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Why there are no spinors for GL(n)

by JohnSt
Tags: spinors
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JohnSt
#1
Jun14-08, 02:53 AM
P: 8
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...
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samalkhaiat
#2
Jun21-08, 09:03 PM
Sci Advisor
P: 888
[QUOTE=JohnSt;1766142]
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
JohnSt
#3
Jun24-08, 04:09 AM
P: 8
Thank you very much. These are well or less-known but nontrivial facts. For example, highly nontrivial is the fact that the double covering of the real general linear group is not a matrix group. I wonder if there is a simple proof with no mention of double-coverings etc

Haelfix
#4
Jun24-08, 08:11 AM
Sci Advisor
P: 1,677
Why there are no spinors for GL(n)

Theres a cute little book called "spin geometry" by Lawson et al that proves the result (chapter 5) by noting there is no faithful finite dimensional representations possible. The double cover is called the metalinear group incidentally.
JohnSt
#5
Jun25-08, 04:18 AM
P: 8
Excuse me, I found no chapter 5 in this book. Paragraph 5 deals with representation, but non a single word had Lawson said about general linear group and its infinite-dimensional spinors.
Haelfix
#6
Jun25-08, 08:33 AM
Sci Advisor
P: 1,677
Eeep, apologies, The PDF I have is evidently a lot different than the published book. Unfortunately the book is checked out of our library so I can't find the appropriate corresponding sections. Eyeballing the google book chapter content, maybe look in the representation section on page 30, or somewhere where they talk about Dirac operators?

Its possible that its not there though, in which case I apologize.
samalkhaiat
#7
Jun25-08, 05:19 PM
Sci Advisor
P: 888
[QUOTE=JohnSt;1777697]
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
humanino
#8
Jun25-08, 05:42 PM
humanino's Avatar
P: 2,828
Quote Quote by samalkhaiat View Post
No, I don't think you will find such a "proof".
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)
samalkhaiat
#9
Jun25-08, 05:54 PM
Sci Advisor
P: 888
[QUOTE=humanino;1779554]
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
JohnSt
#10
Jun26-08, 02:42 AM
P: 8
You are too skilled! One need not "spinor bundles" to prove that it is not possible to extend spinorial representation to a representation of general linear group without enlargring the representation space. It is all about spinors and vectors, not about sections of bundles. Thank you for trying.


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