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A representation of a $C^*$-algebra, $A$, is a pair $(H,\pi)$ where $H$ is a Hilbert space and $\pi$ is a *-homomorphism from $A$ to $B(H)$. A representation is non-degenerate if $\{\pi(a)h:a\in A, h\in H\}$ is dense in $H$. If $A$ is a unital $C^*$-algebra this means $\pi(1)=1$.

I sort of understand what this definition means in the unital case. But I am having trouble understanding the meaning behind the general definition. Most books I have read provide little motivation of this (although think Murphy does say something about $\pi(A)$ acting on $H$).

I was wondering If someone could explain the meaning behind this definition or perhaps link to somewhere that already does

Thanks!

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3 Answers 3

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Though not a full answer, I'll add to Martin Argerami's answer by stating that his example is the only way in which a representation of a C*-algebra can be "degenerate". Indeed, if $\sigma : A\to\mathcal B(H)$ is such that $H_0 := \overline{\sigma(A)H}\neq H$, then letting $H_1 := H_0^\perp$ (so that $H = H_0\oplus H_1$), for any $\eta\in H_0$ and $\xi\in H_1$ then there are $a_n\in A$ and $\eta_n\in H$ such that $\eta = \lim_n\sigma(a_n)\eta_n$, and so for any $a\in A$ we have $$ \begin{aligned} \langle \sigma(a)\xi, \eta\rangle & = \langle\sigma(a)\xi, \lim_n\sigma(a_n)\eta_n\rangle \\ & = \lim_n\langle \xi, \sigma(a^*a_n)\eta_n\rangle \end{aligned} $$ But $\sigma(a^*a_n)\eta_n\in H_0 = H_1^\perp$, and so this limit is equal to zero. In other words, we've just shown that $\sigma(a)\xi\in H_1$ for all $\xi\in H_1$.

On the other hand, we clearly also have $\sigma(a)\xi\in H_0$ (by the very definition of $H_0$), so $\sigma(a)\xi\in H_0\cap H_0^\perp = 0$. Thus, we've just shown that $\sigma(a)$ restricted to $H_1$ must be zero. In other words, $$ \sigma = \rho\oplus 0 $$ where $\rho(a) := \sigma(a)|_{H_0}$ is a non-degenerate representation, and $0$ is the zero representation on $H_1$.

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    $\begingroup$ Great explanation, have an updoot +1. But what is a $C^*$-algebra? $\endgroup$
    – Ltoll
    Mar 24, 2022 at 2:30
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Maybe seeing where it fails helps you understand it. What you want with non-degeneracy is to avoid the following situation: let $A_0\subset B(H_0)$ be a C$^*$-algebra, and let $H=H_0\oplus H_0$ and $A\subset B(H)$ be $$ A=\left\{\begin{bmatrix}a&0\\0&0\end{bmatrix}:\ a\in A_0\right\}. $$ Here $AH=H_0\oplus 0$, so the identity representation is degenerate. Note that this construction can be done even when $A$ is non-unital.

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  • $\begingroup$ You mean $[Id(A)H]$ is not equal to $H$? $\endgroup$
    – math112358
    Sep 19, 2018 at 17:46
  • $\begingroup$ No, of course not. That's why it's "degenerate". $\endgroup$ Sep 19, 2018 at 18:48
  • $\begingroup$ I am confused,when we talk about the representation of a $C^*$ algebra,why we always assume that the reprentation is non-degenerate. $\endgroup$
    – math112358
    Sep 20, 2018 at 2:13
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    $\begingroup$ Why not? What information do you get from writing, say, the algebra $\mathbb C $ as $\{(c,0): \ c\in\mathbb C\} $? $\endgroup$ Sep 20, 2018 at 4:41
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In the non-unital case you still have $\pi(u_\lambda)\xi \to \xi$ for an approximate identity $u_\lambda$. In particular $\pi$ is non-degenerate iff $H$ is cyclic for the representation.

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