# Banach-algebra homeomorphism.

Let $A$ be a commutative unital Banach algebra that is generated by a set $Y \subseteq A$. I want to show that $\Phi(A)$ is homeomorphic to a closed subset of the Cartesian product $\displaystyle \prod_{y \in Y} \sigma(y)$. Moreover, if $Y = \{ a \}$ for some $a \in A$, I want to show that the map is onto.

Notation: $\Phi(A)$ is the set of characters on $A$ and $\sigma(y)$ is the spectrum of $y$.

I tried to do this with the map $$f: \Phi(A) \longrightarrow \prod_{y \in Y} \sigma(y)$$ defined by $$f(\phi) \stackrel{\text{def}}{=} (\phi(y))_{y \in Y}.$$ I don’t know if $f$ makes sense, and I can’t show that it is open or continuous. Need your help. Thank you!

The mapping $f$ is well-defined.

Given each $\phi \in \Phi(A)$, we must have $\phi(a) \in \sigma(a)$ for all $a \in A$. Indeed, as $\phi(a - \phi(a) \cdot \mathbf{1}_{A}) = 0$, we see that $a - \phi(a) \cdot \mathbf{1}_{A}$ is not invertible, or equivalently, $\phi(a) \in \sigma(a)$. Therefore, the mapping $$f: \Phi(A) \longrightarrow \prod_{y \in Y} \sigma(y)$$ as defined above makes sense.

The mapping $f$ is continuous.

Equip $\Phi(A)$ with the weak$^{*}$-topology inherited from $A^{*}$. To prove that $\displaystyle f: \Phi(A) \to \prod_{y \in Y} \sigma(y)$ is continuous, it suffices to show that $p_{y_{0}} \circ f: \Phi(A) \to \mathbb{R}$ is continuous for each $y_{0} \in Y$, where $p_{y_{0}}$ is the projection mapping of $\displaystyle \prod_{y \in Y} \sigma(y)$ onto the $y_{0}$-th coordinate. As $$\forall \phi \in \Phi(A): \quad \left( p_{y_{0}} \circ f \right)(\phi) = \phi(y_{0}),$$ we see that $p_{y_{0}} \circ f$ is simply the mapping $\phi \longmapsto \phi(y_{0})$, which is obviously continuous with respect to the weak$^{*}$-topology on $\Phi(A)$. Therefore, as $y_{0} \in Y$ is arbitrary, it follows that $f$ is continuous.

The mapping $f$ is injective.

Observe that $f$ is injective because any $\phi \in \Phi(A)$ is uniquely determined by its values on a generating set for $A$.

The mapping $f$ is a topological embedding.

To show that $\displaystyle f: \Phi(A) \to \prod_{y \in Y} \sigma(y)$ is a topological embedding, observe firstly that $\Phi(A)$ is weak$^{*}$-compact (as it is a weak$^{*}$-closed subset of $\text{Ball}(A^{*})$, which is weak$^{*}$-compact by the Banach-Alaoglu Theorem) and $\displaystyle \prod_{y \in Y} \sigma(y)$ is Hausdorff. Then use the fact that a continuous injection from a compact space to a Hausdorff space is a topological embedding.

If $A$ is generated by a single element, then $f$ is onto.

Suppose that $Y = \{ y_{0} \}$. Then $$\sigma(y_{0}) = \{ \phi(y_{0}) ~|~ \phi \in \Phi(A) \},$$ which shows that $f: \Phi(A) \to \sigma(y_{0})$ is onto.

• Since $f$ is an injective continuous map from a compact space to a compact Hausdorff space, does not that itself imply that $\Phi(A)$ is homeomorphic to its image under $f$ ? – user44349 Feb 28 '13 at 7:04
• Oops! I didn’t see your comment until just now. Yes, I realized on my own that my original argument was too convoluted. We don’t need $\displaystyle \prod_{y \in Y} \sigma(y)$ to be compact (although it is) in order to prove that $f$ is an embedding. We only need that it is Hausdorff. – Haskell Curry Feb 28 '13 at 7:24

Note that $\Phi (A)$ is compact in the w$^*$-topology. Also, $\prod \sigma(y)$ is compact Hausdorff in the product topology. For the map $f$ you defined, note that $Ker f = \{0\}$ since $Y$ generates $A$.

To prove continuity, take a net $\{\phi_\alpha\}_{\alpha \in I}$ in $\Phi(A)$, such that $\phi_\alpha \rightarrow \phi$ in w$^*$-topology. Then, $\phi_\alpha (y) \rightarrow \phi(y)$ in norm topology, for any $y \in Y$. ------------- (*)

Now consider a basic open set $V$ around the point $\prod_{y \in Y} \phi(y)$. Then, there exists $y_1, y_2, \ldots, y_k \in Y$ such that $V = \prod_{y \in Y} V_y$, where $V_y = \sigma (y)$ for any $y \in Y \setminus \{y_1, y_2, \ldots, y_k\}$ and $V_{y_i} = V_i$ is an open ball in $\sigma (y_i)$ containing $\phi(y_i)$, for $i = 1, 2, \ldots, k$.

Using (*) we get that for each $i$, $\exists$ $\alpha_i$ such that $\phi_{\beta} (y_i) \in V_i$ for any $\beta \geq \alpha_i$. Since the index set $I$ is directed, $\exists$ $\alpha_0 \in I$ such that $\alpha_0 \geq \alpha_i$ for all $i$. Thus for any $\beta \geq \alpha_0$, we have that $\phi_{\beta} (y_i) \in V_i$ for each $i$, and hence $\prod_{y \in Y} \phi_{\beta} (y) \in V$.

Thus, it follows that $\prod \phi_\alpha (y) \rightarrow \prod \phi(y)$ in $\prod \sigma(y)$, i.e, $f$ is continuous.