User torsten asselmeyer-maluga - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-18T05:53:11Z http://mathoverflow.net/feeds/user/7608 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/34334/how-well-can-we-localize-the-exoticness-in-exotic-r4/35208#35208 Answer by Torsten Asselmeyer-Maluga for How well can we localize the "exoticness" in exotic R^4? Torsten Asselmeyer-Maluga 2010-08-11T10:26:27Z 2010-08-14T20:52:34Z <p>The compactness of the 4-manifold is really necessary for the existence of the Akbulut cork. The non-compact case cannot be done with this method. If there is a compact subset in the exotic $\mathbb{R}^4$ determing the exoticness, then one can make a one-point compactifications of the $\mathbb{R}^4$ getting an exotic 4-sphere $S^4$ (constructing a counterexample to the smooth Poince conjecture in dimension 4 (SPC4)).</p> <p>More exact: Let $X$ be an exotic $\mathbb{R}^4$ and $R$ the standard $\mathbb{R}^4$. Let $C\subset X, C'\subset R$ be compact, contractable subsets. Now we assume that the subsets $C,C'$ are like Akbulut corks, i.e. $C$ and $C'$ are homeomorphic but non-diffeomorphic and $X-C$ is diffeomorphic to $R-C'$. The compactification of $X$ and $R$ result in a homotopy $S^4$ homeomorphic to the $S^4$. Then $X-C$ and $R-C'$ are changed to $\hat{X}-C$ and $\hat{R}-C'$ where $\hat{X},\hat{R}$ are homeomorphic to the $S^4$. Then $\hat{X}-C$ and $\hat{R}-C'$ are diffeomorphic (the assumption above) but $\hat{X},\hat{R}$ are not diffeomorphic. Thus we produce a counterexample to SPC4. But there is not such a compact subset (someday I heart the formulation: "the exoticness is located at infinity").</p> <p>Therefore the approach in the 94'paper is correct, the world tube is a non-compact area. No contradiction.</p> http://mathoverflow.net/questions/33841/is-it-true-that-exotic-smooth-r4-cannot-be-diffeomorphic-to-rxn-where-n-is-a-3/35212#35212 Answer by Torsten Asselmeyer-Maluga for Is it true that exotic smooth R^4 cannot be diffeomorphic to RxN, where N is a 3-manifold? Torsten Asselmeyer-Maluga 2010-08-11T10:42:28Z 2010-08-14T20:25:30Z <p>Steven Sievik comment is very important. Following the approach of Munkres and McMillan, then every 4-manifold $N\times\mathbb{R}$ with $N$ a contractable 3-manifold is diffeomorphic to the standard $\mathbb{R}^4$. Therefore the exotic $\mathbb{R}^4$ cannot be splitted like $N \times\mathbb{R}$ and esspecially not like $\mathbb{R}^3 \times\mathbb{R}$. Or, there is no diffeomorphism between the exotic $\mathbb{R}^4$ and $\mathbb{R}^3 \times\mathbb{R}$. But by definition there is a homeomorphism between the exotic $\mathbb{R}^4$ and the standard $\mathbb{R}^4$. Then we have a homeomorphism between the exotic $\mathbb{R}^4$ and $\mathbb{R}^3 \times\mathbb{R}$.</p> <p>In the topological category we have a splitting $\mathbb{R}^3 \times\mathbb{R}$ but not in the smooth category.</p> <p>Addendum: In contrast, every exotic $\mathbb{R}^4$ admits a $C^\infty$ codimension-1 foliation because any non-compact manifold admits one (see the BAMS article of Lawson "Foliations", Corollary 1.2).</p> http://mathoverflow.net/questions/26338/triangulations-of-exotic-4-spheres/35210#35210 Answer by Torsten Asselmeyer-Maluga for Triangulations of exotic 4-spheres Torsten Asselmeyer-Maluga 2010-08-11T10:33:31Z 2010-08-11T10:33:31Z <p>The current status of the smooth Poincare conjecture in dimension 4 is presented in the paper: Michael Freedman, Robert Gompf, Scott Morrison and Kevin Walker "Man and machine thinking about the smooth 4-dimensional Poincare conjecture" in Quantum Topology, Volume 1, Issue 2 (2010), pp. 171–208 (<a href="http://arxiv.org/abs/0906.5177" rel="nofollow">arXiv</a>) Thus the Cappel-Shaneson approach seem to fail by Akbuluts work. Now there is only possible construction via the Gluck twist with a real 2-knot (i.e. knotted 2-sphere), i.e. a knot not coming from a 3-dimensional (classical) knot.</p> http://mathoverflow.net/questions/30632/theoretical-physics-why-not-just-r4/31841#31841 Answer by Torsten Asselmeyer-Maluga for Theoretical physics: Why not just R^4? Torsten Asselmeyer-Maluga 2010-07-14T12:12:15Z 2010-07-16T08:52:18Z <p>The idea: exotic smoothness=matter is fascinating and cannot wiped off by the locality argument. A dynamical process is a submanifold of the spacetime. We found this page accidentally but had parallel ideas. In a recent paper <a href="http://arxiv.org/abs/1006.2230" rel="nofollow">geometrization of matter by exotic smoothness</a> we realize the idea: matter = Casson handle, i.e. we showed that the action of the fermion and gauge fields follow from exotic smoothness by considering the structure of the Casson handle. In dimension 4 one has the special fact that a local change of the 4-manifold can change the smoothness, i.e. from the physical point of view we have a local theory.</p> <p>Now some words about quantization:</p> <p>In a paper <a href="http://arxiv.org/abs/1001.0882" rel="nofollow">Exotic smooth R^4, noncommutative algebras and quantization</a> we found a close relation between codimension-1 foliations and exotic smoothness. Especially I'm interested in the question: given a 4-manifold with boundary, how can I detect exotic smoothness on the boundary? The answer is also strongly related to the exotic R^4 where one has also a kind of localization. Surprisingly we found an answer: on the boundary (i.e. a 3-manifold) one has a codimension-1 foliation and its cobordism class (detected by the Godbillon-Vey invariant) gives you the exotic smoothness class. But by Connes work, foliations are strongly connected to C* algebras and thus we have the link to QFT. In the paper above we construct an example for the exotic R^4 and show that this C* algebra appears by deformation quantization of a Poisson algebra, all details can be read there.</p> http://mathoverflow.net/questions/34334/how-well-can-we-localize-the-exoticness-in-exotic-r4/35208#35208 Comment by Torsten Asselmeyer-Maluga Torsten Asselmeyer-Maluga 2010-08-14T20:32:03Z 2010-08-14T20:32:03Z Thanks for the comments. Of course you are right and I changed the answer. http://mathoverflow.net/questions/33841/is-it-true-that-exotic-smooth-r4-cannot-be-diffeomorphic-to-rxn-where-n-is-a-3/35212#35212 Comment by Torsten Asselmeyer-Maluga Torsten Asselmeyer-Maluga 2010-08-14T20:19:48Z 2010-08-14T20:19:48Z Yes, you are right but I was not sure. I edited the answer. Thanks for the comment. http://mathoverflow.net/questions/30632/theoretical-physics-why-not-just-r4/31841#31841 Comment by Torsten Asselmeyer-Maluga Torsten Asselmeyer-Maluga 2010-07-16T08:32:21Z 2010-07-16T08:32:21Z @Aaron -- exotic smoothness is a part of the path integral independent of the integration over the possible geometries see [[<a href="http://arxiv.org/abs/1003.5506]]" rel="nofollow">arxiv.org/abs/1003.5506]]</a> for the details