# Why does an infinite tensor product depend on some vectors for operator algebras?

I have read that, in the definition of the infinite tensor product of operator algebras such as ${\rm C}^\ast$-algebras and ${\rm W}^\ast$-algebras, every factor in the product is associated with a vector (or a state $s$); and if the vectors (or states) are different, the infinite tensor products formed in this way may also be different. For example, we have non-isomorphic (mainly type III) factors arising as the infinite tensor product of ${\rm I}_2$ factors with different states.

Why are these infinite tensor products different? Comparing with the construction of a finite tensor product, what is the difference?

Here is the definition from Blackadar's book Operator algebras:

• I don't understand the downvotes: it seems that the OP is asking for explanation of Blackadar's remark in III.3.1.2 – Yemon Choi Sep 24 '14 at 10:44
• I vote to close because: 1. this is not a research level question. 2. The OP should not copy the context directly from the text book, he should explain the question according to his own understandings. 3. As he has posted the question on stackexchange, it should not appear on MO. – YHBKJ Sep 26 '14 at 15:22
• In the next page of the book you mention he says why they can be different (III.3.1.4 and III.3.1.7) – Caleb Eckhardt Sep 27 '14 at 23:18
• @Strongart You didn't read the book carefully enough, as pointed out by Eckhardt above, you question is actually trivial! – YHBKJ Sep 28 '14 at 12:38
• @Caleb Eckhardt III.3.1.4 and III.3.1.7 are some examples I have said in the first part "we have nonisomorphism (mainly type III) factors from the infinite tensor product of I2 factors with different states". It is a single example, but I want to find an explaination which is a way to this example. – Strongart Sep 28 '14 at 13:51

The point is the following. If one takes a reduced tensor product of a pair of $W^*$-algebras $A_1,A_2$ with states $\phi_1$, $\phi_2$, the resulting tensor product does not depend on the choice of $\phi_i$ as long as they give rise to equivalent GNS representations of $A_i$ (e.g. the states are faithful). Note, however, that it is not as straightforward as one might think to write an isomorphism between $(A_1,\phi_1)\otimes (A_2,\phi_2)$ and $(A_1,\phi'_1)\otimes (A_2,\phi'_2)$, since in general there is no unitary from $L^2(A_i,\phi_i)$ to $L^2(A_i,\phi'_i)$ which intertwines the actions of $A_i$. For $A_i$ finite-dimensional, there is an invertible linear map (similarity) $S_i : L^2(A_i,\phi_i)\to L^2(A_i,\phi'_i)$ with this property, but it is not isometric. There are analogs of this more general situations, too.
For infinite tensor products, the situation is even more delicate. Indeed, the naive way to prove isomorphism between $\bigotimes_i (A_i,\phi_i)$ and $\bigotimes_i (A_i,\phi'_i)$ even for finite-dimensional $A_i$ runs into the problem that similarities $S_i$ do not give a bounded map between the infinite tensor products of Hilbert spaces: $\bigotimes S_i$ may not exist.
It turns out, for example, that for $A_i\cong M_{2\times 2}$, the algebras of $2\times 2$ matrices and $\phi_i(x) = Tr(d x)$, with $d$ a positive operator with eigenvalues $\lambda_1,\lambda_2$ satisfying $\lambda_1+\lambda_2=1$, $\lambda_1/\lambda_2 = \lambda\in (0,1]$, the infinite von Neumann tensor product $\bigotimes (A_i,\phi_i)$ depends on $\lambda$. The resulting von Neumann algebra is called an ITPFI factor, and is the unique type $III_\lambda$ (if $\lambda\neq 1$) or type $II_1$ (if $\lambda=1$) hyperfinite factor.
• Thanks, S$_i$ maybe is not bounded, that is the key, but does any specific examples of unbounded S$_i$? – Strongart Oct 19 '14 at 6:31
• If $A_i$ are finite-dimensional, $S_i$ are bounded, for each $i$. In the example given in the answer, $\bigotimes_i S_i$ is unbounded even when each $S_i$ is. If $A_i$ are e.g. II$_1$ factors, and $\phi_i(x)=\tau(d_i x)$, $\phi'_i(x)=\tau(d_i' x)$ then you have something like $S_i = d_i^{1/2}(d_i')^{-1/2}$ and this of course may be unbounded even for a fixed $i$. – user60602 Oct 19 '14 at 22:48