# Cohomological obstructions to lift $\pi_0$ of a topological group

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?

• 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. Nov 19, 2015 at 12:50
• 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?
– YCor
Nov 19, 2015 at 16:58
• I do mean a group homomorphism (otherwise the problem becomes trivial). Continuity is no assumption since $\pi_0G$ is discrete. Nov 19, 2015 at 23:15
• @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.
– YCor
May 12, 2016 at 7:56

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.)
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.
• 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. May 13, 2016 at 18:44