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Let $A$ be a unital $C^\star$-algebra and let $a_1,\dots,a_n$ be a finite list of normal elements in $A$ which (together with their adjoints) generate a norm-dense $\star$-subalgebra $B \subset A$. Clearly, if $A$ is finite-dimensional, then every element in $A$ (and hence $B$) has finite spectrum. I am asking for the converse.

Question: Assume that each element of $B$ has finite spectrum. Is it true that $A$ has to be finite-dimensional?

The existence of finitely-generated infinite torsion groups shows that this might be a highly non-trivial problem. In this case one would consider the reduced group $C^\star$-algebra and note that all monomials in the generators of the group and their inverses (which are equal to the adjoints of the generators) would have finite spectrum. However, the generated algebra would still be infinite-dimensional. I do not know of any simpler way to come up with such an example. In this case it is conceivable that the random-walk operator associated with the generating set (which is an element in the real group ring) has infinite spectrum, even though I did not prove this.

Maybe there is also need to consider the spectra of elements in matrices over $B$ (which of course follow to be finite if $A$ is finite-dimensional.) In view of this, I am not only asking for an answer to the question but also for the right question (or a better one) if the answer to the original question is negative.

A stronger assumption would be to assume that $A$ itself consists only of elements with finite spectrum. This case seems much easier to approach and the answer seems to be positive. In fact every infinite-dimensional $C^\star$-algebra should contain an element with infinite spectrum.

Just to get started, a more concrete instance of the question above is:

Question: Let $p_1, p_2$ and $p_3$ be three projections in a $C^*$-algebra with the property that every (non-commutative) polynomial in $p_1, p_2$ and $p_3$ has finite spectrum. Is it true that the projections generate a finite-dimensional algebra?

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I have just learned that Choi-Elliott proved in (ams.org/mathscinet-getitem?mr=1081290) that the set of self-adjoint elements with finite spectrum in certain rotation algebras are dense in the set of all self-adjoint elements. (Interestingly, this result is not explicit, in the sense that there is no concrete $\theta$ for which it happens.) However, from what I understand, it seems unclear whether there is a finitely generated dense sub-algebra such that each element has finite spectrum. –  Andreas Thom Aug 17 '10 at 15:05
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Suppose you have a C$^*$ algebra generated by $g_1$, $g_2$ and $g_3$. Consider the elements $x_1g_1+x_2g_2+x_3g_3$, where the $x_i$ are variables. If the spectra of these elements are the solutions of a polynomial equation in $x_1$, $x_2$ and $x_3$, is the algebra finite dimensional? –  Peter Shor Aug 19 '10 at 23:13
    
I do not know. Beautiful question! Do you want to (or may I) ask this as a separate question? –  Andreas Thom Aug 20 '10 at 9:07
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@Andreas, Why don't you go ahead and ask this as a separate question. I was thinking about this as an approach to your question. One next step (which would be much harder) would to say if the spectra of these elements isn't the solution of a polynomial equation, then one of these elements has an infinite spectrum. I have no idea how to prove this, or even if it's true, but it seems to work for a few examples. –  Peter Shor Aug 20 '10 at 12:54
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Could be; I do not know. I was just talking with Joachim Cuntz and George Elliott about the problem, but they did not mention Richard Kadison. There is a related problem to find an element with infinite spectrum in an infinite dimensional $C^\ast$-algebra, which is a lot easier. Where did you hear that Kadison asked the question? –  Andreas Thom Sep 4 '10 at 21:28

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