P \times Q}(G)$. Composition of 1-morphisms is given by convolution:</p> <p>$* : D_{P \times Q}(G) \times D_{Q \times R}(G) \to D_{P \times R}(G)$</p> <p>(you can probably guess how this is defined using the description you give above). </p> <p>Then your question is a special case of the following question:</p> <p>Q: describe the Grothendieck group of this 2-category (a 1-category).</p> <p>The answer is that Grothendieck group is* what I call the Schur algebroid (for want of a better name). See the beginning of my paper "Singular Soergel bimodules" on the arXiv. (I am only referring to my paper because it is a convenient reference. Certainly these things have been known to experts since at least the early 80's.)</p> <p>If you would prefer not to look at this paper here is a direct answer to your question. For any subset $I \subset S$ of the simple reflections let $\underline{H}_I$ denote the Kazhdan-Lusztig basis element indexed by the longest element of the standard parabolic $W_I$. Then, given $h \in \mathcal{H}\underline{H}_I$ and $h' \in \underline{H}_I \mathcal{H}$ one can define</p> <p>$h*_I h' := \frac{1}{\pi(I)} hh'$</p> <p>where $\pi(I)$ denotes the Poincaré polynomial of $W_I$. (Note that this element really lives in the $\mathbb{Z}[v,v^{-1}]$ form of the Hecke algebra.) This is the class in the Hecke algebra you are looking for.</p> <p>*: Of course this is not true because one has lost the $q$: what I really mean is that one should either consider the split Grothendieck group of semi-simple complexes (or parity sheaves if one prefers) or use an appropriate mixed version.</p> <p>Final note: you talk about Polo's result about arbitrary polynomials $\in 1 + q\mathbb{N}[q]$ occuring as KL-polynomials. I recall that there is a purely combinatorial proof of this result in the literature. Unfortunately I can't remember the title or author, but it shouldn't be difficult to find.</p> http://mathoverflow.net/questions/67831/morita-invariance-of-drinfeld-centre Morita invariance of Drinfeld centre Geordie Williamson 2011-06-15T05:05:15Z 2011-06-15T05:48:30Z <p>Given a monoidal category $M$ one can consider its Drinfeld centre $Z(M)$. Objects of the Drinfeld centre are pairs $(m, \alpha)$ where $m$ is an object and $\alpha$ is an isomorphism $\alpha: - \otimes m \to m \otimes -$ satisfying some "obvious" conditions.</p> <p>A simple and important example of a monoidal category is the category of $G$-equivariant sheaves of $k$-vector spaces on $Y \times Y$ where $Y$ is a finite $G$-set. The monoidal structure is given by </p> <p>$V\otimes W = p_{13*} (p_{12}^*V \otimes p_{23}^* W)$</p> <p>where $p_{ij} : Y \times Y \times Y \to Y \times Y$ denotes the projection map (in a hopefully obvious notation).</p> <p>For example, if $Y$ is a point then one recovers the (tensor) category of representations of $G$. If $Y = G$ then one recovers a monoidal category equivalent vectors spaces graded by $G$. If $G$ is the trivial group then one obtains the tensor category of matrices of vector spaces'' over $Y$.</p> <p>Now there is a result, which I have heard (by Ostrik) called Muerger's Morita invariance of Drinfeld centre''. It should have the consequence that, with $G$ and $Y$ as above:</p> <blockquote> <p>The Drinfeld centre $Z(Sh_G(Y \times Y))$ does not depend on $Y$ up to equivalence.</p> </blockquote> <p>(I guess the baby example of $G$ the trivial group explains the term Morita invariance''.)</p> <p>My question is:</p> <blockquote> <p>Where can I read about these results in the literature?</p> </blockquote> http://mathoverflow.net/questions/46521/formality-of-classifying-spaces Formality of classifying spaces Geordie Williamson 2010-11-18T18:29:21Z 2011-04-05T13:58:57Z <p>Let $G$ be a compact Lie group (or reductive algebraic group over $\mathbb{C}$), and let $BG$ be its classifying space. Fix a prime $p$. Let $\mathcal{A}$ denote the dg algebra of singular cochains on $BG$ with coefficients in a field or characteristic $p$ (or if you prefer the dg algebra of endomorphisms of the constant sheaf). My question is:</p> <blockquote> <p>Is it known for which primes $p$ the dg algebra $\mathcal{A}$ is formal, that is, quasi-isomorphic to a dg algebra with trivial differential?