Hochschild/Cyclic Homology of von Neumann Algebras: Useless? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-18T21:31:36Z http://mathoverflow.net/feeds/question/786 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/786/hochschild-cyclic-homology-of-von-neumann-algebras-useless Hochschild/Cyclic Homology of von Neumann Algebras: Useless? Dave Penneys 2009-10-16T19:28:36Z 2010-08-08T10:59:12Z <p>Hochschild homology gives invariants of (unital) $k$-algebras for $k$ a unital, commutative ring. If we let our algebra $A$ be the group ring $k[G]$ for $G$ a finite group, we get group homology. There are plenty of other connections to homological algebra. If we use cyclic homology, there are connections to geometry and topology involving the Chern character.</p> <p>Von Neumann algebras are complex algebras, so we can take their Hochschild and cyclic homologies. When I have asked experts in the fields of von Neumann algebras and non-commutative geometry about what you get, I usually hear some approximation of the following: "There's also analysis in von Neumann algebras, so I wouldn't expect an algebraic invariant like Hochschild or cyclic homology to tell you anything useful."</p> <p>Although this answer makes some sense, I find it very displeasing and cryptic. Why shouldn't it tell you something? Is there some way to make "it doesn't tell you anything" quantitative? Is there an example of a von Neumann algebra with nontrivial Hochschild or cyclic homology (different from that of the complex numbers)?</p> <p>EDIT: After reading the responses so far, I should specify that I really want to know if there is a $II_1$-factor with nontrivial Hochschild or cyclic (co)homology.</p> http://mathoverflow.net/questions/786/hochschild-cyclic-homology-of-von-neumann-algebras-useless/790#790 Answer by Grétar Amazeen for Hochschild/Cyclic Homology of von Neumann Algebras: Useless? Grétar Amazeen 2009-10-16T19:57:30Z 2009-10-16T19:57:30Z <p>I am certainly not an expert, but I guess that when people say </p> <blockquote> <p>There's also analysis in von Neumann algebras, so I wouldn't expect an algebraic invariant like Hochschild or cyclic homology to tell you anything useful.</p> </blockquote> <p>they mean: "Von Neumann algebras are complex algebras with extra structure, and it is this extra structure that make them interesting. Therefore knowing something about Hochschild cohomology is not very interesting to us because it doesn't tell us anything about this extra structure."</p> <p>It might very well tell you something interesting, I don't know, but it still feels a bit strange to ignore this extra structure.</p> http://mathoverflow.net/questions/786/hochschild-cyclic-homology-of-von-neumann-algebras-useless/799#799 Answer by Yemon Choi for Hochschild/Cyclic Homology of von Neumann Algebras: Useless? Yemon Choi 2009-10-16T20:45:07Z 2009-10-16T20:45:07Z <p>As someone who works on the continuous (bounded) cohomology of Banach algebras: I think the quote is a way of saying "we don't really know". There are certainly questions which start off with extra continuity/boundedness requirements and turn out to be rephraseable in the "purely algebraic" module categories - L^2 cohomology of discrete groups is one, if I remember correctly, Farber and Lueck have written about this.</p> <p>I'm prepared to believe what's said about algebraic modules over II_1 factors, although I worry/suspect that one has to work with modules over a nastier algebra. Is this the case?</p> <p>If you want to <em>compute</em> Hochschild cohomology (with coefficients in the algebra, I guess you mean) then this is just <em>hard</em>. It is not easy to find well-defined projective resolutions of Banach objects (if you pass to some dense subalgebra or submodule then more tools are available, this seems to be the approach adopted in much of NCG a la Connes). </p> <p>In fact, given a non-injective von Neumann algebra M (something like a free group factor will do) then there exists an M-bimodule X, which is a Banach space and on which M acts continuously, and a continuous derivation M --> X which is not inner. Which sort of answers your question, though probably not in the sense you meant...