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- Thread starter JohnSt
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samalkhaiat

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Does anybody know a simple proof of the fact that there are no finite-dimensional extensions of the [tex]\textsl{so(n)}[/tex]-spinor representation to the group of general linear transformations. The proof seems can be based on the well-known fact that when rotated [tex]2\pi[/tex] a spinor transforms [tex]\psi\rightarrow-\psi[/tex]. But i have found no elementary proof.

SO(n)-spinors belong not to the group SO(n) but to its universal covering group which happens to have linear representations (spinors) other than SO(n) representations: SU(2) for SO(3) and SL(2,C) for SO(1,3).

A spinor or double-valued representation of SO(n,m) is by definition a linear representation of Sp(n,m) that cannot be obtained from a representation of SO(n,m).

Like SO(n), the general linear group GL(n) is not simply connected. However [unlike SO(n)], its universal covering group has no linear representations other than GL(n) representations.

See the theorem on page 151 of E. Cartan, "The Theory of Spinors", Dover Edition 1981.

regards

sam

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Haelfix

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Haelfix

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Its possible that its not there though, in which case I apologize.

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samalkhaiat

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I wonder if there is a simple proof with no mention of double-coverings etc

No, I don't think you will find such a "proof". The Following is the only way to define spinors on an oriented manifold M:

One starts from a principal dundle, say a, over M, with total space denoted by E(a). Then one assumes that SO(n) is the structural group of a. A spin-structure on a is (by definition) a pair (b,f) consisting of;

1) a principal bundle b over M, with total space E(b) and structural group identified with Spin(n), i.e., the 2-fold covering of SO(n).

2) a map [itex]f: E(b) \rightarrow E(a)[/itex] such that

[tex]fr_{1} = r_{2}( f \times g )[/tex]

where g is the homomorphism from Spin(n) to SO(n), [itex]r_{1}[/itex] is the right action of Spin(n) and [itex]r_{2}[/itex] is the right action of SO(n).

So, it is all about replacing SO(n) by its 2-fold covering group Spin(n). If this is possible, one then says that M admits a spin-structure;

" The necessary and sufficient condition for an SO(n) bundle to be endowed with a spin-structure is that its second Stiefel-Whitney class index should vanish."

The point is that we can not construct spinors from the metric tensor alone and the GL(n) generators can not be written in terms of Clifford numbers.

regards

sam

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I agree with you that modern language is best suited, and I would strongly recommand to use it as you do. However, I want to point out that spinors were first defined by Cartan, as you know since you have his book, at a time where vector bundles were not even known. So I do believe one could construct a proof without explicit use of fiber bundles (although one will not get away without topological consideration of course, such as double-covering)No, I don't think you will find such a "proof".

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samalkhaiat

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So I do believe one could construct a proof without explicit use of fiber bundles (although one will not get away without topological consideration of course, such as double-covering)

Yes. And I did mention that Cartan proves it on page 151. The OP asked for a "Simple Proof" that avoids the use of the double covering! Such a proof, I believe, does not exists.

regards

sam

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