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:


The same question is here: https://math.stackexchange.com/questions/940238/why-does-infinite-tensor-product-associated-with-some-vectors-in-the-operator-al
 A: 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.  
