# Meaning of non-degenerate representation in $C^*$-algebras

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!

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 $$H_1$$.

• Great explanation, have an updoot +1. But what is a $C^*$-algebra? Mar 24, 2022 at 2:30

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.

• You mean $[Id(A)H]$ is not equal to $H$? Sep 19, 2018 at 17:46
• No, of course not. That's why it's "degenerate". Sep 19, 2018 at 18:48
• I am confused,when we talk about the representation of a $C^*$ algebra,why we always assume that the reprentation is non-degenerate. Sep 20, 2018 at 2:13
• Why not? What information do you get from writing, say, the algebra $\mathbb C$ as $\{(c,0): \ c\in\mathbb C\}$? Sep 20, 2018 at 4:41

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.