# non-degenerate representation of a C*-algebra A is a direct sum of cyclic representations of A.

I am studying chapter 5 of the book Murphy. In the proof of Theorem 5.1.3 is at the bottom of my many questions. I'm Thanks for help in understanding prove. Thank advance.

Theorem 5.1.3

Let $(H,\varphi)$ be a non-degenerate representation of a C*-algebra A. Then it is a direct sum of cyclic representations of A.

Proof : For each $x\in H$, set $H_x=\overline{\varphi(A)x}$. An easy application of Zorn’s Lemma shows that there is a maximal set $\Lambda$ of non-zero elements of H such that the spaces $H_x$ are pairwise orthogonal for $x\in \Lambda$. If $y\in \left(\bigcup_{x\in\Lambda}H_x\right)^\perp$, then for all $x\in\Lambda$ we have $\langle y,\varphi(a^*b)x\rangle=0$, so $\langle \varphi(a)y$,$\varphi(b)x\rangle=0$, and therefore the spaces $H_x$,$\ H_y$ are orthogonal. Observe that since $(H,\varphi)$ is non-degenerate, $y\in H_y$. It follows from the maximality of $\Lambda$ that $y=0$. Therefore, H is the orthogonal direct sum of the family of Hilbert spaces $\{H_x\}_{x\in \Lambda}$. Obviously, these spaces are invariant for $\varphi(A)$, and the restriction representation

$\varphi_x:A\rightarrow B(H_x), a\mapsto \varphi(a)_{H_x}$,

has x as a cycle vector. Since $(H,\varphi)$ is the direct sum of the representations $\{(H_x,\varphi_x)\}_{x\in\Lambda}$ the theorem is proved $\bullet$

set $\Lambda$ In what form?

$(H,\varphi)$ is non-degenerate, $y\in H_y$, why??

Is from $H=⊕H_{x}^{⊥}$ ?? and Why is this form ??

these spaces are invariant for $\varphi(A)$ why??

Let us define $S$ to be the set of subsets $\Lambda$ of $H$ such that the closed subspace $H_x$ is non-zero and the spaces $H_x$ and $H_y$ are orthogonal for all distinct $x,y\in\Lambda$. The maximal $\Lambda$ is such a subset, and the existence of $\Lambda$ follows directly from Zorn's lemma.

The fact that $y\in H_y$ for all $y\in H$ is proved here, in Lemma 2.26. Hence if $y\in(\bigcup_{x\in\Lambda}H_x)^\perp$ and $y$ were non-zero, then $H_y$ would be a non-zero subspace, and as $H_y$ is orthogonal to all $H_x$, we obtain a contradiction of the maximality of $\Lambda$. Hence $y$ must be the zero vector.

I would conclude the proof as follows. Let $P_x$ be the projection onto $H_x$ for all $\xi\in\Lambda$. The above argument shows that $\sum_{x\in\Lambda}P_x=1$ (the sum converging in the strong operator topology), since the $P_x$'s are all orthogonal. Moreover, since $\varphi(a)[\varphi(b)x]=\varphi(ab)x\in\varphi(A)x$, so by continuity we have $\varphi(a)H_x\subseteq H_x$. This shows that $\varphi(a)P_x=P_x\varphi(a)P_x$ and hence $P_x\varphi(a)=P_x\varphi(a)P_x$ by using the $^*$-operation on the former equation and noting that $A=A^*$. Hence $$\varphi(a)P_x=P_x\varphi(a)$$ for all $x\in\Lambda$ and $a\in A$.

By the first part of the proof of Proposition 2.20, we have a unitary map $U\colon H\to K=\bigoplus_{x\in\Lambda}H_x$ given by $$U\xi=(P_x\xi)_{x\in\Lambda}.$$ We can define representations $\varphi_x\colon A\to B(H_x)$ by $\varphi_x(a)=\varphi(a)|_{H_x}$ since $\varphi(a)H_x\subseteq H_x$. Now consider the representation $\varphi'=\bigoplus_{x\in\Lambda}\varphi_x\colon A\to B(K)$ given by $$\varphi'(a)=(\varphi_x(a))_{x\in\Lambda}$$ (where each $\varphi_x(a)$ acts on $H_x$). For all $\xi\in H$ and $a\in A$ it satisfies $$U^*\varphi'(a)U\xi=U^*(\varphi_x(a)P_x\xi)_{x\in\Lambda}\stackrel{\star^1}{=}U^*(\varphi(a)P_x\xi)_{x\in\Lambda}\stackrel{\star^2}{=}U^*(P_x\varphi(a)\xi)_{x\in\Lambda}=\varphi(a)\xi,$$ where $\star^1$ follows from $\varphi(a)H_x\subseteq H_x$ and $\star^2$ follows from the fact that $\varphi(a)$ and $P_x$ commute for all $x\in\Lambda$. This proves that $\varphi$ is equivalent to the direct sum $\varphi'$. Moreover, each $\varphi_x$ is cyclic: note that $x\in H_x$ so that $\varphi_x(a)x=\varphi(a)x$ for all $x\in\Lambda$, and since $\varphi(A)x$ is dense in $H_x$ by construction, "cyclicity" follows.

• dear Bryder thanks a lot
– reza
Jan 6, 2015 at 19:43
• No problem! I love these things. :) Jan 7, 2015 at 8:57