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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!

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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.

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    $\begingroup$ 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 $f$ ? $\endgroup$ – user44349 Feb 28 '13 at 7:04
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    $\begingroup$ 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. $\endgroup$ – Haskell Curry Feb 28 '13 at 7:24
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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.

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