</p> <p>If one restricts the module categories then there is a whole theory of Tor and Ext for Banach modules, due to Helemskii - though it only works on a relatively small class of short exact sequences. However, for von Neumann algebras things are still hard (see work of Christensen, Sinclair, Smith and others).</p> http://mathoverflow.net/questions/786/hochschild-cyclic-homology-of-von-neumann-algebras-useless/1008#1008 Answer by Yemon Choi for Hochschild/Cyclic Homology of von Neumann Algebras: Useless? Yemon Choi 2009-10-18T05:59:55Z 2009-10-18T05:59:55Z <p>Some further thoughts: the most striking results I know of on "purely algebraic cyclic/Hochschild homology" are due to Wodzicki, see e.g.</p> <p>Homological properties of rings of functional-analytic type, Proceedings of the National Academy of Sciences USA 87 (1990), 4910-4911</p> <p>which states that stable C*-algebras have trivial cyclic homology. Obviously this doesn't answer your II_1 factor question...</p> <p>Also: your remark that <em>in some cases, we can ignore the analysis and make the situation a bit simpler</em> confuses me a little. To get anywhere with cyclic or Hochschild homology, we need to do some kind of comparison of resolutions, or construction of contracting homotopies, or something like that. My intuition - but I don't work much on operator algebras, so I could well be wrong here - is that a von Neumann algebra is such a big object we usually can only get a handle on it by looking at suitable subsets which generate its unit ball in the WOT/SOT. So for group von Neumann algebras, one tries to see what's going on for translations, and thence to deduce more general results by exploiting w*-w* continuity; or else use projections and approximation arguments. If we go to a purely algebraic category, then it is no longer sufficient to define things on dense subsets - one really needs a global definition, one really needs to verify that certain putative identities are satisfied by each element of the von Neumann algebra.</p> <p>Sorry if that's a bit waffly. I think my point is that imposing continuity restrictions actually makes things <em>easier</em>, because - intuitively - more things are going to be projective/injective/flat relative to one's restricted class of short exact sequences. This is why, for instance, we know that $H^n_{cb}(M,M)=0$ for any von Neumann algebra M, but why the analogous claim without the 'cb' is open and back-breaking. In a similar vein, if you work in a restricted category then one does indeed get some known instances of homological non-triviality (though at the level of modules, not at the level of cyclic homology):</p> <p>M. E. Polyakov, <a href="http://www.springerlink.com/content/p0u475547t380814/" rel="nofollow">An Example of a Spatially Nonflat von Neumann Algebra</a></p> <p>I should also say that the Hilbert module stuff you mention doesn't really connect to your original question about cyclic (co)homology. It's interesting, and I think more has been done, but it's just <em>different</em> - so if that's what interests you, cyclic and Hochschild homology may be something of a distraction.</p> http://mathoverflow.net/questions/786/hochschild-cyclic-homology-of-von-neumann-algebras-useless/1255#1255 Answer by Dmitri Pavlov for Hochschild/Cyclic Homology of von Neumann Algebras: Useless? Dmitri Pavlov 2009-10-19T18:26:35Z 2009-10-19T18:26:35Z <p>I am surprised that nobody mentioned the book by Sinclair and Smith “Hochschild Cohomology of von Neumann algebras” so far. If I remember it correctly, they claim that nobody knows whether there is a von Neumann algebra with a non-trivial Hochschild cohomology. But the book is 14 years old, so this problem might be solved already.