The spinor bundle $S$ on $\bS^2$

Consider the 2-dimensional sphere $\bS^2$, with its usual orientation, $\bS^2 = \bC \cup \{\infty\} \isom \bCP^1$. The usual spherical coordinates on $\bS^2$ are

$$p = (\sin\th\cos\phi, \sin\th\sin\phi, \cos\th) \in \bS^2.$$

The poles are $N = (0,0,1)$ and $S = (0,0,-1)$. Let $U_N = \bS^2 \setminus \{N\}$, $U_S = \bS^2 \setminus \{S\}$ be the two charts on $\bS^2$. Consider the stereographic projections $p \mapsto z : U_N \to \bC$, $p \mapsto \ze : U_S \to \bC$ given by

$$z := e^{-i\phi} \cot\frac{\th}{2}, \qquad \ze := e^{+i\phi} \tan\frac{\th}{2},$$

so that $\ze = 1/z$ on $U_N \cap U_S$. Write

$$q := 1 + z \bar z = \frac{2}{1 - \cos\th}, \words{and} q' := 1 + \ze\bar\ze = \frac{q}{z\bar z}.$$

The sphere $\bS^2$ has only the ``trivial'' spin structure $\sS = \Ga(\bS^2, S)$, where $S \to \bS^2$ has rank two. Now $S = S^+ \oplus S^-$, where $S^{\pm} \to \bS^2$ are complex line bundles, and these may be (and are) nontrivial. We argue that $S^+ \to \bS^2$ is the ``tautological'' line bundle coming from $\bS^2 \isom \bCP^1$. We know already that

$$\sS^\3 \isom \sS \iff S^* \isom S \Longleftarrow\ (S^+)^* \isom S^-$$

and the converse $S^* \isom S \implies (S^+)^* \isom S^-$ will hold provided we can show that $S^\pm \to \bS^2$ are nontrivial line bundles. (Otherwise, $S^+$ and $S^-$ would each be selfdual, but we know that the only selfdual line bundle on $\bS^2$ is the trivial one, since $H^2(\bS^2,\bZ) \isom \bZ$.)

Consider now the (tautological) line bundle $L \to \bS^2$, where

$$L_z := \set{(\la z_0, \la z_1) \in \bC^2 : \la \in bC}, \... ...}, \qquad L_\infty := \set{(0,\la) \in \bC^2 : \la \in \bC}.$$

In other words, $L_z$ is the complex line through the point $(1,z)$, for $z \in \bC$. A particular local section of $L$, defined over $U_N$, is $\sg_N(z) := (q^{-\shalf}, zq^{-\shalf})$, which is normalized so that $\pairing{\sg_N}{\sg_N} = q^{-1}(1 + \bar zz) = 1$ on $U_N$: this hermitian pairing on $\Ga(\bS^2,L)$ comes from the standard scalar product on $\bC^2$ --each $L_z$ is a line in $\bC^2$.

Let also $\sg_S(\ze) := (\ze q'^{-\shalf}, q'^{-\shalf})$, normalized so that $\pairing{\sg_S}{\sg_S} = 1$ on $U_S$. Now if $z \neq 0$, then

$$\sg_S(z^{-1}) = \biggl( \frac{1}{z\sqrt{q'}}, \frac{1}{\sq... ...}, \frac{z}{\sqrt{q}} \biggr) = (\bar z/z)^{1/2} \sg_N(z).$$

To avoid ambiguity, we state that $(\bar z/z)^{1/2}$ means $e^{-i\phi}$, and also $(z/\bar z)^{1/2}$ will mean $e^{+i\phi}$.

A smooth section of $L$ is given by two functions $\psi^+_N(z,\bar z)$ and $\psi^+_S(\ze, \bar\ze)$ satisfying the relation $\psi^+_N(z, \bar z) \sg_N(z) = \psi^+_S(\ze, \bar\ze)\sg_S(\ze)$ on $U_N \cap U_S$. Thus we argue that

$$\psi^+_N(z, \bar z) = (\bar z/z)^{1/2} \psi_S^+(z^{-1}, \bar z^{-1}) \word{for} z \neq 0,$$

and $\psi_N^+$, $\psi_S^+$ are regular at $z = 0$ or $\ze = 0$ respectively. Likewise, a pair of smooth functions $\psi_N^-,\psi_S^-$ on $\bC$ is a section of the dual line bundle $L^* \to \bS^2$ if and only if

$$\psi^-_N(z, \bar z) = (z/\bar z)^{1/2} \psi_S^-(z^{-1}, \bar z^{-1}) \word{for} z \neq 0.$$

We claim now that we can identify $S^+ \isom L$ and $S^- \isom L^* = L^{-1}$ --here the notation $L^{-1}$ means that $[L^{-1}]$ is the inverse of $[L]$ in the Picard group $H^2(\bS^2,\bZ)$ that classifies $\bC$-line bundles-- so that a spinor in $\bS = \Ga(\bS^2,S)$ is given precisely by two pairs of smooth functions

$$\begin{pmatrix}\psi_N^+(z, \bar z) \ [\jot] \psi_N^-(z, \... ... [\jot] \psi_S^-(\ze, \bar\ze) \end{pmatrix} \word{on} U_S,$$

satisfying the above transformation rules. (The nontrivial thing is that the spinor components must both be regular at the south pole $z = 0$ and the north pole $\ze = 0$, respectively.)

Since $\sS \ox_{\bA} \sS^* \isom \End_{\sA}(\sS) \isom \sB \isom \sA^\8(\bS^2)$ as $\sA$-module isomorphisms (we know that $\sB \isom \sA^\8(\bS^2)$ as sections of vector bundles), it is enough to show that, as vector bundles,

$$\sA^\8(\bS^2) \isom L^0 \oplus L^2\oplus L^{-2}\oplus L^0,$$

where $L^2 = L \ox L$, $L^{-2} = L^* \ox L^*$, and $L^0 = \bS^2 \x \bC$ is the trivial line bundle. It is clear that $\sA^0(\bS^2) = \Coo(\bS^2) = \sA = \Ga(\bS^2, L^0)$; and furthermore, $\sA^2(\bS^2) \isom \sA = \Ga(\bS^2, L^0)$ since $\La^2 T^*\bS^2$ has a nonvanishing global section, namely the volume form $\nu = \sin\th d\th \w d\phi$.

With respect to the ``round'' metric on $\bS^2$, namely,

$$g := d\th^2 + \sin^2\th d\phi^2 = \frac{4}{q^2} (dx^1 \ox dx^1 + dx^2 \ox dx^2),$$

the pairs of $1$-forms $\biggl\{ \dfrac{dz}{q}, \dfrac{d\bar z}{q} \biggr\}$ and $\biggl\{-\dfrac{d\ze}{q'}, -\dfrac{d\bar\ze}{q'} \biggr\}$ are local bases for $\sA^1(\bS^2)$, over $U_N$ and $U_S$ respectively.

Note that the last exercise now justifies the claim that the half-spin bundles were indeed $S^+ \oplus S^- \isom L \oplus L^*$.