Def./Examples of Spin Structure Re Trivial Bundles?

In summary: I think it was H_1(Sg,Z/2)) ; a result I find hard to believe. In summary, a spin structure in a manifold M is equivalent to M admitting a trivialization of the tg.bundle over the 1-skeleton of M. A spin structure (in this layout) is a choice, up to homotopy of a (global) trivialization of the bundle over the 1-skeleton that extends to a trivialization over the 2-skeleton.
  • #1
Bacle
662
1
Hi, All:

I am reading a paper in which , if I understood well, a spin structure in a manifold M

is equivalent to M admitting a trivialization of the tg.bundle over the 1-skeleton of M

( I guess M is assumed to be "nice-enough" so that it is a simplicial complex )

so that this trivialization extends to a trivialization of the 2-skeleton; then , e.g., S^2 (seen as a

2-simplex, with 1-simplices as S^1's ) would not be spin, since a trivialization over any 1-subskeleton S^1

of S^2 cannot extend to a trivialization of the bundle over S^2 --because the

bundle over S^2 is not trivial, . A spin structure (in this layout) is a choice, up to homotopy

of a (global) trivialization of the bundle over the 1-skeleton that extends to a trivialization over the

2-skeleton.


O.K, so far. But now there is a different use of spin structure which is supposedly

equivalent to the one just given: a 4-manifold is spin if every homology 2-class can be

repd. by an embedded sphere; ** BUT ** , while then S^2 is not Spin by the first

def ( bundle over S^2 is not trivial, by , e.g. the Hairy Ball Thm. ), it is spin by

the second def., since its second homology class is S^2 itself --because S^2 is

orientable. Am I missing something here somewhere?

Any Ideas?

Thanks in Advance.
 
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  • #3
I would think that the second condition doesn't apply to S^2 because it's not 4-dimensional :)

Anyway, you have to be careful here. In dimensions greater than two, SO(n) has fundamental group Z/2. Spin(n) is the unique double cover of this. However, this doesn't go through for two dimensions, since the double cover of the circle is itself. The condition that does go through in both cases is that w_2(TM) vanishes.

To see why the condition of passing from the one-skeleton to the two-skeleton doesn't work in two dimensions, let's do the obstruction theory. Fix n>2, and fix a trivialization over the one-skeleton, thought of as a map S^1 -> BSO(n). The obstruction to extending the map over the two-skeleton lies in H^2(S^2, pi_1(BSO(n)))=0, while the obstruction to extending the map as a homotopically trivial map lies in H^2(S^2,pi_2(BSO(n)))=H^2(S^2,pi_1(SO(n)))=H^2(S^2,Z/2), which is precisely w_2. I.e. two SO(n) bundles over S^2 are isomorphic iff they have the same second Stiefel-Whitney class. From Cech cohomology, you see right away that this is precisely the obstruction to lifting the bundle to a Spin(n) bundle, i.e. an SO(n) bundle over S^2 is trivial iff its second Stiefel-Whitney class vanishes iff it's Spin. You also glean from this that the obstruction for any manifold being spin is w_2 being trivial, which is the same thing as the classifying map over the two-skeleton being homotopically trivial.

This all changes if n=2. Here since Spin(2)=S^1, we want to lift the bundle to an S^1 bundle which is compatible with the double cover S^1 -> S^1 (this is still w_2). You'll notice right away that this has nothing to do with the bundle being trivial over the two-skeleton.
 
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  • #4
Thanks, Zhentil:
As you may have noticed, this is not my area of expertise.

Still, is there a connection here with being able to represent homology two-classes

by embedded spheres ( in which case CP^2 would be spin, since H_2(CP^2)=CP^1 , and

CP^1 ~ S^2 --actually they are diffeomorphic )? But, as Niehoff points out, Wikipedia

claims that only CP^(2n+1) is spin . I was just reading some article by Kirby and Friedman

in which , if I had understood well, they were using a certain quadratic Z/2-form

( a quadratic form over Z/2) called the Rokhlin form ; defined on H_1(Sg,Z/2) ;(I think

Sg was an embedded copy of g-torus on the manifold M in question) to determine if a

manifold was spin. The result was then that the manifold was spin if the form was killed

on a symplectic basis {x1,y1,x2,y2,...,x2g,y2g} (this is a basis for the 1-homology

of the embedded g-torus g-torus, in which (xi,yj)=1 if i=j , and (xi,yj)=0 otherwise )

So, basically, if the Rokhlin form was zero on the symplectic basis, then the handles

of the 2-class rep. could be surgered away without affecting the overall homology.

Sorry for running off; the point is it seemed like they were arguing that if the manifoild

was spin, then the handles in the 2-class could be surgered away, and the class would

be represented by an embedded sphere.
 
  • #5
No, there is no way that CP^2 is spin. w_2 is the mod 2 reduction of the first Chern class, which is non-zero for CP^2.
 
  • #6
Thanks, Zhentil; I don't mean to be argumentative, I just want to clarify some things:

so being spin has no relation then with representing homology 2-classes (in a 4-manifold )

by embedded spheres, i.e., with being able to lower the genus of the representatives

for H_2(M^4)?

P.S: thanks for your patience; I am doing way too much stuff and I sometimes skip
on important details; hopefully that will change soon.
 
  • #7
No, it's certainly possible that that's true. I can't prove it off the top of my head, but I just wanted to point out that CP^2 isn't a counterexample, since the higher multiples of CP^1 aren't representable by embedded spheres.
 
  • #8
Zhentil:Sorry for the bother, just to know what B(SO(n)) stands for. Also wondering if you got some
good refs. for this topic.
 
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  • #9
It's no trouble at all :) BSO(n) is the classifying space for SO(n) bundles. Other people might be able to help you more on the references, but I learned what I know from Milnor's characteristic classes, Differential Topology by Hirsch, and Differential Forms and Rational Homotopy Theory by Griffiths and Morgan (this one is the best imo for obstruction theory).
 
  • #10
Also Kirby Calculus by Gompf and Stipsicz is an amazing book. It's a good place to learn, e.g., that a spin structure is equivalent to a trivialization over the one-skeleton that extends over the two-skeleton.
 

1. What is a spin structure?

A spin structure is a mathematical concept used in differential geometry and topology to describe the geometric properties of a space. In simple terms, it is a way of assigning a set of frames (or basis vectors) to each point on a manifold, which allows for the definition of a spinor field.

2. What is a trivial bundle?

A trivial bundle is a type of fiber bundle where the fibers are all identical copies of the same space. In other words, the bundle looks the same at every point and can be thought of as a product space. For example, a cylinder can be thought of as a trivial bundle of circles over a line.

3. How are spin structures and trivial bundles related?

Spin structures are often used to define bundles, and in particular, the spin structure on a manifold determines the type of bundle that can be defined on it. In the case of trivial bundles, the spin structure is always trivial, meaning that it can be extended to a spin structure on the entire bundle.

4. What are some examples of trivial bundles?

Some examples of trivial bundles include the tangent bundle of a manifold, the Möbius strip, and the Klein bottle. In all these cases, the base space (manifold) is the same as the total space, and the fibers are all copies of the same space.

5. Why is the spin structure on trivial bundles considered trivial?

The spin structure on a trivial bundle is considered trivial because it can be extended to a spin structure on the entire bundle. In other words, the spin structure at each point is the same, and it does not change as we move along the base space. This makes it relatively easy to study the spin structure and other geometric properties of these bundles.

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