# Finite dimensional $C^*$-algebras [closed]

Show that if a $C^*$-algebra $A$ is reflexive as a Banach space, then $A$ must be finite dimentional.

I tried to solve it; but, I could not. please help me for this exercise. Thanks a lot!

## closed as off-topic by Thomas, Davide Giraudo, user26857, Marconius, 6005Nov 5 '15 at 3:11

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Lemma 0. Let maximal abelian $$^*$$-subalgebra $$M$$ of $$C^*$$ algebra $$A$$ which is finite dimensional, then (i) $$M$$ is a linear span of finite family of projections $$\{p_1,\ldots,p_n\}\subset A$$ such that $$p_i p_j=0$$. (ii) $$A$$ is finite dimensional.

See exercise 4.6.12 in Fundamentals of the theory of operator algebras. Vol.1. Elementary theory. R. V. Kadison, J. R. Ringrose.

Lemma 1. Any infinite dimensional $$C^*$$-algebra $$A$$ contains selfadjoint element with infinite spectrum.

Since $$A$$ is infinite dimensional by lemma 0, so does its maximal abelian $$^*$$-subalgebra $$M$$. Therefore there is infinite family of pairwise orthogonal projections $$\{p_n:n\in\mathbb{N}\}$$. Now define $$a=\sum\nolimits_{n=1}^\infty 2^{-n} p_n$$. Since $$a p_k=2^{-k}p_k$$ for all $$k\in\mathbb{N}$$, then $$\{2^{-k}:k\in\mathbb{N}\}\subset \sigma(a)$$. Hence $$\sigma(a)$$ is infinite. For details also see this discussion.

Lemma 2. Let $$\Omega$$ be a locally compact space, and $$C_0(\Omega)$$ is reflexive, then $$\Omega$$ is finite.

I'll borrow the main idea from this answer. Assume $$\Omega$$ is infinite, then we have infinite sequence of distinct point $$(\omega_n)_{n=1}^\infty$$. Consider sequence of linear functionals $$\delta_n:C_0(\Omega)\to\mathbb{C}:f\mapsto f(\omega_n)$$ of norm $$1$$ and an isometric operator $$I:\ell_1\to C(\Omega)^*: a\mapsto \sum\nolimits_{n=1}^\infty a_n\delta_n$$. Thus we may regard $$\ell_1$$ as closed subspace of $$C(\Omega)^*$$. Sicne $$C_0(\Omega)$$ is reflexive, then from this answer we know that so does $$C(\Omega)^*$$. Now we see that $$\ell_1$$ is reflexive as closed subspace of reflexive $$C_0(\Omega)^*$$. Contradiction, hence $$\Omega$$ is finite.

Proposition. Any reflxive $$C^*$$-algebra $$A$$ is finite dimensional.

Assume $$A$$ is infinite dimensional, then by lemma 1 we have a positive element $$a\in A$$ with infinite spectrum. Consider commutative $$C^*$$ subalgebra $$C^*(a)$$ generated by $$a$$. Since $$C^*(a)$$ is a closed subspace of reflxive space, then it is also reflexive as closed subspace of reflexive space. From this aswer we know that $$C^*(a)\cong_1 C_0(\sigma(a)\setminus\{0\})$$, hence $$C_0(\sigma(a)\setminus\{0\})$$ is reflexive. By lemma 2 we get that $$\sigma(a)\setminus\{0\}$$ is finite and so does $$\sigma(a)$$. Contradiction, hence $$A$$ is finite dimensional.

• Why is the maximal abelian *-subalgebra infinite? – math112358 Jan 5 at 5:42
• @mathrookie, I fixed the statement of the lemma. Now it must be clear that $M$ is infinite-dimensional – Norbert Jan 6 at 13:53