Hilbert tensor construction: inner product, completion and product bases

[The quotient is a complex vector space] Since $\mathbb{C}$ is commutative, for each $c\in \mathbb{C}$ the $\mathbb{Z}$-linear self-map $L_c$ of $F$ induced by $e_{(u,v)}\mapsto e_{(cu,v)}$ maps every defining generator of $H_0$ to a defining generator: $L_c(e_{(u+u^{\mathrm{\prime }},v)}-e_{(u,v)}-e_{(u^{\mathrm{\prime }},v)})=e_{(cu+c{u}^{\mathrm{\prime }},v)}-e_{(cu,v)}-e_{(c{u}^{\mathrm{\prime }},v)}$ is the first relation at $(cu,c{u}^{\mathrm{\prime }},v)$, the second family is preserved in the same way, and $L_c(e_{(ur,v)}-e_{(u,rv)})=e_{(c(ur),v)}-e_{(cu,rv)}$ is the third relation at $(cu,r,v)$ because $c(ur)=(cu)r$. Hence $L_c[H_0]\subseteq H_0$ and $L_c$ descends to a $\mathbb{Z}$-linear map of $F\mathrm{/}{H}_0$; because $L_c{L}_{c^{\mathrm{\prime }}}=L_{c{c}^{\mathrm{\prime }}}$ and the actions distribute over the abelian group, $F\mathrm{/}{H}_0$ becomes a $\mathbb{C}$-vector space with $c\cdot (u\otimes v)=(cu)\otimes v$; the third relation gives $(cu)\otimes v=u\otimes (cv)$, so the quotient is the $\mathbb{C}$-bilinear tensor product $H\odot K$.

[Well-definedness and sesquilinearity] For fixed $(u_0,v_0)$, put $b_{u_0,v_0}(u,v)=⟨u,u_0{⟩}_H⟨v,v_0{⟩}_K$. By [F3] this is $\mathbb{C}$-bilinear in $u,v$, and $b_{u_0,v_0}(ur,v)=r{b}_{u_0,v_0}(u,v)=b_{u_0,v_0}(u,rv)$. Extend its values on the basis of $F={\mathbb{Z}}^{(H\times K)}$ by ${\tilde{b}}_{u_0,v_0}({\sum }_i{n}_i{e}_{(u_i,v_i)})={\sum }_i{n}_i{b}_{u_0,v_0}(u_i,v_i)$ for $n_i\in \mathbb{Z}$. This is a $\mathbb{Z}$-linear map on $F$, and additivity in each variable and the displayed balancing identity show that it vanishes on all three families of generators of $H_0$ in [F1], hence on $H_0$. It therefore induces an additive functional ${\lambda }_{u_0,v_0}:F\mathrm{/}{H}_0\to \mathbb{C}$. Since ${\tilde{b}}_{u_0,v_0}(L_{c}x)=c{\tilde{b}}_{u_0,v_0}(x)$ for every $x\in F$, the scalar action of step 1.1 makes ${\lambda }_{u_0,v_0}$ complex-linear. For any finite representation $w={\sum }_j{u}_j^{\mathrm{\prime }}\otimes v_j^{\mathrm{\prime }}$, the displayed form is consequently independent of the representation of its first argument $z$, being ${\sum }_j{\lambda }_{u_j^{\mathrm{\prime }},v_j^{\mathrm{\prime }}}(z)$. On free representatives, conjugate symmetry of the factor pairings gives $⟨z,w⟩=\overline{⟨w,z⟩}$ term by term. Independence in the first argument applied to the right-hand side therefore gives independence of the representation of $w$ as well. The descended form is linear in $z$; conjugate symmetry then makes it conjugate-linear in $w$, and proves it is Hermitian.

[Positive definiteness and unique orthonormal coordinates] Let $z={\sum }_{i=1}^n{u}_i\otimes v_i$, let $U_0=\mathrm{span}\{u_i\}$ and $V_0=\mathrm{span}\{v_i\}$, and choose orthonormal bases $e_1,\dots ,e_p$ of $U_0$ and $f_1,\dots ,f_q$ of $V_0$ by [F4]. Write $u_i={\sum }_a{\alpha }_{ia}{e}_a$ and $v_i={\sum }_b{\beta }_{ib}{f}_b$; bilinearity of $\otimes$ (step 1.1) gives $z={\sum }_{a,b}{c}_{ab}{e}_a\otimes f_b$ with $c_{ab}={\sum }_i{\alpha }_{ia}{\beta }_{ib}$. By the defining formula of the Statement and orthonormality, $⟨z,z⟩={\sum }_{a,b,c,d}{c}_{ab}\overline{c_{cd}}⟨e_a,e_c⟩⟨f_b,f_d⟩={\sum }_{a,b}\mathrm{\mid }{c}_{ab}{\mathrm{\mid }}^2\ge 0$. If $z=0$, then this same displayed formula gives ${\sum }_{a,b}\mathrm{\mid }{c}_{ab}{\mathrm{\mid }}^2=⟨z,z⟩=0$, so every $c_{ab}=0$, because a sum of nonnegative real numbers vanishes only if every term does; conversely, if $c_{ab}=0$ for all $a,b$ then bilinearity gives $z={\sum }_{a,b}0\cdot e_a\otimes f_b=0$. Hence $⟨z,z⟩=0$ exactly when $z=0$, and applying the same argument to the difference of two expansions shows that the coefficients relative to orthonormal families are unique.

