Timeline for Is $\pi_2$ algebraic?
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
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| Jun 4, 2014 at 16:37 | comment | added | Tony Pantev | 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 4, 2014 at 16:06 | comment | added | Ben Wieland | I skimmed your papers and found no examples of a group that is not good. Do you have any? How about $Sp_4(\mathbb Z)=\pi_1(A_g)$, the standard example that is not Serre good? Is your condition weaker or stronger? | |
| Jun 4, 2014 at 15:19 | vote | accept | Alex Gavrilov | ||
| Jun 4, 2014 at 15:19 | comment | added | Alex Gavrilov | Tony Pantev@ I am going to accept your answer, because it is much more informative then my own. | |
| Jun 3, 2014 at 14:33 | comment | added | Tony Pantev | 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, 2014 at 11:51 | comment | added | Alex Gavrilov | Thank you. This is quite interesting. But if I got it right, it is by no means clear if fundamental groups of complex surfaces are algebraically good in this sense. | |
| Jun 3, 2014 at 4:19 | comment | added | André Henriques | Thank you. I'm glad I asked, because what I had in mind (as what this notion might have been) was quite different. | |
| Jun 3, 2014 at 3:56 | comment | added | Tony Pantev | 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, 2014 at 3:51 | comment | added | Tony Pantev | 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, 2014 at 3:36 | comment | added | André Henriques | What does "algebraically good" mean? | |
| Jun 3, 2014 at 2:34 | history | answered | Tony Pantev | CC BY-SA 3.0 |