Talk:Tensor product of Hilbert spaces

Notation for the Hilbertian tensor product

I find a little unsatisfactory that the completed tensor product is still denoted by L² ⊗ L² in the section "Examples and applications", giving the strange equation

$L^{2}(X)\otimes L^{2}(Y)=L^{2}(X\times Y).$

In books they sometimes use a hat. And one cannot say that only the completed tensor product will appear in discussions: the space of simple functions is also important.

Also I am not sure that the restriction of separability is needed in

It turns out that the set of linear combinations is in fact dense in L²(X × Y), if L²(X) and L²(Y) are separable

and similarly for L² ⊗ H. Bdmy (talk) 20:16, 30 November 2008 (UTC)

It bothers me too. I suggest ${\widehat {\otimes }}$ for the completed tensor product. For the second issue you raise, the proof I know of this result uses Fubini's theorem, which I believe requires sigma-finite measures (does this imply separability of L^2?), although there are probably still better results. siℓℓy rabbit (talk) 21:01, 30 November 2008 (UTC)

What happens I think is that any fixed element of the Hilbertian tensor product belongs to the product of separable subspaces on both sides, so that the general case should follow from the proof in the separable case. Bdmy (talk) 21:34, 30 November 2008 (UTC)

Well, if φ_i, i∈I, is an orthonormal basis of L²(X), and ψ_j, j∈J is an orthonormal basis of L²(Y), then certainly φ_i⊗ψ_j is an orthonormal basis of $L^{2}(X)\otimes L^{2}(Y)$. What I'm a little unclear on is whether the products φ_i(x)ψ_j(y) span a dense subspace of L²(X×Y). It seems to require the Fubini theorem, which in turn requires σ-finiteness. Now it may be that there is an entirely different sort of argument that there exists an isomorphism between these two spaces (e.g., by showing that they have the same cardinal as Hilbert dimension), but this seems to be a less natural sort of isomorphism. siℓℓy rabbit (talk) 21:57, 30 November 2008 (UTC)

First of all one has to make precise what σ-field is taken on X×Y; if it is the tensor product of σ-fields (what else?), then I think it is OK: the products φ_i(x)ψ_j(y) will span the products $1_{A}(x)1_{B}(y)$. Anyway this is a minor issue that can wait a few days to find a reference for... Bdmy (talk) 22:06, 30 November 2008 (UTC)

Yes, that makes sense. siℓℓy rabbit (talk) 22:13, 30 November 2008 (UTC)

I forgot to answer one of your questions: you may define the product probability measure on the uncountable product P of copies of the two-point space {-1, 1}, each copy equipped with the probability that gives mass 1/2 to each point. This is a probability but the L²(P) is not separable: the projection functions are an uncountable orthogonal system. I think that's all for today. Bdmy (talk) 22:22, 30 November 2008 (UTC)

Inner product?

What guarantees the function defined in the text will be an inner product? I don't see why it must be positive definite. 94.21.187.19 (talk) 14:09, 25 August 2010 (UTC)

weakly Hilbert Schmidt

what does that mean? It's not defined here, nor at Hilbert-Schmidt operator.--155.198.192.75 (talk) 14:20, 25 August 2011 (UTC)

You can find it in Kadison and Ringrose (search Google books and then search inside for "weak Hilbert Schmidt"). Reducing the definition from multilinear to bilinear case and adapting a bit to the notation used in this article the definitions go something like this:

A mapping $\phi :H_{1}\times H_{2}\to C$ is a Hilbert-Schmidt functional if it is a bounded bilinear functional (p. 127). A bounded linear mapping $L:H_{1}\times H_{2}\to K$ is weakly Hilbert-Schmidt if for all $v\in K$ the mapping $\phi _{v}=(u_{1},u_{2})\mapsto \langle L(u_{1},u_{2}),v\rangle$ is a Hilbert-Schmidt functional and $\|\phi _{v}\|\leq M\|v\|$ for some real number $M\geq 0$ (p. 131).

The definition of a Hilbert-Schmidt functional is incorrect. Not only does $\phi :H_{1}\times H_{2}\to C$ need to be bounded and bilinear we require that $\sum _{n,m}|\phi (e_{n},f_{m})|^{2}$ be finite (where $(e_{n})$ and $(f_{m})$ are bases of $H_{1}$ and $H_{2}$ respectively).

I that's ok maybe somebody wants to put this into the article? (ezander) 134.169.77.151 (talk) 13:00, 19 October 2011 (UTC)

I copied it into Hilbert-Schmidt operator. It looks good to me. User:Linas (talk) 12:38, 23 November 2013 (UTC)

However it looks, it is wrong and also not what is written in the book. 131.246.131.151 (talk) 11:08, 27 November 2014 (UTC)

I've put the definition of weakly Hilbert Schmidt functional in this article, it has already bemoved from the Hilbert Schmidt operator article (probably because it was wrong). I hope that everything is correct now. TSBM (talk) 12:06, 8 December 2014 (UTC)

Problematic formula

I am attempting to repair this problematic formula, which also occurs in Hilbert space:

$x^{*}\in H_{1}^{*}\to x^{*}(x_{1})x_{2}$

I know what its trying to say, but, as a formula, the above just doesn't make sense. So, for example ... obviously, the intent is that $x^{*}$ is an element of $H_{1}^{*}$ but then, what the heck does $x^{*}\to x^{*}(x_{1})x_{2}$ mean? The intent seemed to be to use a \mapsto not a \to, so that $x^{*}\mapsto x^{*}(x_{1})x_{2}$. But this doesn't really make sense (because there's no X_2 on the left-hand side) By contrast, a map $x_{1}\otimes x_{2}\mapsto x_{1}^{*}(-)x_{2}$ does make sense, notationally.

The alternative parse is that $H_{1}^{*}\to x^{*}(x_{1})x_{2}$ but this doesn't make much sense either. By contrast, $H_{1}^{*}\to H_{2}$ does make sense.

What we really want to write is that there's a map $H_{1}\otimes H_{2}\to (H_{1}^{*}\to H_{2})$ which $x_{1}\otimes x_{2}\mapsto x_{1}^{*}(-)x_{2}$ for $x_{1}\otimes x_{2}\in H_{1}\otimes H_{2}$ and $x_{1}^{*}(-)x_{2}\in (H_{1}\to H_{2})$

That's why I made the edits that I did, but any better fix would be fine with me. Anyway, the error is prehistoric; it was there when sillyrabbit copied it over in 2008 from another article. I'm surpirsed it hasn never been fixed. User:Linas (talk) 14:06, 23 November 2013 (UTC)

Ohhhh. I get it it was trying to say this:

${\begin{aligned}x_{1}\otimes x_{2}:H_{1}^{*}&\to H_{2}\\x^{*}&\mapsto x^{*}(x_{1})x_{2}\\\end{aligned}}$