bio | website | ma.utexas.edu/users/… |
---|---|---|
location | Austin, TX | |
age | ||
visits | member for | 4 years, 11 months |
seen | yesterday | |
stats | profile views | 1,786 |
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. |