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  Is my Summary of a Spinor Bundle Associated with a String Worldsheet Correct?

+ 1 like - 0 dislike
941 views

I've been having difficulty finding a source that lists all the properties of the spinor bundle of a string worldsheet explicitly, so I've had a go at creating my own description. I'd really appreciate it if someone could tell me if the following is true:


Take the worldsheet to be some 2d pseudo-Riemannian orientable manifold M.

One can associate with each point xM a 2d tangent space TMx. The disjoint union of TMx at all x defines the total space TM of a tangent bundle (TM, πTM, M) whos projection is given by:

πTM:TMM

The worldsheet M is the base space of the tangent bundle and each TMx is a fibre.

Since the tangent space is 2d, the bases that exist in each TMx are 2d also. Since the base space is pseudo-Riemannian, so is the tangent space and the ordered bases (frames) that exist on each TMx are 'pseudo-orthonormal'. This would mean that the bases transform under an O(1,1) group. However, Since the base space M is orientable, so is each TMx and that means that the frames are oriented pseudo-orthonormal and transform under SO(1,1) instead.

This allows the oriented orthonormal frame bundle (a specific sub-class of principal bundle) to be written as (FSO(1,1)(M),πF,M,SO(1,1)), where the projection acts as:

πF:FSO(1,1)(M)M

The fibre Fx of this frame bundle at a point x on M is the set of all frames of TMx at the same point x. Fx is homeomorphic to the gauge group SO(1,1) and is said to be an SO(1,1)-torsor.

However, now one can define a lift of the group SO(1,1) to Spin(1,1). The corresponding frame bundle is now (P,πP,M,SO(1,1)) with projection:

πP:PM

The fibre Px of this frame bundle at a point x on M is the set of all frames of TMx at the same point x. Px is homeomorphic to the gauge group Spin(1,1) and is said to be an Spin(1,1)-torsor.

How can the set of all frames in TMx be homeomorphic to both SO(1,1) and Spin(1,1)?

The spinor bundle can then be defined to be given by (S,πS,M,Δ(1,1)Spin(1,1)), with projection that acts as:

κ:SM

Here S is given by:

S=P×κΔ(1,1)=(P×Δ(1,1))/Spin(1,1)

The fibre is given by Δ(1,1) which is the Hilbert space of all spinor states. Each section of this bundle then corresponds to a particular Majorana-Weyl spinor field configuration on the worldsheet.

This post imported from StackExchange Physics at 2015-03-04 12:46 (UTC), posted by SE-user Siraj R Khan
asked Mar 3, 2015 in Theoretical Physics by Siraj R Khan (105 points) [ no revision ]

1 Answer

+ 1 like - 0 dislike

It is nearly correct. First, Fx is the space of oriented and orthonormal frames of TxM. Furthermore, you gave the correct definition of the Spin(1,1) bundle but confused a little bit its meaning. The Spin(1,1,) bundle is defined as the lift of the SO(1,1) bundle to Spin(1,1). This means we have a principal Spin(1,1)-bundle P and a principal bundle map ϕ:PF which is at the same time a covering (this literally lifts the double covering of SO(1,1) by Spin(1,1) to the bundle picture). For more details see http://en.wikipedia.org/wiki/Spin_structure#Spin_structures_on_Riemannian_manifolds. Now P is not a frame bundle. More precisely, for every oriented, orthonormal frame of TxM there exists two points p and p in the fiber Px which yield this frame by ϕ(p)=ϕ(p). Thus Px is not homeomorphic to the frames on TxM.

This post imported from StackExchange Physics at 2015-03-04 12:46 (UTC), posted by SE-user Tobias Diez
answered Mar 4, 2015 by Tobias Diez (90 points) [ no revision ]
Thanks for the reply Tobias. Wouldn't Fx be the space of oriented and pseudo-orthonormal frames of TxM, just to be a little more precise? (Since TxM has Lorentzian metric signature). And I understand the Spin(1,1)-bundle now, thank you.

This post imported from StackExchange Physics at 2015-03-04 12:46 (UTC), posted by SE-user Siraj R Khan

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