How Does Positivity in C*-Algebras Relate to Their Representations?

Oxymoron
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If I have a faithful nondegenerate representation of a C*-algebra, A:

\pi\,:\,A \rightarrow B(\mathcal{H})

where B(\mathcal{H}) is the set of all bounded linear operators on a Hilbert space. And just suppose that a\geq 0 \in A. How is the fact that a is positive got anything to do with \pi(a) being positive?

Apparantly there is an if and only if relationship!? How does one begin to prove something like a\geq 0 \Leftrightarrow \pi(a) \geq 0 \in B(\mathcal{H})?
 
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Define positive for me again for a C* alg.
 
Let A be a C*-algebra, and a,b \in A. Then a\geq b if a-b has the form c^*c for some c\in A. In particular, a is positive if a = c^*c for some c \in A.
 
Well, one implication is obvious a positive implies pi(a) positive since pi is a *-homomorphism.

Conversely, hmm, well, faithfulness and nondegeneracy must come into it somewhere.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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