</p> </blockquote> <p>I assume / hope that the answer is that this is true if $p$ is not a torsion prime for $G$ (i.e. $p$ arbitrary in types $A$ and $C$, $p \ne 2$ in types $B$, $D$ and $G_2$, $p \ne 2, 3$ in types $F_4$, $E_6$ and $E_7$, and $p \ne 2,3,5$ in type $E_8$.)</p> <p>Note that we know* that $\mathcal{A}$ is formal in characteristic 0.</p> <blockquote> <p>Can one then conclude that it is formal in any characteristic in which the cohomology of $\mathcal{A}$ is torsion free?</p> </blockquote> <p>If so I think this would give the above list of primes.</p> <p>*) for example because $H(BG, \mathbb{Q})$ is a poynomial algebra, and $\mathcal{A}$ admits a graded commutative model using the de Rham complex -- see Bernstein-Lunts "Equivariant sheaves and functors".</p> http://mathoverflow.net/questions/34755/what-is-the-hopf-algebra-structures-in-the-homology-of-the-based-loop-spaces-of/59147#59147 Answer by Geordie Williamson for What is the Hopf algebra structures in the homology of the based loop spaces of $E_7$ and $E_8$? Geordie Williamson 2011-03-22T07:54:31Z 2011-03-22T07:54:31Z <p>This is more of a suggestion than an answer. Let $K$ be a compact Lie group and $G$ its complexification. It is known (see Pressley-Segal) that the based loop space of $K$ and the affine Grassmannian $Gr_G$ of $G$ have the same (co)homology groups. It is a theorem of Ginzburg that the the cohomology ring of $Gr_G$ may be (canonically) identified with the enveloping algebra of the centraliser of a regular nilpotent element $e \in \mathfrak{g}^{\vee}$ (here $\mathfrak{g}^{\vee}$ denotes the Lie algebra of the group $G^{\vee}$ Langlands dual to $G$).</p> <p>One can dualise this to get that the homology of $Gr_G$ can be identified with the functions on <code>$B^{\vee}_e$</code>, where <code>$B^{\vee}$</code> denotes a Borel subgroup of $G^{\vee}$ and $e$ is a regular nilpotent element in the Lie algebra of $B^{\vee}$. This is explained in a very nice paper by Yun and Zhu called "Integral homology of loop groups via Langlands dual groups". They prove that this induces an isomorphism of the corresponding group schemes (with the Hopf algebra structure on the homology of $Gr_G$ inducing the group scheme structure on one side).</p> <p>So this doesn't provide explicit formulas, but it does give you a way to work out the Hopf algebra structure (even over the integers in the simply laced case -- again see Yun-Zhu for this statement) in terms of combinatorics of root systems.</p> <p>If you are interested in torsion phenomena the paper "Some arithmetical results on semi-simple Lie algebras" by Springer might be useful.</p> http://mathoverflow.net/questions/47852/examples-of-mixed-hodge-structures/47906#47906 Answer by Geordie Williamson for Examples of Mixed Hodge Structures Geordie Williamson 2010-12-01T14:12:03Z 2010-12-01T14:12:03Z <p>A.H. Durfee, A naive guide to mixed Hodge theory,'' Singularities, Part 1 (Arcata, Calif., 1981), 313-320, Proc. Sympos. Pure Math., 40, Amer. Math. Soc., Providence, RI, 1983.</p> <p>is quite nice to read, but probably doesn't handle any more cases than you already know.</p> http://mathoverflow.net/questions/29409/is-soergels-proof-of-kazhdan-lusztig-positivity-for-weyl-groups-independent-of-o/35858#35858 Answer by Geordie Williamson for Is Soergel's proof of Kazhdan-Lusztig positivity for Weyl groups independent of other proofs? Geordie Williamson 2010-08-17T11:59:45Z 2010-08-17T11:59:45Z <p>Perhaps I can supplement Jim's answer a little.</p> <p>In the paper "Kazhdan-Lusztig-Polynome und unzerlegbare Bimoduln uber Polynomringen" Soergel shows that there are certain graded indecomposable bimodules over a polynomial ring (now known as Soergel bimodules) which categorify the Hecke algebra. (Note that these are <em>not</em> projective!!)</p> <p>That is, the indecomposable objects are classified (up to shifts and isomorphism) by the Weyl group, and one has an isomorphism between the Hecke algebra and the split Grothendieck group of the category of Soergel bimodules.</p> <p>As a consequence, one obtains a basis for the Hecke algebra which is positive in the standard basis (as follows from the construction of the isomorphism of the Grothendieck group with the Hecke algebra) and has positive structure constants (because it is a categorification).