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Let $G$ be a topological group. Denote the same group with the discrete topology by $G^\delta$ and denote the group of connected components of $G$ by $\pi_0G$. I am interested in the question when we can find a section to the canonical map $G \to \pi_0G$.

An obstruction to do that is the requirement that the map $$H^{\ast}(BG^{\delta}) \leftarrow H^{\ast}(B\pi_0G): \alpha$$ (cohomology with $\mathbb Z$ coefficients) induced by $G^{\delta} \to \pi_0G$ has to be split injective.

(side remark: This is exactly how one proves Morita's theorem: For $g \geq 5$, the mapping class group of a genus $g$ surface can not be realized by diffeomorphisms: Here $G = \text{Diffeo}(S_g)$, $\pi_0G = \text{Mod}_g$ and the class $\kappa_3 \in H^6(BG) = H^6(B\text{Mod}_g)$ lies in the kernel of the map $\alpha$ mentioned above. A reference for this is Morita's book on characteristic classes of surface bundles, page 172.)

My question is about the case where this obstruction vanishes: Does someone know an example of a topological group $G$ where the discussed map $\alpha$ is split injective, but still the map $G \to \pi_0 G$ does not admit a section?

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    $\begingroup$ An example worth trying out is the subgroup $\{a+bi:a^2+b^2=1\}\cup\{aj+bk:a^2+b^2=1\}$ of the quaternions. $\endgroup$ Nov 19, 2015 at 12:50
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    $\begingroup$ When $G$ is not discrete, you maybe mean a group homomorphism lifting $G\to\pi_0(G)$, or possibly require this homomorphism be continuous. Could you be more precise? $\endgroup$
    – YCor
    Nov 19, 2015 at 16:58
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    $\begingroup$ I do mean a group homomorphism (otherwise the problem becomes trivial). Continuity is no assumption since $\pi_0G$ is discrete. $\endgroup$ Nov 19, 2015 at 23:15
  • $\begingroup$ @JensReinhold $\pi_0(G)=G/G_0$ is discrete when $G$ is a Lie group, but not in general. Hence continuity is definitely a nontrivial assumption. $\endgroup$
    – YCor
    May 12, 2016 at 7:56

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Here's a reduction of the problem (if terrible groups are allowed). Suppose $G$ is any discrete group with a normal subgroup $N$.

We can define a topology on $G$ by saying that $U \subset G$ is open if and only if it's the preimage of a subset $S \subset G/N$ (i.e. it is a union of cosets of $N$). This makes $G$ into a topological group. Any coset $gN$ has the indiscrete topology, which makes it contractible (and hence path-connected). As a result, the set of path components can be identified with $G/N$.

Therefore, if you can find any group $G$ with normal subgroup $N$ such that the map $G \to G/N$ does not split, but the map $H^*(G/N) \to H^*(G)$ is a split injection, you can construct an example. (I don't know one off-hand.)

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Probably you know that, by MacLane's classification of non-abelian group extensions, once you fix the outer group action of $\pi_0G$ on the connected component of the identity $G_0$, which is a homomorphism $\pi_0G\rightarrow \operatorname{Out}(G_0)$ induced by conjugation in $G$, the obstruction to algebraically split $G\twoheadrightarrow\pi_0G$ lies precisely in $H^2(\pi_0G,Z(G_0))$ where $Z(G_0)$ is the center of $G_0$ endowed with the action of $\pi_0G$ induced again by conjugation in $G$. Actually, any obstruction is realizable by some group $G$, but all this is discrete, it would be necessary to check how things work if you take into account the topology.

The cohomomlogy group classifies the set of extensions because it's a torsor, but this doesn't mean that there's even a split extension at all with the given outer action.

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  • $\begingroup$ That's not quite correct. There are two obstructions. One is lifting from $Out$ to $Aut$ and the other is the one you mention. Take a group with trivial center. Then your obstruction to extensions by that group always vanishes. But does the universal extension $G\to Aut(G)\to Out(G)$ split? Wikipedia says that it fails for $G=A_6$, although I am nervous. I think that it fails for free groups. $\endgroup$ May 13, 2016 at 18:44
  • $\begingroup$ @BenWieland you're right, the cohomology group classifies, but unlike with abelian extensions, it's not a bijection of pointed sets, it's just a torsor, so there's no trivial extension corresponding to the trivial cohomology class. I'm sorry for the mistake. $\endgroup$ May 16, 2016 at 16:18

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