bio | website | ma.utexas.edu/users/… |
---|---|---|
location | Austin, TX | |
age | ||
visits | member for | 4 years, 8 months |
seen | Mar 26 at 4:00 | |
stats | profile views | 1,727 |
Postdoc at UT Austin.
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. |
Apr 16 |
answered | Equivariant derived category and invariant divisor |
Apr 8 |
comment |
Smooth mixed hodge modules - representations of fundamental group?
My expectation is that there will be no positive answer to your question. The data of a smooth (pure) Hodge module (aka variation of Hodge structure) involves a filtration on the holomorphic sections of the underlying vector bundle. This filtration is not flat, but rather satisfies the Griffiths transversality condition $\nabla \mathcal F^n \subseteq \mathcal F^{n-1}\otimes \Omega^1$. It doesn't look like it is possible to express these data and conditions in terms of the monodromy. |
Mar 29 |
comment |
Why should noncommutative CYs be dgas?
I think of Calabi-Yau as being a property of a dga rather than differential graded being a property of a CY algebra... |
Feb 14 |
answered | regular singularities and comparison isomorphism |
Jan 27 |
answered | Most general “finiteness of de Rham cohomology” statement for holonomic $D$-modules in the algebraic case? |
Dec 2 |
answered | How is a descent datum the same as a comodule structure? |