The (flat) torus

On the $2$-torus $\bT^2 := \bR^2/\bZ^2$, we use the Riemannian metric coming from the usual flat metric on $\bR^2$. Thus, if we regard $\sA = \Coo(\bT^2)$ as the smooth periodic functions on $\bR^2$ with $f(t^1, t^2) \equiv f(t^1 + 1, t^2) \equiv f(t^1, t^2 + 1)$, then $(t^1,t^2)$ define local coordinates on $\bT^2$, with respect to which all Christoffel symbols are zero, namely $\Ga_{ij}^k = 0$, and thus $\nabla = d$ represents the Levi-Civita connection on $1$-forms.

In this case, $n = 2$, $m = 1$ and $2^m = 2$, so we use ``two-component'' spinors; that is, the spinor bundles $S \to \bT^2$ are of rank two. There is the ``untwisted'' one, where $S$ is the trivial rank-two $\bC$-vector bundle, and $\sS \isom \sA^2$. The Clifford algebra in this case is just $\sB = M_2(\sA)$. Using the standard Pauli matrices:

$$\sg^1 := \twobytwo{0}{1}{1}{0}, \qquad \sg^2 := \twobytwo{0}{-i}{i}{0}, \qquad \sg^3 := \twobytwo{1}{0}{0}{-1},$$

we can write the charge conjugation operator as

$$C = -i \sg^2 K$$

where $K$ again denotes (componentwise) complex conjugation.

Notice that $\sg^3$ does not appear in the formula for $\Dslash$; its role here is to give the $\bZ_2$-grading operator: $c(\ga) = \sg^3$ --regarded as a constant function with values in $M_2(\bC)$-- in view of the relation $\sg^3 = -i \sg^1\sg^2$ among Pauli matrices.

On the $3$-torus $\bT^3 := \bR^3/\bZ^3$, where now $n = 3$, $m = 1$ and again $2^m = 2$, we get two-component spinors. Again we may use a flat metric and an untwisted spin structure with $\sS = \sA^2$. The charge conjugation is still $C = -i \sg^2 K$ on $\sS$, so that $C^2 = -1$ also in this $3$-dimensional case. The Dirac operator is now

$$\Dslash = -i (\sg^1 \del_1 + \sg^2 \del_2 + \sg^3 \del_... ...{-i \del_3}{-\del_2 - i\del_1}{\del_2 - i\del_1}{i \del_3}.$$