Spectrum of $C(K) \oplus C(K')$

Question

Let $K$, $K'$ be compact Hausdorff spaces. How does the spectrum of the $C^*$-algebra given by the direct sum $C(K) \oplus C(K')$ look like?

Answer 1

Note that the spectrum of a C$^*$-algebra $A$ is the topological space of all nonzero $*$-homomorphisms $A\to\mathbb C$.

For the direct sum ,the map $\alpha:C(K)\oplus C(K')\to C(K\sqcup K')$ given by $$\alpha(f,g)(k)=\begin{cases} f(k),&\ k\in K\\ \ \\ g(k),&\ k\in K'\end{cases}$$ is a $*$-isomorphism.

For the tensor product $C(K)\otimes C(K')$, consider the map $\beta:C(K)\otimes C(K')\to C(K\times K')$ induced by by $$\beta(f\otimes g)(k,k')=f(k)g(k')$$and extended by linearity. It is straightforward that $\beta$ is a $*$-homomorphism. That $\beta$ is onto follows from the Stone-Weierstrass Theorem: the $*$-algebra $$ \text{span}\,\{(x,y)\longmapsto f(x)g(y):\ f\in C(K),\ g\in C(K')\} $$ separates points, so it is dense; as the image of a $*$-homomorphism is closed, $\beta$ is onto. For injectivity, if $\beta(\sum_jf_j\otimes g_j)=0$, we have $\sum_jf_jg_j=0$. Let $h_1,\ldots,h_m\subset\{f_1,\ldots,f_n\}$ be a basis for the span of $f_1,\ldots,f_n$. Then there exist coefficients $c_{jr}$ with $f_j=\sum_rc_{jr}h_r$. We obtain $$ 0=\sum_jf_jg_j=\sum_j\sum_rc_{jr}h_rg_j=\sum_r\left(\sum_jc_{jr}g_j\right)\,h_r. $$ So for any $y\in K'$ we obtain $$ 0=\sum_jf_jg_j)y_=\sum_j\sum_rc_{jr}h_rg_j(y)=\sum_r\left(\sum_jc_{jr}g_j(y)\right)\,h_r. $$ The linear independence then gives $\sum_jc_{jr}g_j(y)=0$ for all $y$, so $$ \sum_jc_{jr}g_j=0. $$ Now $$ \sum_jf_j\otimes g_j=\sum_j\left(\sum_rc_{jr}h_r\right)\otimes g_j =\sum_r h_r\otimes \left(\sum_jc_{rj}g_j\right)=0. $$ Thus $\beta$ is injective. So $\beta $ is a $*$-isomorphism that gives us $C(K)\otimes C(K')\simeq C(K\times K')$.