Lie Groups – Gelfand-Naimark and Peter-Weyl for the Unitary Group

c-star-algebraslie-algebraslie-groupsnoncommutative-geometrytopological-groups

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?

Best Answer

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 $(\omega_k,\omega_k^{-1}I_k)$.

Related Question