What are "good" examples of spin manifolds? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T09:22:43Z http://mathoverflow.net/feeds/question/80081 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/80081/what-are-good-examples-of-spin-manifolds What are "good" examples of spin manifolds? Otis Chodosh 2011-11-04T19:15:16Z 2011-11-06T18:38:05Z <p>I'm trying to get a grasp on what it means for a manifold to be spin. My question is, roughly:</p> <blockquote> <p>What are some "good" (in the sense of illustrating the concept) examples of manifolds which are spin (or not spin) (and why)?</p> </blockquote> <hr> <p>For comparison, I'd consider the cylinder and the mobius strip to be "good" examples of orientable (or not) bundles.</p> <hr> <p>I've read the answers to <a href="http://mathoverflow.net/questions/66681/classical-geometric-interpretation-of-spinors" rel="nofollow">http://mathoverflow.net/questions/66681/classical-geometric-interpretation-of-spinors</a> which are helpful, but I'd like specific examples (non-examples) to think about. </p> http://mathoverflow.net/questions/80081/what-are-good-examples-of-spin-manifolds/80082#80082 Answer by Kofi for What are "good" examples of spin manifolds? Kofi 2011-11-04T19:40:32Z 2011-11-06T18:01:42Z <p>If $M$ is a spin manifold, then any submanifold of codimension 1 is also a spin manifold. This yields a lot of examples, for example, that $S^n$ is spin etc.</p> <p>(I may not have understood your point completely.)</p> <p>Edit: As pointed out in the comments, one has to demand as well that the submanifold is orientable.</p> http://mathoverflow.net/questions/80081/what-are-good-examples-of-spin-manifolds/80089#80089 Answer by Johannes Ebert for What are "good" examples of spin manifolds? Johannes Ebert 2011-11-04T22:08:22Z 2011-11-04T22:08:22Z <p>A simply connected $4$-manifold is spin iff all embedded oriented surfaces have even self-intersection number or, equivalently, if the quadratic form $H_2 (M;Z) \to Z$ induced by the intersection form takes even values. This is by the following string of arguments:</p> <ol> <li><p>$M$ is spin iff $w_2 (TM)=0$.</p></li> <li><p>$w_2 (TM)=0$ iff the linear form $H_2 (M; Z/2) \to Z/2$, $a \mapsto \langle w_2 (TM);a\rangle$ is null.</p></li> <li><p>Any class $a \in H_2 (M;Z)$ can be represented as the fundamental class of an embedded oriented surface $F \subset M$.</p></li> <li><p>$w_2 (TM)|_F = w_2 (\nu_F)$ by the product formula for Stiefel-Whitney classes and because $F$ is spin.</p></li> <li><p>$w_2 (\nu_F)$ is the mod $2$ reduction of the Euler class of the normal bundle of $F$.</p></li> <li><p>$\langle [F]; \chi(\nu_F) \rangle$ is the self-intersection number of $F$, or equivalently, the value of the quadratic form at $[F]$.</p></li> </ol> <p>Now you should play a bit with $4$-manifolds and might get a feeling for the spin condition.</p> http://mathoverflow.net/questions/80081/what-are-good-examples-of-spin-manifolds/80090#80090 Answer by Ryan Budney for What are "good" examples of spin manifolds? Ryan Budney 2011-11-04T22:15:23Z 2011-11-04T23:34:02Z <p>There's the traditional obstruction-theoretic perspective. Orientability means the tangent bundle trivializes over a 1-skeleton. Dually you could think of that as saying the complement of a co-dimension $2$ subcomplex has a trivial tangent bundle. </p> <p>So admitting a spin structure is the same, but it will be the tangent bundle trivializes over a 2-skeleton, dually the complement of a co-dimension three subcomplex admits a trivial tangent bundle. </p> <p>A surface is orientable if and only if it contains no Moebius bands -- a regular neighbourhood of any simple closed curve must be a cylinder. In higher dimensions this translates into a manifold being orientable if and only if it contains no twisted bundles $D^{n-1} \rtimes S^1$, i.e. regular neighbourhoods of simple closed curves are diffeomorphic to $D^{n-1} \times S^1$. </p> <p>For spin structures there's something very similar. Of course, a surface admits a spin structure if and only if it is orientable. It's a more interesting notion in higher dimensions. The statement there is the manifold is orientable, and if you take a regular neighbourhood of any surface in the manifold, then it has a trivial tangent bundle. So manifolds like $\mathbb RP^3$ are perfectly valid spin manifolds -- $\mathbb RP^3$ contains $\mathbb RP^2$ but the total space of its normal bundle has a perfectly trivializable tangent bundle. Technically, the condition is a little stronger than that -- you can trivialize the tangent bundle of the complement of a co-dimension $3$ subset. So not only can you trivialize the total spaces of normal bundles of surfaces, but even the regular neighbourhoods of unions of surfaces. </p> <p>So if you want a manifold that isn't spin, the archetype would be a vector bundle over a surface so that the total space does not have a trivializable tangent bundle. Take the $D^2$-bundle over $S^2$ with Euler Class $\chi$. I think this happens if and only if $\chi$ is even. I suppose you have more entertaining examples when dealing with the regular neighbourhood of a 2-complex that isn't itself a manifold. </p> <p>edit: Milnor's "Spin structures on manifolds" in L'Enseignement Mathematique Vol 9 (1963) is an excellent reference for most of the above. I don't believe he goes into all the descriptions above since I think he wants to keep the article simple. The Poincare duality interpretation above is a very standard mode of thinking that's employed throughout much of low-dimensional topology. Kirby's book on 4-manifolds is a nice place to look for this material. Specifically, R. Kirby "The topology of 4-manifolds" Springer-Verlag (1989). A more modern reference would be Gompf and Stipsicz, but again I don't think they use all the above descriptions. Milnor and Stasheff's "Characteristic Classes" describes most of the basic constructions involved above, in the obstruction theory section. In a couple of months I'll be putting up a paper on the arXiv that gives some very combinatorial ways of describing spin and spin^c-structures on manifolds (mostly for computer implementation). I hope that will be a good reference, too! But the paper is still unreadable. </p> http://mathoverflow.net/questions/80081/what-are-good-examples-of-spin-manifolds/80231#80231 Answer by AndrĂ© Henriques for What are "good" examples of spin manifolds? AndrĂ© Henriques 2011-11-06T18:38:05Z 2011-11-06T18:38:05Z <p>If you know about Steenrod operations, here's a very convenient characterization:</p> <blockquote> <p>A manifold $M$ is <i>Spin</i> iff its Poincare duality in $H^*(M,\mathbb Z/2)$ is compatible with $Sq^1$ and $Sq^2$.</p> </blockquote> <p>Similarly, <i>oriented</i> manifolds are those whose Poincare duality in $H^*(M,\mathbb Z/2)$ is compatible with $Sq^1$.</p> <p>The story continues: <i>String</i> manifolds have a Poincare duality in $H^*(M,\mathbb Z/2)$ that is compatible with $Sq^1$, $Sq^2$ and $Sq^4$ (but now, that's no longer an if and only if). My paper <a href="http://arxiv.org/abs/0810.2131" rel="nofollow">http://arxiv.org/abs/0810.2131</a> with Chris Douglas and Mike Hill describes all that in detail and provides many concrete examples.</p>