Spin$^c$ Dirac operators on the 2-sphere

We know that finitely generated projective modules over the $C^*$-algebra $A = C(\bS^2)$ are of the form $p A^k$, where $p = [p_{ij}]$ is an $k \x k$ matrix with elements in $A$, such that $p (= p^2 = p^*)$ is an orthogonal projector, whose rank is $\tr p = p_{11} +\cdots+ p_{kk}$. To get modules of sections of line bundles, we impose the condition that $\tr p = 1$, so that $p A^k$ is an $A$-module ``of rank one''. It turns out that it is enough to consider the case $k = 2$ of $2 \x 2$ matrices.

After stereographic projection, we can replace $\vec n$ by $f(z) := \dfrac{n_1 - in_2}{1 - n_3}$. where $z = e^{-i\phi} \cot\frac{\th}{2}$ is allowed to take the value $z = \infty$ at the north pole. Then $f$ is a continuous map from the Riemann sphere $\bC \cup \{\infty\} = \bCP^1$ into itself. If two projectors $p$ and $q$ are homotopic --there is a continuous path of projectors $\set{p_t : 0 \leq t \leq 1}$ with $p_0 = p$ and $p_1 = q$-- then they give the same class $[p] = [q]$ in $K^0(\bS^2)$; and this happens if and only if the corresponding maps $\vec n$, or functions $f(z)$, are homotopic.

Let $\sE_{(m)} = p_m A^2$ and $\sE_{(-m)} = p_{-m} A^2$, where

$$p_m(z) = \frac{1}{1 + z^m\bar z^m} \twobytwo{z^m\bar z^m}{... ...1}{1 + z^m\bar z^m} \twobytwo{z^m\bar z^m}{\bar z^m}{z^m}{1},$$

with the obvious definition (¿what is it?) for $z = \infty$.

For $m \in \bZ$, $m \neq 0$, we redefine $\sE_{(m)} := p_m \sA^2$ with $\sA = \Coo(\bS^2)$; so that $\sE_{(m)}$ now denotes smooth sections over a nontrivial line bundle on $\bS^2$. We can identify each element of $\sE_{(m)}$ with a smooth function $f_N\: U_N \to \bC$ for which there is another smooth function $f_S\: U_S \to \bC$, such that

$$f_N(z) = (\bar z/z)^{m/2} f_S(z^{-1}) \word{for all} z \neq 0. \eqno (\ul{m})$$

Here, as before, $(\bar z/z)$ means $e^{i\phi}$ in polar coordinates.

To get all the structures on $\bS^2$, we twist the spinor module $\sS$ for the spin structure, namely $\sS = \sE_{(1)} \oplus \sE_{(-1)}$, by the rank-one module $\sE_{(m)}$. On the tensor product $\sS \ox_\sA \sE_{(m)}$ we use the connection

$$\nb^{S,m} := \nb^S \ox 1_{\sE_{(m)}} + 1_\sS \ox \nb^{(m)}.$$

The sign of a selfadjoint operator $D$ on a Hilbert space is given by the relation $D =: F \vert D\vert = F (D^2)^{1/2}$, where we put $F := 0$ on $\ker D$. Thus $F$ is a bounded selfadjoint operator such that $1 - F^2$ is the orthogonal projector whose range is $\ker D$. When $\ker D$ is finite-dimensional, $1 - F^2$ has finite rank, so it is a compact operator.

An even Fredholm module over an algebra $\sA$ is given by:

1. a $\bZ_2$-graded Hilbert space $\sH = \sH^0 \oplus \sH^1$;
2. a representation $a \mapsto \pi(a) = \twobytwo{\pi^0(a)}{0}{0}{\pi^1(a)}$ of $\sA$ on $\sH$ by bounded operators that commute with the $\bZ_2$-grading;
3. a selfadjoint operator $F = \twobytwo{0}{F^-}{F^+}{0}$ on $\sH$ that anticommutes with the $\bZ_2$-grading, such that $F^2 - 1$ and $[F,\pi(a)]$ are compact operators on $\sH$, for each $a \in \sA$.

We can extend the twisted Dirac operator $\Dslash_m$ to a selfadjoint operator on $\sH = \sH^0 \oplus \sH^1$, where $\sH^0$ and $\sH^1$ are two copies of the Hilbert space $L^2(\bS^2,\nu)$ where $\nu = 2i q^{-2} dz d\bar z$. We define $\pi^0(a) = \pi^1(a)$ to be the usual multiplication operator of a function $a \in \Coo(\bS^2)$ on this $L^2$-space.