# 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 $(e_{\lambda })_{\lambda \in \Lambda }$ is an orthonormal basis of $H_{1}$ and $(f_{\mu })_{\mu \in \mathrm {M} }$ is an orthonormal basis of $H_{2}$. Then $(e_{\lambda }\otimes e_{\mu })_{(\lambda ,\mu )\in \Lambda \times \mathrm {M} }$ 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 $\epsilon >0$. 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
    $\left\|\sum _{k=1}^{m_{j}}\alpha _{j,k}e_{\lambda _{j,k}}-f_{j}\right\|<\epsilon$ and $\left\|\sum _{k=1}^{l_{j}}\beta _{j,k}f_{\mu _{j,k}}-g_{j}\right\|<\epsilon$.
Then note that by the triangle inequality,
    $\left\|\sum _{j=1}^{n}f_{j}\otimes g_{j}-\sum _{j=1}^{n}\left(\sum _{k=1}^{m_{j}}\alpha _{j,k}e_{\lambda _{j,k}}\right)\otimes \left(\sum _{k=1}^{l_{j}}\beta _{j,k}f_{\mu _{j,k}}\right)\right\|\leq \sum _{j=1}^{n}\left\|f_{j}\otimes g_{j}-\left(\sum _{k=1}^{m_{j}}\alpha _{j,k}e_{\lambda _{j,k}}\right)\otimes \left(\sum _{k=1}^{l_{j}}\beta _{j,k}f_{\mu _{j,k}}\right)\right\|$.
Now fix $j\in [n]$. Then by the triangle inequality,
    ${\begin{aligned}\left\|f_{j}\otimes g_{j}-\left(\sum _{k=1}^{m_{j}}\alpha _{j,k}e_{\lambda _{j,k}}\right)\otimes \left(\sum _{k=1}^{l_{j}}\beta _{j,k}f_{\mu _{j,k}}\right)\right\|&\leq \left\|f_{j}\otimes g_{j}-\left(\sum _{k=1}^{m_{j}}\alpha _{j,k}e_{\lambda _{j,k}}\right)\otimes g_{j}\right\|+\left\|\left(\sum _{k=1}^{m_{j}}\alpha _{j,k}e_{\lambda _{j,k}}\right)\otimes g_{j}-\left(\sum _{k=1}^{m_{j}}\alpha _{j,k}e_{\lambda _{j,k}}\right)\otimes \left(\sum _{k=1}^{l_{j}}\beta _{j,k}f_{\mu _{j,k}}\right)\right\|\\&=\left\|f_{j}-\left(\sum _{k=1}^{m_{j}}\alpha _{j,k}e_{\lambda _{j,k}}\right)\right\|\|g_{j}\|+\left\|\sum _{k=1}^{m_{j}}\alpha _{j,k}e_{\lambda _{j,k}}\right\|\left\|g_{j}-\left(\sum _{k=1}^{l_{j}}\beta _{j,k}f_{\mu _{j,k}}\right)\right\|\\&\leq \epsilon \left(\|g_{j}\|+\left\|\sum _{k=1}^{m_{j}}\alpha _{j,k}e_{\lambda _{j,k}}\right\|\right).\end{aligned}}$
In total, we obtain that
    $\left\|\sum _{j=1}^{n}f_{j}\otimes g_{j}-\sum _{j=1}^{n}\left(\sum _{k=1}^{m_{j}}\alpha _{j,k}e_{\lambda _{j,k}}\right)\otimes \left(\sum _{k=1}^{l_{j}}\beta _{j,k}f_{\mu _{j,k}}\right)\right\|\leq \epsilon \sum _{j=1}^{n}\left(\|g_{j}\|+2\|f_{j}\|\right)$
(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 $(\lambda ,\mu )\neq (\lambda ',\mu ')$. Then
    $\langle e_{\lambda }\otimes f_{\mu },e_{\lambda '}\otimes f_{\mu '}\rangle =\langle e_{\lambda },e_{\lambda '}\rangle \langle f_{\mu },f_{\mu '}\rangle =0$.
Similarly, the above expression evaluates to $1$ when $\lambda =\lambda '$ and $\mu =\mu '$. Hence, $(e_{\lambda }\otimes e_{\mu })_{(\lambda ,\mu )\in \Lambda \times \mathrm {M} }$ does constitute an orthonormal basis of $H_{1}\otimes H_{2}$. $\Box$