ΦA×ΦB is Homeomorphic to ΦA⊗pB

# For Commutative Banach Algebras A and B over C - Φ_{A}×Φ_{B} is Homeomorphic to Φ_{A⊗pB}

Proposition: Let $\mathfrak{A}$ and $\mathfrak{B}$ be commutative Banach algebras over $\mathbf{C}$ with unit (though this is not a necessary requirement). Then $\Phi_{\mathfrak{A}} \times \Phi_{\mathfrak{B}}$ is homeomorphic to $\Phi_{\mathfrak{A} \otimes_p \mathfrak{B}}$. |

*Recall that a map $T$ between two topological spaces is a homeomorphism if it is a bijection such that $T$ and $T^{-1}$ are continuous.*

**Proof:**Let $f \in \Phi_{\mathfrak{A}}$ and let $g \in \Phi_{\mathfrak{B}}$. Let $T[f, g] : \mathfrak{A} \times \mathfrak{B} \to \mathbf{C}$ be defined for all $x \in \mathfrak{A}$ and all $y \in \mathfrak{B}$ by $T[f, g](x, y) = f(x)g(y)$. Then $T$ is a bilinear map, and so by the theorem on The Existence of a Linear Map σ on X⊗Y to Z that Matches a Bilinear Map on X×Y to Z page, there exists a unique linear map $f \square g : \mathfrak{A} \otimes \mathfrak{B} \to \mathbf{F}$ such that:

\begin{align} \quad [f \square g](x \otimes y) = f(x)g(y) \quad \forall x \in \mathfrak{A}, \: \forall y \in \mathfrak{B} \end{align}

- Furthermore, $f \square g$ is actually a bounded linear operator on $\mathfrak{A} \otimes \mathfrak{B}$ (via the projective tensor norm) with $\| f \square g \| \leq 1$, since for all $u = \sum_{i} x_i \otimes y_i \in \mathfrak{A} \otimes \mathfrak{B}$ we have that:

\begin{align} \quad | ([f \square g](u) | &= \left | [f \square g] \left ( \sum_{i} x_i \otimes y_i \right ) \right | \\ &= \left | \sum_{i} [f \square g](x_i \otimes y_i) \right | \\ &= \left | \sum_{i} f(x_i)g(y_i) \right | \\ & \leq \sum_{i} \| f(x_i) \| \| g(y_i) \| \\ & \leq \underbrace{\| f \| \| g \|}_{\leq 1 \: \mathrm{since \:} f \in \Phi_{\mathfrak{A}}, \: g \in \Phi_{\mathfrak{B}}} \sum_{i} \| x_i \| \| y_i \| \\ & \leq \sum_{i} \| x_i \| \| y_i \| \end{align}

- Since the above inequality holds for all representations of $u$ we have that $| [f \square g](u) | \leq p(u)$ for all $u \in \mathfrak{A} \otimes \mathfrak{B}$, so $\| f \square g \| \leq 1$. Furthermore, each $f \square g$ is multiplicative since if $u = \sum_{i} x_i \otimes y_i$ and $v = \sum_{j} x_j' \otimes y_j'$ then:

\begin{align} \quad [f \square g](uv) = [f \square g] \left ( \sum_{i} x_i \otimes y_i \sum_{j} x_j' \otimes y_j' \right ) &= [f \square g] \left ( \sum_{i,} \sum_{j} x_ix_j \otimes y_iy_j \right ) \\ &= \sum_{i} \sum_{j} [f \square g](x_ix_j \otimes y_jy_j) \\ &= \sum_{i} \sum_{j} f(x_ix_j') g(y_iy_j') \\ &= \sum_{i} \sum_{j} f(x_i)f(x_j')g(y_i)g(y_j') \\ &= \sum_{i} f(x_i)g(y_i) \sum_{j} f(x_j')g(y_j') \\ &= \sum_{i} [f \square g](x_i \otimes y_i) \sum_{j} [f \square g](x_j' \otimes y_j') \\ &= [f \square g] \left ( \sum_{i} x_i \otimes y_i \right ) [f \square g] \left ( \sum_{j} x_j' \otimes y_j' \right ) \\ &= [f \square g](u) [f \square g](v) \end{align}

- So $f \square g : \mathfrak{A} \otimes \mathfrak{B} \to \mathbf{F}$ is a multiplicative linear functional on $\mathfrak{A} \otimes \mathfrak{B}$ such that $| [f \square g](u)| \leq p(u)$ for all $u \in \mathfrak{A} \otimes \mathfrak{B}$, so $f \square g$ can be UNIQUELY extended to a multiplicative linear functional on $\mathfrak{A} \otimes_p \mathfrak{B}$, i.e., $f \square g$ can be uniquely extended to a member of $\Phi_{\mathfrak{A} \otimes_p \mathfrak{B}}$.

