How Do Spinors Fit in With Differential Geometry

[lavinia]

The complicated detail was worked out by Karoubi in 1968[*] I summarize his results here:
If an orientable bundle E has structure group SO(n,m), one can always reduce the structure group to its maximal compact subgroup S(O(n) × O(m)). This reduction is equivalent to the choice of a maximal positive definite sub-bundle E^{+} < E, which allows us to write E as the direct sum, E = E^{+} \oplus E^{-} , where E^{-} = (E^{+})^{\perp} is negative definite. Then, one can show that the SO(n, m) bundle E admits \mbox{Spin} (n, m) structure if and only if w_{2} (E^{+}) = w_{2} (E^{-}).

I would like to see the proof. It seems that the condition only guarantees that the second Stiefel-Whitney class of the bundle is zero which means that it has a Spin(p+q) structure not a Spin(p,q) structure. For that I would think the second Whitney class of each summand must be zero.

* Roughly, the universal classifying space for SO(p,q) structures ought to be BSO(p) x BSO(q) up to homotopy equivalence so that its second Z2 cohomology would be generated by the two universal Stiefel-Whitney classes(Kunneth formula) of the two factors and the condition that the manifold admits a Spin(p,q) structure should imply that the induced map on the second Z2 cohomology group is zero.

One can define a Spin structure as a cohomology class in $H^1$(SO(p) bundle;Z2) whose restriction to $H^1$(Fiber;Z2) is surjective for each fiber,F. (For details see http://retro.seals.ch/cntmng?pid=ensmat-001:1963:9::66 )

It follows that the transgression mapping $t: H^1$(Fiber) $-> H^2$(Manifold) is zero.
However, in the universal bundle the transgression mapping is injective. So by a diagram chasing argument the classifying map for the bundle must map the universal second Stiefel-Whitney class to zero.

For an SO(p,q) bundle the argument is the same except the cohomology coefficients are Z2xZ2 rather than Z2 and there are now two universal second Stiefel Whitney classes, one for each factor.

 

[pellis]

.
You might have thought that this system (save for the ribbon) is in fact a vector. The concept of a vector in school is taught as something that has direction and magnitude. This is not true. A vector can describe much more. Something with direction and magnitude (plus parity number of wraps) is a spinor. A vector is a much more complex object.
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Interesting thread on a perennial topic.

Could you please expand on the ways in which you reckon that a vector "is a much more complex object"? (I realize that your use of "complex" here is in the natural language rather than mathematical sense).

 

[pellis]

Could you please expand on the ways in which you reckon that a vector "is a much more complex object"? .

Further: Having looked at the various uses of the term vector in maths ( http://en.wikipedia.org/wiki/Vector_(mathematics_and_physics) ) it seems that for almost every entry one can replace the word vector by the word spinor, e.g. vector field, vector space, vector calculus, vector bundle etc. all have their equivalents as spinor field, spinor space, spinor calculus, spinor bundle. The more specific terms like Darboux vector or wave vector refer to physical situations represented by vectors, just as there are specific applications of spinors, e.g pulse synthesis in medical NMR spectroscopy.

Can you still support your contention that a vector "is a much more complex object" than a spinor, please, haael?

Regards - pellis

 

[lavinia]

But once you introduce spinors, it seems that the geometric way of understanding things fails. People are reduced to saying "a spinor is something that transforms in such-and-such a way under rotations" again.The mathematically sophisticated way of describing them in terms of representations of the group of rotations (or Lorentz boosts) don't really provide much more understanding of what they really "are".

Is there some geometric way of understanding spinors that is more in line with the "tangent to a parametrized path" understanding of a vector?

While this doesn't really answer your question I find these ideas helpful.

- First of all, on a general vector bundle, there is no natural geometric way to interpret the vectors other than algebraically as elements of a vector space at each point of the manifold.

When one wants to take directional derivatives, or covariant derivatives, one must have tangent vectors and it is the tangent bundle which is naturally geometric. One can have some generalized ideas of geometry on vectors bundles using a connection but intuition goes away quickly and differentiation is still done with respect to tangent vectors.

- When a manifold is oriented its structure group can be reduced to SO(n) from O(n). An orientation is not a property of a vector but rather of the entire vector space.

By analogy one can think of a Spin structure as an "orientation" of an oriented manifold.

A manifold is orientable when its first Stiefel-Whitney class is zero and then it has two orientations ,one for each element of $H^0(M;Z2)$

Similarly, if an oriented manifold has a zero second Stiefel Whitney class then it has one Spin structure for each element of $H^1(M;Z2)$.
This can be seen from the exact cohomology sequence (with Z2 coefficients),

$0 ->H^1$(Manifold) ->$H^1$(SO(p) Principal bundle) ->$H^1$(SO(p)) ->$H^2$(Manifold)

In the case that the second Stiefel-Whitney class is zero the last term on the right may be taken to be zero. Since $H^1(SO(p)) = Z2$
the sequence becomes

$0 ->H^1$(Manifold) -> $H^1$(SO(p) Principal bundle) -> Z2 -> 0

So each element of $H^1$(Manifold) corresponds to a Spin structure on the bundle of oriented orthonormal frames.