Ytukyg

Suppose that $G$ is a compact abelian group. I denote by $G_0$ the connected component of the identity in $G$.

If $G_0$ is open in $G$ (equivalently $G/G_0$ is finite) is it true that $Gcong G_0times G/G_0$?

I also assume that $G/G_0$ is equipped with the quotient topology and that everything is Hausdorff.

*If the title confuses you, the statement above is equivalent to whether the short exact sequence $0rightarrow G_0rightarrow G rightarrow G/G_0rightarrow 0$ splits.

Yes, this is true. The case for Lie groups is pretty easy (and I explained it here). For the general case, one can prove this using the fact compact groups are inverse limits of Lie groups.

Let $G$ be a compact abelian group, and denote $A=G/G^0$. By a corollary of Peter-Weyl, every identity neighbourhood contains a subgroup which is co-Lie (i.e. the quotient by it is a Lie group). Thus we can form a net of subgroups $N_alpha$, ordered by reverse inclusion (so $alpha geq beta$ if and only if $N_beta subseteq N_alpha$), such that $N_alphasubseteq G^0$ for all $alpha$ (sine $G^0$ is open) and such that $bigcap N_alpha={1}$. Denote $G_alpha=G/N_alpha$ and denote by $p_alpha:Gto G_alpha$ the natural quotient map.

Since each $p_alpha$ is open, continuous and surjective, the subgroup $p_alpha(G^0)$ is open and connected, and hence is the connected component of $G_alpha$. Since $N_alphasubseteq G^0$, we have by the third isomorphism theorem (for topological groups) that $G_alpha/G_alpha^0=(G/N_alpha)/(G^0/N_alpha)cong G/G^0 = A$. Thus, by the case for Lie groups, we know $G_alpha cong G_alpha^0times A_alpha$ for $A_alphacong A$.

Denote $A'_alpha = p_alpha^{-1}(A_alpha)$ and $A'=bigcap A'_alpha$. I will show that $A'$ is mapped isomorphically onto $A$ via the quotient $q:Gto G/G^0$, which shows the sequence does indeed split.

First let us see that it is mapped injectively into $A$ via $q$. We have, for every $alpha$, $$p_alpha(A'_alphacap G^0)subseteq p_alpha(A'_alpha)cap p_alpha(G^0)=A_alphacap G_alpha^0={1}.$$ Thus $A'_alphacap G^0subseteq N_alpha$ for every $alpha$, so $bigcap A'_alphacap G^0subseteqbigcap N_alpha={1}$. This precisely mean that the intersection of $A'=bigcap A'_alpha$ with $ker q=G^0$ is trivial, i.e. $q$ mapps $A'$ injectively into $A$.

It remains to show $q(A')=A$. I think this is pretty clear. For every $alpha$ we know $q(A'_alpha)=A$. To see this, first note that $p_alpha(A'_alphacdot G^0)$ contains both $p_alpha(A'_alpha)=A_alpha$ and $p_alpha(G^0)=G_alpha^0$ and hence is all of $G_alpha=G_alpha^0times A_alpha$, and therefore $A'_alphacdot G^0cdot ker p_alpha = G$, but $ker p_alpha = N_alpha subseteq G^0$, so $A'_alphacdot G^0=G$. Since $ker q= G^0$ we have $q(A'_alpha)=A$. But by compactness this implies $q(A')=A$ (since if $ain A$ then for every $alpha$ there is $a_alphain A'_alpha$ such that $q(a_alpha)=a$; this net has some converging subnet which converges to some $a'in A$, and by continuity $q(a')=a$).

This answer turned out a lot longer than I expected... Using the language of inverse limits it would have been a lot shorter, I think, if you know how to prove $Gcong varprojlim G_alpha$ in a natural way and not just that every neighbourhood contains a co-Lie subgroup.