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I was reading a paper which mentioned without proof that every infinite-dimensional $C$* algebra has an infinite-dimensional commutative $C$* subalgebra.

Thinking about it for 10 minutes, I didn't see an immediately proof. It is sufficient to construct an element with infinite spectrum, but I don't see how to construct such an element.

Moreover, if one takes infinitely many noncommuting infinite and co-infinite projections on a Hilbert space, and takes the $C$* algebra generated by those projections, I don't see a reason why this contains an infinite-dimensional commutative $C$* subalgebra. (Clearly this is not a proven counterexample.)

Is there an easy proof of this fact?

Marten Wortel
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    Maybe http://math.stackexchange.com/questions/83101/the-spectrum-of-normal-operators-in-c-algebras will help. – Tom Cooney Feb 10 '15 at 19:23

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Marten, this is true by an observation due to Kaplansky, I believe, which asserts that an infinite-dimensional C*-algebra contains a self-adjoint element with infinite spectrum. See Ex. 4.6.12 in

R. V. Kadison and J. R. Ringrose, Fundamentals of the Theory of Operator Algebras, Vol. I, Elementary Theory, Pure and Applied Math., Vol. 100 Academic Press, New York (1983).

Now, apply the spectral theorem to such an element $x$. Using continuous functional calculus, you will get that $C^*(x)=C_0(\sigma(x))$, which is infinite-dimensional as $\sigma(x)$ is an infinite compact set.

Tomasz Kania
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