</p> http://mathoverflow.net/questions/786/hochschild-cyclic-homology-of-von-neumann-algebras-useless/1263#1263 Answer by Yemon Choi for Hochschild/Cyclic Homology of von Neumann Algebras: Useless? Yemon Choi 2009-10-19T19:13:29Z 2009-10-25T21:09:56Z <p>This is in response to Dmitri's remark. The reason I didn't bring up Sinclair &amp; Smith's book (which is where I first started trying to learn Hochschild cohomology) is that it deals with <strong>continuous cochains</strong> with coefficients in the algebra. I understood DP's original question as being about purely algebraic cyclic cohomology, which involves not-necessarily continuous cochains taking values in the dual of the algebra (not the algebra itself). I hope this addresses your "surprise". FWIW, I'm more interested in the Sinclair &amp; Smith setting myself, but I don't think that's what DP was asking about - though I may have misunderstood.</p> <p>And yes, there is still no example of a von Neumann algebra M for which H^n(M,M) is nonzero for some n > 1; while the vanishing or otherwise of H^2(L(F_2),L(F_2)) is still unknown...</p> http://mathoverflow.net/questions/786/hochschild-cyclic-homology-of-von-neumann-algebras-useless/12664#12664 Answer by Claude Schochet for Hochschild/Cyclic Homology of von Neumann Algebras: Useless? Claude Schochet 2010-01-22T17:14:09Z 2010-01-22T17:14:09Z <p>The question is not well-posed. There are various versions of cyclic theory (for instance) which differ according to continuity conditions that are assumed. In Connes' original IHES papers he deals with both discrete (useful for arbitrary rings) and topological (useful in the $C^\infty$-setting.) </p> <p>The basic problem is that cyclic theory is very very sensitive. Consider the following example (using Connes' topological cyclic theory.) Let M denote a compact smooth manifold with a smooth foliation. Then there are three operator algebras that you can associate with the situation;</p> <p>a) $C(M)$, the $C^\ast$-algebra of continuous complex functions on $M$.</p> <p>b) $C^\infty (M)$, the smooth functions on $M$.</p> <p>c) $C_\tau ^\infty (M)$, the continuous functions on $M$ which are smooth in the leaf directions.</p> <p>The cyclic cohomology of these three rings are all different typically (e.g. for the Kronecker flow on the torus). </p> <p>a) leads to measures on $M$;</p> <p>b) leads to deRham cohomology on $M$;</p> <p>c) leads to "tangential cohomology" on $M$ (cf. my book with Cal Moore).</p> <p>Hope this helps.</p> <p>CS</p> http://mathoverflow.net/questions/786/hochschild-cyclic-homology-of-von-neumann-algebras-useless/34913#34913 Answer by Andreas Thom for Hochschild/Cyclic Homology of von Neumann Algebras: Useless? Andreas Thom 2010-08-08T10:53:30Z 2010-08-08T10:59:12Z <p>There is important work by Alain Connes and Dimitri Shlyakhtenko (<a href="http://www.ams.org/mathscinet/search/publdoc.html?arg3=&amp;co4=AND&amp;co5=AND&amp;co6=AND&amp;co7=AND&amp;dr=all&amp;pg4=AUCN&amp;pg5=TI&amp;pg6=ALLF&amp;pg7=ALLF&amp;pg8=ET&amp;r=1&amp;review_format=html&amp;s4=shlyakhtenko%20and%20connes&amp;s5=&amp;s6=&amp;s7=&amp;s8=All&amp;vfpref=html&amp;yearRangeFirst=&amp;yearRangeSecond=&amp;yrop=eq" rel="nofollow">see here</a>). They come up with a definition of $\ell^2$-homology for finite von Neumann algebras and define numerical invariants called $\ell^2$-Betti numbers for finite von Neumann algebras. This approach builds on the more classical theory of $\ell^2$-invariants developed by Atiyah, Cheeger-Gromov and also Lück. So far, there are no really interesting computations. </p> <p>However, it seems that this homology group (or some variant of it) is more likely to be able to detect the differences among free group factors. Of course, this is only speculation. There is also a cohomological picture (<a href="http://www.ams.org/mathscinet/search/publdoc.html?arg3=&amp;co4=AND&amp;co5=AND&amp;co6=AND&amp;co7=AND&amp;dr=all&amp;pg4=AUCN&amp;pg5=TI&amp;pg6=ALLF&amp;pg7=ALLF&amp;pg8=ET&amp;review_format=html&amp;s4=thom&amp;s5=&amp;s6=&amp;s7=&amp;s8=All&amp;vfpref=html&amp;yearRangeFirst=&amp;yearRangeSecond=&amp;yrop=eq&amp;r=12&amp;mx-pid=2399103" rel="nofollow">see here</a>) which boils down (in dimension one) to a study of derivations with values in the algebra of affiliated operators. Unfortunately, this more algebraic approach has not been very successful so far.</p>