# How Do Spinors Fit in With Differential Geometry

#### Ben Niehoff

Science Advisor
Gold Member
Thank you. I guess all those buzz words don't apply to the humble torus, no spinors on a torus?
You can have spinors on the torus, as well as every orientable Riemann surface.

#### Spinnor

Gold Member
You can have spinors on the torus, as well as every orientable Riemann surface.
That is cool! Thank you. You can't know too many facts about spinors and now I know 2 more!

The Wiki article on spinors has a table summary that lists "objects" that exist for a given signature of Euclidean space (which I guess are different types of spinors)?

Can there be different signatures on a torus that also give rise to different types of spinors?

http://en.wikipedia.org/wiki/Spinor#Summary_in_low_dimensions

Thanks for the help!

.

#### lavinia

Science Advisor
Gold Member
Spin structure in indefinite signature is significantly more complicated, if I recall some classes in string theory from grad school. I think the trick was that one had to split the orientable bundle E into a diirect sum of a positive definite part and a negative definite part, and then the obstruction to having a spin structure was that the difference of the second Stieffel Whitney class applied to both pieces had to vanish. I would be incapable of proving this piece of memory though.
The reason, I think, is that if the SO(n,m) bundle is (n,m)-orientable, that is, the structure group can be reduced to the connected component of the identity, then the bundle splits into a Whitney sum of two oriented sub-bundles and a Spin(n,m) structure reduces to a Spin structure on each summand. It follows that the second Stiefel Whitney class of each summand must be zero.

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#### lavinia

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While I don't know anything about Spin geometry, in order to answer the original question which was how do Spinors play a role in Differential Geometry, perhaps this may be helpful to you. I am paraphrasing a section of the introduction to the book "Spin Geometry" by Lawson and Michelson.

The tangent bundle of a manifold has structure group the general linear group. If the structure group can be lifted to the universal covering group, the manifold is a Spin manifold. Sadly, the universal covering group has no finite dimensional representations other than those obtained by projecting onto the general linear group and then following the projection with a representation of GL(n). So from this point of view Spin structures do not provide any new geometry.

However with a Riemannian metric on an orientable manifold, one can reduce the structure group to SO(n), and the universal cover, Spin(n), does in fact have independent finite dimensional representations. This means that the metric allows one to construct new vector bundles which do not exist on manifolds that do not have a Spin structure.

These "bundles of Spinors" depend on the choice of metric and thus should be expected to be linked with the geometry of the manifold.

Interestingly, the authors say that mathematicians are trying to formulate a "spinor calculus" analogous to tensor calculus but at the time of the writing of the book, it has not been well defined.

I imagine that since the Spin(n) principle bundle covers the principle SO(n) bundle that a connection on the SO(n) bundle lifts to a connection on the principal Spin(n) bundle. This connection together with a finite dimensional representation of the Spinor group will produce a covariant differentiation on the induced Spinor bundle.

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#### samalkhaiat

Science Advisor
Spin structure in indefinite signature is significantly more complicated, if I recall some classes in string theory from grad school. I think the trick was that one had to split the orientable bundle E into a diirect sum of a positive definite part and a negative definite part, and then the obstruction to having a spin structure was that the difference of the second Stieffel Whitney class applied to both pieces had to vanish. I would be incapable of proving this piece of memory though.
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 $w_{2} (E^{+}) = w_{2} (E^{-})$.

[*] M. Karoubi (1968). "Algèbres de Clifford et K-théorie". Ann. Sci. Éc. Norm. Sup.1 (2): 161–270.

#### lavinia

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Gold Member
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 [Broken])

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.

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#### pellis

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

Science Advisor
Gold Member
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.

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