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Let $A$ be a unital C$^*$-algebra. We claim that the projections are precisely the extreme points of $(A_+)_1$. I've found a few references for this, but I'm confused about a step in the proofs.

Let $p = \frac{a+b}{2}$ with $p$ a projection and $\|a\| = \|b\| = 1$, with $a, b$ positive. To reduce to the commutative case, we just need to show that $a$ commutes with $p$.

Since $0 \le \frac{a}{2} = p - \frac{b}{2} \le p$, Conway (in his operator theory book) immediately concludes that $a$ commutes with $p$. In Jesse Peterson's von Neumann algebras notes, he first conjugates the above inequalities by $(1-p)$ to see that $0 \le (1-p)a(1-p) \le 0$, and then he says that the commutativity of $a$ with $p$ follows.

I'm struggling to see why commutativity follows in either of these references. Certainly $(1-p)x(1-p) = 0$ need not imply that $x$ commutes with $p$, so we must be leveraging positivity or the inequality $\frac{a}{2} \le p$ somehow. I'm just not seeing it. Any thoughts?

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You have that $a\geq0$, so $a=c^*c$ for some $c$. Then since $(1-p)a(1-p)$, we get $$ 0=(1-p)c^*c(1-p)=[c(1-p)]^*c(1-p). $$ If follows that $c(1-p)=0$, and so $$a(1-p)=c^*c(1-p)=0.$$ Thus, $a=pa$. Since $a$ is selfadjoint, $ap=(pa)^*=a^*=pa$.


It is important to notice that the above is precisely the proof that if $A\in M_n(\mathbb C)$ is positive and $A_{kk}=0$, then $A_{kj}=A_{jk}=0$ for all $j$.

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  • $\begingroup$ Of course. Thanks so much. $\endgroup$ – Josh Keneda Dec 10 '16 at 5:39

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