Non-commutative geometry from von Neumann algebras? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T08:13:16Z http://mathoverflow.net/feeds/question/3150 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/3150/non-commutative-geometry-from-von-neumann-algebras Non-commutative geometry from von Neumann algebras? Dave Penneys 2009-10-28T21:48:34Z 2010-03-08T13:29:03Z <p>The Gelfand transform gives an equivalence of categories from the category of unital, commutative $C^*$-algebras with unital $*$-homomorphisms to the category of compact Hausdorff spaces with continuous maps. Hence the study of $C^*$-algebras is sometimes referred to as non-commutative topology.</p> <p>All diffuse commutative von Neumann algebras acting on separable Hilbert space are isomorphic to $L^\infty[0,1]$. Hence the study of von Neumann algebras is sometimes referred to as non-commutative measure theory.</p> <p>Connes proposed that the definition of a non-commutative manifold is a spectral triple $(A,H,D)$. From a $C^*$-algebra, we can recover the "differentiable elements" as those elements of the $C^*$-algebra $A$ that have bounded commutator with the Dirac operator $D$. </p> <p>What happens if we start with a von Neumann algebra? Does the same definition give a "differentiable" structure? Is there a way of recovering a $C^*$-algebra from a von Neumann algebra that contains the "differentiable" structure on our non commutative measure space? This would be akin to our von Neumann algebra being $L^\infty(M)$ for $M$ a compact, orientable manifold (so we have a volume form). Or are von Neumann algebras just "too big" for this?</p> <p>One of the reasons I am asking this question is Connes' spectral characterization of manifolds (<a href="http://arxiv.org/abs/0810.2088v1" rel="nofollow">arXiv:0810.2088v1</a>) which shows we get a "Gelfand theory" for Riemannian manifolds if the spectral triples satisfy certain axioms. Connes starts with the von Neumann algebra $L^\infty(M)$ instead of the $C^*$-algebra $C(M)$.</p> http://mathoverflow.net/questions/3150/non-commutative-geometry-from-von-neumann-algebras/3152#3152 Answer by Yemon Choi for Non-commutative geometry from von Neumann algebras? Yemon Choi 2009-10-28T22:10:13Z 2009-10-28T22:10:13Z <p>You definitely need some extra structure on your von Neumann algebra, but I'm not quite sure what you're asking for. Intuitively I would think that just as different topological spaces share the same measure space structure, trying to extract NC-topological information out of a von Neumann algebra is going to need extra structure. (For instance, no one does topological K-theory of von Neumann algebras as far as I know.)</p> <p>I see that on page 7 of that Connes paper, he shows that the WOT-closure A'' does remember the original algebra A <em>if</em> extra data are given (the Dirac operator and its interaction with A). </p> <p>Although it's probably not what you want: if you're looking at group von Neumann algebras and looking at the "geometry of the dual group", then the original group can be recovered from a suitable coproduct on the original (Hopf) von Neumann algebra. This is vaguely on the lines of Weil's theorem that "essentially" recovers a locally compact group and its Haar measure from a measurable group.</p> http://mathoverflow.net/questions/3150/non-commutative-geometry-from-von-neumann-algebras/3279#3279 Answer by Dmitri Pavlov for Non-commutative geometry from von Neumann algebras? Dmitri Pavlov 2009-10-29T15:57:22Z 2009-10-29T15:57:22Z <p>First let me note that one does not need additional conditions of diffuseness and separability in the statement of Gelfand-Neumark theorem for commutative von Neumann algebras. In fact, the category of commutative von Neumann algebras is contravariantly equivalent to the category of localizable measurable spaces. The definition of measurable space includes the σ-ideal of null sets (i.e., sets of measure 0) as an additional datum. Localizable here means one of several equivalent properties (Radon-Nikodym, Riesz representation, completeness of Boolean algebra of measurable sets etc.). It is an extremely natural property in the sense that one cannot prove a single interesting theorem in measure theory without requiring the measure space to be localizable.</p> <p>As for the main question, I doubt that Connes' definition makes any sense in the world of von Neumann algebras. Look at the commutative case. The category of commutative von Neumann algebras (thus the category of localizable measurable spaces) is contravariantly (resp. covariantly) equivalent to the category of hyperstonean spaces and hyperstonean maps (a subset of continuous maps) by Gelfand-Neumark duality. Connes' definition in the commutative case equips a compact Hausdorff topological space with a structure of an oriented Riemannian smooth manifold. But hyperstonean spaces cannot have any smooth structures because they are not locally Euclidean.</p> <p>Yemon Choi's suggestion sounds more reasonable to me, i.e., one needs to introduce additional data to recover the original topological space from its hyperstonean version (i.e., recover a topological space from its measurable space).</p> http://mathoverflow.net/questions/3150/non-commutative-geometry-from-von-neumann-algebras/3286#3286 Answer by Matthew Daws for Non-commutative geometry from von Neumann algebras? Matthew Daws 2009-10-29T16:48:23Z 2009-10-29T16:48:23Z <p>I guess a vaguely related idea is what happens for Locally Compact Quantum Groups (a la the Kustermans + Vaes axiomatisation). Here you can define everything using von Neumann algebras (A von Neumann algebra with a normal coproduct which has left and right invariant weights: the commutative case is L^\infty(G) with the left and right Haar measures). However, it's then part of the theory that you always get a (unique) C*-algebra version of the object (in the commutative case, you recover C_0(G)). Conversely, starting the the C*-algebra axiomatisation (which is more complicated) you can build a von Neumann algebra. Thus, here, the C* and von Neumann worlds are equivalent.</p> <p>This is exactly what Dmitri (and Yemon) mention: it's a process that allows you to cover the topological structure of the group G from knowing it's measure space L^\infty(G) (as long as you have these extra bits of data floating about).</p> http://mathoverflow.net/questions/3150/non-commutative-geometry-from-von-neumann-algebras/3293#3293 Answer by Urs Schreiber for Non-commutative geometry from von Neumann algebras? Urs Schreiber 2009-10-29T17:25:35Z 2009-10-29T17:25:35Z <p>People did consider the notion of <a href="http://adsabs.harvard.edu/abs/2000math.....12233B" rel="nofollow">vonNeumann spectral triple</a>.</p> <p>(also referenced at <a href="http://ncatlab.org/nlab/show/spectral+triple" rel="nofollow">nLab:spectral triple</a>)</p> http://mathoverflow.net/questions/3150/non-commutative-geometry-from-von-neumann-algebras/17469#17469 Answer by Jan Jitse Venselaar for Non-commutative geometry from von Neumann algebras? Jan Jitse Venselaar 2010-03-08T13:29:03Z 2010-03-08T13:29:03Z <p>Alain Connes proves in the paper "A unitary invariant in Riemannian Geometry" ( <a href="http://arxiv.org/abs/0810.2091" rel="nofollow">http://arxiv.org/abs/0810.2091</a> ) that all information on a Riemannian geometry can be decoded from the spectrum of a Dirac operator and the relative position of two commutative von Neumann algebras. I don't know how you would translate this into a statement about noncommutative spaces however, I'll have to look at the paper in more detail.</p>