Cohomology group $H^n(G, \mathbb{R}/\mathbb{Z})$ as a continuous group or a Lie group?

Question

Is there any example of Borel cohomology group $H^n(G, \mathbb{R}/\mathbb{Z})$ for any $G$ such that $H^n(G, \mathbb{R}/\mathbb{Z})$ is a continuous group? Such as a Lie group?

Most of the examples I know, are $H^n(G, \mathbb{R}/\mathbb{Z})$ are simply $\mathbb{Z}_N$ or $\mathbb{Z}$ and/or products of cyclic groups. For example, I know that $H^2(SO(3), \mathbb{R}/\mathbb{Z})=\mathbb{Z}_2$, $H^3(SO(3), \mathbb{R}/\mathbb{Z})=\mathbb{Z}$. Let us consider the case that $H^n(G, \mathbb{R}/\mathbb{Z})$ for $n>0$.

p.s. Najib Idrissi's comment raises a good point that $H^0(G, \mathbb{R}/\mathbb{Z})=\mathbb{R}/\mathbb{Z}=U(1)$. This example is a trivial counter example. So let us consider and focus on $n>0$ case only. Many thanks.

p.s.2 - I mean the Borel group cohomology studied in, for example here in condensed matter phyiscs and here in a field theory language. Please look at their TABLEs.

Answer 1

Okay, I think I at least know what you mean now. Let me list the five things you could have meant by $H^n(G, \mathbb{R}/\mathbb{Z})$ (none of which have anything to do with Borel-Moore homology). Below I will write $U(1)$ instead of $\mathbb{R}/\mathbb{Z}$.

1. The group cohomology of $G_{\delta}$ with coefficients in $U(1)_{\delta}$, where $G_{\delta}$ denotes $G$ with the discrete topology. This is equivalently the cohomology of the classifying space $BG_{\delta} \cong K(G_{\delta}, 1)$ with coefficients in $U(1)_{\delta}$.
2. The cohomology of $BG$ with coefficients in $U(1)_{\delta}$. Presumably this is what Najib Idrissi thought you meant.
3. The cohomology of $BG_{\delta}$ with coefficients in $U(1)$, where $U(1)$ retains its usual topology. This is equivalently the set of homotopy classes of maps from $BG_{\delta}$ to $B^n U(1) \cong B^{n+1} \mathbb{Z}$, or also equivalently $H^{n+1}(BG_{\delta}, \mathbb{Z})$. This one was probably the least likely.
4. The cohomology of $BG$ with coefficients in $U(1)$, where $U(1)$ retains its usual topology. This is equivalently the set of homotopy classes of maps from $BG$ to $B^n U(1) \cong B^{n+1} \mathbb{Z}$, or also equivalently $H^{n+1}(BG, \mathbb{Z})$.
5. The Borel cohomology, which is the cohomology of a cochain complex constructed like the usual cochain complex computing group cohomology, except that the cocycles $G^n \to U(1)$ are required to be Borel measurable.

(Maybe I should've just known that you meant the last thing, but I didn't know that it existed before googling it carefully. This is not a thing that mathematicians learn about by default.)

If I'm reading this paper correctly (in particular the proof of Corollary 1.14), then in this case options #4 and #5 are isomorphic. In any case this agrees with the computations you've given already (see this MO question for the integral cohomology of $BSO(3)$). But I think the five options are genuinely different from each other in general.

In any case, if I'm right that options #4 and #5 are isomorphic in this case, then I believe that for a compact connected Lie group $G$,

$H^n(BG, U(1)) \cong H^{n+1}(BG, \mathbb{Z})$ is always a finitely generated abelian group, hence a product of $\mathbb{Z}$s and cyclic groups.

So the answer to the question is no. Incidentally, $H^0(BG, U(1)) \cong H^1(BG, \mathbb{Z})$ is not $U(1)$; instead it is the group $\text{Hom}(\pi_0(G), \mathbb{Z})$. So that's not a counterexample either.