# Does a $C^*$ subalgebra of the centralizer of a unitary representation always contain the unit?

I am studying a theorem in Folland's "Course in Abstract Harmonic Analysis" where the following ingredients/assumptions are needed:

$G$ a locally compact group, $\pi$ a unitary representation of $G$ on a separable Hilbert space $\mathcal{H}$ and $\mathcal{B}$ a weakly closed commutative $C^*$ subalgebra of $C(\pi)$, where $C(\pi)$ is the centralizer of $\pi$, i.e., the space of bounded operators on the representation space $\mathcal{H}_\pi$ that commute with $\pi(x)$ for every $x \in G$.

We know that $C(\pi)$ is a weakly closed unital $C^*$ algebra of operators. In the course of the proof, there is the statement that one can assume that $\mathcal{B}$ contains the identity map $I$, i.e., the unit of $C(\pi)$. My question now is, why this assumption can be made. Is there something I am missing? Thanks for any input.

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I think it will help if you state where this is being used because whether or not you can make that assumption probably depends on what is being proved. –  Eric O. Korman Mar 17 '13 at 21:39
Allen, can you please state the theorem and reproduce its proof here? Not everyone has Folland’s book, and no one reading your question will understand the context in which $\mathcal{B}$ appears. –  Haskell Curry Mar 18 '13 at 2:56
A rather drastic edit was proposed by @user67291. Are you the same user? –  Dennis Gulko Mar 18 '13 at 13:54
Yes, I am the same user. I registered a new account. Sorry for the confusion. –  Allen Mar 18 '13 at 19:04