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I was trying to solve exercises (4) on Page 59 of the book "A short course on spectral theory", William Avreson.

Let $A$ be a Banach star-algebra. A representation $\pi\in$rep$(A,H)$ is said to be irreducible if the only closed $\pi(A)$-invariant subspaces of $H$ are the trivial ones $\{0\}$ and $H$. Show that $\pi$ is irreducible if and only if the commutant of $\pi(A)$ consists of scalar multiples of the identity operator.

I know the irreducible representation may have another definition. Namely, if any projection commutes with $\pi(A)$ then $\pi$ is irreducible. But the question is how to prove its commutant consists of multiples of identity if given it is irreducible? (This one is much stronger.)

Martin Argerami
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Chen Ke
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1 Answers1

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Let $P$ be a projection in the commutant $\pi(A)'$. Then $PH$ is a subspace, invariant for $\pi(A)$. Since $\pi$ is irreducible, $PH$ is either $0$ or $H$, which implies that $P=0$ or $P=I$.

The above shows that the only projections in $\pi(A)'$ are $0$ and $I$. As a von Neumann algebra is the norm-closed span of its projections, $\pi(A)=\mathbb C I$.

Martin Argerami
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  • That is quick and smart. Thank you! But before this section, the author didn't talk much about von Neumann algebra. I guess there might be a more fundamental way to solve it. – Chen Ke Mar 26 '16 at 23:04
  • Arveson's book assumes that the reader knows the Spectral Theorem. Usually (for instance Theorem IX.2.2 in Conway's A Course in Functional Analysis), the Spectral Theorem includes the fact that $T\in M'$ if and only if all of its spectral projections are in $M'$. Thus, it is immediate that $M'$ is the closed span of its projections. – Martin Argerami Mar 26 '16 at 23:52
  • Here is a dirtier argument without the spectral theorem, and using facts about functional calculus from Arveson's book. $$\ $$ In section 1.2, you have the fact that when you calculate the polar decomposition $T=U|T|$ for $T\in \pi(A)'$ (any von Neumann algebra, for that matter), then $U^U,UU^\in\pi(A)'$, and these are the range projections of $T^*$ and $T$.

    So, if the only projections are $0$ and $I$, every nonzero operator in $\pi(A)'$ is onto; then every selfadjoint is invertible. (continued)

     – Martin Argerami Mar 27 '16 at 00:22
  • By functional calculus (section 1.1), if $T\in\pi(A)'$ is selfadjoint and $f_1(t)=\max{t,0}$, $f_2(t)=-\min{t,0}$, let $T_+=f_1(T)$, $T_-=f_2(t)$. Then $T=T_1-T_2$, with both $T_1,T_2$ positive. As $T_1T_2=0$ (by functional calculus, because $f_1(t)f_2(t)=0$), we have $\text{ran}, T_2\subset\ker T_1$. But $\text{ran}, T_2$ can only be $0$ or $H$, so $T_2=0$, or $T_1=0$. This implies that any selfadjoint has spectrum either contained in $(-\infty,0]$ or in $[0,\infty)$. This forces $T$ to be a scalar multiple of the identity. – Martin Argerami Mar 27 '16 at 00:23
  • Sorry I don't understand about the last statement. why $T$ is forced to be multiple of identity if it is self-adjoint and has a non-positive( or negative) spectrum? And why if any self-adjoint operator in $\pi(A)'$ is multiple of identity, then so is any other operator in $\pi(A)'$. – Chen Ke Mar 27 '16 at 03:12
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    Because if you add a multiple of the identity to $T$, the spectrum shifts. If the spectrum is not a point, you can add some constant so that it will have some positive part and some negative part; my argument shows that's impossible, so the spectrum necessarily has a single point. For your second question, every operator is a linear combination of selfadjoints. – Martin Argerami Mar 27 '16 at 06:25