The Algebraic Tensor Product of Two Normed Algebras

Proposition: Let $\mathfrak{A}$ and $\mathfrak{B}$ be algebras over $\mathbf{F}$. Then there exists a unique product of $\mathfrak{A} \otimes \mathfrak{B}$ for which $\mathfrak{A} \otimes \mathfrak{B}$ is both an algebra and has the property that $(x_1 \otimes y_1)(x_2 \otimes y_2) = x_1x_2 \otimes y_1y_2$ for all $x_1, x_2 \in \mathfrak{A}$ and for all $y_1, y_2 \in \mathfrak{B}$.

• Proof: Let $x_1 \in \mathfrak{A}$ and let $y_1 \in \mathfrak{B}$. Then the map $\mathfrak{A} \times \mathfrak{B} \to \mathfrak{A} \otimes \mathfrak{B}$ defined for all $x_2 \in \mathfrak{A}$ and all $y_2 \in \mathfrak{B}$ by $(x_2, y_2) \to x_1x_2 \otimes y_1y_2$ is bilinear map.

• To see this, fix $x_2 \in \mathfrak{A}$ and let $y_2, y_2' \in \mathfrak{B}$, $\alpha \in \mathbf{F}$. Then:

(1)

\begin{align} \quad (x_2, y_2 + y_2') = x_1x_2 \otimes y_1(y_2 + y_2') = x_1x_2 \otimes [y_1y_2 + y_1y_2'] = x_1x_2 \otimes y_1y_2 + x_1x_2 \otimes y_1y_2' = (x_2, y_2) + (x_2, y_2') \end{align}

(2)

\begin{align} \quad (x_2, \alpha y_2) = x_1x_2 \otimes y_1(\alpha y_2) = x_1x_2 \otimes (\alpha y_1 y_2) = \alpha (x_1x_2 \otimes y_1y_2) = \alpha (x_2, y_2) \end{align}

• So for each fixed $x_2 \in \mathfrak{A}$ the map $y_2 \to x_1x_2 \otimes y_1y_2$ is linear. Similarly it can be shown that for each fixed $y_2 \in \mathfrak{B}$ the map $x_2 \to x_1x_2 \otimes y_1y_2$ is linear. Thus this map is bilinear.

• 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 bilinear map $T[x_1, y_2] : \mathfrak{A} \otimes \mathfrak{B} \to \mathfrak{A} \otimes \mathfrak{B}$ such that for all $x_2 \in \mathfrak{A}$ and all $y_2 \in \mathfrak{B}$:

(3)

\begin{align} \quad T[x_1, y_1](x_2 \otimes y_2) = x_1x_2 \otimes y_1y_2 \quad (*) \end{align}

• Note that the map $(x_1, y_1) \to T[x_1, y_1]$ is also a bilinear map from $\mathfrak{A} \times \mathfrak{B}$ to $\mathcal{L}(\mathfrak{A} \otimes \mathfrak{B})$. Using the theorem above again, there exists a unique linear map $\sigma : \mathfrak{A} \otimes \mathfrak{B} \to \mathcal{L}(\mathfrak{A} \otimes \mathfrak{B})$ such that:

(4)

\begin{align} \quad \sigma(x_1, x_2) = T[x_1, y_1] \end{align}

• So define the product on $\mathfrak{A} \otimes \mathfrak{B}$ for all $u, v \in \mathfrak{A} \otimes \mathfrak{B}$ by $(u, v) \to \sigma(u)(v)$. There are a few things to check.

