Let $A$ be a subhomogeneous C$^{*}$-algebra (i.e., there is a finite upper bound on the size of the irreducible representations of $A$). Let $\hat{A}$ denote its spectrum. I heard of a result that states that

If $(\pi_{n})$ is a sequence in $\hat{A}$, then $(\pi_{n})$ can converge to at most finitely many points.

I'd like to know why this statement is true.

For instance, if we look at the algebra

$$ B:=\left\{f\in C([0,1],M_{2}(\mathbb{C})):\exists \lambda,\mu,\nu\in\mathbb{C}\text{ with }f(0)=\begin{pmatrix}\lambda & 0\\ 0 & 0\end{pmatrix}\text{ and }f(1)=\begin{pmatrix}\mu & 0\\ 0 & \nu\end{pmatrix}\right\} $$ it is clear that there is a sequence of $2\times 2$ irreducible representations converging to two $1\times 1$ irreducible representations (the point evaluations converging to either the upper-left or bottom-right evaluation at $1$). But there is not sequence of $2\times 2$ irreducible representations converging to $\lambda,\mu$, and $\nu$, which makes sense to me because we cannot stick all three of them down the diagonal, as we are in $M_{2}(\mathbb{C})$.

If $A$ is homogeneous, then $\hat{A}$ is Hausdorff, so the sequence can converge to at most one point in this case. I guess, in the subhomogeneous case the result has something to do with the fact that there are only finitely many choices for the dimensions of the irreducible representations.

I also know that each subspace $\hat{A}_{n}:=\{\pi\in\hat{A}:\dim\pi=n\}$ is Hausdorff, but I'm not sure how to put this together. Any help is appreciated.


1 Answer 1


See JMG Fell, The Dual Spaces of C$^*$-Algebras, Trans. Amer. Math. Soc. 94 (1960), 365-403, Corollary 1 on p. 388:

Let $A$ be a C$^*$-algebra with dual space $\hat{A}$. Let $T^i$ be a net of elements of $\hat{A}$, all of dimension equal to or less than the integer $n$; and let $S^1, \ldots S^r$ be distinct elements of $\hat{A}$ such that $T^i\to_i S^k$ for each $k$. Then $$\sum_{k=1}^r \dim S^k\le n.$$


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