Let $G$ and $H$ be two compact Lie groups with isomorphic Lie algebras $\frak{h} \simeq \frak{g}$, but which are non-isomorphic as topological spaces. From the isomorphism assumption it (should) follows that we have a bijection between the irreducible representations of $G$ and $H$. From this it should also follow that the Peter–Weyl decomposition of the space matrix coefficient functions of both groups are isomorphic as $\frak{g}$-modules. Now the completion of the space of matrix coefficients of both groups to their spaces of continuous functions cannot be isomorphic since $G$ and $H$ are not isomorphic as topological spaces. So where does this failure of isomorphism come from? Does it come from the algebra structure of the space of matrix coefficients? Or might it come from the completion being different, i.e. the $\|\cdot\|\_{\infty}$ norm is different in both cases.


Edit: Following the comments below, the example I am thinking about is $U\_2$ and $SU\_2 \times U\_1$. Does $SU\_2 \times U\_1$ have more irreducible representations than $U\_2$?



Peter–Weyl decomposition for compact Lie groups with isomorphic Lie algebras
It's just not true that having isomorphic Lie algebras implies a bijection between the irreducibles (presumably you mean a bijection compatible with the isomorphism between the Lie algebras). For example when $G = SU(2), H = SO(3)$ only half of the irreducibles of $SU(2)$ come from irreducibles of $SO(3)$.


Continuing, the natural double cover $SU(2) \to SO(3)$ identifies the algebra $C(SO(3))$ of continuous functions on $SO(3)$ with a subalgebra of $C(SU(2))$ (exactly the subalgebra of "even functions" with respect to $-1 \in SU(2)$) and in the Peter-Weyl decomposition this subalgebra corresponds to the irreducibles of $SU(2)$ which descend to irreducibles of $SO(3)$ ("integer spin"), while there's a whole other subspace of odd functions corresponding to the irreducibles which don't ("half-integer spin").


Generally, if $G$ and $H$ are compact connected Lie groups related by a covering map $p : G \to H$ then they fit into a short exact sequence


$$1 \to Z \to G \to H \to 1$$


where $Z = \text{ker}(p)$ is a finite central subgroup of $G$, which when $G$ is simply connected can be identified with $\pi\_1(H)$. When $G = SU(2), H = SO(3)$ we have $Z = \{ \pm 1 \}$. The pullback $p^{\ast} : \text{Rep}(H) \to \text{Rep}(G)$ then identifies the representations of $H$ with the representations of $G$ on which $Z$ acts trivially; meanwhile there must be other representations of $G$ on which $Z$ doesn't act trivially (by Peter-Weyl). This pullback is compatible with the pullback $p^{\ast} : C(H) \to C(G)$ on continuous functions, which identifies $C(H)$ with the subalgebra of $C(G)$ on which $Z$ acts trivially (the generalized "even subalgebra").


