bio | website | math.stanford.edu/~rmbellov |
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location | ||
age | ||
visits | member for | 4 years, 6 months |
seen | 3 hours ago | |
stats | profile views | 1,756 |
Dec 23 |
comment |
Fiber functors to derived categories
Is it really true that $G$ linearly reductive implies $\delta=0$? For example, consider a cts homomorphism $\Gamma\rightarrow G(k)$ where $\Gamma$ is a procyclic group ($k$ could be discrete or have an interesting topology). Then the functor $V\mapsto V^\Gamma$ may not be exact, but the group cohomology $H^i(\Gamma,V)$ is likely computed by the complex $V\rightarrow V$, where the map is $\gamma-1$ for a topological generator $\gamma$ of $\Gamma$. If $V\mapsto V^\Gamma$ is exact, it looks to me like I get 2 $G$-torsors (one for $H^0$, one for $H^1$), but if not, I don't think $\delta=0$. |
Dec 22 |
awarded | Nice Question |
Oct 7 |
awarded | Yearling |
Jun 25 |
awarded | Citizen Patrol |
Jan 15 |
answered | the dual abelian scheme |
Oct 7 |
awarded | Yearling |
Oct 4 |
awarded | Nice Question |
Sep 30 |
accepted | Are D_dR and D_st “potentially comparable”? |
Sep 30 |
comment |
Are D_dR and D_st “potentially comparable”?
Great, thanks! |
Sep 29 |
asked | Are D_dR and D_st “potentially comparable”? |
Jul 3 |
awarded | Enlightened |
Jul 3 |
awarded | Nice Answer |
Jul 2 |
comment |
Some help in digesting a paragraph in the introduction of Deligne/Rapoport's “Les Schemas de Modules de Courbes Elliptique”
Yes, we're thinking of $(\mathbf{Z}/n\mathbf{Z})^2$ as a group scheme. Over any base scheme $S$, the scheme structure on $(\mathbf{Z}/n\mathbf{Z})_S^2$ is just $n^2$ disjoint copies of $S$. |
Jul 2 |
awarded | Enlightened |
Jul 2 |
awarded | Nice Answer |
Jul 2 |
answered | Some help in digesting a paragraph in the introduction of Deligne/Rapoport's “Les Schemas de Modules de Courbes Elliptique” |
May 9 |
awarded | Nice Answer |
Jan 20 |
asked | Fiber functors to derived categories |
Dec 25 |
comment |
Deformation of ordinary p-divisible groups via Grothendieck-Messing
I think you're implicitly asserting that the map $\omega_{G^m}\rightarrow D(G_0)(W(k))$ coming from $G'$ is the same as the composition $\omega_{G^m}\rightarrow \omega_G\rightarrow D(G_0)(W(k))$. |
Dec 24 |
comment |
Shape of snowflakes
It's not the same as assuming no molecules attach at the same time, because in your random model you're effectively adjusting the probability of attachment based on the size of the snowflake. If anything, you want to adjust the probabilities based on the geometry --- I would guess that available vertices deep "inside" the snowflake would have lower probabilities of attachment. |