Norm 1 projection in a $C^*$-algebra

Question

On a Hilbert space, it is known that projections have a norm greater than 1 (except $0$), and that those of norm 1 are exactly the orthogonal projections. Most of the proofs I have seen use the Hilbert space, either a specific vector in it, or $\ker p$, or ${\rm im\,}P$.

I was wondering if we could prove this in a general $C^*$-algebra (when there are non trivial projections), not using an underlying Hilbert space (given by a GNS construction).

For example, the implication "orthogonal projection implies norm 1 (or 0)" works in a general $C^*$-algebra because $$\|x\| = \|x^2\| = \|x^{*} x\| = \|x\|^2 \Rightarrow \|x\| \in \{0,1\}$$

I have not been able to prove the converse without using a Hilbert space. So I have a norm 1 projection $x$, I am pretty sure I need to work inside the $C^*$-algebra $C(x)$ generated by $x$ and its adjoint.

The fact that $\|x\| = \underset{y \in C(x), \|y\|=1}\sup \|xy\| = \underset{y \in C(x), \|y\|=1}\sup \|yx\|$ (attained for $y = x^*$) might be useful. Any help would be appreciated.

@AndréS.: how to prove that a norm-one idempotent is selfadjoint. — Martin Argerami, Apr 15 '18 at 16:33
It seems implied by your question that you don't want this approach, but you could take the universal representation of the $C^*$-algebra and use an argument mentioned in the first paragraph to obtain the result. — Aweygan, Apr 16 '18 at 04:31
I have a question regarding the answer. How do you know that there are no negative numbers in the spectrum of $a$ ? — Philip, Jun 06 '19 at 16:17
If you have a new question, please ask it by clicking the Ask Question button. Include a link to this question if it helps provide context. - From Review — Chain Markov, Jun 06 '19 at 17:28

Martin Argerami · Accepted Answer · 2020-04-13T02:30:40.200

6

We have $\|x\|=1$, $x^2=x$. Write $x=a+ib$, with $a,b$ selfadjoint. The equality $x^2=x$ is now $$ a=a^2-b^2, \ \ \ b=ab+ba. $$ From the first equality, $a^2-a=b^2\geq0$. So $a^2\geq a$, which tells us that $\sigma(a)\subset \{0\}\cup[1,\|a\|]$. In particular, $\|a\|\geq1$ (the case $a=0$ is easily discarded because it forces $b=0$). But $\|a\|\leq\|x\|=1$ (because $a=\frac{x+x^*}2$), so $\|a\|=1$. Now $\sigma(a)\subset\{0,1\}$, so $a$ is a projection. From $a=a^2-b^2$, we get $b^2=0$, so $b=0$. Thus $x=a$ is selfadjoint.

edited Apr 13 '20 at 02:30

answered Apr 15 '18 at 16:52

Martin Argerami

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Took me a minute: $||b||^2=||bb^*||=||b^2||=0$, right. – David C. Ullrich Apr 15 '18 at 16:56
That is eactly what I was looking for, thank you very much. – Ugo Massariol Apr 15 '18 at 17:04
@David: that's probable neater than the way I think about it. If $b^2=0$, its spectrum is ${0}$ and then so is the spectrum of $b$. Being selfadjoint, that's enough. – Martin Argerami Apr 15 '18 at 17:09
@Ugo: glad I could help. I hope you'll stay in the site. – Martin Argerami Apr 15 '18 at 17:14

Norm 1 projection in a $C^*$-algebra

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