The problem asks whether, given von Neumann factors ( M ) and ( N ), the isomorphism ( L^\infty[0,1] \mathbin{\bar\otimes} M \cong L^\infty[0,1] \mathbin{\bar\otimes} N ) implies ( M \cong N ).

Key Steps:

1. Center of the Tensor Product: For von Neumann algebras ( A ) and ( B ), the center of the von Neumann tensor product ( A \mathbin{\bar\otimes} B ) is ( Z(A) \mathbin{...

Yes. In fact, for any diffuse abelian von Neumann algebra A (in particular A = L∞[0,1]), if M and N are factors and A ⊗ M ≅ A ⊗ N, then M ≅ N. No separability assumption on M or N is needed.

Here is a short proof.

• Let A = L∞[0,1]. Since M and N are factors, the center of A ⊗ M (resp. A ⊗ N) is A ⊗ 1. Thus any ∗-isomorphism θ: A ⊗ M → A ⊗ N maps the center onto the center and hence induces an a...

An "Exercise" on von Neumann Algebra Tensor Product

This is an interesting question about whether the isomorphism of tensor products implies the isomorphism of the original von Neumann factors when the separability assumption is removed.

Let me work through this step by step:

When and are von Neumann factors, the tensor products and $L^\infty[0,1] \bar{...

Yes. In fact one can replace L∞[0,1] by any diffuse abelian von Neumann algebra A :

If A is diffuse abelian and M , N are factors, then
A \bar⊗ M ≅ A \bar⊗ N  implies M ≅ N.

Separability of the pre-duals is not needed.
(The argument uses characters of the abelian algebra; they are not normal, so no measurable-field or direct-integral machinery is required.)

Proof.

1. Characters of A....

Yes – the factor that appears in the homogeneous direct integral L∞[0,1] ⊗̄ M completely determines the isomorphism class of the integral, no matter how big (or non–separable) the factors are. Hence, if

  L∞[0,1]  ⊗̄  M   ≅   L∞[0,1]  ⊗̄  N                                    (1)

then M ≅ N. A convenient way to see this is to use the standard central– (or direct–integral–) decomposition of ...

This is an excellent question. The statement is indeed true, even without the assumption of separable preduals. The proof avoids the machinery of direct integrals (disintegration theory), which is typically developed under separability assumptions, and instead relies on a more algebraic characterization of the factors within the tensor product.

The core idea is to characterize the factor insi...

<think> I need to tackle this problem about von Neumann algebra tensor products. The question is: if \( L^\infty[0,1] \mathbin{\bar{\otimes}} M \) is isomorphic to \( L^\infty[0,1] \mathbin{\bar{\otimes}} N \), does it follow that \( M \) is isomorphic to \( N \)? Here, \( M \) and \( N \) are von Neumann factors, and the tensor product is the spatial one.

First, recall that the bar over the tens...

This is indeed a subtle question, and you're right to be cautious. The answer is no in general - there exist non-isomorphic factors and such that .

Here's why this fails and what we know:

The Separable Case

As you noted, when and have separable preduals, the answer is yes. This follows from the disintegrat...

Yes – nothing “mystical’’ happens in the non‐separable case: tensoring a factor with the diffuse abelian algebra A=L^∞[0,1] is cancellative. In fact one shows that

A⊗̄M and A⊗̄N

are canonically W^*–bundles over the standard measure space [0,1], with every fibre isomorphic to M (resp. to N). Any *–isomorphism

φ:A⊗̄M → A⊗̄N

must carry the centre onto the centre, hence induces a meas...