Let $X$ be a smooth complex projective surface. Is the Hurewicz image $h(\alpha)\in H_2(X)$ of a homotopy class $\alpha\in\pi_2(X)$ algebraic?
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5$\begingroup$ Not in general: if $X$ is simply connected, the natural map from $\pi _2$ to $H_2$ is an isomorphism, hence any class in $H_2$ comes from $\pi _2$. $\endgroup$– abxCommented Jun 2, 2014 at 11:52
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4$\begingroup$ But what's a counterexample? $\endgroup$– Tom GoodwillieCommented Jun 2, 2014 at 11:55
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1$\begingroup$ K-3 surface works. $\endgroup$– mehCommented Jun 4, 2014 at 4:07
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2 Answers
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A slightly better variant of this question is to ask: is the Hurewitz image of $\pi_{2}(X)$ in $H_{2}(X)$ a sub Hodge structure? This is in fact an old question of Philippe Eyssidieux. In section 4.3 of this paper we proved that if the fundamental group of $X$ is nice enough (more precisely if $\pi_{1}(X)$ is algebraically good) , then $\text{im}\left[ \pi_{n}(X) \to H_{n}(X)\right]$ is a sub Hodge structure for all $n$.
answered Jun 3, 2014 at 2:34
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2$\begingroup$ What does "algebraically good" mean? $\endgroup$ Commented Jun 3, 2014 at 3:36
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5$\begingroup$ 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. $\endgroup$ Commented Jun 3, 2014 at 3:51
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2$\begingroup$ 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. $\endgroup$ Commented Jun 3, 2014 at 3:56
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2$\begingroup$ 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? $\endgroup$ Commented Jun 4, 2014 at 16:06
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3$\begingroup$ 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. $\endgroup$ Commented Jun 4, 2014 at 16:37
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This question is simple, and with some suggestion I can answer it myself. As abx pointed out, for a simply connected surface $h$ is an isomorphism, by the Hurewicz theorem. There are simply connected surfaces with non-algebraic homology classes in $H_2$, for example K3 surfaces.