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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 $x \in M$ a 2d tangent space $TM_{x}$. The disjoint union of $TM_{x}$ at all $x$ defines the total space $TM$ of a tangent bundle ($TM$, $\pi_{TM}$, $M$) whos projection is given by:

\begin{equation} \pi_{TM}: TM \rightarrow M \end{equation}

The worldsheet $M$ is the base space of the tangent bundle and each $TM_{x}$ is a fibre.

Since the tangent space is 2d, the bases that exist in each $TM_{x}$ are 2d also. Since the base space is pseudo-Riemannian, so is the tangent space and the ordered bases (frames) that exist on each $TM_{x}$ 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 $TM_{x}$ 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 $(F_{SO(1,1)}(M), \pi_{F}, M, SO(1,1))$, where the projection acts as:

\begin{equation} \pi_{F}: F_{SO(1,1)}(M) \rightarrow M \end{equation}

The fibre $F_{x}$ of this frame bundle at a point $x$ on $M$ is the set of all frames of $TM_{x}$ at the same point $x$. $F_{x}$ 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 $(P, \pi_{P}, M, SO(1,1))$ with projection:

\begin{equation} \pi_{P}: P \rightarrow M \end{equation}

The fibre $P_{x}$ of this frame bundle at a point $x$ on $M$ is the set of all frames of $TM_{x}$ at the same point $x$. $P_{x}$ 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 $TM_{x}$ be homeomorphic to both $SO(1,1)$ and $Spin(1,1)$?

The spinor bundle can then be defined to be given by $(S, \pi_{S}, M, \Delta_{(1,1)} Spin(1,1))$, with projection that acts as:

\begin{equation} \kappa: S \rightarrow M \end{equation}

Here $S$ is given by:

\begin{equation} S = P \times_{\kappa} \Delta_{(1,1)} = (P \times \Delta_{(1,1)})/Spin(1,1) \end{equation}

The fibre is given by $\Delta_{(1,1)}$ 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.

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It is nearly correct. First, $F_x$ is the space of oriented and orthonormal frames of $T_x M$. 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 $\phi: P \to F$ 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 $T_xM$ there exists two points $p$ and $p'$ in the fiber $P_x$ which yield this frame by $\phi(p) = \phi(p')$. Thus $P_x$ is not homeomorphic to the frames on $T_xM$.

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  • $\begingroup$ Thanks for the reply Tobias. Wouldn't $F_{x}$ be the space of oriented and pseudo-orthonormal frames of $T_{x}M$, just to be a little more precise? (Since $T_{x}M$ has Lorentzian metric signature). And I understand the $Spin(1,1)$-bundle now, thank you. $\endgroup$ – Siraj R Khan Mar 4 '15 at 10:49
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    $\begingroup$ You are welcome. Yes of course, it is pseudo-orthonormal. I was just too lazy to write it ;-). But note that the same argumentation goes through for the Riemannian case and in arbitrary dimension (with the obvious modifications of the groups, i.e. you get a $Spin(n)$ bundle). $\endgroup$ – Tobias Diez Mar 4 '15 at 13:32

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