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My question is based on the following vague belief, shared by many people: It should be possible to use von Neumann algebras in order to define the cohomology theory TMF (topological modular forms) in the same way one uses Hilbert spaces in order to define topological K-theory. More precisely, one expects hyperfinite type $\mathit{III}$ factors to be the analogs of (separable) infinite dimensional Hilbert spaces.

Now, here is a fundamental difference between Hilbert spaces and type $\mathit{III}$ factors: The category of Hilbert spaces has two monoidal structures: direct sum $\oplus$, and tensor product $\otimes$, and both of them preserve the property of being an infinite dimensional Hilbert space.

In von Neumann algebras, the tensor product of two hyperfinite type $\mathit{III}$ factors is again a hyperfinite type $\mathit{III}$ factor, but their direct sum isn't (it's not a factor).

Hence my question: are there other monoidal structures on the category of von Neumann algebras that I might not be aware of? More broadly phrased, how many ways are there of building a new von Neumann algebra from two given ones, other than tensoring them together?

Ideally, I would need something that distributes over tensor product, and that preserves the property of being a hyperfinite type $\mathit{III}$ factor... but that might be too much to ask for.

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So what's so bad about relaxing "factor" to "direct sums of finitely many factors"? I rarely see anywhere that factor-ness is really necessary theoretically (this is closely related to the fact that people often assume that the unit object in a tensor category is simple, but the theory mostly works fine without that assumption). –  Noah Snyder May 5 '10 at 16:38
Non-factors have very few automorphisms: Assuming $M$ and $N$ are factors, an automorphism of $M\oplus N$ either preserves $M$ and $N$, or permutes the two subalgebras. This is in sharp contrast with the situation in Hilbert spaces. –  André Henriques May 5 '10 at 16:59
So there's some theorem about type III factors which tells you that you never need to look at bimodules and can instead always look at endomorphisms. I don't know how that theorem is proved, but does it break down in the non-factor case? If so could allowing bimodules as morphisms help you here? –  Noah Snyder May 5 '10 at 17:02
Even with bimodules as morphisms, $M\oplus N$ has very few automorphisms. An invertible $(M\oplus N)$-$(M\oplus N)$-bimodule is always of one of the following two sorts: 1) a direct sum of an invertible $M$-$M$-bimodule and an invertible $N$-$N$-bimodule or 2) a direct sum of an invertible $M$-$N$-bimodule and an invertible $N$-$M$-bimodule. –  André Henriques May 5 '10 at 17:48
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4 Answers

up vote 4 down vote accepted

The category of von Neumann algebras W* admits a variety of monoidal structures of three distinct flavors.

(1) W* is complete and therefore you have a monoidal structure given by the categorical product.

(2a) W* is cocomplete and therefore you also have a monoidal structure given by the categorical coproduct.

(2b) I suspect that there is also a “spatial coproduct”, just as there is a categorical tensor product and a spatial tensor product (see below). The spatial coproduct should correspond to a certain central projection in the categorical coproduct. Perhaps the spatial coproduct is some sort of coordinate-free version of the free product mentioned in Dmitri Nikshych's answer.

(3a) For any two von Neumann algebras M and N consider the functor F from W* to Set that sends a von Neumann algebra L to the set of all pairs of morphisms M→L and N→L with commuting images. The functor F preserves limits and satisfies the solution set condition, therefore it is representable. The representing object is the categorical tensor product of M and N.

(3b) There is also the classical spatial tensor product. I don't know any good universal property that characterizes it except that there is a canonical map from (3a) to (3b) and its kernel corresponds to some central projection in (3a). Perhaps there is a nice description of this central projection.

Since your monoidal structure is of the third flavor and you don't want a monoidal structure of the first flavor, I suggest that you try a monoidal structures of the second flavor. I suspect that the spatial coproduct of two factors is actually a factor. You are lucky to work with factors, because in the commutative case 2=3, in particular 2a=3a and 2b=3b.

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Thank you Dmitri for your nice survey. (1) is the direct sum. (2a) and (3a) will typically be non-separable, and with very big center: I don't like those features. (2b) sounds interesting. Does such a construction actually exist? A priori, the free product depends on a chosen state. Can it be made not to depend on the state? Equivalently: does varying the states on A and B change the isomorphism type of the diagram A --> A*B <-- B? –  André Henriques Aug 4 '10 at 13:54
I'll think about this issue. RIght now I can only conjecture that the situation might be similar to the one of Connes' fusion, which was originally defined using a choice of weight, but later a weight-independent construction was found. –  Dmitri Pavlov Aug 5 '10 at 2:15
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Let me flesh out my comment above a little bit. Rather than answering your question I'm going to try to convince you that you shouldn't be asking this question. In particular, I claim that assuming that a von Neumann algebra is a factor is typically unmotivated and the right assumption is that it have finite dimensional center. Type III hyperfinite von Neumann algebras with finite dimensional center are perfectly happy and have a tensor product and a direct sum with the right properties.

