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Why there are no spinors for GL(n)by JohnSt
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#1
Jun1408, 02:53 AM

P: 8

Does anybody know a simple proof of the fact that there are no finitedimensional 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 wellknown fact that when rotated [tex]2\pi[/tex] a spinor transforms [tex]\psi\rightarrow\psi[/tex]. But i have found no elementary proof...



#2
Jun2108, 09:03 PM

Sci Advisor
P: 888

[QUOTE=JohnSt;1766142]
A spinor or doublevalued 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 


#3
Jun2408, 04:09 AM

P: 8

Thank you very much. These are well or lessknown 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 doublecoverings etc



#4
Jun2408, 08:11 AM

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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.



#5
Jun2508, 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 infinitedimensional spinors.



#6
Jun2508, 08:33 AM

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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. 


#7
Jun2508, 05:19 PM

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P: 888

[QUOTE=JohnSt;1777697]
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 spinstructure 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 2fold 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 2fold covering group Spin(n). If this is possible, one then says that M admits a spinstructure; " The necessary and sufficient condition for an SO(n) bundle to be endowed with a spinstructure is that its second StiefelWhitney 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 


#8
Jun2508, 05:42 PM

P: 2,828




#9
Jun2508, 05:54 PM

Sci Advisor
P: 888

[QUOTE=humanino;1779554]
regards sam 


#10
Jun2608, 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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