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Let A be a prime $C^{*}$ algebra and $e= \alpha^{-1} c^{*}c$ for some $c \in A$ which satisfies $(c^{*}c)^{2}=\alpha c^{*}c$. Then $e^{2}=e=e^{*}$ and $eAe=\mathbb Ce$ so $eAe$ may be identified with $\mathbb C$.

Endow $Ae$ with an inner product $<ae,be>=eb^{*}ae$. Then the inner product norm coincides with the original norm on $Ae$ and is therefore complete. The corresponding Hilbert space is denoted by $H$.

Define $ \pi:A\longrightarrow B(H)$ by $ \pi(a) \xi= a \xi, \ a \in A, \ \xi \in H$. We must prove that $\pi$ is an irreducible and faithful representation. That it is a representation is straightforward. How do we prove that it is irreducible and faithful?

Arundhathi
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  • Is the following proof for showing that $\pi$ is faithful correct? Suppose $\pi(a)\xi=0 \forall \xi \in H$. i.e. $a\xi=0 \forall \xi \in H$. i.e. $axe=0 \forall x \in A$. Suppose $a\neq 0$. Then the (left)ideal generated by $a$ is non-zero. The (left)ideal generated by e is also non zero, so $\exists x \in A$ such that $axe \neq 0$ since A is a prime $C^{*}$ algebra. This gives us a contradiction. So $a=0$. – Arundhathi Jan 17 '14 at 06:49
  • But this only proves that its faithful right? Also, isnt the fact that left ideal generated by $a$ is non zero redundant? – Nasibabuba Jan 17 '14 at 07:05

1 Answers1

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  1. Faithfulness: suppose that $a\in A$ with $a\ne0$. As $e\ne0$, by primality there exists $b\in A$ with $abe\ne0$. That is, $\pi(a)be\ne0$ and so $\pi(a)\ne0$.

  2. Irreducibility: Let $P\in B(H)$ be a projection with $P\pi(a)=\pi(a)P$ for all $a\in A$. Let $K=PH\subset H$. Define $$ I=\{b\in A:\ be\in K\},\ \ J=\{b\in A:\ be\in K^\perp\}. $$ The fact that $P$ commutes with all $\pi(a)$ implies that $I,J$ are left-ideals: if $b\in I$, $a\in A$, then $abe=\pi(a)be=\pi(a)Pbe=P\pi(a)be\in K$ (it is trivial to verify that $I,J$ are subspaces of $A$).

If $Ie=0$, then $K=0$ and so $P=0$. If $Ie\ne0$, there exists $b\in I$ with $be\ne0$. Let $d\in J$, $a\in A$. Then $ad\in J$, and $ade\in K^\perp$. So $eb^*ade=0$ by the orthogonality of $K$ and $K^\perp$. As this happens for every $a\in A$, we conclude by the primality of $A$ that $de=0$ (as we are assuming that $be\ne0$). As $d$ was an arbitrary element of $Je$, we conclude that $Je=0$, i.e. $K^\perp=0$, so $P=I$.

We have shown that $P$ is either $0$ or $I$, so $\pi(A)$ is irreducible.

Martin Argerami
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