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Postdoc at UT Austin.


Jun
21
comment Representation Theory of $U(N)$
@André: Fair point! I started to write a more detailed answer, then remembered that I had other things to do... probably this answer should have been a comment.
Jun
20
answered Representation Theory of $U(N)$
Jun
18
answered Equivariant Derived Category
Jun
1
awarded  Popular Question
May
3
comment Characterizations of regular holonomic D-modules
Have you looked in Bjork's book? I think most of this stuff is there.
Apr
9
awarded  Nice Answer
Mar
7
answered Relations between functors in a recollement
Mar
6
comment Relations between functors in a recollement
Note that if you have a recollement with the property that $i_R q_L = 0 = i_L q_R$, then the category $\mathbb D$ splits as an orthogonal sum of $\mathbb D^0$ and $\mathbb D^1$; i.e. there are no Homs in either direction. This is certainly not the case for sheaves on a locally closed decomposition of a space...
Mar
6
comment Relations between functors in a recollement
Maybe I am getting confused here, but it is not true that $j^\ast i_\ast = 0$ (where $i$ is the open embedding and $j$ is the closed complement - opposite to my usual convention!). For example if you take the direct image of the constant sheaf under the open embedding $i:\mathbb R - \{0\} \hookrightarrow \mathbb R$, then the stalk $j^\ast i_\ast \mathbb Z_{\mathbb R - \{0\}}$ is 2-dimensional. Similarly, $j^!i_!$ is not zero in general.
Nov
22
comment Is there a cotangent bundle of a stable $\infty$-category?
You may complain that this is more like the tangent bundle than the cotangent, but I think some kind of Koszul duality should relate sheaves on this odd tangent bundle with sheaves on the cotangent bundle (at least up to shifts by 2).
Nov
22
comment Is there a cotangent bundle of a stable $\infty$-category?
Not sure if this is relevant or not, but the categorical Hochschild homology (CHH) of a monoidal category seems to be a reasonable candidate for part 1 (no idea about part 2). By CHH I mean the category $\mathcal C \otimes{\mathcal C \times \mathcal C^{op}} \mathcal C$ associated to a monoidal category $(\mathcal C, \otimes)$. In the case $\mathcal C = QC(X)$, for X a scheme $CHH(\mathcal C) = QC(LX)$, where $LX = X\times_{X\times X} X$ is the derived loopspace (by a result of Ben-Zvi-Francis-Nadler). The derived loopspace is the same as the odd tangent complex by HKR.
Sep
24
awarded  Autobiographer
Sep
2
awarded  Revival
Jul
20
awarded  Yearling
Jul
6
awarded  Enlightened
Jul
2
awarded  Curious
Jun
15
comment in which sense is a mixed Hodge structure an extension of pure ones?
You are not missing anything. The term iterated extension means that (in your example): $W_1$ is an extension of $Gr_1$ by $W_0$, $W_2$ is an extension of $Gr_2$ by $W_1$ etc. So $H$ is "built up" from pure modules, in a similar way that a finite group is built as an iterated extension of finite simple groups.
May
27
comment The general Smith homomorphism in bordism
strung : string ?
May
25
comment Equivariant Sheaves, Local system
More naively, what you are asking for is the analogue of the statement that a 1-dimensional representation $\rho: G \to \mathbb C^\times$ of a group is a class function. The condition $m^\ast L = L \boxtimes L$ just says (roughly) that the stalk $L_{gh}$ at $gh$ is identified with $L_g \otimes L_h$. Thus we have $L_{ghg^{-1}} \simeq L_g \otimes L_h \otimes L_{g^{-1}} \simeq L_g \otimes L_{g^{-1}} \otimes L_h \simeq L_h$, which is (roughly) what it means to be $G$ equivariant. To get rid of the "roughly" you will need to express the above in terms of diagrams...
May
20
comment What's the relationship between these two isomorphisms involving G and T?
I don't know how to answer your question, but let me try to clarify one thing that I probably said hastily in a talk some time. Of course, there is no equivalence (or even a map) of stacks $T/W \to G/G$ - the stabilizers are all wrong, even if the orbit spaces agree. Let us write $N=N_G(T)$. You do have a map of stacks $BN \to BG$, which gives a map on loop spaces $N/N \to G/G$ ($BN$ is what you write as $BT/W$). Inside $N/N$ you have a copy of $T/N = (T/T)/W$ which also maps to $G/G$. Not sure what else to say yet.