Talk:Continuous functional calculus

Commutative only?
The article invokes the Gelfand representation of the C* algebra, but that only applies for commutative algebras. Seems to me that maybe what is meant is the algebra generated by x, which, since generated by a normal element, will be commutative. Is it so? -lethe talk [ +] 17:44, 6 April 2006 (UTC)


 * I thought this was clear. The article states:
 * The proof of this fact is almost immediate from the Gelfand representation: it suffices to assume A is the C*-algebra of continuous functions on some compact space X and define


 * $$ \pi(f) = f \circ x. $$


 * Uniqueness follows from application of the Stone-Weierstrass theorem.
 * However, to reduce risk of confusion, maybe we should put in "without loss of generality, we may assume A is commutative since the C*-algebra generated by a single normal element is commutative.--CSTAR 18:22, 6 April 2006 (UTC)
 * I'm still trying to decide whether I find it confusing or not. The article doesn't make clear what algebra A is.  Is it the algebra generated by x?  What if I take A to be some larger commutative algebra?  Then X will not be the spectrum of x.  Also, this composition fx is bothering me.  A priori, it's not clear that the composition makes sense, since, as they're defined, the codomain of x (as a function on X) is all of C, while the domain of f is only sp(x).  I guess the range of x turns out to coincide with the spectrum so that the composition does actually make sense.  But maybe this needs to be made clear to the reader?  -lethe talk [ +] 18:46, 6 April 2006 (UTC)


 * One way to deal with this technically (altough not in the article) is to note that the functional calculus does not depend on the choice of commutative C*-algebra A containing x.--CSTAR 19:03, 6 April 2006 (UTC)