Suppose $X$ and $Y$ are compact metric spaces. Let $\varphi\colon C(X)\to M\_{n}(C(Y))$ be any $\*$-homomorphism. If $\pi$ is an irreducible representation of $M\_{n}(C(Y))$, then $\pi$ is unitarily equivalent to a point evaluation $\textrm{ev}\_{y}$. The $\*$-homomorphism $\textrm{ev}\_{y}\circ\varphi\colon C(X)\to M\_{n}(\mathbb{C})$ is a representation of $C(X)$. Since it's a finite-dimensional representation, we can find a unitary $u\_{y}\in M\_{n}(\mathbb{C})$ and a set of points $X\_{y}=\{x^{y}\_{1},\ldots,x^{y}\_{n}\}\subset X$ such that for all $f\in C(X)$,
$$
(\varphi\circ f)(y)=(\textrm{ev}\_{y}\circ\varphi)(f)=u\_{y}
\begin{pmatrix}
f(x^{y}\_{1}) & 0 & \cdots & 0\\
0 & f(x^{y}\_{2}) & \cdots & 0\\
\vdots & \vdots & \ddots & \vdots\\
0 & 0 & \cdots & f(x^{y}\_{n})
\end{pmatrix}
u\_{y}^{\*}.
$$


**My question is:**



> 
> Does the Hausdorff distance of the set $\{x\_{1}^{y},\ldots,x\_{n}^{y}\}$ vary continuously in $y$? More precisely, if $y\_{m}\to y$ in $Y$, does $D\_{m}:=\max\_{1\leq i\leq n}\min\_{1\leq j\leq n}d\_{X}(x^{y}\_{i},x^{y\_{m}}\_{j})\to 0$ as $m\to \infty$?
> 
> 
> 


This question is motivated by the simpler case where $\varphi$ gives rise to continuous functions $\lambda\_{1},\ldots,\lambda\_{n}:Y\to X$ satisfying
$$
\varphi(f)=\begin{pmatrix}
f\circ\lambda\_{1} & 0 & \cdots & 0\\
0 & f\circ\lambda\_{2} & \cdots & 0\\
\vdots & \vdots & \ddots & \vdots\\
0 & 0 & \cdots & f\circ\lambda\_{n}
\end{pmatrix}.
$$


In this case for each $y\in Y$, $X\_{y}=\{\lambda\_{1}(y),\ldots,\lambda\_{n}(y)\}$, and the result holds.


I asked this question some weeks ago on stack exchange, but did not get an answer: <https://math.stackexchange.com/questions/2787579/closeness-of-points-in-the-irreducible-decomposition-of-a-c-algebra-repres>



Closeness of points in the irreducible decomposition of a C$^{*}$-algebra representation
Yes, it varies continuously. To see this, suppose it did not, i.e., ${\rm max}\_i {\rm min}\_j d(x^y\_i, x^{y\_m}\_j) \not\to 0$. Passing to a subsequence, we can assume that ${\rm min}\_j d(x^y\_i, x^{y\_m}\_j)$ is bounded away from $0$ for some fixed value of $i$. Now find $f \in C(X)$ such that $f(x\_i) = 1$ and $f(x^{y\_m}\_j) = 0$ for all $m$ and $j$. Then ${\rm ev}\_{y\_m}\circ \phi(f) = 0$ for all $m$ but ${\rm ev}\_y\circ \phi(f) \neq 0$. But $\phi(f) \in M\_n(C(Y)) \cong C(Y; M\_n)$ so that $y \mapsto {\rm ev}\_y(\phi(f))$ must be a continuous map from $Y$ into $M\_n$, which is a contradiction.


