# Tensor Product: Hilbert Spaces

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Problem

Given Hilbert spaces.

In general, their algebraic tensor product isn't complete: $$\mathcal{H}\hat{\otimes}\mathcal{K}=\mathcal{H}\otimes\mathcal{K}\iff\dim\mathcal{H}<\infty\lor\dim\mathcal{K}<\infty$$ How to prove this from scratch?

Attempt

Choose orthonormal bases: $$\mathcal{S}\otimes\mathcal{T}:=\{\sigma\otimes\tau:\sigma\in\mathcal{S},\tau\in\mathcal{T}\}$$

One obtains some candidates: $$\sigma_k\otimes\tau_l\in\mathcal{S}\otimes\mathcal{T}:\quad\sum_{kl=1}^\infty\frac{1}{kl}\sigma_k\otimes\tau_l\quad\sum_{k=1}^\infty\frac{1}{kl}\delta_{kl}\sigma_k\otimes\tau_l$$ However the former one drops out: $$\sum_{kl=1}^\infty\frac{1}{kl}\sigma_k\otimes\tau_l=\left(\sum_{k=1}^\infty\frac{1}{k}\sigma_k\right)\otimes\left(\sum_{l=1}^\infty\frac{1}{l}\tau_l\right)$$ So it is not obvious at all wether the latter one works out!

Reference

Build-up on: Vector Spaces: Tensor Product

• `Obviously, it is not complete yet.' Certainly it is complete because finite dimensional. – Tomek Kania Dec 1 '14 at 10:50
• Come on, that was a typo, you could have known that! – C-Star-W-Star Dec 1 '14 at 11:23
• So what is your question? Could you explain it? – Yan kai Dec 1 '14 at 16:24
• Really it is not a good idea to modify your question so as to render obsolete any answer(s) thoughtfully provided. Better is to ask a new question. – user43208 Dec 23 '14 at 16:41

That element you want to form is just an elementary tensor $x\otimes y$ in the algebraic tensor product $\mathcal H\otimes\mathcal H$. Then you want to have sums of those guys, and then limits of them.

• Aaaah ok, so is it like this?: So if I would instead consider formally $\sum_{k,l}a_{k,l}e_k\otimes e_l$ then they're not necessarily of the form $\sum_{k,l}x_ky_le_k\otimes e_l$. – C-Star-W-Star Dec 1 '14 at 23:05
• Yes. But a Hilbert space is in particular a complete Banach space. So you would have to argue that the formal space constructed like that is a complete Banach space with the norm induced by the inner product. It doesn't look simpler than taking the algebraic tensor product and completing it. – Martin Argerami Dec 1 '14 at 23:26
• Sorry for the late come-back. I got my question now: Why is the algebraic tensor product of infinite dimensional Hilbert spaces incomplete yet: $\mathcal{H}\otimes\mathcal{H}\subsetneq\mathcal{H}\hat{\otimes}\mathcal{H}$ More specifically, can one construct a sequence in a general or in a special Hilbert space algebraic tensor product whos limit is certainly not contained: $z_n:=\sum_{1\leq k\leq K(n)}\sum_{1\leq l\leq L(n)}\alpha_n(i,j)s_n(i)\otimes s_n(j):\quad\|z_m-z_n\|\to0,z_n\nrightarrow z$ – C-Star-W-Star Dec 3 '14 at 10:42
• I hope you don't mind if I unaccept your answer, do you? Still I gave you: (+1) – C-Star-W-Star Dec 19 '14 at 17:59

Let us choose orthonormal bases $\{e_i: i \in I\}$ of $\mathcal H$ and $\{f_j : j\in J \}$ of $\mathcal K$.

Assume $\mathcal K$ is finite dimensional, thus $J$ is finite. Let $x\in \mathcal H \hat\otimes \mathcal K$ and write it as $$x=\sum_{j\in J} x_j \otimes f_j \quad \text{with}\quad x_j =\sum_{i\in I} (e_i\otimes e_j, x) \, e_i\,,$$ where $(\,\cdot\,,\,\cdot\,)$ is the scalar product to be chosen linear in the second argument. This shows that $x\in \mathcal H\otimes \mathcal K$.

Conversely, if both are infnite dimensional, we can identify a subset of $I$ and $J$ with $\mathbb N$. Then $$y=\sum_{n\in \mathbb N} \frac1{n!}e_n\otimes f_n \in \mathcal H\hat\otimes \mathcal K.$$ Show that $y\not \in \mathcal H\otimes \mathcal K$.

• Oh I forgot to answer: That is precisely I'm looking for. I tried similar constructions but couldn't figure it out. Can you explain why: $\sum_{n\in\mathbb{N}}\frac{1}{n!}e_n\otimes f_n\notin\mathcal{H}\otimes\mathcal{K}$ – C-Star-W-Star Jan 3 '15 at 16:06
• because if $x\in \mathcal H\otimes \mathcal K$ it is finite sum of elementary tensors not infinite. (sorry for my weak english in advanced) – R.N Sep 3 '15 at 15:09
• I linked to this question and answer here math.stackexchange.com/q/2157955 as I can't prove the last part. – Tom Collinge Feb 23 '17 at 15:08