quantum field theory - Is there a 2D manifold on which the Dirac equation has a zero mode?

The two-dimensional (2D) Dirac equation $(\sigma_1iD_1+\sigma_2 iD_2)\psi=E\psi$ admits zero mode ($E=0$) solutions on a non-trivial gauge background, such as the zero mode at the core of a U(1) gauge flux of $\pi$. I am wondering if zero modes of Dirac equation also arise on a non-trivial gravitational background (curved space), given the analogy between the gauge curvature and the gravitational curvature. So here is my question: on which 2D closed manifold does the Dirac equation has a zero mode?

My first try is to consider the sphere $S^2$, which has non-trivial curvature. But I found this paper (http://arxiv.org/abs/hep-th/0111084v1) claiming that there is no zero mode for Dirac fermions on the sphere. So I am wondering if there is any simple example of 2D closed manifold that supports Dirac fermion zero modes.

Answer

First, on torus with both periodic boundary condition, there are two zero modes, basically constant spinor fields, and "2" is because: the spinor representation of $\operatorname{SO}(2)$ is two dimensional. Or you can think "2" as one left spinor field and one right hand spinor field. However, you will immediately see that the zero modes are not "robust". If we decide to choose other 3 types of spin structure, there is no zero mode. In this sense, the zero mode itself is not robust. Instead, one can prove, the Dirac Index, basically "zero mode" of left hand minus "zero mode" of right hand is a topological invariants (c.f. Atiyah-Singer(AS) Index theorem, e.g., see wikipedia: https://en.wikipedia.org/wiki/Atiyah-Singer_index_theorem, applying to Dirac operator).

The AS index theorem for Dirac operator relates Dirac Index to curvature terms, more precisely,

\begin{eqnarray} \operatorname{Index}=\int_M \hat{A}(TM) ch(V) \end{eqnarray}

Roughly speaking, A-hat(sometimes called A-roof, see: https://en.wikipedia.org/wiki/Genus_of_a_multiplicative_sequence#.C3.82_genus) genus is "gravitational" contribution, $\hat{A}=1-\frac{1}{24}p_1+\ldots$. At dimension 2, A is trivial, so the only contribution comes from Chern character (see https://en.wikipedia.org/wiki/Chern_class#The_Chern_character) $ch(V)=1+F+\ldots$, of complex vector bundle $V$. So this implies, if you have a "magnetic monopole" inside your torus, there must be a zero mode. Because the AS index theorem implies, although a pair of zero modes can be lifted, the single one is always there.

Let us explain it in your example more detaily. Take your torus and parametrize it by two coordinates: $\theta \in [0,2\pi]$ with period $2\pi$ and $t \in [0,1]$ with period $1$. The spinor representation of $SO(2)$ is two dimensional, e.g., pauli matrices $\gamma^\theta=\sigma_2$ and $\gamma^t=\sigma_1$. Chiral element $\gamma^c=i\gamma^t \gamma^\theta = -\sigma_3$ Therefore, we can write down the Dirac operator explicitly:

$$\mathcal{D}=i \gamma^j \partial_j =\left( \begin{array}{c c} 0 & i\partial_t+ \partial_\theta \\ i\partial_t- \partial_\theta & 0 \end{array} \right)= \left( \begin{array}{cc} 0 & D \\ D^\dagger & 0 \end{array} \right)$$

where $D=i\partial_t+\partial_\theta: \Gamma^+ \rightarrow \Gamma^-$ and $D^\dagger=i\partial_t-\partial_\theta: \Gamma^- \rightarrow \Gamma^+$. And $\Gamma^+ \oplus \Gamma^- = \Gamma(\mathcal{S},T^2)$. We can also twist the spinor bundle by "U(1) gauge fields" $\mathcal{A}$, with vector representations forming a vector bundle over $T^2$, $V\rightarrow T^2$. The total bundle will be $\mathcal{S}\otimes V$, and Dirac operator is modified, by replacing $i\partial_j$ by covariant derivative $i\partial_j - A_j$. For example, we set a uniform magnetic field with total flux $2\pi$ by "Landau gauge": $A_\theta(\theta,t)=-t$ and $A_t(\theta,t)=0$. Therefore, the twisted Dirac operator is:

\begin{eqnarray} \mathcal{D}=\left( \begin{array}{cc} 0 & i\partial_t-it+ \partial_\theta \\ i\partial_t+it- \partial_\theta & 0 \end{array} \right)= \left( \begin{array}{cc} 0 & D_A \\ D^\dagger_A & 0 \end{array} \right) \end{eqnarray}

where $D_A=i\partial_t-it+\partial_\theta$ and $D_A^\dagger=i\partial_t+it-\partial_\theta$. We solve the Dirac equation $\mathcal{D}\psi=0$ on spinor fields on torus with the boundary condition: \begin{eqnarray} \psi(\theta,t)=\psi(\theta+2\pi,t), \quad \psi(\theta,t+1)=e^{i\theta}\psi(\theta,t) \end{eqnarray}

The solution space of $\mathcal{D}\psi=0$ can be decomposed into left hand zero modes $D_A\psi_L=0$ and right hand zero modes $D^\dagger_A\psi_R=0$, and index is the number of left hand zero modes $n_L=\operatorname{dim}(\operatorname{ker} D_A)$ minus right hand zero modes $n_R=\operatorname{dim}(\operatorname{ker} D^\dagger_A)$. Now assume $\psi_L \in \operatorname{ker} D_A$, by modes decomposition $$\psi_L=\sum_n c_n(t) e^{in\theta}, \quad \frac{dc_n(t)}{dt}-t c_n(t)+n c_n(t)=0 $$ so the t dependence of the coefficient $c_n(t)$ will be Gaussian type $c_n(t)=c_n e^{\frac{(n-t)^2}{2}}$, and constant $c_n$ will be fixed by boundary condition: $$c_{n+1}(t+1)=c_n(t)$$ therefore, $c_n=c$ for all $n$ and solution:

\begin{eqnarray} \psi_L(\theta,t)=c \sum_n \exp\left(\frac{(n-t)^2}{2}+in\theta \right) \end{eqnarray} however, the summation is not renormalizable, so we conclude $n_L=\operatorname{ker} D_A=0$. Similarly, we can solve $D^\dagger_A \psi_R=0$ for $\psi_R$, and the solution:

\begin{eqnarray} \psi_R(\theta,t)=c \sum_n \exp \left(-\frac{(n-t)^2}{2}+in\theta \right) \end{eqnarray} this wavefunction is normalizable, therefore $n_R=1$. So Index$=0-1=-1$. To verify the AS index theorem, we notice the integral $\int_M \hat{A}(TM) ch(E)$ at 2d reduces to $\frac{1}{2\pi} \int_{M} \mathcal{F}=-1={\rm Index}$ (Explanation of "$-1$" in the integral: we choose left-right handness by $\gamma^c=i\gamma^t \gamma^\theta$, so we define volume form to be $dt \wedge d\theta$, therefore, $\mathcal{F}= d (A_{\theta} (\theta,t) d \theta )=- dt \wedge d\theta$)

This example is attributed to Atiyah: Eigenvalues of Dirac operator, for a similar but different purpose.

On the other hand, if you wish to have a purely "gravitational zero mode", you need to find a 4k (k integer) dimensional manifold with nontrivial $\hat{A}$. A simple and famous example with nonzero pontryagin number is $K3$ (https://en.wikipedia.org/wiki/K3_surface) at 4 dimension, which is also a spin manifold.