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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.

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    $\begingroup$ Comment to the question (v1): If the Dirac operator has a zero-mode, and if the metric has Euclidean/positive signature, then the scalar curvature cannot be positive everywhere, cf. the Lichnerowicz vanishing theorem. This rules out e.g. the sphere $S^2$. $\endgroup$ – Qmechanic Mar 24 '15 at 22:47
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    $\begingroup$ On torus with both periodic boundary condition, there are two zero modes, basically constant spinor fields, and "2" because the spinor representation of SO(2) is two dimensional. Or you can think 2 as one left and one right hand. However, you can immediately see this is not "stable", because if we change spin structure to the other 3 type, there is no zero modes. In this sense, the zero mode itself is not stable. Instead, one can prove, the Dirac Index, basically "zero mode" of left hand - "zero mode" of right hand is a topological invariants. And the index theorem relates that to $\endgroup$ – Yingfei Gu Dec 11 '15 at 5:49
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    $\begingroup$ curvature terms, more precisely, $\int ch(V)\hat{A}$, the chern character with A hat genus. At dimension 2, A have to be c-number 1, and only contribution comes from $ch(V)$, say, gauge curvature F. So if my reasoning above is correct, this implies, if we have a monopole inside any 2-manifold, there must be a zero mode some where. Because the index theorem implies there is precisely one more left handed zero modes comparing to right handed. Although a pair of zero modes can be lifted, the single one is always there. $\endgroup$ – Yingfei Gu Dec 11 '15 at 5:54
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    $\begingroup$ One amusing example can demonstrate this, and has condensed matter interpretation I believe is the torus with a magnetic monopole inside. On torus, you can write the dirac equation with complex variables, and the zero modes corresponding to holomorphic and anti-holomorphic functions, for left and right handed respectively. Now, with appearance of the magnetic monopole, the problem becomes analogous to Landau level problem. I am not sure about the physical interpretation, but one can concretely solve the equation by specially chosen gauge, and the solutions are theta functions $\endgroup$ – Yingfei Gu Dec 11 '15 at 6:01
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    $\begingroup$ @EverettYou Yes, chirality in even dimensions is always possible to define, simply by looking at the spinor representations. In 2d, the only reason we need gauge field to make an nonzero index is because the A hat genus is trivial. While in 4d, A hat genus could have nontrivial contribution from $p_1$, then at 4d, the index is formally "index= gauge + gravitational". $\endgroup$ – Yingfei Gu Dec 11 '15 at 17:23
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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.

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  • $\begingroup$ May I know why the index of $\mathcal{D}$ is the number of left hand zero modes $n_L=dim(\textrm{ker }D_A)$ minus right hand zero modes $n_R=dim(\textrm{ker }D^{\dagger}_A)$. $\endgroup$ – Mtheorist Apr 6 '17 at 9:42
  • $\begingroup$ @Mtheorist this is the definition of analytic index $\endgroup$ – Yingfei Gu Jul 20 '18 at 2:14
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There seems to be some confusion. In the referenced paper there is no temporal direction, and even though there is curvature there, it has nothing to do with gravity (there is no time). Where as when you compare with $2D$ gauge theory, one of those $2$ dimensions is temporal in deed. So if you wanted to carry the analogy you should consider two dimensions, one of which is temporal and ask about whether there are any zero modes for a specific manifold. The answer there is that in $(1+1)$ dimensions gravity is non dynamical and the spin connection vanishes, so you cannot make an analogy with the gauge theories in $(1+1)$ dimensions.

Your question is valid, but your analogy isn't.

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  • $\begingroup$ Do you mean that Dirac fermion can never have zero mode on any 2D manifold with Euclidian metric signature? $\endgroup$ – Everett You Mar 26 '15 at 5:20
  • $\begingroup$ That's not what I meant, this is why I said that your question is valid. $\endgroup$ – Ali Moh Mar 26 '15 at 5:22
  • $\begingroup$ Thanks. So do you have an answer of the question, if we restrict to the Euclidian metric signature? $\endgroup$ – Everett You Mar 26 '15 at 7:06

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