Examples of noncommutative analogs outside operator algebras? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T19:00:41Z http://mathoverflow.net/feeds/question/7250 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/7250/examples-of-noncommutative-analogs-outside-operator-algebras Examples of noncommutative analogs outside operator algebras? Dave Penneys 2009-11-30T07:46:14Z 2012-06-18T14:45:24Z <p>Theo's <a href="http://mathoverflow.net/questions/7095" rel="nofollow">question</a> made me wonder if there are other "noncommutative analogs" outside of operator algebras. Some noncommutative analogs from operator algebras include:</p> <ul> <li>A $C^\ast$-<a href="http://en.wikipedia.org/wiki/C%2A-algebra" rel="nofollow">algebra</a> is a noncommutative <a href="http://en.wikipedia.org/wiki/Topological%5Fspace" rel="nofollow">topological space</a> (cf. the <a href="http://en.wikipedia.org/wiki/Gelfand%5Ftransform" rel="nofollow">Gelfand transform</a>).</li> <li>The <a href="http://en.wikipedia.org/wiki/Multiplier%5Falgebra" rel="nofollow">multiplier algebra</a> of a nonunital $C^\ast$-algebra is the noncommutative <a href="http://en.wikipedia.org/wiki/Stone%2DCech%5Fcompactification" rel="nofollow">Stone-Cech compactification</a>.</li> <li>A <a href="http://en.wikipedia.org/wiki/Spectral%5Ftriple" rel="nofollow">spectral triple</a> is a noncommutative <a href="http://en.wikipedia.org/wiki/Manifold" rel="nofollow">manifold</a> (add some extra data to the spectral triple to get a noncommutative <a href="http://en.wikipedia.org/wiki/Riemannian%5Fmanifold" rel="nofollow">Riemannian manifold</a> cf. arXiv:0810.2088).</li> <li>A <a href="http://en.wikipedia.org/wiki/Von%5FNeumann%5Falgebra" rel="nofollow">von Neumann algebra</a> is a noncommutative <a href="http://en.wikipedia.org/wiki/Measure%5Fspace" rel="nofollow">measure space</a>.</li> </ul> <p>Are there any other good examples? If you know more in operator algebras, that's great too.</p> <p>EDIT: these algebras should be considered as various functions spaces for noncommutative spaces as per @Yemon's answer. I'm going to leave the above text as is unless there are requests for another edit.</p> http://mathoverflow.net/questions/7250/examples-of-noncommutative-analogs-outside-operator-algebras/7251#7251 Answer by Yemon Choi for Examples of noncommutative analogs outside operator algebras? Yemon Choi 2009-11-30T08:11:34Z 2009-11-30T08:11:34Z <p>No answers, I'm afraid, but some remarks.</p> <p>If I can indulge one of my pet peeves: a $C^*$-algebra is IMHO <em>not</em> a noncommutative topological space. What it might be, depending on how deeply one has drunk from the Well of Wisdom, is the <em>coordinate ring</em> of a NC topological space (the correspondences are contravariant). In fact, if one actually tries to do any NCG then this kind of reversal of arrows gets used all over the place, e..g K<sup>*</sup> of $C(X)$ being K<sub>*</sub> of $X$, cyclic cohomology of algebras really being cyclic homology of spaces, and so forth. There might be a good categorical explanation of why a good category of "spaces" should be self dual, but I don't remember seeing one.</p> <p>To be slightly more serious, I am made rather uncomfortable by the abandon with which the NC dictionary gets brandished. (Connes' axioms for spectral triples seem, from my limited reading, to be really quite subtle; the fact they work well hides a lot of work that goes into choosing the right definition.) It seems to me a bit easy to take some class of commutative algebras associated to a geometric or topological widget, widen the class to include noncommutative algebras, and then claim to be doing noncommutative widgetry. That's not meant to downplay the very real and interesting work done by several groups; merely to caution that (IMHO) there's a lot of hidden work that has to be done to make the <em>right</em> definitions.</p> <p>OK, one suggestion. As I think I said on the other post/thread, there is a book by Nik Weaver where he tries to develop an approach to spaces of Lipschitz functions on metric spaces, which will allow one to do some kind of "analysis on noncommutative metric spaces". Unfortunately I can't remember at the moment what connection, if any, this has with Rieffel's program of quantum metric spaces. The basic idea is very similar to that used by Connes et al, but with potentially broader application since we are not imposing any (NC) smooth structure.