- On the other hand, if $\chi \in \Phi_{\mathfrak{A} \otimes_p \mathfrak{B}}$ then let $f : X \to \mathbf{F}$ and $g : Y \to \mathbf{F}$ be defined as:

\begin{align} \quad f(x) &= \chi (x \otimes 1_{\mathfrak{B}}) \quad \forall x \in \mathfrak{A} \\ \quad g(y) &= \chi(1_{\mathfrak{A}} \otimes y) \quad \forall y \in \mathfrak{B} \end{align}

- Observe that $f$ linear and further $f$ is multiplicative since for all $x_1, x_2 \in \mathfrak{A}$ we have that:

\begin{align} \quad f(x_1x_2) = \chi(x_1x_2 \otimes 1_{\mathfrak{B}}) = \chi((x_1 \otimes 1_{\mathfrak{B}})(x_2 \otimes 1_{\mathfrak{B}})) = \chi(x_1 \otimes 1_{\mathfrak{B}})\chi(x_2 \otimes 1_{\mathfrak{B}}) = f(x_1)f(x_2) \end{align}

- Also observe that $g$ is linear and further $g$ multiplicative since for all $y_1, y_2 \in \mathfrak{B}$ we have that:

\begin{align} \quad g(y_1y_2) = \chi(1_{\mathfrak{A}} \otimes y_1y_2) = \chi((1_{\mathfrak{A}} \otimes y_1)(1_{\mathfrak{A}} \otimes y_2)) = \chi(1_{\mathfrak{A}} \otimes y_1) \chi(1_{\mathfrak{A}} \otimes y_2) = g(y_1)g(y_2) \end{align}

- So $f \in \Phi_{\mathfrak{A}}$ and $g \in \Phi_{\mathfrak{B}}$.

- Furthermore we have that $\chi = f \square g$ since for all $u \in \mathfrak{A} \otimes_p \mathfrak{B}$ with $u = \sum_{i} x_i \otimes y_i$ we have that:

\begin{align} \quad \chi (u) = \chi \left ( \sum_{i} x_i \otimes y_i \right )= \sum_{i} \chi (x_i \otimes y_i) = \sum_{i} \chi((x_i \otimes 1_{\mathfrak{B}})(1_{\mathfrak{A}} \otimes y_i)) = \sum_{i} \chi(x_i \otimes 1_{\mathfrak{B}}) \chi(1_{\mathfrak{A}} \otimes y_i) &= \sum_{i}f(x_i)g(y_i) \\ &= \sum_{i} [f \square g](x_i \otimes y_i) \\ &= [f \square g] \left ( \sum_{i} x_i \otimes y_i \right ) \\ &= [f \square g](u) \end{align}

- So $\chi(u) = [f \square g](u)$ for all $u \in \mathfrak{A} \otimes_p \mathfrak{B}$. Thus $\chi = f \square g$. So for all $\chi \in \Phi_{\mathfrak{A} \otimes_p \mathfrak{B}}$ there exists $f \in \Phi_{\mathfrak{A}}$ and $g \in \Phi_{\mathfrak{B}}$ with $\chi = f \square g$.

- Let $T : \Phi_{\mathfrak{A}} \times \Phi_{\mathfrak{B}} \to \Phi_{\mathfrak{A} \otimes_p \mathfrak{B}}$ be defined for all $(f, g) \in \Phi_{\mathfrak{A}} \times \Phi_{\mathfrak{B}}$ by $T(f, g) = f \square g$. Then $T$ is injective and surjective by the remarks made above. We further claim that $T$ is a homeomorphism of $\Phi_{\mathfrak{A}} \times \Phi_{\mathfrak{B}}$ to $\Phi_{\mathfrak{A} \otimes_p \mathfrak{B}}$.

**1. Showing that $T$ is continuous:**It can be shown that $T$ is weak-* continuous, and since weak-* continuity implies continuity, we have that $T$ is continuous.

**2. Showing that $T^{-1}$ is continuous:**Since $T : \Phi_{\mathfrak{A}} \times \Phi_{\mathfrak{B}} \to \Phi_{\mathfrak{A} \otimes_p \mathfrak{B}}$ is continuous, in order to show that $T^{-1}$ is continuous it is sufficient to show that $\Phi_{\mathfrak{A}} \times \Phi_{\mathfrak{B}}$ and $\Phi_{\mathfrak{A} \otimes_p \mathfrak{B}}$ are both compact.

- First, from the proposition on The Carrier Space of a Commutative Banach Algebra page, since $\mathfrak{A}$ is a commutative Banach algebra WITH unit, $\Phi_{\mathfrak{A}}$ is compact. Similarly, since $\mathfrak{B}$ is a commutative Banach algebra WITH unit, $\Phi_{\mathfrak{B}}$ is compact. By Tychonoff's theorem, the product $\Phi_{\mathfrak{A}} \times \Phi_{\mathfrak{B}}$ is compact.

- Secondly, the spaces $\Phi_{\mathfrak{A}}$ and $\Phi_{\mathfrak{B}}$ are compact, they are complete and totally bounded. Furthermore, $\Phi_{\mathfrak{A}} \times \Phi_{\mathfrak{B}}$ is complete and totally bounded, and hence compact.

- Thus $T$ is a continuous bijection from the compact space $\Phi_{\mathfrak{A}} \times \Phi_{\mathfrak{B}}$ to the compact space $\Phi_{\mathfrak{A} \otimes_p \mathfrak{B}}$, and so $T^{-1}$ is continuous.

- Thus we conclude that $T$ is a homeomorphism of $\Phi_{\mathfrak{A}} \times \Phi_{\mathfrak{B}}$ onto $\Phi_{\mathfrak{A} \otimes_p \mathfrak{B}}$. $\blacksquare$