In particular, what I'd like to claim (but which I'm not going to try to prove) is that if you took your recent paper replaced "factor" everywhere with "von Neumann algebra" then dualizability would force you to restrict to the case of finite dimensional centers, but not to factors.

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Here's why using non-factors looks like a bad idea to me (recall that my goal is to construct the cohomology theory TMF). So there's an obvious functor from the category of vN algebras with finite dimensional center to Sets. It sends an algebra to its set of minimal central projections. Now, it is well known that the cohomology theory associated to (finite) sets is the sphere spectrum. So a cohomology theory built from non-factors will probably have a map to the sphere spectrum. But TMF is known not to have a map to the sphere spectrum. That's why I don't want non-factors. –  André Henriques May 9 '10 at 18:52
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There is a construction of free product of von Neumann algebras due to Voiculescu. It is very popular among operator algebraists. Definitions can be found, e.g., in Lance Barnett, Free Product Von Neumann Algebras of Type III. I wonder if this helps.

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Suppose that there was such a category. Then, all objects would isomorphic to $R$ anyway and the question is how sum and tensor product act on the morphisms. The natural choice of morphisms is the set of endomorphisms of the hyperfinite factor. Now, pick a unital inclusion $\iota :R \oplus R \subset R$ and and isomorphism $\mu \colon R \cong R \otimes R$ and define $$(\phi \oplus \psi)(x) := \iota(\phi(x) \oplus \psi(x)) \quad \mbox{and} \quad (\phi \otimes \psi)(x) := \mu( (\phi \otimes \psi)(\mu^{-1}(x))).$$

The question is now how sum and tensor product behave. This of course requires a bit of work (and may depend on the choice of $\mu$ and $\iota$).

However, let us look more concretely. Assume for the moment that $R = \otimes_{n \in {\mathbb N}} M_2 {\mathbb C}$ (with some fixed state on $M_2 {\mathbb C}$). Then, it seems that the inclusion ${\mathbb C}^{\oplus 2} \subset M_2 {\mathbb C}$ in the first factor gives some $\iota$ and a bijection between ${\mathbb N}$ and ${\mathbb N} \times {\mathbb N}$ yields some map $\mu$. It seems straightforward that this should give a bimonoidal category with the required distributivity isomorphisms.

Added August 15 to incorporate Andre's remarks:

The choice of $\iota$ is basically a choice of an infinite projection $p \in R$ and isomorphisms $\phi : R \to pRp$ and $\psi : R \to (1-p)R(1-p)$. Since all projections in $R$ are equivalent (we are in type $III$) i.e. $p \sim 1-p$, $\phi(p) \sim p$ and $\psi(p) \sim p$ the sum should be coherently associative. I think that it is a first step to provide an (inner) isomorphism of $R$ which maps the triple $p, \psi(p),\psi(1-p)$ to $\phi(p),\phi(1-p),1-p$. I have to really think about it, but each triple is a triple of pairwise orthogonal projections in $R$ which sums up to one. I think, that up to unitary conjugacy, there is only one such triple (in the type $III$-setting).

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I'll rephrase your construction in more invariant terms. Given M and N, you construct a new factor M[+]N as follows. First, you pick a Morita equivalence between M and N (given isos M=R and N=R, that's what you get from \iota). That Morita eq. can be interpreted as a choice of 2x2 matrix algebra, with upper-left and lower-right corners given by M and N respectively. Then define M[+]N to be that 2x2 matrix algebra. It comes with two nonunital inclusions M --> M[+]N and N --> M[+]N. I would be surprised if this construction was (coherently) associative (M [+] N) [+] P =?= M [+] (N [+] P). Is it? –  André Henriques Aug 15 '10 at 0:15
Maybe there is a way of picking \iota so that the operation [+] is coherently associative... Is there one? Can one then also arrange it so that [+] distributes over (x) ? It would be great if this worked... –  André Henriques Aug 15 '10 at 7:05
My reply was to long to for a comment. I edited the original answer instead. –  Andreas Thom Aug 15 '10 at 10:10
I think that there is still some misunderstanding. The $\sum$-operation applied to two automorphisms would not be an automorphisms. (This is analogous to the fact that the sum of two line-bundles is not a line-bundle.) However, as far as I understand, your construction on the level of algebras would always yield that the sum of two automorphisms is an automorphism. –  Andreas Thom Aug 15 '10 at 10:52
Picking an inner automorphism of R that maps the triple p,psi(p),psi(1−p) to phi(p),phi(1−p),1−p is always possible. But picking one that satisfies the pentagon axiom is much more tricky... But maybe it's possible. –  André Henriques Aug 15 '10 at 22:16
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