</p> <p>Soergel conjectures that this basis is in fact the Kazhdan-Lusztig basis, which would imply positivity in general.</p> <p>Up until now there are only two cases when one can verify Soergel's conjecture:</p> <ul> <li>when one has some sort of geometry (in which case one can show that the indecomposable Soergel bimodules are the equivariant intersection cohomology of Schubert varieties, and then use old results of Kazhdan and Lusztig). This shows Soergel's conjecture for Coxeter groups associated to Kac-Moody groups (in particular finite and affine Weyl groups).</li> <li>when the combinatorics is very simple (i.e. for dihedral groups (Soergel) or universal Coxeter groups (Fiebig, Libedinsky)).</li> </ul> <p>Hence, up until now there are no examples where Soergel's conjecture has yielded positivity when it was not known by other means. Also note that in the vast majority of cases, the proof using Soergel bimodules is strictly more complicated than the geometry proof, as one needs an extra step to get from geometry to Soergel bimodules.</p> <p>Soergel's conjecture would however have more far reaching consequences than a proof that Kazhdan-Lusztig polynomials have positive coefficients. For example it provides a natural "geometry" for arbitrary Coxeter groups. For example, generalising some sort of Soergel bimodules to complex reflection groups would yield a natural setting for the study of "spetses" (unipotent characters associated to complex reflection groups).</p> <p>One should also note that Dyer has developed a very similar conjectural world associating commutative algebra categories to Coxeter groups. He instead considers modules over the dual nil Hecke ring (which is the analogue of the cohomology of a flag variety), and has many nice results and conjectures. (Much of his work considers more general orders than the Bruhat order, and so will probably come in handy soon ...!)</p> <p>While I am at it I should mention Peter Fiebig's theory of Braden-MacPherson sheaves on moment graphs. This is (in a sense made precise in one of Peter's papers) a local version of Soergel bimodules, and hence many questions become more natural on the moment graph.</p> <p>Finally, one should mention the recent work of Elias-Khovanov and Libedinsky, which give generators and relations for the monoidal category of Soergel bimodules for certain Coxeter groups. (Elias-Khovanov in type A and Libedinsky in right-angled type.) These are very interesting results, but it is unclear to what extent they can be used to attack Soergel's conjecture.</p> http://mathoverflow.net/questions/31266/which-statements-in-section-5-of-bbd-will-fail-if-we-consider-mathbbq-l-adic/31283#31283 Answer by Geordie Williamson for Which statements in section 5 of BBD will fail if we consider $\mathbb{Q}_l$-adic sheaves there? Geordie Williamson 2010-07-10T10:34:16Z 2010-07-10T10:34:16Z <p>Most of the statements go through. This is discussed in some detail at the start of Chapter 4. They also give a list of things that one has to be careful of when using other coefficients.</p> <p>In particular, I think Proposition 5.1.15 stays true with coefficients in $\mathbb{Q}_{\ell}$. This should follow because all functors commute with the extension of scalars functor from <code>$\mathbb{Q}_{\ell}$</code> to <code>$\overline{\mathbb{Q}}_{\ell}$</code>.</p> http://mathoverflow.net/questions/22699/morphisms-between-pure-complexes-of-sheaves/23257#23257 Answer by Geordie Williamson for Morphisms between pure complexes of sheaves Geordie Williamson 2010-05-02T13:52:30Z 2010-05-02T13:52:30Z <p>Dear Mikhail,</p> <p>I had been hoping someone else would attempt to answer this question, as I have been wondering very similar things lately. (In fact I drove myself crazy for about a month last year trying to work out some solution to what you are asking in the second paragraph.)