# Infinite tensor product of states

Tensor products of finite number of different objects are always well described in the literature. However, the situation of infinite tensor products seems to be much tougher. Even in the simplest case, the infinite tensor product of Hilbert spaces, we have to be patient to find a good references (apart from the original article by von Neumann). Can one explain or give the references to how the infinite tensor products of infinitely many (normal) states is defined. Is there an analogue to the "stabilizing sequence" and all but finite number of states are e.g. vector states?

If we consider an infinte tensor product of von Neumann algebras, it's predual seem to correspond to the space which consists of infinite tensor product of states, right?

Last question is a bit vague, but I want to ask it. Say that we have the predual of the infinte tensor product of the same von Neumann algebra $M$. We take the state $\omega \in M_{*}$, can we view $\omega^{\otimes_{\sigma-\text{weak}} n}$ for some $n \in \mathbb{N}$ an element of the predual of the infinite tensor product of $M$ ? When the limit of this expression makes sense I mean $\lim_{n \to \infty} \omega^{\otimes n}$, for instance when it is an element of the predual of the infinite tensor product of $M$ and how does it possibly look.

I'm studying "INFINITE TENSOR PRODUCTS OF VON NEUMANN ALGEBRAS" by Nakagami, so maybe by the end I will be able to answer my question, but still I will be grateful for any help.

• Did you already check out: mathoverflow.net/questions/11767/infinite-tensor-products – Suvrit Mar 23 '14 at 15:36
• Yes, I looked there. They describe more algebraic setting and I'm sure it is possible to find a nice reference (book) which would answer my question directly. The article which I mentioned (in my question) seems to be alright, however it would be better to see something more recent. – Glacier Mar 23 '14 at 15:46

## 1 Answer

If you're working at the level of Hilbert spaces, I think the more usual procedure nowadays is to fix a distinguished unit vector $u_i$ in each Hilbert space $H_i$, then define the tensor product to be the Hilbert space generated by all elementary tensors $\bigotimes v_i$ such that $v_i \in H_i$ for all $i$ and $v_i = u_i$ for all but finitely many $i$. This is a direct limit of the tensor products over finite sets of indices in a fairly obvious way. In physical models the $u_i$ would typically be ground states. This definition seems more usable than the old "full" infinite tensor product defined by von Neumann which is typically nonseparable. I give details in Section 2.5 of my book Mathematical Quantization.

It seems that the infinite tensor product defined by Nakagami in general will not act faithfully on the Hilbert space defined above. I am not sure what to make of this fact. An analogous construction of an infinite tensor product of von Neumann algebras can be given on my Hilbert space tensor product. (Say each $M_i$ acts on $H_i$, then each $M_i$ acts on $\bigotimes H_i$ and we let $\bigotimes M_i$ be the von Neumann algebra generated by all the $M_i$ within $B(\bigotimes H_i)$. But since my Hilbert space is just a piece of the full tensor product, I expect my von Neumann algebra tensor product to be contained in Nakagami's. So this gets kind of interesting!

I suppose the predual of $\bigotimes M_i$ as I am defining it will be a direct limit of the maximal tensor products of finite sets of preduals of the $M_i$, with the embedding maps given by tensoring with the vector states coming from the $u_i$.

For your last question, in my setup this certainly works if $\omega$ is the vector state coming from $u_i$, and in Nakagami's definition you'll be okay as long as $\omega$ is any vector state. So it seems like you want to be sure $M$ is in standard form to begin with, so that every normal state is a vector state, and then Nakagami's construction should do what you want. (Otherwise I imagine it doesn't.)

• Concretely, given a ${\rm II}_1$ factor $M$, is there a state on $M^{\otimes \infty}$ generating a ${\rm II}_1$ factor? – Sebastien Palcoux Aug 3 '14 at 5:14
• And, is there a state generating a ${\rm III}_1$ factor? In this case, what's the core? (I think to the example $M=L(\mathbb{F}_2)$). – Sebastien Palcoux Aug 3 '14 at 5:16