[Completion and the norm of elementary tensors] By steps 2.1 and 3.1, $(H\odot K,⟨\cdot ,\cdot ⟩)$ is a complex inner product space with induced norm $\mathrm{\parallel }z\mathrm{\parallel }=⟨z,z{⟩}^{1\mathrm{/}2}$ by [F3]. Its norm completion $(\widehat{H\odot K},\kappa )$ exists by [F2], and by [F2] it carries a unique inner product extending the given one whose induced length is the completion norm; with it $\widehat{H\odot K}$ is a Hilbert space and $\kappa$ is a linear isometry with dense image. The norm of an elementary tensor is $\mathrm{\parallel }\kappa (u\otimes v){\mathrm{\parallel }}^2=⟨u\otimes v,u\otimes v⟩=⟨u,u{⟩}_H⟨v,v{⟩}_K=\mathrm{\parallel }u{\mathrm{\parallel }}^2\mathrm{\parallel }v{\mathrm{\parallel }}^2$, hence $\mathrm{\parallel }\kappa (u\otimes v)\mathrm{\parallel }=\mathrm{\parallel }u\mathrm{\parallel }\mathrm{\parallel }v\mathrm{\parallel }$. To prove the completion uniqueness asserted above, let $(Y,\iota )$ and $(Z,j)$ be completions. The rule $\iota (z)\mapsto j(z)$ is a well-defined linear isometry on the dense subspace $\iota [H\odot K]$. For $y\in Y$, choose $z_n\in H\odot K$ with $\iota (z_n)\to y$ and define $U(y):={\lim }_{n}j(z_n)$; the isometry makes $(j(z_n))$ Cauchy and makes this limit independent of the approximating sequence. Thus $U$ is a linear isometry with $U\circ \iota =j$. Its image is closed because $Y$ is complete and contains the dense set $j[H\odot K]$, so it is onto; any two such continuous extensions agree on a dense subspace and hence everywhere.

[Orthonormal family of products] Let $(e_a{)}_{a\in A}$, $(f_b{)}_{b\in B}$ be orthonormal with closed linear spans $H$ and $K$. For pairs $(a,b)\mathrm{\ne }(a^{\mathrm{\prime }},b^{\mathrm{\prime }})$ at least one index differs, so $⟨\kappa (e_a\otimes f_b),\kappa (e_{a^{\mathrm{\prime }}}\otimes f_{b^{\mathrm{\prime }}})⟩=⟨e_a,e_{a^{\mathrm{\prime }}}{⟩}_H⟨f_b,f_{b^{\mathrm{\prime }}}{⟩}_K=0$, and for equal pairs the value is $1$; hence the products are orthonormal in $\widehat{H\odot K}$.

[Density of the product span] Fix $\epsilon >0$ and $z={\sum }_{i=1}^n{u}_i\otimes v_i\in H\odot K$. By [F5] applied to $(e_a)$ and to $(f_b)$ there are finite sets $A_i\subseteq A$, $B_i\subseteq B$ with $\mathrm{\parallel }{u}_i-{\sum }_{a\in A_i}⟨u_i,e_a⟩e_a\mathrm{\parallel }<\epsilon \mathrm{/}(n(\mathrm{\parallel }{v}_i\mathrm{\parallel }+1))$ and $\mathrm{\parallel }{v}_i-{\sum }_{b\in B_i}⟨v_i,f_b⟩f_b\mathrm{\parallel }<\epsilon \mathrm{/}(n(\mathrm{\parallel }{u}_i\mathrm{\parallel }+1))$; write $u_i^{\mathrm{\prime }},v_i^{\mathrm{\prime }}$ for the finite sums and $z^{\mathrm{\prime }}={\sum }_i{u}_i^{\mathrm{\prime }}\otimes v_i^{\mathrm{\prime }}$. Then $z-z^{\mathrm{\prime }}={\sum }_i((u_i-u_i^{\mathrm{\prime }})\otimes v_i+u_i^{\mathrm{\prime }}\otimes (v_i-v_i^{\mathrm{\prime }}))$, and by the triangle inequality, $\mathrm{\parallel }{u}_i^{\mathrm{\prime }}\mathrm{\parallel }\le \mathrm{\parallel }{u}_i\mathrm{\parallel }$ (Bessel, [F5]) and the norm formula of step 4.1, $\mathrm{\parallel }z-z^{\mathrm{\prime }}\mathrm{\parallel }\le {\sum }_i(\mathrm{\parallel }{u}_i-u_i^{\mathrm{\prime }}\mathrm{\parallel }\mathrm{\parallel }{v}_i\mathrm{\parallel }+\mathrm{\parallel }{u}_i\mathrm{\parallel }\mathrm{\parallel }{v}_i-v_i^{\mathrm{\prime }}\mathrm{\parallel })<2\epsilon$. Hence every element of the dense subspace $\kappa [H\odot K]$ is approximated by finite sums $\sum c_{ab}\kappa (e_a\otimes f_b)$, so the closed linear span of the product family contains $\kappa [H\odot K]$ and therefore coincides with $\widehat{H\odot K}$; by [F5] the family is a complete orthonormal system. If $A$ and $B$ are countable, then $A\times B$ is countable and the closed span of a countable family is separable, so $\widehat{H\odot K}$ is separable.