The Algebraic Tensor Product of Two Normed Algebras

The Algebraic Tensor Product of Two Normed Algebras

Proposition: Let $X$ and $Y$ be algebras over $\mathbf{F}$. Then there exists a unique product of $X \otimes Y$ for which $X \otimes Y$ is an algebra and with the property that $(x_1 \otimes y_1)(x_2 \otimes y_2) = x_1x_2 \otimes y_1y_2$ for all $x_1, x_2 \in X$ and for all $y_1, y_2 \in Y$.
  • Proof: Let $x_1 \in X$ and let $y_1 \in Y$. Then the map $X \times Y \to X \otimes Y$ defined for all $x_2 \in X$ and all $y_2 \in Y$ by $(x_2, y_2) \to x_1x_2 \otimes y_1y_2$ is bilinear map.
  • To see this, fix $x_2 \in X$ and let $y_2, y_2' \in Y$, $\alpha \in \mathbf{F}$. 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 $x_2 \in X$ the map $y_2 \to x_1x_2 \otimes y_1y_2$ is linear. Similarly it can be shown that for each fixed $y_2 \in Y$ the map $x_2 \to x_1x_2 \otimes y_1y_2$ is linear. Thus this map is bilinear.
(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 $X \times Y$ to $\mathrm{L}(X \otimes Y)$. Using the theorem above again, there exists a unique linear map $\sigma : X \otimes Y \to \mathrm{L}(X \otimes Y)$ such that:
(4)
\begin{align} \quad \sigma(x_1, x_2) = T[x_1, y_1] \end{align}
  • So define the product on $X \otimes Y$ for all $u, v \in X \otimes Y$ 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 $x_1, x_2 \in X$ and let $y_1, y_2 \in Y$. 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 $\sigma : X \otimes Y \to \mathrm{L}(X \otimes Y)$ is unique, and each $T[x_1, x_2] : X \otimes Y \to X \otimes Y$ is unique, we see that this product $(u, v) \to \sigma (u)(v)$ is unique.
  • 2. Showing that the product on $X \otimes Y$ defined above is algebra product on $X \otimes Y$: Let $u, v, w \in X \otimes Y$ 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 $X$ and $Y$ 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 $X \otimes Y$ is an algebra product on $X \otimes Y$. Thus $X \otimes Y$ is an algebra over $\mathbf{F}$. $\blacksquare$
Proposition 2: Let $X$ and $Y$ be normed algebras over $\mathbf{F}$. Then the algebraic tensor product $X \otimes Y$ is a normed algebra with the operations of addition, scalar multiplication, and a unique product on $X \otimes Y$ with the property that for all $x_1, x_2 \in X$ and $y_1, y_2 \in Y$ 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 $X \otimes Y$ is algebra and that $p$ is a norm on $X \otimes Y$ (as a normed linear space). All that remains to show is that $p(uv) \leq p(u)p(v)$ for all $u, v \in X \otimes Y$.
  • Let $u, v \in X \otimes Y$ 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 $X \otimes Y$ 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 $X \otimes Y$.
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