</p> <p>I can't answer everything but here is a start:</p> <p>Write $K = \overline{\mathbb{Q}_{\ell}}$. One already sees the problems with what you are asking for $X_0 = Spec \mathbb{F}_q$. Then the category of constructible $K$-sheaves on $X_0$ is equivalent to the category of finite dimensional $K$-representations of $Gal(\mathbb{F}/\mathbb{F}_q)$, the absolute Galois group of $\mathbb{F}_q$. (The absolute Galois group is generated topologically by Frobenius, and so this is the same as giving a finite dimensional $K$-vector space together with an endomorphism.) [See BBD 5.1.11 for a statement, I think this is explained in Milne, but don't have it at the moment.]</p> <p>Now, a pure sheaf on $X_0$ is pure of weight $i$ if all the eigenvalues of Frobenius are algebraic integers all of whose complex conjugates over $\mathbb{C}$ have the same absolute value $q^{i/2}$.</p> <p>Note that here we already see that over $X_0$ a pure sheaf does not need to be semi-simple. Indeed, there is no reason why Frobenius should act semi-simply. (This is one example of what de Cataldo and Migliorini are talking about in Remark 3.1.9 after the Theorem 3.1.8.) I think it is part of the standard conjectures that Frobenius acts semi-simply on the $\ell$-adic cohomology of smooth projective varieties, which as I understand it, is still not known.</p> <p>I don't know what you mean when you ask:</p> <blockquote> <p>Does the splitting of part 1 (only!) really requires $\overline{\mathbb{Q}_{\ell}}$-coefficients?</p> </blockquote> <p>As to your main question, I think that the above example shows that this is too much to hope. Without working over $X_0$ one cannot define what it means to be mixed, and without going to $X_0$ one can't expect the same ext vanishing.</p> <p>I recently discovered your work on "weight structures" and found it very interesting. I guess you are asking the above, because you would like to argue that one gets a weight structure in the setting of $\overline{\mathbb{Q}_{\ell}}$-sheaves.</p> <p>There is one setting where I think that one really does get a weight structure. This is in the (at least formally) very similar world of "mixed Hodge modules" on complex varieties. There one has the desired ext vanishing from the outset.</p> http://mathoverflow.net/questions/18301/decomposition-theorem-and-blow-ups/18317#18317 Answer by Geordie Williamson for Decomposition theorem and blow-ups Geordie Williamson 2010-03-15T22:37:57Z 2010-03-16T09:41:48Z <p>In this example we have $p : X \to Y$ and we may assume, wlog, that $X$ is isomorphic to the total space of the normal bundle to the surface, and $p$ is the contraction of the zero section.</p> <p>Then, by the Deligne construction, $IC(Y) = \tau_{\le -1} j_* \mathbb{Q}[3]$, where $j : Y^0 \hookrightarrow Y$ is the inclusion of the smooth locus (which is isomorphic to $X^0$ the complement of the zero section in $X$).</p> <p>In order to work this out, we can use the Leray-Hirsch spectral sequence</p> <p>$E_2^{p,q} = H^p(S) \otimes H^q(\mathbb{C}^*) \Rightarrow H^{p+q}(X^0)$</p> <p>this converges at $E_3$ and we get that the degree 0, 1 and 2 parts of the cohomology of $X^0$ is given by the <em>primitive</em> classes in $H^i(S)$ for $i = 0, 1, 2$. Note that this is everything in degrees 0 and 1, but in degree two the primitive classes form a codimension one subspace $P_2 \subset H^2(S)$.</p> <p>The Deligne construction above, gives us that <code>$IC(Y)_0 = H^0(S)[3] \oplus H^1(S)[2] \oplus P_2[1]$</code>.</p> <p>(This is a general fact: whenever you take a cone over a smooth projective variety, the stalk of the intersection cohomology complex at 0 is given by the primitive classes with respect to the ample bundle used to embed the variety. This follows by exactly the same arguments given above.)</p> <p>Then the decomposition theorem gives</p> <p>$p_* \mathbb{Q} = \mathbb{Q}_0[1] \oplus ( IC(Y) \oplus H^3(S) ) \oplus H^4(S)[-1]$.</p> <p>EDIT: fixed typos pointed out by Chris.</p> http://mathoverflow.net/questions/17886/why-would-one-expect-a-derived-equivalence-of-categories-to-hold/18010#18010 Answer by Geordie Williamson for Why would one expect a derived equivalence of categories to hold? Geordie Williamson 2010-03-12T20:51:48Z 2010-03-14T17:14:26Z <p>Another motivation is more basic. For example, in the modular representation theory of finite groups it is often the case that one has two blocks $A$ and $B$ of two different groups and one suspects (for example) that both blocks have the same number of simple modules (an example of this is given by Alperin's conjecture).