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

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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?

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Contrary to what has been written above, diffeomorphisms do not act on spinors, since not every diffeomorphism preserves the spin structure.  The spin structure depends only on the conformal class of the metric, hence you have to restrict yourself to conformal diffeomorphisms; that is, diffeomorphisms $\varphi: M \to M$, such that $\varphi^*g = e^{2\sigma} g$ for some function $\sigma$, where $(M,g)$ is the (pseudo)riemannian spin manifold in question.

At the infinitesimal level, this means that one can define the Lie derivative of a spinor along a conformal Killing vector field, but not along just any vector field.  This Lie derivative was defined in the 1972 PhD thesis of Yvonne Kosmann-Schwarzbach, supervised by André Lichnerowicz who had earlier defined the Lie derivative of a spinor along a Killing vector field.

If $K$ is a Killing vector field, so that $\mathcal{L}_K g = 0$, then the Lie derivative (also called $\mathcal{L}_K$) on spinors obeys the following properties:

• $[\mathcal{L}_{K_1},\mathcal{L}_{K_2}] = \mathcal{L}_{[K_1,K_2]}$
• $\mathcal{L}_K (f \psi ) = K(f) \psi + f \mathcal{L}_{K}\psi$
• $[\mathcal{L}_K, \nabla_X] = \nabla_{[K,X]}$
• $\mathcal{L}_K ( X \cdot \psi) = [K,X] \cdot \psi + X \cdot \mathcal{L}_K \psi$

where $K,K_1,K_2$ are Killing vector fields, $\psi$ is a spinor field, $f$ a smooth function and $X$ an arbitrary vector field.

It follows from the latter two properties above that $\mathcal{L}_K$ commutes with the Dirac operator and hence that the Lie derivative along a Killing vector $K$ of a harmonic spinor is again harmonic.  (Harmonic spinors are those annihilated by the Dirac operator.)  By the first property of  $\mathcal{L}_K$ it follows that the space of harmonic spinors is a representation of the Lie algebra of isometries of $(M,g)$ and hence also of the connected component of the group of isometries.

answered Jun 17, 2014 by (2,315 points)

So, long story short, they are representations under conformal isometries, an only conformal isometries right?

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EDIT of the EDIT: one of the main assertion below (the existence of an action of diffeomorphisms on the space of spinors) is WRONG. (composing by a diffeomorphism of course does not provide a new section of the spinor bundle if the spinor bundle is not preserved by the diffeomorphism). See the answer of  José Figueroa-O'Farrill for a CORRECT one.

.

Every diffeomorphism $f$ of $B$ naturally acts on the space of spinors of $B$ by composition: if $\psi(x)$ is a spinor then $\psi(f(x))$ is another spinor. But in general, $f$ does not act on the space of zero modes of the Dirac operator because the action of $f$ does not in general commute with the action of the Dirac operator. But it is the case if $f$ is an isometry. Indeed, as by defintion an isometry preserves the metric, the action induced by an isometry commutes with the covariant derivative and so with the Dirac operator.

EDIT: as explained above, the space of spinors on $B$ is a representation of the group of diffeomorphisms of $B$ (a diffeomorphism acts on a spinor by composition). In particular, it is also a representation of the group of isometries of $B$ (isometries are just special diffeomorphisms). So in order to prove that the space of spinor zero-modes is a representation of the group of isometries, it is enough to show that it is a stable subspace of the space of spinors for the action of the group of isometries. So if $\psi$ is a spinor zero mode and $f$ an isometry, we have to show that $f.\psi$ is a zero mode (where $f. \psi$ denotes the image of $\psi$ by the action of $f$: $(f.\psi)(x)=\psi(f(x))$), i.e. that $D(f.\psi)=0$ where $D$ is the Dirac operator. But if we know that $D$ and $f$ commute, we have $D(f . \psi)= f.(D \psi)=0$ because $D\psi=0$ because $\psi$ is a zero-mode of $D$.

answered Jun 14, 2014 by (5,120 points)
edited Jun 18, 2014 by 40227

I am just beginning to work with spinors and I would greatly appreciate if you could give me a (more or less) step by step explanation of why the fact that the Dirac operator and an isometry commute makes the 4-dimensional  spinor corresponding to a zero mode of the compact dirac operator a representation of the isometry group.

I liked you answer but I have a final question. The spinor on B is a rep of the isometry group. But if I consider a spinor of the FULL space, that is a tensor product of a normal spinor and a zero mode on B, is the 4 spinor a rep of the isometry group on its own(and if this is so under what operator) or must I consider the whole $\phi\otimes\psi$ as a rep of the isometry group?

By the way, I haven't seen your EDIT just till today and I had started a bounty on stack.exchange. if you answer, its yours.

thanks for the honesty

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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
answered Jun 13, 2014 by (215 points)

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