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I'm trying to get a grasp on what it means for a manifold to be spin. My question is, roughly:

What are some "good" (in the sense of illustrating the concept) examples of manifolds which are spin (or not spin) (and why)?


For comparison, I'd consider the cylinder and the mobius strip to be "good" examples of orientable (or not) bundles.


I've read the answers to Classical geometric interpretation of spinors which are helpful, but I'd like specific examples (non-examples) to think about.

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The manifold $X$ is spin iff the loop space $LX$ is orientable. But I don't think this will help. If you look at $spin^c$, almost complexe manifold $M$ is $spin^c$. All of the 4 dimention manifold is $spin^c$. So if more over, $det^(1/2) TM$ exist, $M$ is spin. –  shu Nov 4 '11 at 20:02
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Projective spaces are good examples and non-examples. Similarly, you can consider discrete quotients of spheres. Then depending on the group you quotient by, the resulting space is (or not) spin. –  José Figueroa-O'Farrill Nov 4 '11 at 20:13
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@Otis: See a preprint by Stephan Stolz and Peter Teichner (web.me.com/teichner/Math/Surveys_files/MPI.pdf), where they prove that M is spin if and only if LM is fusion-orientable and M is string if and only if LM is fusion-spin. –  Dmitri Pavlov Nov 5 '11 at 10:48
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@Otis:For the spin^c and spin , the reference is the lovely book write by Morgan. press.princeton.edu/titles/5866.html. –  shu Nov 5 '11 at 21:02
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@shu, Otis: The fusion condition mentioned by Dmitri Pavlov is essential - a real Enriques surface is not spin, but its loop space is orientable. –  Konrad Waldorf Nov 7 '11 at 8:55

4 Answers 4

up vote 26 down vote accepted

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.

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.

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 $D^{n-1} \times S^1$.

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.

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.

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.

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Thanks! This is enlightening. Can you suggest a reference for this? –  Otis Chodosh Nov 4 '11 at 23:14

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:

  1. $M$ is spin iff $w_2 (TM)=0$.

  2. $w_2 (TM)=0$ iff the linear form $H_2 (M; Z/2) \to Z/2$, $a \mapsto \langle w_2 (TM);a\rangle$ is null.

  3. Any class $a \in H_2 (M;Z)$ can be represented as the fundamental class of an embedded oriented surface $F \subset M$.

  4. $w_2 (TM)|_F = w_2 (\nu_F)$ by the product formula for Stiefel-Whitney classes and because $F$ is spin.

  5. $w_2 (\nu_F)$ is the mod $2$ reduction of the Euler class of the normal bundle of $F$.

  6. $\langle [F]; \chi(\nu_F) \rangle $ is the self-intersection number of $F$, or equivalently, the value of the quadratic form at $[F]$.

Now you should play a bit with $4$-manifolds and might get a feeling for the spin condition.

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If you know about Steenrod operations, here's a very convenient characterization:

A manifold $M$ is Spin iff its Poincare duality in $H^*(M,\mathbb Z/2)$ is compatible with $Sq^1$ and $Sq^2$.

Similarly, oriented manifolds are those whose Poincare duality in $H^*(M,\mathbb Z/2)$ is compatible with $Sq^1$.

The story continues: String 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 http://arxiv.org/abs/0810.2131 with Chris Douglas and Mike Hill describes all that in detail and provides many concrete examples.

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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.

(I may not have understood your point completely.)

Edit: As pointed out in the comments, one has to demand as well that the submanifold is orientable.

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this is false: see Ryan Budney's answer (RP^2 in RP^3). Maybe this is true if the submanifold is codimension 1 and 2-sided? –  Ian Agol Nov 5 '11 at 21:31
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It's true so long as the normal bundle to the codimension 1 submanifold is trivial. So if both manifolds involved are orientable then it's true. –  Vitali Kapovitch Nov 6 '11 at 1:38

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