User mkouboi - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T05:35:24Z http://mathoverflow.net/feeds/user/20005 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/130950/groupoids-vs-action-groupoids/130973#130973 Answer by Mkouboi for Groupoids vs. action groupoids Mkouboi 2013-05-17T15:34:03Z 2013-05-17T15:34:03Z <p>I might be wrong but at first glance, I would say "yes" to <em>Question 1</em>. The reason is the following. If a groupoid $\Gamma \rightrightarrows Y$ acts by automorphisms on a groupoid $A\rightrightarrows X$, one can form the semi-direct product groupoid $\Gamma\ltimes A$ with unit space $X$ and morphisms $(\Gamma\ltimes A)_1$ elements $(\gamma,a)\in \Gamma\times A$ such that $\gamma a$ makes sense. Now, if $A\cong G\ltimes X$ is connected, and if $\Gamma$ is a group acting compatibly on $G$ and $X$, then $\Gamma\ltimes (G\ltimes X)\rightrightarrows X$ is isomorphic to $(\Gamma\ltimes G)\ltimes X \rightrightarrows X$. Moreover, any connected $\Gamma$-groupoid is in particular connected and hence in the form $G\ltimes X$. By construction of the group $G$, $\Gamma$ acts compatibly on it and $X$, and $\Gamma\ltimes A$ is a action $\Gamma$-groupoid. </p> http://mathoverflow.net/questions/128831/infinity-categories-vs-kan-complexes Infinity-categories vs Kan complexes Mkouboi 2013-04-26T15:08:46Z 2013-05-01T11:05:49Z <p>Hi all,</p> <p>It is known (cf. Lurie's book "Higher Topos Theory", for instance) that higher ($\infty$-) category, in particular topological higher ($\infty$-) groupoids are "better" defined as <em>weak Kan complexes</em>, aka <em>quasi-categories</em>. Let me recall that a simplicial space $X_\bullet$ is <em>weak Kan</em> if every map $\Lambda_i^n\longrightarrow X$ (where $\Lambda_i^n$ is the <em>horn</em>) can be extended to a map $\Delta^n\longrightarrow X$ for all $0&lt; i &lt; n$. </p> <p>My problem is the following. Let us consider the <em>fundamental</em> $(\infty,0)$-groupoid $\pi_{\leq \infty}X$ of a nice space $X$: $X$ is the set of objects, morphisms between objects are maps $f:[0,1]\longrightarrow X$, $2$-morphisms are homotopies between morphisms, and so on. I am quite confused about how does one, "geometrically", see $\pi_{\leq \infty}X$ as a simplicial space; for an element of $\pi_2X$, for instance, has four edges, and hence four different face maps. </p> <p>Many thanks! </p> http://mathoverflow.net/questions/30511/ebook-readers-for-mathematics/88999#88999 Answer by Mkouboi for eBook readers for mathematics Mkouboi 2012-02-20T08:04:47Z 2012-02-20T08:04:47Z <p>Hi, I have been using the new Kindle (dimensions: 166 mm x 114 mm x 8.7 mm) for a week; here are my observations. It does read mathematics texts written in pdf, hence arXiv papers. I noticed only one issue which concerns margins and the font size. Indeed, most mathematics textbooks and papers have big margins, and then not look good on the Kindle. There is however the possibility of zooming, but as mentioned above, this is very disturbing since then one can no longer see the entire page one is reading. Another alternative is to convert pdf or djvu to e-reader formats such as mobi, rtf, etc. by using a program such as <em>Calibre</em> (for Linux distribution); but then one gets some ugly extraterrestrial maths formulas.</p> <p>Here is the solution I have been adopting so far to read arXiv papers: <strong>I download the .tex files, and fiddle about with the margins and font size; more precisely, the reasonable font size is 12pt, I cut almost all the margins and enlarge the lines (this is useful especially when one deals with big formulas or diagrams) by adding "\usepackage[paperwidth=16cm, paperheight=20cm,top=0.5cm, bottom=0.5cm, left=0.5cm, right=0.5cm]{geometry}"</strong>. The result I get is much better, and I can read without trouble ... just imagine one of the printed two pages on an A4 sheet.</p> http://mathoverflow.net/questions/88460/group-extensions-and-actions-on-categories/88538#88538 Answer by Mkouboi for Group extensions and actions on categories Mkouboi 2012-02-15T18:46:01Z 2012-02-15T18:46:01Z <p>Such extensions are what D. Conduché called <em>crossed 2-modules</em> (in French: modules croisés de longueur 2) in his 1983 paper: <em>Modules croisés généralisés de longueurs 2</em>, J. Pure and Appl. Alg. Vol. 34, Issues 2-3 (1984), pp. 155-178. My guess is that an action of $G$ on the category of $H$-sets amounts to what the same author called "<em>non-abelian pre-crossed complex</em>" (in French: complexe pré-croisé non abélien) in the same paper. The equivalence you talked about is then Théorème 2.6 of the above mentioned reference.