</p> <p>Now, suppose that $A$ and $B$ are Morita equivalent. Then not only do $A$ and $B$ have the same number of simple modules, but specifying a Morita equivalence gives a bijection between the simple modules.</p> <p>Often, however, $A$ and $B$ are not Morita equivalent, but rather derived equivalent. The Grothendieck groups of $D^b(A)$ and $D^b(C)$ have a basis given by the classes of the simple modules of $A$ and $B$. A derived equivalence induces an isomorphism between the Grothendieck groups, and hence $A$ and $B$ have the same number of simple modules if they are derived equivalent. Note now, however, that the derived equivalence does not induce a bijection between simple modules, because simple modules need not correspond under the derived equivalence.</p> <p>As an example of this approach is Broué's abelian defect group conjecture (which predicts a derived equivalence between certain $A$ and $B$). It implies Alperin's conjecture (in the abelian defect case), but provides a structural reason for the equality (and also implies much more structural results about characters, "perfect isometries" ...).</p> <p>Hence, one may search for a derived equivalence to give a structure explanation for various concrete numerical equalities.</p> <p>(I think another example of this is given by Bezrukavnikov's use of perverse coherent sheaves on the nilpotent cone to explain some numerical equivalences observed by Vogan, but I don't know much about this.)</p> http://mathoverflow.net/questions/17821/steinberg-representations-of-finite-groups-of-lie-type/17865#17865 Answer by Geordie Williamson for Steinberg Representations of Finite Groups of Lie Type Geordie Williamson 2010-03-11T14:11:11Z 2010-03-11T14:55:22Z <p>I don't think so. Let $T$ be an $F$-stable torus. A character of $T^F$ is in <em>general position</em> if its stabiliser under $N_G(T)/T$ is trivial. I assume by <em>generic</em> you mean "obtained by Deligne-Lusztig induction from a character in general position". (These are exactly the characters which appear in MacDonald's conjecture, and are therefore "generic".)</p> <p>In this setting the Steinberg character is the opposite of generic. It appears, for example, when one induces the trivial character from a split torus (and I think it occurs in the Deligne-Lusztig induction from <em>any</em> $F$-stable torus, but am not sure). For example, in $SL_2$ the (Harish-Chandra = Deligne-Lusztig) induction of the trivial character yields $1 + St$ and Deligne-Ludztig induction of the trivial character from the non-split torus yields $1 - St$.</p> http://mathoverflow.net/questions/15072/how-good-an-approximation-to-the-equivariant-derived-category-is-given-by-the-gra/15208#15208 Answer by Geordie Williamson for how good an approximation to the equivariant derived category is given by the Grassmannian filtration of the classifying space? Geordie Williamson 2010-02-13T20:49:26Z 2010-02-13T20:49:26Z <p>Something related (but not exactly what you are asking) is covered in some detail in Bernstein-Lunts (around 2.2.3).</p> <p>Call a map $\pi : P \to X$ $n$-acyclic if for any <em>sheaf</em> $F$ the truncated adjunction morphism $F \to \tau_{\le n} \pi_* \pi^* F$ is an isomorphism.</p> <p>Given an $n$-acyclic $G$-equivariant map $P \to X$ where $P$ has free $G$-action one has a map $p: P/G \to X/G$ (keeping your notation).</p> <p>Then, if $F$ and $G$ have cohomology sheaves concentrated in an interval $I$ with $|I| &lt; n$ then the natural map from $Hom (F, G) \to Hom(p^* F, p^* G)$ is an isomorphism. (Here, in contrast to your usage in the question, $Hom$ means only degree zero homomorphisms).</p> <p>Two comments:</p> <ul> <li>(depending on your definition) $Hom(F,G)$ is <em>defined</em> to be $\varprojlim Hom(p_m^* F, p_m^* G)$ and so the statement "is injective" is a bit misleading!</li> <li>in geometric representation theory the objects on $pt/G$ that are being considered are often direct sums of equivariant constant sheaves (eg if one takes the equivariant intersection cohomology of a projective variety) in which case your description of the kernel works just fine!