</p> http://mathoverflow.net/questions/7250/examples-of-noncommutative-analogs-outside-operator-algebras/7254#7254 Answer by Konstantin Slutsky for Examples of noncommutative analogs outside operator algebras? Konstantin Slutsky 2009-11-30T09:00:39Z 2009-11-30T09:00:39Z <p>There are quantum groups which are non-commutative analogs of locally compact topological groups (non-commutative in the sense of topology, not in the sense of group structure). <a href="http://en.wikipedia.org/wiki/Quantum%5Fgroup" rel="nofollow">Here</a> is Wikipedia article about them.</p> http://mathoverflow.net/questions/7250/examples-of-noncommutative-analogs-outside-operator-algebras/7260#7260 Answer by Dave Penneys for Examples of noncommutative analogs outside operator algebras? Dave Penneys 2009-11-30T11:37:24Z 2009-11-30T11:37:24Z <p>I remembered a few more from operator algebras:</p> <p>a trace is like a noncommutative integral. This has lead to:</p> <ul> <li>The Schatten $p$-ideals</li> <li>If $M$ is a $II_1$-factor, then one forms the noncommutative $L^P$-spaces (see this <a href="http://en.wikipedia.org/wiki/Lp%5Fspace" rel="nofollow">link</a>) $L^p(M)$ as the completion of $M$ with the norm $$tr(|x|^p)^{1/p}$$ for $1\leq p &lt;\infty$ and as $M$ with the operator norm for $p=\infty$. Then we get the duality $L^1(M)^\ast=L^\infty(M)$, and since the trace is finite, we have, as in the finite measure space case, that $$M=L^\infty(M)\subset L^2(M)\subset L^1(M).$$ I believe we have $L^p(M)\subset L^q(M)$ if $p>q$, but I haven't checked it. Is it also true that $L^p(M)^*=L^q(M)$ if $p^{-1}+q^{-1}=1$? I have never checked this either.</li> </ul> http://mathoverflow.net/questions/7250/examples-of-noncommutative-analogs-outside-operator-algebras/7269#7269 Answer by Peter Arndt for Examples of noncommutative analogs outside operator algebras? Peter Arndt 2009-11-30T13:09:36Z 2009-11-30T13:09:36Z <p><a href="http://en.wikipedia.org/wiki/Quantale" rel="nofollow">Quantales</a> are another sort of noncommutative space.</p> <p>A general remark: It was said in two other answers that "a good category of spaces should be self dual". If by "self dual" you mean "equivalent to its opposite category", then this should rather not be the case for a category of spaces. Dual cats to cats of spaces are rather of algebraic character, a resonable definition for being "of algebraic character" is being "locally presentable". But a theorem of Gabriel/Ulmer states that if the opposite of a locally presentable cat is locally presentable again, then the cat is preorder (i.e. we are in a trivial case) - this can be seen as a mathematical statement reflecting the duality between algebra and geometry</p> http://mathoverflow.net/questions/7250/examples-of-noncommutative-analogs-outside-operator-algebras/7294#7294 Answer by Hugh Thomas for Examples of noncommutative analogs outside operator algebras? Hugh Thomas 2009-11-30T17:11:42Z 2009-11-30T17:11:42Z <p>There is a notion of non-commutative resolution of a singularity (in the sense of commutative algebraic geometry). See section 5 of <a href="http://arxiv.org/abs/math/0206295" rel="nofollow">this ICM talk</a> by Bondal and Orlov. </p> <p>The following is a simple example. </p> <p>Let $G$ be a finite subgroup of $SL_n$ acting on $\mathbb C^n$ in the usual way. So $G$ also acts on the coordinate ring $R=\mathbb C[x_1,\dots,x_n]$. If you like, you can consider $R^G$, the invariant subring. This ring might be singular, so you might want to resolve the singularity. </p> <p>A nice noncommutative alternative to doing so is to consider $R*G$, the skew group ring. The elements of $G$ form a free $R$-basis additively, just like for the usual group ring. However, the multiplication rule is given by:</p> <p>$$(r*g)(s*h)=(r\, g(s))*(gh)$$</p> <p>The skew group ring is a "non-commutative crepant resolution" of the singularity. See <a href="http://arxiv.org/abs/math/0211064" rel="nofollow">this paper</a> by Michel van den Bergh, of which the above is Example 1.1. </p> <p>Part of its "resolution-ness" is the fact that $R*G$ has finite homological dimension (which is a non-commutative way to think of smoothness). </p> http://mathoverflow.net/questions/7250/examples-of-noncommutative-analogs-outside-operator-algebras/7528#7528 Answer by John McCarthy for Examples of noncommutative analogs outside operator algebras? John McCarthy 2009-12-02T00:14:10Z 2009-12-02T00:36:08Z <p>There exist a large family of noncommutative spaces that arise from the quantum matrices. These purely algebraic objects q-deform the coordinate rings of certain varieties. For example, take quantum SL(2), this is the algebra $\mathbb{C} &lt; a,b,c,d\ >$ quotiented by the ideal generated by $$ab - qba, ~~ ac - qca, ~~ bc - cb, ~~ bd - qdb, ~~ cd - qdc, ~~ ad - da - (q - q^{-1})bc\$$ and the "q-det" relation $$ad -qbc - 1.