The spin connection $\nabla^S$ over $\bS^2$

Given any local orthonormal basis of $1$-forms $\{E_1,\dots,E_n\}$, we can compute Christoffel symbols with all three indices taken from this basis, by setting $\Hat{\Ga}_{\mu\al}^{\bt} := (E_\mu)^i \Tilde{\Ga}_{i\al}^{\bt}$, or equivalently, by requiring that

$$\nabla_{E_\mu} E_\al =: \Hat{\Ga}_{\mu\al}^\bt E_\bt$$

for $\mu,\al,\bt = 1,2,\dots,n$. (This works because the first index is tensorial).

This yields the local orthonormal bases $E_1 := \half q \del/\del x^1$, $E_2 := \half q \del/\del x^2$ for vector fields, and dually $\th^1 = (2/q) dx^1$, $\th^2 = (2/q) dx^2$ for $1$-forms. However, since $\bS^2 = \bCP^1$ is a complex manifold, it is convenient to pass to ``isotropic'' bases, as follows. We introduce
$$\begin{align*} E_+ &:= E_1 - iE_2 = q \dd{z}, & \th^+ &:= \half(\th^1 + i\th... ...\bar z}, & \th^- &:= \half(\th^1 - i\th^2) = \frac{d\bar z}{q}. \end{align*}$$

The Clifford action on spinors is given (over $U_N$, say) by $\ga^1 := \sg^1 = \twobytwo{0}{1}{1}{0}$ and $\ga^2 := \sg^2 = \twobytwo{0}{-i}{i}{0}$. The $\bZ_2$-grading operator is given by

$$\chi := (-i) \sg^1 \sg^2 = \sg^3 = \twobytwo{1}{0}{0}{-1}.$$

The spin connection is now specified by

$$\nabla^S_{E_\pm} := E_\pm - \quarter \Hat{\Ga}_{\pm\al}^\bt \ga^\al \ga_\bt.$$

A similar expression is valid over $U_S$, by replacing $z,\bar z,q$ by $\ze,\bar\ze,q'$ respectively, and by changing the overall $(-i)$ factor to $(+i)$. This formal change of sign is brought about by the local coordinate transformation formulas induced by $\zeta = 1/z$. (Here is an instance of the ``unique continuation property'' of $\Dslash$: the local expression for the Dirac operator on any one chart determines its expressions on any overlapping chart, and then by induction, on the whole manifold.)