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visits | member for | 5 years, 7 months |
seen | 2 mins ago | |
stats | profile views | 3,039 |
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 |
Jun 3 |
comment |
Is $\pi_2$ algebraic?
Now given a finitely generated group $\pi$, we say that $\pi$ is algebraically good if the schematization of the Eilenberg-Mclane space $K(\pi,1)$ is $K(\pi^{alg},1)$, i.e. if the schematization has no higher homotopy groups. Algebraically good groups are not easy to come by. In the paper we prove that free groups, fundamental groups of compact surfaces, and fundamental groups of Artin neighborhoods are algebraically good. We also give a lenghty discussion and characterization of goodness. |
Jun 3 |
comment |
Is $\pi_2$ algebraic?
It is a tricky notion. Given a homotopy type $X$ one can construct its complex schematization $(X\otimes \mathbb{C})^{sch}$. By definition this is a higher stack on the etale site of schemes, which has the property that its fundamental group is the pro-algebraic completion of $\pi_{1}(X)$ and every finite dimensional representation $V$ of $\pi_{1}(X)$ gives rise to a coherent sheaf on the schematization, so that the cohomology of the local system $V$ on $X$ is naturally isomorphic to the cohomology of the coherent sheaf on the schematization. The schematization exists by a theorem of Toen. |
Jun 3 |
answered | Is $\pi_2$ algebraic? |
Apr 6 |
awarded | Enlightened |
Apr 6 |
awarded | Nice Answer |
Apr 3 |
answered | On a question motivating Lurie's treatment of formal moduli problems |