This question Tensor Product: Hilbert Spaces looks for an example to show that the algebraic tensor product of infinite dimensional Hilbert spaces is not complete. The answers suggest "choose orthonormal bases $\{e_i: i \in I\}$ of $\mathcal H$ and $\{f_j : j\in J \}$ of $\mathcal K$" ...."Then $$y=\sum_{n\in \mathbb N} \frac1{n!}e_n\otimes f_n \in \mathcal H\otimes_{HS}\mathcal K. $$ Show that $y\not \in \mathcal H\otimes_{Alg} \mathcal K$."

I'm having a problem with the last part. Clearly $y$ is an infinite sum of pure tensors, but $I, J$ are not algebraic bases and I don't see that it would immediately follow that $y\not \in \mathcal H\otimes_{Alg} \mathcal K$. Pure tensors in $H\otimes_{Alg} \mathcal K$ can have infinite components, for example (I think)..... $$h=\sum_{n\in \mathbb N} \frac1{n}e_n \in \mathcal H$$ $$ k=\sum_{m\in \mathbb N} \frac1{m}f_m \in \mathcal K$$
$$h \otimes k = \sum_{n\in \mathbb N} \sum_{m\in \mathbb N}\frac1{n}\frac1{m}e_n \otimes f_m \in \mathcal H \otimes_{alg} K $$

So, can anyone please give a proof of either this or a different example ?


I will not give a solution to the question, but rather try to clarify the point on which you seem to be confused.

The elements $h,k$ you defined exist not by an algebraic argument, but by an analytic one, by which I mean that in general given a vector space $V$ and a basis $\{v_i\}$, in general one cannot take infinite sums of elements of the basis and get a well-defined element of $V$. However, a Hilbert space is complete and the sequences of partial sums for $h$ and $k$ are Cauchy, and thus converge.

Now, the elements of $\mathcal{H}\otimes\mathcal{K}$ are finite sums of pure tensors. You have to show that it is not possible to write $y$ as such a finite sum.

  • $\begingroup$ Thanks. Would such a proof come from a contradiction around a finite system of linear equations ? $\endgroup$ – Tom Collinge Feb 23 '17 at 15:32

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