User danny ruberman - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T20:20:09Z http://mathoverflow.net/feeds/user/3460 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/130078/the-classifying-space-of-a-gauge-group/130114#130114 Answer by Danny Ruberman for The classifying space of a gauge group Danny Ruberman 2013-05-08T20:09:45Z 2013-05-08T20:09:45Z <p>A nice proof of a related proposition is given in Section 5.1 of Donaldson-Kronheimer's book, The Geometry of 4-manifolds. See Prop. 5.1.4, which gives the homotopy type of the classifying space for the based gauge group. This classifying space is related to the one you asked about by the so-called base-point fibration; you might have to do a little work to get the Atiyah-Bott proposition.</p> http://mathoverflow.net/questions/129833/classification-of-higher-dimensional-manifolds/129842#129842 Answer by Danny Ruberman for Classification of higher dimensional manifolds Danny Ruberman 2013-05-06T14:11:12Z 2013-05-06T14:11:12Z <p>There is a series of papers from the mid-1960's by C.T.C. Wall on classification of highly connected smooth manifolds, starting with Wall, C. T. C., Classification of (n−1)-connected 2n-manifolds. Ann. of Math. vol. 75 1962 163–189. I think this paper answers your question.</p> <p>The Math Review of this paper by Kervaire starts, "This paper is an application of almost everything known in differential topology to the problem of classifying (n−1)-connected differential 2n-manifolds under diffeomorphism." The classification of 2-connected 6-manifolds to which you refer is actually a later result in this series.</p> <p>Wall addresses the implications of his results to PL classification; presumably, later developments about topological manifolds would give the classification up to homeomorphism. There are probably more modern' ways of formulating Wall's results, as the paper was written in the early days of surgery theory. It would be a worthwhile exercise to compare the way in which classification results are done these days with the methods of the original paper.</p> http://mathoverflow.net/questions/126444/linkage-between-singularities-of-algebraic-varieties-and-continued-fractions/126463#126463 Answer by Danny Ruberman for Linkage between singularities of algebraic varieties and continued fractions Danny Ruberman 2013-04-04T00:49:21Z 2013-04-04T00:49:21Z <p>Continued fractions appear naturally in the resolution of quotient singularities of surfaces (and presumably in higher dimensions as well). From a topological point of view, a neighborhood of the singularity is the cone on a lens space L(p,q), and a particular continued fraction for q/p gives an explicit piece of a smooth complex surface with the same boundary. This is explained very nicely in the notes, "Differentiable Manifolds and Quadratic Forms" by Hirzebruch, Neumann, and Koh, and presumably in many algebraic geometry texts.</p> http://mathoverflow.net/questions/115054/lefschetz-duality-for-twist-coefficient/115079#115079 Answer by Danny Ruberman for Lefschetz duality for twist coefficient Danny Ruberman 2012-12-01T15:40:28Z 2012-12-01T15:40:28Z <p>Poincaré-Lefschetz duality for twisted coefficients is fundamental to surgery theory, and for the universal' case of $Z[\pi_1(X)]$ coefficients is discussed in Chapter 2 of Wall's book, Surgery on Compact Manifolds. As in the case of integral coefficients, the "half die half alive" principle (suitably phrased) holds in arbitrary dimensions. </p> <p>For the setting which seems to interest you, ie duality with coefficients in a representation, you might look at Milnor's "A duality theorem for Reidemeister torsion", Annals 76 (1962), or alternately his survey "Whitehead torsion" (BAMS 72, 1966). There, the "half die half alive" principle shows up (written in a multiplicative way) as the statement that the torsion of a boundary factors as the torsion of the manifold times its conjugate (in a certain sense). A nice application is the famous Fox-Milnor condition on the Alexander polynomial of a slice knot. This approach using duality with twisted coefficients was greatly extended to the setting of the so-called twisted torsion by Kirk and Livingston (two papers in Topology Vol. 38, pp. 635-661 and 663-671, 1999).</p> http://mathoverflow.net/questions/114430/topology-of-the-universal-spinor-field-bundle/114472#114472 Answer by Danny Ruberman for Topology of the Universal Spinor Field Bundle Danny Ruberman 2012-11-26T01:32:26Z 2012-11-26T01:32:26Z <p>I think that much of what you want to know can be summarized in the question: how do you compare spin bundles for different metrics. This question, at least in the Riemannian setting, is treated with some care in the paper:</p> <p>Bourguignon, Jean-Pierre; Gauduchon, Paul, Spineurs, opérateurs de Dirac et variations de métriques. Comm. Math. Phys. 144 (1992), no. 3, 581–599. </p> <p>This paper is the jumping-off point for Maier's paper ([3] in your citations). I haven't looked at Bourguignon-Gauduchon for a while, but I believe that it provides a way of identifying spin bundles for nearby metrics; this identification then tells you what the topology of the space E should be, and provide local trivializations for the bundle. The paper goes much further, and actually shows how the Dirac operators compare for different metrics, on the basis of this comparison of spinor bundles.