</li> </ul> http://mathoverflow.net/questions/1675/how-to-do-computations-using-the-decomposition-theorem-for-perverse-sheaves/1694#1694 Answer by Geordie Williamson for How to do Computations Using the Decomposition Theorem for Perverse Sheaves Geordie Williamson 2009-10-21T17:19:43Z 2009-12-19T09:43:46Z <p>To supplement Ben's answer, basically every aspect of the decomposition theorem is hard.</p> <p>To give you a simple example of something which is implied by the decomposition theorem but is far from trivial is the following statement: given a proper smooth map of smooth varieties f : X -> Y the direct image of the constant sheaf splits as a direct sum of local systems. Note that this implies (but is stronger than) the degeneration of the Leray-Serre spectral sequence for the fibration. This answers to some extent your question "what is so special about algebraic varieties" because Leray-Serre just doesn't degenerate in general.</p> <p>I think the situation has been cleared up considerably by the work of de Cataldo and Migliorini which (IMHO) is the first genuinely geometric proof of the decomposition theorem.</p> <p>One might think of the "smooth map" case above as the "easiest case" (and indeed it does have an easier proof). However de Cataldo and Migliorini point out that in fact the "easiest case" is the case of a semi-small map, for which the decomposition theorem can be deduced from the non-degeneracy of certain bilinear forms. In a difficult work, they deduce the general case by reducing to this case by induction on the "defect of semi-smallness" (how far away a map is from being semi-small) and by taking hyperplane sections to reduce this defect.</p> <p>An excellent informal survey about the decomposition theorem, with lots of wonderful examples can be found in <em><a href="http://arxiv.org/abs/0712.0349" rel="nofollow">The decomposition theorem, perverse sheaves and the topology of algebraic maps</a></em> by de Cataldo and Migliorini.</p> <p>Note that there are really three statements in the decomposition theorem, all of which are hard:</p> <ol> <li>the direct image is the sum of its perverse cohomology groups;</li> <li>each perverse cohomology is a direct sum of IC extensions of a local system;</li> <li>each local system is semi-simple.</li> </ol> <p>As is often the case in mathematics, a nice way to learn why the decomposition theorem is hard is to go to situations when it fails. This occurs when one takes perverse sheaves with coefficients in positive characteristic (or even Z). Daniel Juteau, Carl Mautner and I have written a survey called "Perverse sheaves and modular representation theory" which contains lots of examples of the failure of the decomposition theorem. (Note that all of 1), 2) and 3) above can fail!)</p> http://mathoverflow.net/questions/1832/homfly-and-homology-also-superalgebras/1842#1842 Answer by Geordie Williamson for HOMFLY and homology; also superalgebras Geordie Williamson 2009-10-22T07:58:22Z 2009-10-22T07:58:22Z <p>I can clarify your initial questions, but can't help with things super!</p> <p>Normally, as you point out, Khovanov-Rozansky homology refers to a bigraded theory whose Euler characteristic is sl(n) homology (which is a specialisation of the HOMFLYPT polynomial). This was originally defined by Khovanov and Rozansky in "Matrix factorizations and link homology". In "Matrix factorizations and link homology II" Khovanov and Rozansky define a triply graded theory whose Euler characteristic is the HOMFLYTPT polynomial and Khovanov later showed that theory has a nice construction using using Soergel bimodules. This is often also referred to Khovanov-Rozansky homology, but the correct name seems to be "triply graded link homology".</p> <p>I think the super situation in general is much less well understood, at least from the representation theoretic perspective.</p> http://mathoverflow.net/questions/1039/explicit-direct-summands-in-the-decomposition-theorem/1686#1686 Answer by Geordie Williamson for Explicit Direct Summands in the Decomposition Theorem Geordie Williamson 2009-10-21T17:01:49Z 2009-10-21T17:01:49Z <p>I agree with Ben that the question is a little confusing.</p> <p>There are two possible questions:</p> <p><em>How do you calculate direct summands of the direct image when we start with an IC (not necessarily the constant sheaf) on the source?