Topological Modules/Topological tensor products

Tensor product of Hilbert spaces

Proposition (tensor product of orthonormal bases is orthonormal basis of tensor product):

Let $H_1,H_2$ be Hilbert spaces, and suppose that $(e_{\lambda }{)}_{\lambda \in \mathrm{\Lambda }}$ is an orthonormal basis of $H_1$ and $(f_{\mu }{)}_{\mu \in \mathrm{M}}$ is an orthonormal basis of $H_2$. Then $(e_{\lambda }\otimes e_{\mu }{)}_{(\lambda ,\mu )\in \mathrm{\Lambda }\times \mathrm{M}}$ is an orthonormal basis of $H_1\otimes H_2$.

Proof: Let any element

$\sum _{j=1}^n{f}_j\otimes g_j$

of $H_1\otimes H_2$ be given; by definition, each element of $H_1\otimes H_2$ may be approximated by such elements. Let $ϵ>0$. Then by definition of an orthonormal basis, we find $m_j,l_j$ for $j\in [n]$ and ${\alpha }_{j,k},{\beta }_{j,k},{\lambda }_{j,k},{\mu }_{j,k}$ for $j\in [n]$ and then $k\in [m_j]$ resp. $[l_j]$ such that

$\Vert \sum _{k=1}^{m_j}{\alpha }_{j,k}{e}_{{\lambda }_{j,k}}-f_j\Vert <ϵ$ and $\Vert \sum _{k=1}^{l_j}{\beta }_{j,k}{f}_{{\mu }_{j,k}}-g_j\Vert <ϵ$.

Then note that by the triangle inequality,

$\Vert \sum _{j=1}^n{f}_j\otimes g_j-\sum _{j=1}^n\left(\sum _{k=1}^{m_j}{\alpha }_{j,k}{e}_{{\lambda }_{j,k}}\right)\otimes \left(\sum _{k=1}^{l_j}{\beta }_{j,k}{f}_{{\mu }_{j,k}}\right)\Vert \le \sum _{j=1}^n\Vert f_j\otimes g_j-\left(\sum _{k=1}^{m_j}{\alpha }_{j,k}{e}_{{\lambda }_{j,k}}\right)\otimes \left(\sum _{k=1}^{l_j}{\beta }_{j,k}{f}_{{\mu }_{j,k}}\right)\Vert$.

Now fix $j\in [n]$. Then by the triangle inequality,

$\begin{array}{cc}\Vert f_j\otimes g_j-\left(\sum _{k=1}^{m_j}{\alpha }_{j,k}{e}_{{\lambda }_{j,k}}\right)\otimes \left(\sum _{k=1}^{l_j}{\beta }_{j,k}{f}_{{\mu }_{j,k}}\right)\Vert & \le \Vert f_j\otimes g_j-\left(\sum _{k=1}^{m_j}{\alpha }_{j,k}{e}_{{\lambda }_{j,k}}\right)\otimes g_j\Vert +\Vert \left(\sum _{k=1}^{m_j}{\alpha }_{j,k}{e}_{{\lambda }_{j,k}}\right)\otimes g_j-\left(\sum _{k=1}^{m_j}{\alpha }_{j,k}{e}_{{\lambda }_{j,k}}\right)\otimes \left(\sum _{k=1}^{l_j}{\beta }_{j,k}{f}_{{\mu }_{j,k}}\right)\Vert \\ & =\Vert f_j-\left(\sum _{k=1}^{m_j}{\alpha }_{j,k}{e}_{{\lambda }_{j,k}}\right)\Vert \Vert g_j\Vert +\Vert \sum _{k=1}^{m_j}{\alpha }_{j,k}{e}_{{\lambda }_{j,k}}\Vert \Vert g_j-\left(\sum _{k=1}^{l_j}{\beta }_{j,k}{f}_{{\mu }_{j,k}}\right)\Vert \\ & \le ϵ(\Vert g_j\Vert +\Vert \sum _{k=1}^{m_j}{\alpha }_{j,k}{e}_{{\lambda }_{j,k}}\Vert ).\end{array}$

In total, we obtain that

$\Vert \sum _{j=1}^n{f}_j\otimes g_j-\sum _{j=1}^n\left(\sum _{k=1}^{m_j}{\alpha }_{j,k}{e}_{{\lambda }_{j,k}}\right)\otimes \left(\sum _{k=1}^{l_j}{\beta }_{j,k}{f}_{{\mu }_{j,k}}\right)\Vert \le ϵ\sum _{j=1}^n(\Vert g_j\Vert +2\Vert f_j\Vert )$

(assuming that the given sum approximates $f_j$ well enough) which is arbitrarily small, so that the span of tensors of the form $e_{\lambda }\otimes f_{\mu }$ is dense in $H_1\otimes H_2$. Now we claim that the basis is orthonormal. Indeed, suppose that $(\lambda ,\mu )\ne ({\lambda }^{\prime },{\mu }^{\prime })$. Then

$⟨e_{\lambda }\otimes f_{\mu },e_{{\lambda }^{\prime }}\otimes f_{{\mu }^{\prime }}⟩=⟨e_{\lambda },e_{{\lambda }^{\prime }}⟩⟨f_{\mu },f_{{\mu }^{\prime }}⟩=0$.

Similarly, the above expression evaluates to $1$ when $\lambda ={\lambda }^{\prime }$ and $\mu ={\mu }^{\prime }$. Hence, $(e_{\lambda }\otimes e_{\mu }{)}_{(\lambda ,\mu )\in \mathrm{\Lambda }\times \mathrm{M}}$ does constitute an orthonormal basis of $H_1\otimes H_2$. $◻$