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Jan
13 |
answered | Looking for a good exposition - Rees Construction |
Nov
13 |
reviewed | Approve Irrationality of $ \pi e, \pi^{\pi}$ and $e^{\pi^2}$ |
Nov
12 |
reviewed | Approve Compactness of cadlag martingales w.r.t. to the point-wise topology |
Oct
16 |
awarded | Yearling |
Jun
25 |
awarded | Enlightened |
Jun
25 |
awarded | Nice Answer |
Mar
27 |
awarded | Custodian |
Mar
27 |
reviewed | Approve One question about iteration on groups |
Mar
6 |
awarded | Enlightened |
Mar
6 |
awarded | Nice Answer |
Feb
20 |
comment |
Can Enriques Surfaces have non-trivial TWISTED Fourier-Mukai partners?
My guess is that it will be impossible to do it with a genuine surface. One might be able to check this directly by looking at the classification of surfaces, listing all possible twisted Hodge structures, and then checking that the required lattice $E_{8}(-2)\oplus H(2)\oplus H$ can never appear in a twist of a surface. I have not done this carefully but it seems doable. You basically have to rule out K#s and rational elliptic surfaces. |
Feb
19 |
answered | Can Enriques Surfaces have non-trivial TWISTED Fourier-Mukai partners? |
Oct
16 |
awarded | Yearling |
Oct
2 |
awarded | Enlightened |
Oct
2 |
awarded | Nice Answer |
Aug
28 |
comment |
Are linear algebraic groups rigid?
The group law on a unipotent algebraic group deforms non-trivially in general. For instance by scaling the symplectic form you get a one parameter deformation of the three dimennsional Heisenberg group (upper triangular $3\times 3$ matrices with ones on the diagonal) to the additive group of a $3$ dimensional vector space. |
Aug
17 |
comment |
FIltrations on a vector bundle on a curve
I must be missing something, we have rank two (semi-)stable bundles of arbitrary negative degree. So you can not bound the degree of the line sub-bundles from below universally. Do you want a bound that depends on $g$, $n$, and the degree? |
Jun
4 |
comment |
Is $\pi_2$ algebraic?
In general it is neither. Serre's goodness says we have an isomorphism of cohomologies of a discrete group with coefficients in all modules and the cohomologies of the pro-finite completion. Our condition is the same but we have the pro-algebraic completion and coefficients which are locally compact vector spaces. I am pretty sure that if a group $\Gamma$ is an arithmetic lattice in a semisimple group of rank $\geq 1$, then the super-rigidity theorem will imply that $\Gamma$ is Serre's good if and only it is algebraically good. In particular $Sp_{4}(\mathbb{Z})$ is not algebraically good. |
Jun
3 |
comment |
Is $\pi_2$ algebraic?
Fundamental groups of algebraic surfaces are completely general. By Lefschetz hyperplane section theorem every fundamental group of a smooth projective variety is the fundamental group of a smooth algebraic surface. So asking if the fundamental groups of algebraic surfaces have a certain property is the same as asking if all fundamental groups of projective varieties have that property. But you are right: fundamental groups of algebraic surfaces are not good in general and it is hard to decide when they are. |
Jun
3 |
awarded | Nice Answer |