</em></p> <p>In this case I think the direct image need not be perverse. In which case you are in the general situation of describing the direct summands of a semi-simple complex, for which you need to know the characters of all IC's, or be very clever.</p> <p><em>How do you calculate the local systems occurring in the direct image of the constant sheaf?</em></p> <p>Ben describes what happens above. Semi-small means that the fibre over each stratum of the base has dimension bounded by half the codimension of the stratum (a number I will call d). The local system is then given by local system of 2d^th cohomology of the fibre.</p> <p>In the semi-small situation this is beautiful: the 2d^th cohomology of the fibre is zero if the stratum isn't relevant, and has a basis given by fundamental classes of irreducible components if the stratum is relevant.</p> <p>Note that the fundamental group of the stratum acts on the irreducible components of the fibre via monodromy, and this is precisely the local system that you get. (As an aside, this explains why the local systems are semi-simple, even though the fundamental group might be infinite: the representation always factors through the permutation group on the irreducible components.)</p> <p>I first learnt this in the beautiful article "The Hard Lefschetz Theorem and the topology of semismall maps" by de Cataldo and Migliorini.</p> http://mathoverflow.net/questions/129218/algebraic-stratifications-of-g-varieties Comment by Geordie Williamson Geordie Williamson 2013-04-30T18:03:46Z 2013-04-30T18:03:46Z yes, this is true and Braden is the perfect person to ask as ulrich says! If all you want is that the categories are preserved by the six operations then this is not difficult to see using equivariant sheaves (see related comments at the beginning of &quot;tilting exercises&quot; by Beilinson, Bezrukavnikov, Mirkovic). http://mathoverflow.net/questions/128029/decomposition-of-c-kazhdan-lusztig-basis-element-associated-to-longest-word-in-s Comment by Geordie Williamson Geordie Williamson 2013-04-19T05:43:44Z 2013-04-19T05:43:44Z I think that this would be very difficult in general, as it depends on choices of reduced expression. Ben Elias has found a nice answer for dihedral groups in terms of representations of sl_2. However you are asking about $S_n$, so these results probably aren't so useful. http://mathoverflow.net/questions/125252/is-this-intersection-complex-a-sheaf/125272#125272 Comment by Geordie Williamson Geordie Williamson 2013-03-23T08:41:00Z 2013-03-23T08:41:00Z Another point of view which might be useful: If $X = X_1 \times X_2$ then $IC(L_1 \boxtimes L_2) = IC(L_1) \boxtimes IC(L_2)$ and so one can reduce to the case of a line. http://mathoverflow.net/questions/117978/small-maps-extension-of-ic-sheaves-and-bm-homology Comment by Geordie Williamson Geordie Williamson 2013-01-05T19:30:10Z 2013-01-05T19:30:10Z I think it is correct that the total spaces of the two resolutions involved are isomorphic, however as you point out the isomorphism does not commute with the projection to $X$. (In the general case I guess it should be true that all small resolution $X' -&gt; X$ have equal motives over $X'$ (point counts of all fibres agree, total cohomology agrees etc.), however this is still a long way away from being isomorphic as varieties.) http://mathoverflow.net/questions/117978/small-maps-extension-of-ic-sheaves-and-bm-homology Comment by Geordie Williamson Geordie Williamson 2013-01-04T10:31:28Z 2013-01-04T10:31:28Z Hello again! With regards to your question 3, the Atiyah flop (see wikipedia) gives two inequivalent resolutions of the singularity $xw = yz$ in $C^4$. (This singularity occurs in a Schubert variety in the Grassmannian of 2-planes in 4 space.) http://mathoverflow.net/questions/117978/small-maps-extension-of-ic-sheaves-and-bm-homology Comment by Geordie Williamson Geordie Williamson 2013-01-03T20:59:26Z 2013-01-03T20:59:26Z Hi Dragos, it seems that your definition in the second paragraph is for a semi-small map rather than a small map? http://mathoverflow.net/questions/116348/morphisms-between-verma-modules Comment by Geordie Williamson Geordie Williamson 2012-12-16T02:50:07Z 2012-12-16T02:50:07Z Good question! I