A "group" analogy of the Gelfand Naimark theorem

Question

Are there two compact Hausdorff topological spaces $X,Y$ which are not homeomorphic spaces but the following two groups are isomorphic groups? $C(X, \mathbb{H}^2)$ and $C(Y,\mathbb{H}^2)$ where $\mathbb{H}^2$ is the Poincare upper half plane whose group structure is the semi direct product of $\mathbb{R}$ by $\mathbb{R}^+$ via the standard action of $\mathbb{R}^+$ on additive group $\mathbb{R}$.

Answer 1

Yes, this is true. Let us fix a compact Hausdorff space $X$ and write $G=C(X,\mathbb{H}^2)$. Let $A=[G,G]$; note that $A$ consists of all $f:X\to\mathbb{H}^2$ whose second coordinate is always $1$, and so from now on we will identify it with the ordinary additive group $C(X,\mathbb{R})$. Each element of $G$ acts on $A$ by conjugation; let $R$ be the subring of $\operatorname{End}(A)$ generated by all these conjugations. Note that the conjugation action of $g\in G$ is just multiplication by the first coordinate of $G$, considered as a positive element of $C(X,\mathbb{R})$. So $R$ is naturally isomorphic to the subring of $C(X,\mathbb{R})$ generated by the positive elements. But every $f\in C(X,\mathbb{R})$ is a difference of positive elements (say, $f=(|f|+1)-(|f|-f+1)$), so $R\cong C(X,\mathbb{R})$.

So, we can canonically (functorially with respect to isomorphisms) construct the ring $C(X,\mathbb{R})$ from the group $C(X,\mathbb{H}^2)$. By Gelfand duality, this ring determines $X$ up to homeomorphism (explicitly, $X$ is canonically homeomorphic to its space of maximal ideals with the Zariski topology). So, the group $C(X,\mathbb{H}^2)$ determines $X$ up to homeomorphism.