$$ where $q$ is some complex number. Clearly, when $q=1$ we get back the coordinate ring of $SU(2)$. Other members of this family include quantum $SU(n)$, for $n >2$, quantum $U(2)$, quantum $\mathbb{CP}^n$, quantum Grassamnnian space, quantum spheres .....</p> <p>The basic quantum matrices originally arose from the work of Lenigrad physicists on the inverse scattering problem, and so has a completely different origin to operator algebraic structures. While it is possiple to put norms and involutions on all these algebraic objects and complete them to $C^*$-algebras, they are still very interesting and well studied in their own right. For more information see Shahn Majid's book: A quantum groups primer, or the paper <a href="http://arxiv.org/abs/hep-th/9410039" rel="nofollow">link text</a>.</p> http://mathoverflow.net/questions/7250/examples-of-noncommutative-analogs-outside-operator-algebras/7555#7555 Answer by David Bar Moshe for Examples of noncommutative analogs outside operator algebras? David Bar Moshe 2009-12-02T04:49:30Z 2009-12-02T04:49:30Z <p>Star products of symbols are cosidered as noncommutative analogs of the Poisson algebra of functions (even though, the Poisson algebra itself is noncommutative). For example, the Moyal product (See the Wikipedia entry), is a star product acting on functions on the cotangent bundle of a manifold. There is also the Berezin star product on Kaehler manifold. These products have a correspondence limit (hbar-->0), where the product approaches the usual Poisson bracket, see for example Bordemann,Meinrenken, and Schlichenmaier in arXiv hep-th/930914. Many authors consider the star products as a part of noncommutative geometry (Alain Connes - Noncommutative geometry), for an application, see for example a thesis by Andreas Sykora in <a href="http://edoc.ub.uni-muenchen.de/2893/1/Sykora_Andreas.pdf" rel="nofollow">http://edoc.ub.uni-muenchen.de/2893/1/Sykora_Andreas.pdf</a></p> http://mathoverflow.net/questions/7250/examples-of-noncommutative-analogs-outside-operator-algebras/99903#99903 Answer by NN for Examples of noncommutative analogs outside operator algebras? NN 2012-06-18T14:36:37Z 2012-06-18T14:36:37Z <p>Some non-commutative analogs in lattice theory. </p> <p>von Neumann's coordinatization theorem is the non-commutative analogue of Stone's definitional equivalence between Boolean algebras (complemented distributive lattices) and Boolean rings (associative rings with 1 where all elements are idempotent). </p> <p>The Baer - Inaba - Jonnsson - Monk coordinatization theorem gives the non-commutative analogue of direct products of finite chains (Łukasiewicz propositional logics); for this one uses not the usual formulation of the theorem (which coordinatizes primary lattices with modules over a artinian ring where one-sided ideals are two-sided and form a chain), but a reformulation that gives a true equivalence between the lattice (of submodules of the module) and the ring (of endomorphisms of the module); this way one has a true analogue of the (dual) equivalence between commutative C^*-algebras and compact Hausdorff spaces (or better, their lattice of open sets). </p> <p>One has a common generalization of the two cases above: equivalence between lattices of subobjects and rings of endomorphisms for finitely presented modules (of geometric dimension at least 3) over a "auxiliary" ring which is WQF (weakly quasi Frobenius, a.k.a. IF, injectives are flat). The categories of finitely presented modules over such auxiliary rings are exactly (up to equivalence) the abelian categories with an object which is injective, projective and finitely generates and finitely cogenerates every object.</p> <p>The ultimate generalization is G.Hutchinson's coordinatization theorem, a correspondence between arbitrary abelian categories and modular lattices with 0 where each element can be doubled and intervals are projective (in lattice theory, sense i.e. the classical projective geometry meaning) to initial intervals. When one looks at this theorem together with the Freyd - Mitchell embedding theorem, one has that three languages are fully adequate and equivalent ways to do linear algebra: (1) the usual language of sums and products (modules over associative rings); (2) the language of category theory (abelian categories); (3) the old fashioned language of synthetic geometry of incidence (joins and meets in suitable modular lattices).