</p> http://mathoverflow.net/questions/96924/reference-request-embedded-morse-theory/96958#96958 Answer by Danny Ruberman for Reference request: embedded Morse theory Danny Ruberman 2012-05-15T01:07:45Z 2012-05-15T01:07:45Z <p>The main set of ideas that you want to learn is the following description of an m-dimensional manifold Y sitting in $R^n$, in such a way that the standard height function is a Morse function when restricted to Y. Thus this function, say f, gives a handle decomposition of Y; as you pass a critical point of f|Y, you add an index k handle to Y. Simultaneously, you add a (k + n -m -1)-handle to the complement of Y. This is described in reasonable detail in Section 6.2 of the book of Gompf and Stipsicz, "An Introduction to 4-manifolds and Kirby Calculus". </p> <p>I don't know the original source for this description; I learned it more or less as folklore. The informal explanation that Kirby used to give of this involved sitting in a bathtub (I think the person in the bathtub was supposed to be Y) and watching the topology of the water change as it passed various critical points. (An alternate version, helpful for thinking about knots, was a wire in a bucket of water being filled up.) A nice example to think about is how to build a handle decomposition (or Heegaard splitting) of a knot complement. For instance, for a 2-bridge knot, you should see a handle decomposition with a 0-handle, two 1-handles, and a single 2-handle. </p> <p>As a side remark, the term relative Morse theory, as I understood it, has to do with the study of the Morse function on a manifold Y induced by a Morse function on a larger manifold. In this form, it was extensively studied (in the PL case) in the 60's, in order to give results on embeddings. For instance, various theorems of the form "concordance implies isotopy" in high codimension are proved in this way.</p> http://mathoverflow.net/questions/95154/handlebody-decomposition-of-an-open-4-manifold/95167#95167 Answer by Danny Ruberman for Handlebody decomposition of an open 4-manifold Danny Ruberman 2012-04-25T15:48:24Z 2012-04-25T15:48:24Z <p>There are not that many explicit handlebody pictures of exotic open 4-manifolds, because they get awfully complex in short order. The ones that I know of are in work of Žarko Bižaca from the mid-90's. I think you probably can work out this particular case by hand. You don't really want to use Quinn's theorem for this, because it is not exactly constructive. On the other hand, Freedman's construction produces this manifold in a somewhat explicit manner. </p> <p>Here is a sketch. Start with a 4-ball, and then attach a (+1) framed 2-handle along a trefoil knot. If you've chosen the correct trefoil, the boundary of the resulting manifold is the Poincare homology sphere, say P. Then Freedman tells you that P is the boundary of a (topological) contractible manifold W, which you glue on to make the exotic $CP^2$, commonly known as Ch (for Chern). The reason that you don't need Quinn is that the construction of W is done by making a manifold $W'$ that is proper homotopy equivalent to $S^3 \times [0,\infty)$, and then using the proper h-cobordism theorem to recognize that the end of $W'$ is homeomorphic to $S^3 \times (1,\infty)$, from which you see that you can compactify $W'$ to a manifold by adding in a point.</p> <p>In the case at hand, you can be more explicit. One way (this is what happens in Freedman's original paper) is to consider the building block $P \times I$, and then do (spin) surgery on a circle to kill the fundamental group, resulting in a compact manifold $P'$. The embedding theorems of Freedman find a (topological, locally flat) wedge of spheres in $P'$, which can be surgered out to give a compact, simply-connected homology cobordism from $P$ to itself. Stacking infinitely many of these gives $W'$. Presumably, although I've never done this, you can use techniques of Bizaca to build a handlebody picture of $W'$ from this description.</p> <p>An alternate approach, which might be more amenable to drawing pictures, would come from Freedman's older paper, A Fake $S^3 \times R$. In this paper, which precedes his disk embedding theorem (but has many of the basic ideas, including reimbedding techniques) he constructs what I've called $W'$ by embedding Casson handles. There is also a Bourbaki exposition of this paper by Siebenmann that is helpful in trying to read it. </p> http://mathoverflow.net/questions/89921/compressibility-of-seifert-surface-after-0-surgery compressibility of Seifert surface after 0-surgery Danny Ruberman 2012-03-01T02:29:54Z 2012-03-01T03:35:31Z <p>Gabai's solution of the Property R conjecture shows that a minimal genus Seifert surface of a knot, capped off in the 0-framed surgery along that knot, is of minimal genus in its homology class. In particular, it is incompressible in the 0-surgered manifold. On the other hand, there may be incompressible Seifert surfaces for the knot that are not of minimal genus. (For example many pretzel knots bound incompressible surfaces of arbitrarily high genus.) Presumably, there may be such a (non-minimal genus) incompressible surface that becomes compressible in the 0-surgered manifold. Does anyone know an example of this phenomenon?