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For purposes of my own, I'm interested in constructing connected spaces, without recourse to geometric realisation or the like, that have non-trivial homotopy groups in dimension 1 and 2 and are not products of Eilenberg-MacLane spaces. This rules out obvious constructions using crossed modules (at least, obvious to me).

One idea is this: take a representation $\rho:\Gamma_g \to SO(3)$ of $\Gamma_g = \pi_1(\Sigma_g)$, the fundamental group of a compact, connected, orientable surface or genus $g$. Then form the associated sphere bundle $X=\widetilde{\Sigma_g} \times_\rho S^2 \to \Sigma_g$.

Then, unless my calculation is wrong, $X$ is a 2-type space with $\pi_2(X) = \mathbb{Z}$, $\pi_1(X) = \Gamma_g$ and $k$-invariant $a\in H^3(\Gamma_g,\mathbb{Z})$.

So my question is,

are there any non-trivial representations $\rho$ or cohomology classes $a$?

and secondarily,

would a non-trivial representation $\rho$ give rise to a non-trivial $k$-invariant in the above situation?

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We have that $H^\ast(\Gamma_g,\mathbb Z)=H^\ast(\Sigma_g,\mathbb Z)$ and the latter is $0$ in degrees $\geq3$ so $a=0$ for trivial reasons. –  Torsten Ekedahl Aug 17 '10 at 7:40
Also your calculation is wrong; there are lots of higher homotopy groups of $S^2$ so $X$ is not a $2$-type. –  Torsten Ekedahl Aug 17 '10 at 8:57
Is the space to be a 2-type? If so then it surely corresponds to a crossed module as crossed modules model 2-types, so I presume David does not want a 2-type, just that there are non-trivial homotopy groups in dimensions 1 and 2. –  Tim Porter Aug 17 '10 at 12:16
Hi Tim, Thorsten. I thought I wanted a 2-type, but now that you mention it of course this isn't (I don't know why I forgot about the higher homotopy groups of S^2 - stupid of me!). But that is not so important. I guess I just wanted a nice geometric example of a space with nontrivial pi_1,pi_2. –  David Roberts Aug 18 '10 at 3:32

2 Answers 2

As Torsten Ekedahl mentioned, the resulting object will not be a 2-type. For this, you would instead want to construct $\mathbb{CP}^\infty$-bundles over your surface. These are classified by homomorphisms Γg → ℤ/2, determining the action of $\pi_1$ on $\pi_2$, together with an associated cohomology element (which is zero for the dimension reasons stated in the comments above). There are therefore (ℤ/2)2g different 2-types with these homotopy groups.

If you really did want to construct S2-bundles, it turns out that (again because surfaces are low-dimensional) these are in bijective correspondence with 2-types under the map that sends such a bundle to its second Postnikov stage. As a result, these are again determined solely by how the fundamental group acts on $\pi_2$. Only the trivial bundle arises from an SO(3)-action because SO(3) acts trivially on the first homotopy group of S2.

There are many distinct representations of the group Γg (e.g. g=1!) and they will obviously not necessarily give rise to distinct bundles. One reason for this is that the associated bundle only depends on the path-component of the representation $\rho$ in the space of all SO(3)-actions (or homeomorphism-actions, or self-homotopy-equivalence-actions).

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I guess that replacing an SO(3) action by an O(3) action would not improve things much... :S –  David Roberts Aug 18 '10 at 3:34

For the first part of your question, there are faithful representations $\rho: \Gamma_g \to SO(3)$. One way to see this is by taking an arithmetic realization of the surface $\Sigma_g$, giving a faithful rep. $\rho':\Gamma_g \to PSL(2,R)$. Then a non-trivial Galois conjugate of this representation will give a faithful rep. $\rho:\Gamma_g \to SO(3)\subset PSL(2,C)$.

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