character on commutative $C^*$-algebras involutive?

Question

Let $A$ be a commutative $C^*$-algebra, $0\neq f:A\to\mathbb{C}$ a linear, multiplicative map (a so called character). Is $f(a^*)=f(a)^*$ for all $a\in A$?

I think so, since f can be seen as a point-evaluation map ev$_x$ on $C_0(X)$ for a loccaly compact Hausdorff space X, where $x\in X$. And this point-evaluation maps preserve *. So f must be involutive. Is it correct?

Demophilus · Answer 1 · 2017-06-20T01:55:52.597

2

$A$ doesn't even have to be commutative, here's a short outline to how to prove this: Let $A$ be any unital $C^*$-algebra. Consider any character $f:A \to \mathbb{C}$. It's sufficient to prove that $f(x) \in \mathbb{R}$ for all self-adjoint $x \in A$. You can prove that the spectrum of $x$ is real. Because $f(x) \in \sigma(x)$, we must have $f(x) \in \mathbb{R}$. Now if $A$ is non-unital, you can simply extend $f:A \to \mathbb{C}$ to $\bar{f}:A_1 \to \mathbb{C}$ where $A_1$ is its unitisation. It then follows by the previous result.

As a side note, here's a way to prove $\sigma(x)$ is real when $x$ is self-adjoint without using characters. First we prove the spectra of a unitary element $u$ is contained in $S^1$. Note that $\lVert u \rVert^2 = \lVert u^* u \rVert =1$ so $r(u) \leq 1$ where $r$ denotes the spectral radius. We also have that if $\lambda \in \sigma(u)$, $\lambda^{-1} \in \sigma(u^*)$. So $\lvert \lambda \rvert \leq 1$ and $\lvert \lambda \rvert \geq 1$, from which we conclude that $\lvert \lambda \rvert=1$. Now if $x$ is self-adjoint, by the holomorphic calculus, $e^{ix}$ is unitary. Then by the spectral mapping theorem, $e^{i\sigma(x)} = \sigma(e^{ix}) \subset S^1$. So $\sigma(x) \subset \mathbb{R}$.

As for a reference, I'm pretty sure this argument (or a very similar one) is made in Murphy's $C^*$-algebras and Operator Theory.

edited Jun 20 '17 at 01:55

answered Jun 19 '17 at 23:46

Demophilus

4,170

The only way I know to show that the spectrum of a self-adjoint element is real is using this fact about characters. Could you explain how to do it without that? – Prahlad Vaidyanathan Jun 20 '17 at 01:27
@PrahladVaidyanathan I can, are you familiar with the Riesz calculus/Holomorphic calculus? – Demophilus Jun 20 '17 at 01:34
Sure. If possible, please include a reference or, if it is short enough, just edit your post and add it in. – Prahlad Vaidyanathan Jun 20 '17 at 01:35
@PrahladVaidyanathan I added the argument. – Demophilus Jun 20 '17 at 01:59
Thank you. I don't know why it is $f(x)\in\sigma(x)$. For continuous functions $f\in C(X)$ on a compact Hausdorff space $X$ it is $\sigma(f)=f(X)$. How is it for a character? – Jun 20 '17 at 07:24
Note that if $f$ is a character and $y \in A$ is an invertible element, we have $1 = f(\mathbb{1}) = f(y^{-1}) f(y)$ so $f(y) \neq 0$. Now we have $f( x-f(x) \mathbb{1}) = 0$, so $x-f(x) \mathbb{1}$ can't be invertible. – Demophilus Jun 20 '17 at 10:28
I understand, thank you. – Jun 20 '17 at 12:23
You're welcome! – Demophilus Jun 20 '17 at 12:25

score 2 · Answer 2 · answered Jun 20 '17 at 01:26

2

Here is another proof. As mentioned by Demophilus, one can assume that $A$ is unital. Furthermore, it suffices to show that $f(a) \in \mathbb{R}$ if $a = a^{\ast}$.

Now, we know that $\|f\| = 1$, so if $t\in \mathbb{R}$, then $$ |f(a+it)|^2 \leq \|a+it\|^2 = \|(a-it)(a+it)\| = \|a^2+t^2\| \leq \|a^2\|+t^2 $$ So if $f(a) = \alpha + i\beta$, then $$ |\alpha|^2 + (\beta+t)^2 \leq \|a^2\| + t^2 $$ Hence, $$ |\alpha|^2 + 2t\beta \leq \|a^2\| \quad\forall t\in \mathbb{R} $$ Let $t\to \infty$ to conclude that $\beta = 0$, so $f(a) \in \mathbb{R}$.

answered Jun 20 '17 at 01:26

Prahlad Vaidyanathan

32,330

Thank you. I have two questions: (1) How do you get $|\alpha|^2 + (\beta+t)^2 \leq |a^2| + t^2$? If I consider $f(a)=\alpha +i\beta$, then $|f(a)|^2=\alpha^2+\beta^2\le |a|^2$. (2) Where do you use that $a^*=a$? – Jun 20 '17 at 07:15
(1)$ |\alpha|^2 + (\beta+t)^2 = |f(a+it)|^2$, not $|f(a)|^2$. (2) $a=a^{\ast}$ is used in the fact that $|a+it|^2 = |(a-it)(a+it)|$. In general, you would get $(a^{\ast}-it)$ in the RHS. – Prahlad Vaidyanathan Jun 20 '17 at 10:19
I understand, thank you – Jun 20 '17 at 12:22

character on commutative $C^*$-algebras involutive?

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