</p> <p>In these equivalences, the lattices are the (pointless, noncommutative) spaces, and the rings are the rings of coordinates or functions over the space (I am not considering the distinction between equivalences and dual equivalences because I am more interested in structures, with their unique concept of isomorphism attached, rather than more general morphisms, which depend upon the particular way to define a structure. But the complementarity between the structural and categorical views, where neither subsumes the other, is another long theme).</p> <p>Since all these ideas have their origin in von Neumann works about continuous geometries and rings of operators, I now explain the relation with operator algebras.</p> <p>First note that the usual definition of pointless topological space as complete Heyting algebra is not a true generalization of the "topological space" concept: they are a true generalization of sober spaces, but to generalize topological spaces one must consider pairs: a complete boolean algebra (which in the atomic case is the same thing as a set) with a complete Heyting subalgebra (the lattice of open sets). To obtain the non-commutative analogue, complete boolean algebras are generalized to meet-continuous geometries, and algebras of measurable sets are replaced with suitable structures (projection ortholattices of von Neumann algebras) which are embedded in the meet-continuous geometries in the same way as a right nonsingular ring is embedded in its maximal ring of right fractions (a regular right self-injective ring).</p> <p>A meet continuous geometry is a complete lattice which is modular, complemented and meet distubutes over increasing joins (not arbitray joins, like Heyting algebras). These structures were introduced by von Neumann and Halperin in 1939; they are a common generalization of (possibly reducible) continuous geometries (the subcase where join distrubutes over decreasing meets) and (possibly reducible and infinite dimensional) projective geometries (the atomic subcase). For them one has a dimension and decomposition theory much like the one for continuous geometries and rings of operators (the theory of S.Maeda, in its last version of 1961, is sufficient; one does not need the 2003 theory by Wehrung and Goodearl). One can define the components of various types, in particular I_1 (the boolean component, i.e. classical logic), the I_2 component (the 2-distibutive component i.e. subdirect product of projective lines; physically these are "spin factors" and quantum-logically it is the non-classical component which nonetheless has non-contextual hidden variables), the I_3 nonarguesian component (subdirect product of projective nonarguesian planes i.e. irreducible projective geometries that cannot be embedded in larger irreducible projective geometries; quantum-logically this means that interacion with other components is only possible classically, without superposition). Once these bad low dimensional components are disregarded, von Neumann coordinatization theorem gives a equivalence between the meet-continuos geometries and the right self-injective von Neumann regular rings. So meet-continuous geometries are pointeless quantum (i.e. non-commutative) sets in the same way as complete Boolean algebras are (commutative) pointless sets. Regular rings are the coordinate rings of these quantum sets in the same way as (commutative, regular) rings of step functions (with values in a field) are the ring equivalent of a boolean algebra (classical propositional logic); the important new fact is that in the "truly non-commutative case" the regular ring is uniquely and canonically determined by the lattice (in the distributive case, on the contrary, it is not: one can use step functions with values in any field, and one can change the field with the point; commutative [resp. strongly] regular rings are the subrings of direct products of [skew] fields which are stable for the generalized inverse operation).