# The eigenfunctions of vanishing eigenvalues of Dirac operator

Consider the eigenfunctions to the operator $i\gamma^\mu D_\mu$: $$i\gamma^\mu D_\mu\Psi_i=\lambda_i\Psi_i.$$ Because $\{\gamma^5,\gamma^\mu\}=0$, we know $$i\gamma^\mu D_\mu\gamma^5\Psi_i=-\lambda_i\gamma^5\Psi_i.$$ Thus non-vanishing eigenvalues will always occur in pairs of opposite sign. And for eigenfunction of $\gamma^5$: $$\gamma^5\Psi_i=\chi_i\Psi_i,$$ it must be an eigenfunction of zero eigenvalue of $i\gamma^\mu D_\mu$ since $$\lambda_i\chi_i\Psi_i=i\gamma^\mu D_\mu\chi_i\Psi_i=i\gamma^\mu D_\mu\gamma^5\Psi_i=-\lambda_i\gamma^5\Psi_i=-\lambda_i\chi_i\Psi_i. \,(\chi_i=\pm 1)$$

But now there is a statement that all the eigenfunction of vanishing eigenvalues can be chosen to be eigenfunctions of $\gamma^5$. Equivalently, the dimension of eigenspace of $\gamma^5$ equals the dimension of the zero-eigenspace of $i\gamma^\mu D_\mu$. How to prove this statement?

If $\psi$ is a zero mode, then so is $\gamma_5\psi$. But then $\psi_{L,R} =(1\pm \gamma_5)\psi$ are also zero modes. This means that all zero-modes can be chosen to be eigenstates of $\gamma_5$. Non-zero modes are not eigenstates of $\gamma_5$, because $\psi$ and $\gamma_5\psi$ have different eigenvalues. We conclude that the number of zero modes is equal to to the number of eigenstates of $\gamma_5$.

There are a some important less obvious results. Let $n^0$ be the total number of zero modes, and $n_{L,R}^0$ be the number of left/right handed zero modes. We just saw that $n^0=n_L^0+n_R^0$. The most important result in gauge theories is the Atiyah-Singer index theorem which implies that $$Q=n_L^0-n_R^0,$$ where $Q$ is the topological charge. Finally, at least in the semi-classical regime, the index theorem is saturated by configuration with $Q=n_L^0$ for $Q>0$, and $Q=-n_R^0$ for $Q<0$.

The kernel ${\rm ker}{\cal D}$ for the Dirac operator ${\cal D}$ (which we assume is defined self-adjointly) is diagonalizable wrt. $\gamma^5$ (which we assume is an involution and anticommute with ${\cal D}$), because (among other things):

1. the kernel ${\rm ker}{\cal D}$ for the Dirac operator is an invariant subspace of $\gamma^5$, $$\gamma^5 {\rm ker}{\cal D}~\subseteq ~{\rm ker}{\cal D}.$$

2. It is enough to consider the restriction $\gamma^5_|$ to ${\cal D}$.

3. $\gamma^5_|$ is an involution, and hence diagonalizable, $${\rm ker}{\cal D}~=~{\rm ker}(\gamma^5_|+1)\oplus{\rm ker}(\gamma^5_|-1) .$$

• Well, you certainly provided a good answer, but a bit sketchy... – Wein Eld Nov 5 '16 at 15:31
• I updated the answer. – Qmechanic Nov 5 '16 at 15:50