• 1. Showing that $(x_1 \otimes y_1)(x_2 \otimes y_2) = x_1x_2 \otimes y_1y_2$: Let $x_1, x_2 \in \mathfrak{A}$ and let $y_1, y_2 \in \mathfrak{B}$. Let $u = x_1 \otimes y_1$ and let $v = x_2 \otimes y_2$. Then $(u, v) \to \sigma(u)(v)$ tells us that:

(5)

\begin{align} \quad (x_1 \otimes y_1)(x_2 \otimes y_2) = uv = \sigma (x_1 \otimes y_1)(x_2 \otimes y_2) = T[x_1, y_1](x_2 \otimes y_2) \overset{(*)} = x_1x_2 \otimes y_1y_2 \end{align}

• Moreover, since $\sigma : \mathfrak{A} \otimes \mathfrak{B} \to \mathcal{L}(\mathfrak{A} \otimes \mathfrak{B})$ is unique, and each $T[x_1, x_2] : \mathfrak{A} \otimes \mathfrak{B} \to \mathfrak{A} \otimes \mathfrak{B}$ is unique, we see that this product $(u, v) \to \sigma (u)(v)$ is unique.

• 2. Showing that the product on $\mathfrak{A} \otimes \mathfrak{B}$ defined above is algebra product on $\mathfrak{A} \otimes \mathfrak{B}$: Let $u, v, w \in \mathfrak{A} \otimes \mathfrak{B}$ and let $\alpha \in \mathbf{F}$. Write:

(6)

\begin{align} \quad u = \sum_{i=1}^{m} x_i \otimes y_i \quad , \quad v = \sum_{j=1}^{n} x_j' \otimes y_j' \quad , \quad w = \sum_{k=1}^{o} x_k'' \otimes y_k'' \end{align}

• We want to show that $u(vw) = (uv)w$, $u(v + w) = uv + uw$, and $(\alpha u)v = \alpha (uv) = u(\alpha v)$. All three of these properties follow from the fact that $\mathfrak{A}$ and $\mathfrak{B}$ are algebras.

• For associativity we have that:

(7)

\begin{align} \quad u(vw) = \left ( \sum_{i=1}^{m} x_i \otimes y_i \right ) \left ( \sum_{j=1}^{n} \sum_{k=1}^{o} x_j'x_k'' \otimes y_j'y_k'' \right ) = \sum_{i=1}^{m} \sum_{j=1}^{n} \sum_{k=1}^{o} x_ix_j'x_k'' \otimes y_iy_j'y_k'' = \left ( \sum_{i=1}^{m} \sum_{j=1}^{n} x_ix_j' \otimes y_iy_j' \right ) \left ( \sum_{k=1}^{o} x_k'' \otimes y_k'' \right ) = (uv)w \end{align}

• For distributivity we have that:

(8)

\begin{align} \quad \quad u(v + w) = \left ( \sum_{i=1}^{m} x_i \otimes y_i \right ) \left ( \sum_{j=1}^{n} x_j' \otimes y_j' + \sum_{k=1}^{o} x_k'' \otimes y_k'' \right ) = \sum_{i=1}^{m} \sum_{j=1}^{n} x_ix_j' \otimes y_iy_j' + \sum_{i=1}^{m} \sum_{k=1}^{o} x_ix_k'' \otimes y_iy_k'' = uv + uw \end{align}

• And for the last property we have that:

(9)

\begin{align} \quad \quad (\alpha u)v = \left ( \alpha \sum_{i=1}^{m} x_i \otimes y_i \right ) \left ( \sum_{j=1}^{n} x_j' \otimes y_j' \right ) = \left ( \sum_{i=1}^{m} \alpha x_i \otimes y_i \right ) \left ( \sum_{j=1}^{n} x_j' \otimes y_j' \right ) = \sum_{i=1}^{m} \sum_{j=1}^{n} \alpha x_ix_j' \otimes y_iy_j' = \alpha \sum_{i=1}^{m} \sum_{j=1}^{n} x_ix_j' \otimes y_iy_j' = \alpha (uv) \end{align}

• And also:

(10)

\begin{align} \quad \quad \alpha (uv) = \alpha \sum_{i=1}^{m} \sum_{j=1}^{n} x_ix_j' \otimes y_iy_j' = \sum_{i=1}^{m} \sum_{j=1}^{n} \alpha x_ix_j' \otimes y_iy_j' = \sum_{i=1}^{m} \sum_{j=1}^{n} x_i (\alpha x_j') \otimes y_i y_j' = \left ( \sum_{i=1}^{m} x_i \otimes y_i \right ) \left ( \sum_{j=1}^{n} \alpha x_j' \otimes y_j' \right ) &= \left ( \sum_{i=1}^{m} x_i \otimes y_i \right ) \alpha \left ( \sum_{j=1}^{n} x_j' \otimes y_j' \right ) \\ &= u(\alpha v) \end{align}

