# Index Theorem for the Twisted Dirac Operator

In the Mirror Symmetry monograph (http://www.claymath.org/library/monographs/cmim01c.pdf), on page 297, the index theorem is used for a two-dimensional twisted Dirac operator. Below equation 13.37, it is claimed that the number of $\psi_-$ zero modes is equal to the number of $\overline{\psi}_+$ zero modes, and the number of $\overline{\psi}_-$ zero modes is equal to the number of ${\psi}_+$ zero modes. How does one show that this is true?

In addition, on page 811, a similar problem is considered for a two-dimensional surface with boundary. Here it is claimed in equation 39.213 that the index of the twisted Dirac operator is $$\textrm{Index }\mathcal{D}=\#[(\psi_-,\overline{\psi}_+)\textrm{ zero modes}]-\#[(\overline{\psi}_-,{\psi}_+)\textrm{ zero modes}].$$ However, as far as I understand there is no well-defined index for $\mathcal{D}$, but only well-defined indices for its chiral or antichiral parts $D$ and $\overline{D}$, where $$\mathcal{D}=\bigg(\begin{array}{cc} 0 & D \\ \overline{D} & 0 \\ \end{array} \bigg).$$ What exactly does equation 39.213 mean? Would it be correct to interpret $\textrm{Index }\mathcal{D}$ as $\textrm{Index }\overline{D}$?

First, we need a spin structure to define the spinor bundle. The index theorem does not care which one we take, so we may take even spinors to be $(0,0)$-forms and odd spinors to be $(0,1)$ forms. Both bundles are trivial on $T^2$, so we may take functions as spinors of both parities. Then the positive part of the Dirac operator is $\not\!\partial_+=\partial_{\bar z}$, and the negative is its adjoint, which becomes $\not\!\partial_-=\partial_z$ on functions.
Now twist with $E$ and the connection $A$ to get twisted Dirac operators $\not\!\partial_+^E=D_{\bar z}$ and $\not\!\partial_-^E=D_z$ in their notation. Equation (13.37) follows from Riemann-Roch (or Atiyah-Singer, if you prefer). Taking adjoint bundles is the same as doing a complex conjugation if the operators are compatible with a Hermitian metric, so it maps $E\otimes S_\pm$ to $E^*\otimes S_\mp$, preserving the connections. In particular, it maps $\psi_\pm$ zero modes to $\bar\psi_\mp$ zero modes and vice versa.
For your second question, regard the operator $\mathcal D$ as the formal difference of the operators $\not\!\partial^E$ and $\not\!\partial^{E^*}$. This operator only has an index if we specify suitable boundary conditions; take those of (39.212). Then the index (39.213) is indeed well-defined. The easiest way to see this is to consider a twisted Dirac operator on the double of $\Sigma$, and that is explained in some detail after (39.213). Because the double has no more boundary, you can use Atiyah-Singer (or Riemann-Roch, if you prefer) to see what the index is.
• Thank you for your answer. I am sorry but I do not understand exactly what you mean by formal difference of the operators. Do you mean that $\textrm{Index }\mathcal{D}=\textrm{Index }\overline{D}-\textrm{Index }{D}=2\textrm{ Index }\overline{D}$? – Mtheorist Apr 7 '17 at 7:20
• @Mtheorist I mean, consider $\operatorname{ind}(\not\!\partial^E_+-\not\!\partial^{E^*}_+)=\operatorname{ind}(\not\!\partial^E_+\oplus\not\!\partial^{E^*}_-)$. But please note that the individual operators $\not\!\partial^E_+$ and $\not\!\partial^{E^*}_-$ do not admit local elliptic boundary conditions. For each of them, you could use the nonlocal APS boundary conditions, which are too complicated here. But the operator $\mathcal D$ admits local boundary conditions like those of (39.212). By "local" I mean pointwise on $\partial\Sigma$. In particular, $\operatorname{ind}(D)$ does not exist. – Sebastian Goette Apr 8 '17 at 4:30
• So the RHS of equation 39.213 really means number of $\psi_-$ OR $\overline{\psi}_+$ zero modes minus number of $\overline{\psi}_-$ OR $\psi_+$ zero modes, correct? – Mtheorist Apr 8 '17 at 6:13
• @Mtheorist No, it means, the number of pairs of $\psi_-$ and $\bar\psi_+$ zero modes matching up at the boundary as in (39.212) minus the number of pairs of $\bar\psi_-$ and $\psi_+$ zero modes matching up at the boundary. Without the boundary condition, there are infinitely many of each. For example take $\Sigma=D^2\subset\mathbb C$ and $E$ trivial. Then every holomorphic function on $D^2$ is a $\bar\!\partial_+=\partial_{\bar z}$ zero mode. – Sebastian Goette Apr 8 '17 at 13:19