# Functional calculus of a projection in a C*-algebra

Let $A$ be a C*-algebra and $p$ be a projection in $A$. Let $f$ be a continuous function defined on the spectrum of $p$. In particular let $f(\lambda)=\lambda^{-\frac{1}{2}}$. What will be the adjoint of $f(p)$?

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What does $0^{-1/2}$ mean? Your $f$ does not seem to be defined on the spectrum of $p$ unless $p$ is the identity. If $f$ actually is defined on the spectrum of $p$, then your question can be answered by noting that continuous functional calculus is a $*$-homomorphism. First, do you know what the spectrum of a projection is? –  Jonas Meyer Jan 20 '12 at 22:41
sorry for this, the spectrum of a projection is {0,1}. I wanted to ask the question for a self adjoint element.. Thanks for the answer –  Micheal Jan 20 '12 at 23:21
So what does $0^{-1/2}$ mean? –  Jonas Meyer Jan 21 '12 at 3:12
In some contexts, e.g., the theory of spectral triples, there's a customary "hack" in the case that $0 \in \sigma(a)$ is isolated, by setting $f(a) := \tilde{f}(a)$, where $$f(\lambda) := \begin{cases} f(\lambda), \quad &\lambda \neq 0,\\ 0, \quad &\lambda = 0. \end{cases}$$ In this case, this would yield $f(p) := \tilde{f}(p) = 0(1-p) + 1^{-1/2}p = p$. –  Branimir Ćaćić May 23 '13 at 21:45

If $a$ is a normal element of a C*-algebra and $f$ is a continuous $\mathbb C$-valued function on the spectrum of $a$, then the adjoint of $f(a)$ is $\overline{f}(a)$. In particular, $f(a)$ is self-adjoint if $f$ is real-valued.

$f(\lambda)=\lambda^{-1/2}$ doesn't make sense on the spectrum of a projection unless it is the identity, because, as you indicated in a comment, the spectrum is $\{0,1\}$. This $f$ could be defined (up to choosing branches) continuously on the spectrum of an element if the spectrum is contained in a simply connected domain that doesn't include $0$.

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