3,002 reputation
917
bio website ma.utexas.edu/users/…
location Austin, TX
age
visits member for 4 years, 3 months
seen yesterday

Postdoc at UT Austin.


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?
Nov
5
comment spectrum of an induced algebra
Ah, you got there first...
Oct
8
revised A question about flag variety of $SL(n,\mathbb{C})$
Changed SL(2,C) to SL(n,C) in the first line.
Oct
1
answered For G a Lie group, can I make sense of G/G as a derived manifold in a nice way?
Sep
5
comment Cobordism and finite sheeted covers of manifolds
Well, for the double cover $S^2 \to \mathbb RP^2$, $S^2$ is not cobordant to 2 copies of $\mathbb RP^2$. But maybe you wanted oriented manifolds?
Sep
4
awarded  Civic Duty