5

Let $A$ be a unital $C^*$ algebra. If $z\in A$ is invertible, then so is $z^*$ and $z^*z$ and, furthermore, $z^*z$ is positive, so we can define using the functional calculus $|z|=\sqrt{z^*z}$. My book then claims that $|z|$ is invertible with inverse $\sqrt{(z^*z)^{-1}}$. Why is $|z|*|z|^{-1}=1$? To me this looks like trying to say that $\sqrt{z^*z}*\sqrt{(z^*z)^{-1}}=\sqrt{(z^*z)(z^*z)^{-1}}=\sqrt1=1$, but I don't see why you can pull the product inside the square root like you can for reals (I don't know if this is actually how to prove the claim).

I don't see how some of the basic properties about continuous functions pass through the functional calculus and still hold inside of $A$.

Jake
  • 1,137
  • 1
    Probably $a$ and $b$ have to commute, if you want to deduce that $\sqrt a \sqrt b = \sqrt {ab}$. –  May 31 '16 at 18:02
  • Maybe if you're looking to have that $\sqrt{ab}=\sqrt{ba}$, but otherwise I don't see why $ab=ba$ is important. – Jake May 31 '16 at 18:06
  • It also think it is not always true that $ab$ is positive (or normal), if $a$ and $b$ are. –  May 31 '16 at 18:12
  • edited for clarity – Jake May 31 '16 at 18:19

3 Answers3

2

By definition, the (continuous) functional calculus associated to a normal operator $x$ is a continuous $*$-homomorphism $\varphi:C(\sigma(x))\to A$ sending the inclusion function $\sigma(x)\to\mathbb{C}$ to $x$, where $\sigma(x)$ is the spectrum of $x$. In your case, $x=z^*z$, and $|z|$ is defined as $\varphi(f)$ where $f$ is the square root function. Similarly, $|z|^{-1}$ is $\varphi(g)$, where $g$ is the function $g(t)=1/\sqrt{t}$ (which in $C(\sigma(x))$ since $0\not\in\sigma(x)$). So $|z|\cdot|z|^{-1}=\varphi(f)\varphi(g)=\varphi(fg)$. But the pointwise product of the functions $f$ and $g$ is just the constant function $1$, so $\varphi(fg)=\varphi(1)=1$.

More generally, the entire point of the functional calculus is that you can do everything "like you can for reals" (as long as you are only applying functions to a single normal operator, or more generally a commutative $*$-subalgebra), because addition and multiplication in $C(\sigma(x))$ are just pointwise addition and multiplication of ordinary numbers.

(If your definition of $|z|$ and/or $|z|^{-1}$ is not via the functional calculus on $x$, you can easily verify that the above functional calculus definitions meet the criteria of whatever other definition you have and thus coincide.)

Eric Wofsey
  • 330,363
0

Note that $$\lvert z \rvert ^{-2} = ( \lvert z \rvert ^2)^{-1} = (z^*z)^{-2} $$ Taking square roots gives the result.

0

Let $a=z^*z$ and $b=(z^*z)^{-1}=z^{-1}(z^*)^{-1}=z^{-1}(z^{-1})^*$. Then $a$ and $b$ are positive commuting elements for which $ab=1=ba$. Let $\sqrt{a}$ and $\sqrt{b}$ be the positive square roots of $a$ and $b$. Then $\sqrt{a}\sqrt{b}=\sqrt{b}\sqrt{a}$ because $\sqrt{a}$ and $\sqrt{b}$ are limits of polynomials in $a$ and $b$, respectively. Hence, $\sqrt{a}\sqrt{b}=\sqrt{b}\sqrt{a}$ is positive and $$ (\sqrt{a}\sqrt{b})^2 =ab=1. $$ Therefore $$ (\sqrt{a}\sqrt{b}-1)(\sqrt{a}\sqrt{b}+1)=0. $$ Because $\sqrt{a}\sqrt{b}$ is positive, then $\sqrt{a}\sqrt{b}+1$ is invertible. Hence $\sqrt{a}\sqrt{b}-1=0$, which is what you wanted to show: $$ \sqrt{a}^{-1}=\sqrt{b} \\ \sqrt{z^*z}^{-1} = \sqrt{(z^*z)^{-1}}. $$

Disintegrating By Parts
  • 87,459
  • 5
  • 65
  • 149