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 
\begin{equation}
\textrm{Index }\mathcal{D}=\#[(\psi_-,\overline{\psi}_+)\textrm{ zero modes}]-\#[(\overline{\psi}_-,{\psi}_+)\textrm{ zero modes}].
\end{equation}
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
\begin{equation}
\mathcal{D}=\bigg(\begin{array}{cc}
   0 & D \\
   \overline{D} & 0 \\
  \end{array} \bigg).
\end{equation}
 What exactly does equation 39.213 mean? Would it be correct to interpret $\textrm{Index }\mathcal{D}$ as $\textrm{Index }\overline{D}$? 
 A: 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.