</p> http://mathoverflow.net/questions/54898/what-is-the-right-definition-of-real-von-neumann-algebra/83652#83652 Answer by Mkouboi for What is the right definition of "real von Neumann algebra"? Mkouboi 2011-12-16T18:43:37Z 2011-12-16T18:43:37Z <p>Hi, this is intended as a comment on Jon's comment, but I still lack MO reputation to leave comments; sorry for that. I believe what is mentioned in Li's book is wrong; the right statement should be "<em>a complex $C^\ast$-algebra is the complexification of a real one if and only if it has an involutory ${}^\ast$-antiautomorphism</em>" (here an example by V. Jones of a von Neumann algebra antiautomorphic to itself but without involutory antiautomorphisms: <a href="http://www.mscand.dk/article.php?id=2523" rel="nofollow">http://www.mscand.dk/article.php?id=2523</a>). </p> <p>Conversely, one can study real $C^\ast$-algebras in terms of their complexifications: <em>e.g.</em>, say that $A$ is a real von Neumann algebra if $A\otimes_{\mathbb{R}}\mathbb{C}$ is von Neumann, and so on.</p> http://mathoverflow.net/questions/83363/is-there-a-nice-application-of-category-theory-to-functional-complex-harmonic-ana/83618#83618 Answer by Mkouboi for Is there a nice application of category theory to functional/complex/harmonic analysis? Mkouboi 2011-12-16T13:38:51Z 2011-12-16T13:45:50Z <p>Of course the equivalence of categories mentioned by Eric A. Bunch is true only for commutative $C^\ast$-algebras. There however is a quite similar result for a wider category of non-commutative $C^\ast$-algebras: a continuous-trace $C^\ast$-algebra $A$ with Hausdorff spectrum $X$ is isomorphic to the $C^\ast$-algebra $\Gamma_0(X,\mathcal{A})$ of continuous sections vanishing at $\infty$ of some continuous $C^\ast$-bundle $\mathcal{A}\to X$; the latter is actually a Dixmier-Douady bundle, in the sense that it has typical fiber the algebra $\mathbb{K}(H)$ of compact operators on some separable Hilbert space. This construction yields an equivalence between the category of continuous-trace $C^\ast$-algebras with Hausdorff spectrum and pairs $(X,\mathcal{A})$. </p> <p>When it comes to study $K$-theory of such $C^\ast$-algebras, the above equivalence may be very useful, for if $A$ corresponds to the pair $(X,\mathcal{A})$, then $K_\ast(A)$ is isomorphic to the twisted $K$-theory ${}^{\mathcal{A}}K^\ast(X)$, which in the finite-dimensional case may be interpreted in terms of geometric objects (twisted vector bundles). For instance, let $G$ be a compact Hausdorff group. Then the dual space of $G$ is homeomorphic to the the spectrum $X$ of the $C^\ast$-algebra $C^\ast(G)$ (which is actually continuous-trace). Now, for $k\in \mathbb{N}$, let $X_k$ be the (open) subspace of $X$ consisting of (equivalence classes of) irreducible representations of $G$ of rank $k$. Then, for each $k$, there is an Azumaya bundle $\mathcal{A}_k\to X_k$ (<em>i.e.</em> a Dixmier-Douady bundle of finite dimension), and there is an isomorphism $K_\ast(C^\ast(G))\cong \bigoplus_k {}^{\mathcal{A}_k} K^\ast(X_k)$ (cf. <a href="http://front.math.ucdavis.edu/0201.5207" rel="nofollow">http://front.math.ucdavis.edu/0201.5207</a> for a generalization of this example).</p> http://mathoverflow.net/questions/130950/groupoids-vs-action-groupoids/130973#130973 Comment by Mkouboi Mkouboi 2013-05-18T16:08:46Z 2013-05-18T16:08:46Z @Mikhail: A groupoid isomorphism from $A\rightrightarrows X$ to $G\ltimes X\rightrightarrows X$ induces a $\Gamma$-action on the group bundle $G\ltimes X$. One obtains the $\Gamma$-action on $G$ by identifying $G$ with the fibres $(G\ltimes X)_x=(G\ltimes X)^x=(G\ltimes X)^x_x$. http://mathoverflow.net/questions/128831/infinity-categories-vs-kan-complexes/129009#129009 Comment by Mkouboi Mkouboi 2013-04-28T19:23:00Z 2013-04-28T19:23:00Z Thanks, Carnahan, for this very interesting answer which, combined with Ronnie's intuitive bi-groupoid view point, helped me so much. http://mathoverflow.net/questions/128831/infinity-categories-vs-kan-complexes/128863#128863 Comment by Mkouboi Mkouboi 2013-04-27T17:35:00Z 2013-04-27T17:35:00Z @Ronnie:Thanks for the reference! http://mathoverflow.net/questions/128831/infinity-categories-vs-kan-complexes Comment by Mkouboi Mkouboi 2013-04-26T19:00:50Z 2013-04-26T19:00:50Z @Zhen: Sure! thanks. http://mathoverflow.net/questions/124404/phd-in-operator-algebras-and-non-commutative-geometry/124410#124410 Comment by Mkouboi Mkouboi 2013-03-13T15:41:52Z 2013-03-13T15:41:52Z Here are some places in Europe with very good and reasonably big research groups in operator algebras and non-commutative geometry: - Paris Jussieu (G. Skandalis,M. Hilsum, A. Zuk, E. Blanchard, etc.); - Orl&#233;ans (V. Lafforgue, J. Renault, etc.); - Metz (J.-L. Tu, M. Benameur, H. Oyono, etc.); - Marseille (Puschnig, etc.); - Muenster (J. Cuntz, S. Echterhoff, etc.); - Goettingen (R. Meyer,etc.); - Coppenhagen (R. Nest, Kirchberg, etc.). http://mathoverflow.net/questions/72690/are-bundle-gerbes-bundles-of-algebras/72756#72756 Comment by Mkouboi Mkouboi 2012-03-19T08:46:16Z 2012-03-19T08:46:16Z @Andr&#233;: I'm not sure what you say is true. The reason is the following: stable isomorphism classes of bundle gerbes over a space X form an abelian group bg(X) which is isomorphic to the group of Morita equivalence classes of (separable) continuous trace C*-algebras with spectrum X. Torsion elements in bg(X) correspond exactly to those continuous trace C*-algebras with typical finite-dimensional fibers.