Consider the compact Lie groups $U(l)$ (the unitary group) and $U(1) \times SU(l)$ for some natural number $l$. Both the groups have the same Lie algebra $\frak{gl}\_l$. Which means that they both have the "same" dense Peter-Weyl subalgebra $PW$ of their $C^\*$-algebra of continuous functions.


The Gelfand-Naimark theorem says that the $C^\*$-algebras of $U(l)$ and $U(1) \times SU(l)$ are non-isomorphic.


How to resolve the fact that both $C^\*$-algebras are completions of the same algebra PW? Is it the case that both spaces have different norms, and or algebra structure, on PW? I guess this must be what is happening?



Gelfand-Naimark and Peter-Weyl for the unitary group
There were silly errors in my comments, so let me start again. You have a canonical surjection $U(1) \times SU(k) \to U(k)$ given by scalar multiplication, so that any representation of $U(k)$ lifts to a representation of $U(1) \times SU(k)$; the question is whether or not every representation of $U(1) \times SU(k)$ descends to a $U(k)$.


Now, the kernel of this canonical surjection is $\langle (\omega\_k,\omega\_k^{-1}I\_k)\rangle \cong \mathbb{Z}\_k$, where $\omega\_k$ is a primitive $k$th root of unity. Then, for any integer $n$ not divisible by $k$, the map $\pi\_n : U(1) \times SU(k) \to U(1)$ given by $\pi\_n(z,u) := u^n$ is a representation of $U(1) \times SU(k)$ that doesn’t descend to $U(k)$, since $\pi\_n(\omega\_k,\omega\_k^{-1}I\_k) = \omega\_k^n \neq 1$.


In terms of Peter–Weyl algebras, what’s going on is that the Peter–Weyl algebra $\mathcal{O}(U(k))$ of $U(k)$ can be identified with the proper $\ast$-subalgebra of $\mathbb{Z}\_k$-invariants in the Peter–Weyl algebra $\mathcal{O}(U(1)\times SU(k))$ of $U(1) \times SU(k)$, where $\mathbb{Z}\_k$ acts through translation by the central element $(\omega\_k,\omega\_k^{-1}I\_k)$.