• Thus the product defined on $\mathfrak{A} \otimes \mathfrak{B}$ is an algebra product on $\mathfrak{A} \otimes \mathfrak{B}$. Thus $\mathfrak{A} \otimes \mathfrak{B}$ is an algebra over $\mathbf{F}$. $\blacksquare$

Proposition 2: Let $\mathfrak{A}$ and $\mathfrak{B}$ be normed algebras over $\mathbf{F}$. Then the algebraic tensor product $\mathfrak{A} \otimes \mathfrak{B}$ is a normed algebra with the operations of addition, scalar multiplication, and a unique product on $\mathfrak{A} \otimes \mathfrak{B}$ with the property that for all $x_1, x_2 \in \mathfrak{A}$ and $y_1, y_2 \in \mathfrak{B}$ we have that $(x_1 \otimes y_1)(x_2 \otimes y_2) = x_1x_2 \otimes y_1y_2$, and with the projective tensor norm $p$.

• Proof: We have already proven that $\mathfrak{A} \otimes \mathfrak{B}$ is algebra and that $p$ is a norm on $\mathfrak{A} \otimes \mathfrak{B}$ (as a normed linear space). All that remains to show is that $p(uv) \leq p(u)p(v)$ for all $u, v \in \mathfrak{A} \otimes \mathfrak{B}$.

• Let $u, v \in \mathfrak{A} \otimes \mathfrak{B}$ with $u = \sum_{i=1}^{m} x_i \otimes y_i$ and with $v = \sum_{j=1}^{n} x_j' \otimes y_j'$. Then we have that:

(11)

\begin{align} \quad uv &= \left ( \sum_{i=1}^{m} x_i \otimes y_i \right ) \left ( \sum_{j=1}^{n} x_j' \otimes y_j' \right ) \\ &=\sum_{i=1}^{m} \sum_{j=1}^{n} (x_i \otimes y_i)(x_j' \otimes y_j') \\ &= \sum_{i=1}^{m} \sum_{j=1}^{n} x_ix_j' \otimes y_iy_j' \end{align}

• Therefore we have that:

(12)

\begin{align} \quad \sum_{i=1}^{m} \sum_{j=1}^{n} \| x_i x_j' \| \| y_i y_j' \| \leq \sum_{i=1}^{m} \sum_{j=1}^{n} \| x_i \| \| x_j' \| \| y_i \| \| y_j' \| = \sum_{i=1}^{m} \| x_i \| \| y_i \| \sum_{j=1}^{n} \| x_j' \| \| y_j' \| \end{align}

• Note that if $A$ is the collection of representations of $uv$ of the form $\sum_{i=1}^{m} \sum_{j=1}^{n} x_ix_j' \otimes y_iy_j'$ with $u = \sum_{i=1}^{m} x_i \otimes y_i$ and $v = \sum_{j=1}^{n} x_j' \otimes y_j'$ and if $B$ is the collection of all representations of $uv$ as elements of $\mathfrak{A} \otimes \mathfrak{B}$ then $A \subseteq B$, and so by properties of infimum $p(uv)$ is less than or equal to any sum of the form $\sum_{i=1}^{m} \sum_{j=1}^{n} x_ix_j' \otimes y_iy_j'$ above.

• Also, since the above inequality holds for all representations of $u$ as $\sum_{i=1}^{m} x_i \otimes y_i$ and all representations of $v$ as $\sum_{j=1}^{n} x_j' \otimes y_j'$, we conclude that $p(uv) \leq p(u)p(v)$.

• Therefore the projective tensor norm is an algebra norm on $\mathfrak{A} \otimes \mathfrak{B}$.