I think it could be simply that the Dirac operator is invariant under isometries, so if $\phi$ is an isometry and $\psi$ a solution to $$D\psi = 0,$$ then $\phi^* \psi$ is also a solution, where $\phi^*$is pullback. Then it would be similar to how harmonic functions $f$ on the sphere -- $\nabla^2 f = 0$ -- come in representations of the rotation group, the $Y^l_m$.

In more detail if $\phi$ is a diffemorphism, that in coordinates takes the form $y^\mu = y^\mu(x^\nu)$ (not a tensor expression), and $v^\mu$ is a vector field, then we can define a vector field $$(\phi_* v^\mu)(\phi(p)) = \frac{\partial y^\mu}{\partial x^\nu} v^\nu(p)$$
called the pushforward of $v^\mu$. Naturally we can pushforward any tensor, in particular the metric. By definition $\phi$ is an isometry if $$(\phi_* g_{\mu\nu})(\phi(p))= g_{\mu\nu}(\phi(p)).$$

This means that if we have any tetrad (also known as a vierbein or a frame), that is a set of vector fields $e_a^\mu$ such that $e_{a\mu} e^\mu_b = \eta_{ab}$ for some symmetric matrix $\eta_{ab}$ with signature $+---$, it is pushed forward to another tetrad. I let $\eta_{ab}$ be general because in spinor problems it is more natural to use a **null tetrad** $$\eta_{ab} = \begin{pmatrix} 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & -1 \\ 0 & 0 & -1 & 0\end{pmatrix}.$$
Since $\eta_{ab}$ has zeros on the diagonal all the tetrad vectors are null. It is well known (see for example *Spinors and space-time* or the Newman-Penrose paper) that to every null tetrad corresponds exactly two bases for two-spinors, called **dyads**, say $(o^A, \iota^A)$ and $(-o^A, \iota^A)$. Thus at least for isometries connected to the identity, the pushforward of tetrads lifts to a pushforward of dyads. (However when the isometry group isn't simply connected this might not be continuous globally, but I think it doesn't matter here, since we can consider isometries close to the identity, which will take us to Lie algebra representations, and then we integrate them, and discard the representations that require passing to the simply connected cover.)

Since we can pushforward dyads we can pushforward two-spinors (by linearity), since we can pushforward two-spinors we can pushforward Dirac spinors. $\newcommand{\Dslash}{\!\not D}$ In particular for a Dirac spinor $\psi$, $\Dslash\psi$ is of course also a Dirac spinor, so $$\beta = \phi_* (\Dslash\psi) = \phi_* (\Dslash \phi^* \phi_* \psi) $$ makes sense, where $\phi^*$ as the inverse of $\phi_*$ so the second equality is just inserting the identity. Now $\phi_* \Dslash \phi^*$ defines a differential operator, it is the transformed Dirac operator under the isometry $\phi$. But since the Dirac operator is defined by the metric and $\phi$ preserves the metric, this must be just the Dirac operator again. (You can probably make this argument more convincing.)

Thus we have established that $$\beta = \Dslash (\phi_*\psi). $$ In particular if $\beta = 0$, so that $\psi$ is a zero mode for the Dirac operator, then $\tilde{\psi} = \phi_* \psi$ is also a solution. Thus the isometry group (or at least its Lie algebra) acts on zero modes.

This post imported from StackExchange Physics at 2014-06-17 07:48 (UCT), posted by SE-user Robin Ekman