# Why zero modes of the internal Dirac operator must be in representations of the isometry group of the compact space

Imagine a manifold $\mathbb{R}^{1,3}\times{}B$ where $B$ is a compact group-manifold with isometry group $U(1)\times{}SU(2)\times{}SU(3)$.

Let's consider the Dirac equation for a massless Spinor field.

$iD_{1+n}\Psi=0$

and separating the compact space part with the 4-space part

$i(D_4+D_{compact})\Psi=0$

we see that the compact dirac operator might be regarded as a mass operator for the 4 dimensional spinor.

Imagine we are interested in solutions whose mass is zero, that is, solutions whose eigenvalues of the internal Dirac operator is zero.

I am trying to understand the impossibility of arranging these zero mode fermions on complex representations of the isometry group, but for that first I need to understand why the fermions obtained by dimensional reduction must form a representation (be it real or complex) of my isometry group.

This is asserted in this paper (Witten 1981). On the third paragraph of the ninth page it is said

If the ground state is a product of four dimensional Minkowski space with one of the $M^{pqr}$ (for our purposes this just means a compact group manidold with the desired isometry group), then the zero modes of the Dirac operator in the internal sace will automatucally form multiplets of $U(1)\times{}SU(2)\times{}SU(3)$ since this is the symmetry of the internal space.

Could someone please be more explicit of why this is so?

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.

• thanks, looks fine. Nonetheless, this is a little beyond my mathematical comfort zone but I think I can understand it after some coffees. I'll tell ya. – Yossarian Jun 17 '14 at 14:08
• The gist of it is: a symmetry operation (isometry) can also act on a spinor field. Since the Dirac operator is defined by things that are unchanged by the symmetry operations, it commutes with them. Thus we can generate new zero modes by acting on them with isometries. It's the same principle as if you rotate a harmonic function you get a new harmonic function (so you get the $Y^l_m$:s), because the Laplacian is rotationally invariant. – Robin Ekman Jun 19 '14 at 2:21
• didn't you get the bounty? – Yossarian Jun 24 '14 at 8:40
• I don't think so. – Robin Ekman Jun 24 '14 at 13:51
• @silvrfück did you click the red button to award the bounty, within 24 hours of the bounty ending? If not, see this. – David Z Jun 26 '14 at 2:27