</p> http://mathoverflow.net/questions/14530/4-genus-of-a-2-bridge-link/68299#68299 Answer by Danny Ruberman for 4-genus of a 2-bridge link Danny Ruberman 2011-06-20T16:38:47Z 2011-06-20T16:48:36Z <p>(This is really a comment on the answer relating to concordance order.) </p> <p>Since your p is even, then your 2-bridge knot is actually a link. So, while it makes sense to ask if it's a slice or ribbon link, asking about its concordance order doesn't make sense. </p> http://mathoverflow.net/questions/7155/famous-mathematical-quotes/12600#12600 Answer by Danny Ruberman for Famous mathematical quotes Danny Ruberman 2010-01-22T03:06:01Z 2010-01-22T03:06:01Z <p>I once read, in an autobiographical piece, what the author said to his high-school teacher upon graduation; my recollection is:</p> <p>"Poincaré has written that geometry is the art of making a correct argument from incorrectly drawn figures. For you, sir, it is the opposite."</p> <p>I would love to know the correct quote, and an accurate source. I've seen a version attributed to Poincaré, but couldn't verify that.</p> http://mathoverflow.net/questions/129833/classification-of-higher-dimensional-manifolds/129842#129842 Comment by Danny Ruberman Danny Ruberman 2013-05-08T14:44:50Z 2013-05-08T14:44:50Z In that case, a topological classification for highly-connected 2n-manifolds <i>with given intersection form</i> with n even should be straightforward from the surgery exact sequence and the computation of the homotopy groups of G/TOP. Kreck's paper gives a good guide, or you could try Ranicki chapter 13 (eg example 13.26 for a sense of what the answer would look like). For existence of manifolds with given intersection form, look at the last chapter (plumbing) of Browder's book, Surgery on simply-connected manifolds. http://mathoverflow.net/questions/129833/classification-of-higher-dimensional-manifolds/129842#129842 Comment by Danny Ruberman Danny Ruberman 2013-05-07T13:40:14Z 2013-05-07T13:40:14Z A more careful answer to the original question would be that there is not really going to be a good classification, especially when n is even, as Allen remarks. That's because the simplest homotopy invariant is the intersection form, and symmetric unimodular forms still defy classification; cf. Milnor-Husemoller's book on the subject. Perhaps Allen could clarify his question, which asks about classification up to homeomorphism; this is probably realistic (or maybe easy) for fixed intersection form. But up to diffeomorphism it's a more complicated story. http://mathoverflow.net/questions/129321/does-a-topological-manifold-have-an-exhaustion-by-compact-submanifolds-with-bou/129331#129331 Comment by Danny Ruberman Danny Ruberman 2013-05-02T17:50:19Z 2013-05-02T17:50:19Z Here's an easier (in the sense of quoting fewer results) argument in dimension 4. There is nothing to prove for compact manifolds. In the same paper (Ends of Maps III: Dimensions 3 and 4, JDG 17 (1982)) in which he proved the existence of handle structures and gave transversality results, Quinn proved that non-compact 4-manifolds are smoothable. Hence your favorite method for smooth manifolds will work. In real terms, this isn't any easier, since this argument and the one Ricardo gives depend on essentially the same set of ideas. http://mathoverflow.net/questions/122356/how-to-see-the-quaternionic-hopf-map-generates-the-stable-3-stem Comment by Danny Ruberman Danny Ruberman 2013-02-20T18:52:06Z 2013-02-20T18:52:06Z Chris: Ryan's argument isn't circular. Starting with a Heegaard splitting of M (from a triangulation or Morse function), you get M as surgery on a link. This already shows $\Omega_3$ is trivial. An algorithm of S. Kaplan encodes a spin structure on M in terms of a &quot;characteristic sublink&quot; of this link, and shows how to find a bounding spin manifold by eliminating the characteristic sublink. This route seems easier to me than the AHSS, but as you say this depends on one's perspective. http://mathoverflow.net/questions/120612/trichotomies-in-mathematics/120752#120752 Comment by Danny Ruberman Danny Ruberman 2013-02-06T01:41:21Z 2013-02-06T01:41:21Z A connected sum is a satellite knot, albeit in a somewhat trivial way, cf. <a href="http://en.wikipedia.org/wiki/Satellite_knot" rel="nofollow">en.wikipedia.org/wiki/Satellite_knot</a>. So there's no need to restrict to prime knots for this trichotomy to work. http://mathoverflow.net/questions/119711/proper-use-of-the-word-stereographic Comment by Danny Ruberman Danny Ruberman 2013-01-24T01:32:03Z 2013-01-24T01:32:03Z I've got to agree with Lee; after all, nobody seems to worry too much about the use of the word volume' to describe measure in n dimensions for $n&gt;3$. http://mathoverflow.net/questions/111409/twisted-bundle-definition/111411#111411 Comment by Danny Ruberman Danny Ruberman 2012-11-03T22:48:13Z 2012-11-03T22:48:13Z This is the correct mathematical answer; the sociological answer is that twisted' usually just means non-trivial. As Ryan says, in this setting there's only one non-trivial bundle, so the usage is not ambiguous.