Example of a non-algebraic $\ell^2$-function in two variables Let's call an $\ell^2$-function $\mathbb{N} \times \mathbb{N} \to \mathbb{C}$ algebraic if it is in the image of the natural algebra homomorphism $\ell^2(\mathbb{N}) \otimes \ell^2(\mathbb{N}) \to \ell^2(\mathbb{N} \times \mathbb{N})$, where on the left hand side we consider the usual, non-completed tensor product. In other words, $f(m,n)$ is algebraic iff it may be written as $\sum_{i=1}^{k} g_i(m) h_i(n)$ for some $k \in \mathbb{N}$ and $\ell^2$-functions $g_i,h_i$. Probably there are abstract reasons for the existence of non-algebraic functions. But I would like to know an explicit example of an $\ell^2$-function together with a concise and complete proof that it is not algebraic. For example:
Question. Can you give a proof that the $\ell^2$-function $(n,m) \mapsto \dfrac{1}{2^{n \cdot m}}$ is not algebraic?
 A: Colin McQuillan's answer is terse, correct, but difficult to understand for typographical reasons: There should be some letters below the arrows.
I shall elaborate using the function $f(m,n):=2^{-mn}$ suggested by the OP as example. For any given $p\in{\mathbb N}$ the ($p\times p)$-matrix $$F_p:=\bigl [f(m,n)\bigr]_{1\leq m\leq p, \ 1\leq n\leq p}$$
is essentially a Vandermonde matrix and therefore has rank $p$, whatever $p$.
On the other hand, for any two functions $g,\ h\in\ell^2$ the tensor product $t:=g\otimes h$ is defined by $t(m,n):=g(m)\ h(n)$. The corresponding matrix
$$T_p:=\bigl [t(m,n)\bigr]_{1\leq m\leq p, \ 1\leq n\leq p} =\bigl [g(m)\ h(n)\bigr]_{1\leq m\leq p, \ 1\leq n\leq p}$$
has rank $\ (\leq )\ 1$, whatever $p$,  since all its rows are multiples of the vector $(h(1),h(2),\ldots,h(p))$. If we are given $k\geq 1$ such pairs $g_i$, $h_i$ and put $q:=\sum_{i=1}^k g_i\otimes h_i$ then the matrix 
$$Q_p:=\bigl [q(m,n)\bigr]_{1\leq m\leq p, \ 1\leq n\leq p}$$
is a sum of $k$ rank-one matrices and therefore has rank $\ \leq k$, whatever $p$. It follows that $Q_p\ne F_p$ as soon as $p>k$; whence the function $f$ of the example is not "algebraic" as defined by the OP. 
A: An algebraic function has bounded rank (either as a tensor, or in this case just as a matrix). So let $f(n,m)$ be $1/2^n$ if $n=m$ and zero otherwise.
For all $p$ the composition $\{1,\cdots,p\}\times\{1,\cdots,p\}\to \mathbb{N}\times \mathbb{N}\to \mathbb{C}$ is a $p\times p$ matrix of full rank. If $f$ is algebraic then this matrix is just $\sum_{i=1}^k g_i(m)h_i(n)$ which has rank at most $k$.
