If $Lat(\mathcal{A})$ is trivial then $\mathcal{A}'$ consists of scalars.

Question

This is related to Exercise 3 of Section 2.5 of Arveson's book on spectral theory. For those who are interested, we were asked to show the following

$\mathcal{A}$ is a Banach *-algebra. $\pi:\mathcal{A}\to B(\mathcal{H})$ is a *-homomorphism (a *-representation). If $\pi(\mathcal{A})$ has no nontrivial invariant subspaces ($\pi$ is irreducible), then the commutants of $\pi(\mathcal{A})$ are scalars.

I can do this one if I am allowed to use the Double Commutant Theorem and the fact von Neumann algebras are generated by projections. (If $\pi(\mathcal{A})$ has no nontrivial subspaces then $\pi(\mathcal{A})'$, a von Neumann algebra has no projections hence it consists of the constants. If $\pi(A)'$ consists scalars then the double commutant the is $B(\mathcal{H})$. Thus $\pi(\mathcal{A})$ is strongly dense in $B(\mathcal{H})$ but this implies $\pi(\mathcal{A})$ has no nontrivial subspaces since $B(\mathcal{H})$ acts transitively on $\mathcal{H}$.)

But I am pretty sure this problem can be done in a more elementary way since Arveson has not yet started to talk about von Neumann algebras. And the problem boils down to the title of this post.

Can someone give a more elementary method? I am especially not comfortable with the von Neumann algebras are generated by projections part.

Thanks!

Answer 1

$\def\A{\mathcal A}\def\im{\operatorname{im}}$ Let $T\in\pi(\A)'$ be nonzero. Fix $A\in\A$; then, for any $h\in\overline{\mbox{ran}(T)}$, there is a sequence $\{h_n\}$ with $h=\lim Th_n$, and $$ \pi(A)h=\pi(A)\lim_nTh_n=\lim_n\pi(A)Th_n=\lim_nT\pi(A)h_n\in\overline{\mbox{ran}(T)}. $$ This shows that $\overline{\mbox{ran}(T)}$ is invariant for $\pi(\A)$, and thus $\mbox{ran}(T)$ is dense in $H$. But $T^*\in\pi(\A)'$ too, so $$ \ker T=(\mbox{ran}\,T^*)^\perp=\{0\}. $$ Then $T$ is injective. We conclude that every nonzero element in $\pi(\A)'$ is injective. This implies that $\pi(\A)'$ consists of scalar multiples of the identity, as we show in the next paragraph.

Indeed, fix $T\in\pi(\A)'$ selfadjoint. Then $C^*(T)$ is an abelian C$^*$-algebra (considered by Arveson in 2.2); by Gelfand's Theorem (2.2.4), $C^*(T)$ is isomorphic to $C(\sigma(T))$. Also (Corollary 2 in the same section), the spectrum as an element of $C^*(T)$ is the same than as an element of $\pi(\A)'$. Now, if $\sigma(T)$ has more than one point, we can always construct functions with two eigenvalues (by, say, making it constantly zero on a neighbourhood of one of the points, and constantly one around the other), say zero and one. This would construct a non-injective nonzero operator, which is impossible; so this shows that $\sigma(T)$ consists of a single point, i.e. $T$ is a scalar multiple of the identity. For arbitrary $T\in\pi(\A)'$, we write $T=(T+T^*)/2+i(T-T^*)/2i$, a linear combination of selfadjoints; by the previous argument, both the real and the imaginary part are scalar, so $T$ is scalar.

Answer 2

$\def\A{\mathcal A}\def\im{\operatorname{im}}$Let $T \in \pi(\A)'$. Let $\lambda \in \sigma(T)$, then $\ker(T-\lambda) \ne \{0\}$ or $\im (T-\lambda) \ne H$.

If $\ker (T-\lambda) \ne \{0\}$, we have $\ker(T-\lambda) = H$, as $\ker (T-\lambda)$ is $\pi(\A)$-invariant and closed. Thus $T = \lambda$.

If $\overline{\im (T-\lambda)} \ne H$, we are also done (as this subspace is again closed and invariant, giving $\im(T- \lambda) = \{0\}$, so $T = \lambda$.

So we are left with the case $\ker (T -\lambda) = \{0\}$, $\im (T -\lambda)$ dense. Now $\ker (T^*-\bar\lambda) = \bigl(\im (T-\lambda)\bigr)^\bot = \{0\}$, which gives (as $T^*\in \pi(\A)'$, $\pi(\A)$ being a $^*$-subalgebra) $T^* = \bar\lambda$, hence $T = \lambda$.