remember talking to Soergel about this a while ago. If memory serves me right he told me that this is no longer true on partial flag varieties, but unfortunately I can't remember where the first counterexample occurs. Also I have often wondered about torsion in the cohomology of these intersections, but I haven't gotten beyond wondering... (some motivation for why one might care is in the section on R-varieties in <a href="http://arxiv.org/abs/1209.3760" rel="nofollow">arxiv.org/abs/1209.3760</a>). http://mathoverflow.net/questions/114075/de-rham-isomorphism-without-integration Comment by Geordie Williamson Geordie Williamson 2012-11-21T18:07:06Z 2012-11-21T18:07:06Z @Donu: Yes, of course you are right ... http://mathoverflow.net/questions/114075/de-rham-isomorphism-without-integration Comment by Geordie Williamson Geordie Williamson 2012-11-21T17:39:54Z 2012-11-21T17:39:54Z Isn't this what sheaf theory was born to do?! Both the de Rham complex and resolution by the sheaf of singular cochains are resolutions of the constant sheaf which are acyclic for global sections, and hence both compute the cohomology of the constant sheaf. http://mathoverflow.net/questions/113849/can-an-nonzero-ic-sheaf-have-zero-hypercohomology/113857#113857 Comment by Geordie Williamson Geordie Williamson 2012-11-19T21:04:33Z 2012-11-19T21:04:33Z @Ben: When you say &quot;non-trivial monodromy&quot; I guess you mean invariants = coinvariants = 0. Of course local systems with unipotent monodromy on $\mathbb{C}^*$ have cohomology. http://mathoverflow.net/questions/113849/can-an-nonzero-ic-sheaf-have-zero-hypercohomology/113857#113857 Comment by Geordie Williamson Geordie Williamson 2012-11-19T21:02:37Z 2012-11-19T21:02:37Z More generally one can take any $X$ and a proper and generically finite map $f : X -&gt; X$. Then $\mathbb{H}(f_*\mathbb{Q}_X) = H^*(X)$ and so every summand of $f_*\mathbb{Q}_X$ except for $\mathbb{Q}_X$ (an IC by the decomposition theorem) has vanishing cohomology. http://mathoverflow.net/questions/110962/smoothness-properties-of-the-springer-fiber/110967#110967 Comment by Geordie Williamson Geordie Williamson 2012-11-12T07:40:02Z 2012-11-12T07:40:02Z &quot;In particular, the Kazhdan--Lusztig cell representations of the Hecke algebra are the Springer representations.&quot; One has to be careful here. This would be implied by the irreducibility of the characteristic variety of the intersection cohomology D-module, which is known to be false. See &quot;Geometric construction of crystal bases&quot; by Kashiwara and Saito, in particular Section 1.2. http://mathoverflow.net/questions/109661/polarizable-variations-of-mixed-hodge-structures Comment by Geordie Williamson Geordie Williamson 2012-10-23T08:22:47Z 2012-10-23T08:22:47Z Isn't your 2) (3.2.8) in &quot;Introduction to Mixed Hodge modules&quot;? As Donu points out, the second paragraph after Theorem 2.2 gives that the indecomposable polarisable Hodge modules are IC complexes. http://mathoverflow.net/questions/110149/action-of-frobenius-on-the-tale-cohomology-of-the-variety-of-borel-subgroups-for Comment by Geordie Williamson Geordie Williamson 2012-10-20T19:29:53Z 2012-10-20T19:29:53Z Hi Jay! I thought typically your $\BC_u$ are (geometrically) simply connected. Hence their fundamental groups over $\mathbb{F}_p$ will simply be equal to the fundamental group of $Spec\mathbb{F}_p$ and any local system will be pulled back from a local system on $Spec\mathbb{F})_p$. In this case I would guess that $H^*_c(\BC_u,\pi^*V)$ would just be $H^*_c(\BC_u)\otimes V$ (where $V$ is a local system on $Spec\mathbb{F}_p$ aka continuous rep of $Gal(Spec\mathbb{F}_p)$). http://mathoverflow.net/questions/109885/second-homotopy-groups-of-3-complexes-and-fenns-spiders Comment by Geordie Williamson Geordie Williamson 2012-10-18T07:38:41Z 2012-10-18T07:38:41Z @Ronnie: by fundamental 2-groupoid I mean the 2-groupoid with objects points, 1-morphisms paths between points, and 2-morphisms paths between paths up to homotopy. (A truncation of the fundamental $\infty$-groupoid). I am a complete novice at these things (hence the question), and am sure you are right that other models are easier to work with. Thank you for the link to the review, which looks interesting.