</p> <p>The above "propositional logics" are without the negation operator; on the other hand, the projection ortholattices of von Neumann algebras are complete orthomodular lattices with sufficiently many completely additive probability measures and sufficiently many internal simmetries (von Neumann said that the strict logic of orthocomplementation and the probability logic of the states uniquely determine each other by means of his symmetry axioms in his characterization of finite factors as continuos geometries with a transition probability. One should also note how much more physically meaningful are von Neumann axioms when compared with the "modern" ones based on Soler's theorem, but this is another large topic). Using Gleason's theorem (and as always in absence of the bad low-dimensional components) one obtains an equivalence between von Neumann's "rings of operators" (i.e. <em>real</em> von Neumann algebras, or their self-adjoint part, real JBW-algebras) and their projection ortholattices (the normal measures on the ortholattice give the predual of the ring of operators). One can see these logics inside a meet-continuos geometry by equipping the geometry with a linear orthogonality relation which has for each element a maximum orthogonal element (pseudo-orthocomplementation in part analogous to "external" in a stonean topological space, in part anti-analogous as it happens with Lowere closure when compared to Kuratowski closure). The regular ring of the lattice is the ring generated by all complementary pairs in the lattice (which are the idempotents of the ring: kernel and image) with the relations corresponding to the partial operation e+f-ef which is defined on idempotents whenever fe=0 (at the lattice level this partial operation is implemented with disjoint join of the images and co-disjoint meet of the kernels); in the case associated to a "ring of operators", this regular ring is the ring of maximal right quotients of the von Neumann algebra, and conversely the algebra is recovered from the lattice with orthogonality by taking the subring generated by orthogonal projections (idempotents whose kernel and image are orthogonal); by a theorem of Berberian the algebra is ring generated by its self-adjoint idempotents, and there is clearly at most one involution on the algebra which fixes such generators.</p> <p>Hence, in summary, in absence of bad low dimensional components (whose exclusion is physically meaningful, see their meanings above) one has equivalences between the following concepts:</p> <p>(0) right self-injective regular rings with a suitable additional structure (to associate a orthogonal projection onto the closure of the image to any element) (1) real von Neumann algebras (2) real JBW-algebras (the Jordan algebra of self-adjoint operators i.e. observables) (3) the effect algebra (of operators with spectrum in [0,1]) i.e. unsharp quantum logic (4) the projection ortholattice (the sharp quantum logic) (5) the pointeless quantum set (meet-continuous geometry) with a suitable orthogonality (6) the convex compact set of normal states</p> <p>Any of the above is a adequate starting point for quantum foundation since all the other points of view can be canonically recovered. </p> <p>All this is restricted to the level of non-commutative measure spaces; a locally compact topological space is something more precise (like a C^* algebra when compared to a von Neumann algebra), and a (Riemannian) metric space is something still more precise (and in particular a differentiable structure, which can be seen as an equivalence class of riemannian structures: note that a isometry for the geodesic metric between complete Riemannian manifolds is automatically differentiable, so the differentiable structure must be definable from the geodesic metric, and infact Busemann and Menger had such a explicit definition of the tangent spaces from the global metric. The topological structure is then another equivalence class of metrics, for another weaker equivalence).</p> <p>Given the equivalence (0)--(6) above, one can note that all the concepts which are used in Connes definition of spectral triples, and analogues structures, can be seen equivalently from each of the above points of view. In this way, all of the above points of view are a possible starting point for non-commutative geometry.</p> <p>[Sorry for the too long post and for my bad pseudo-english language]</p> http://mathoverflow.net/questions/7250/examples-of-noncommutative-analogs-outside-operator-algebras/99904#99904 Answer by Yulia Kuznetsova for Examples of noncommutative analogs outside operator algebras? Yulia Kuznetsova 2012-06-18T14:45:24Z 2012-06-18T14:45:24Z <p>In a recent paper Alexei Pirkovskiy defines a class of Frechet algebras such that the subclass of commutative algebras corresponds to Stein complex manifolds (under a technical assumption of of finite embedding dimension). This may be viewed as noncommutative complex analysis. Link: <a href="http://arxiv.org/abs/1204.4936" rel="nofollow">http://arxiv.org/abs/1204.4936</a></p>