# Dirac Operator on Riemann Surface

Let $\Sigma$ be a Riemann surface with canonical bundle $K$ and hermitian metric $h$. A spin bundle is a line bundle $L$ that squares to $K$. More concretely, given a trivializing atlas $\{ \phi_i : U_i \rightarrow \mathbb{C} \}$ with holomorphic transition maps $f_{ij}$, the transition maps for the canonical bundle are $f^{-1}_{ij}$, and for the spin bundle $\pm \sqrt{f^{-1}_{ij}}$.

The Dirac operator is defined by composing $\bar \partial : C^\infty(L) \rightarrow C^\infty(L \otimes \bar K )$ with an isomorphism $L \otimes \bar K \simeq \bar L$ induced by the hermitian metric $h$.

My question is: What is this isomorphism exactly? What does it look like in the coordinate charts?

The key is the hermitian metric $h$. If you analyze it carefully, as a tensor, $h$ is a section of the line bundle $K \otimes \bar{K} \cong \bar{K} \otimes K$. By the way, when I write $\otimes$ I mean $\otimes_{\mathbb{C}}$ implying that it is the tensor product linear with respect to the complex numbers, not just the reals. As the section $h$ is a positive definite and it never becomes zero, it is the square of a another section, $\sigma$ which is a smooth section of the bundle $L \otimes \bar{L}$. In other words, $h = (\sigma)^2$ (in sophisticated symbols $h = \sigma \otimes \sigma)$.
Define the line bundle map $\phi : \bar{L} \to L \otimes\bar{K}$ as follows: for any smooth section $\hat{s}$ of $\bar{L}$ consider the assignment $$\varphi \, : \, \hat{s} \, \mapsto \, \sigma \otimes\hat{s}$$ that is $\varphi(\hat{s}) = \sigma \otimes\hat{s} = \sigma\, \hat{s} \,$. Now, $\sigma \otimes\hat{s}$ is a smooth section of the line bundle $\left(L \otimes \bar{L} \right)\otimes \bar{L} = L \otimes \left( \bar{L} \otimes \bar{L}\right) = L \otimes \overline{\left( {L} \otimes {L}\right)} = L \otimes \bar{K}$ because ${L} \otimes {L} = K$. Clearly $\varphi$ is a map that maps fibers to fibers and is complex linear on the fibers. Because $\sigma$ is never zero, it is injective and hence surjective. Consequently, it is an isomorphism. So your Dirac operator is $$D = \varphi^{-1} \circ \bar{\partial } \, : \, C^{\infty}(L) \, \mapsto \, C^{\infty}(\bar{L})$$ $$D \, s = \varphi^{-1}(\bar{\partial} \, s)$$ for $s \in C^{\infty}(L)$. Furthermore, you can take the section $\sigma^{-1}$ which is a smooth section of $(L\otimes\bar{L})^{-1} = L^{-1}\otimes\bar{L}^{-1}$ and then $\varphi^{-1}(\lambda) = \sigma^{-1} \otimes \lambda = \sigma^{-1} \,\lambda$. Hence, the Dirac operator can be written as $$D \, s = \sigma^{-1} \otimes (\bar{\partial} \, s) = (\sigma^{-1} \otimes \bar{\partial}) \, s = (\sigma^{-1} \, \bar{\partial}) \, s.$$
Now, in coordinate charts everything looks simpler. Indeed, let $s$ be a spinor, i.e. a smooth section of the bundle $L$ and let $$(\phi^{-1}_j)^* s = s(z,\bar{z}) \, \sqrt{dz}$$ in a chart $\phi_j : U_i \to \mathbb{C}.$ Then the hermitian metric looks like $$(\phi^{-1}_j)^* h = h(z,\bar{z})\, dz \, d\bar{z}$$ and consequently $$(\phi^{-1}_j)^* \sigma = \sigma(z,\bar{z}) \, \sqrt{dz \, d\bar{z}} = \sigma(z,\bar{z}) \, \sqrt{|dz|^2} = \sigma(z,\bar{z}) \, |dz|$$ where $$h(z,\bar{z}) = \big( \sigma(z,\bar{z}) \big)^2$$ because $h(z,\bar{z})$ is positive function, so the square root is well defined, i.e. $\sigma(z,\bar{z}) = \sqrt{h(z,\bar{z})}$. Thus, your Dirac operator acts as follows $$\big(D\,s\big)(z,\bar{z}) \, \sqrt{d\bar{z}} = \frac{1}{\sqrt{h(z,\bar{z})}} \frac{\partial s}{\partial \bar{z}}(z,\bar{z}) \, \sqrt{d\bar{z}}$$ because \begin{align} \big(D\,s\big)(z,\bar{z}) \, \sqrt{d\bar{z}} &= D\Big( s(z,\bar{z}) \, \sqrt{dz} \Big) = \frac{1}{\sqrt{h(z,\bar{z}) |dz|}} \, \frac{\partial s}{\partial \bar{z}}(z,\bar{z}) \, d\bar{z} \, \sqrt{d{z}} \\ &= \frac{1}{\sqrt{h(z,\bar{z})} \, \sqrt{dz}\, \sqrt{d\bar{z}}}\, \frac{\partial s}{\partial \bar{z}}(z,\bar{z}) \, d\bar{z} \, \sqrt{d{z}}\\ &= \frac{1}{\sqrt{h(z,\bar{z})}}\, \frac{\partial s}{\partial \bar{z}}(z,\bar{z}) \, \frac{d\bar{z} \, \sqrt{d{z}}}{ \sqrt{dz}\, \sqrt{d\bar{z}}} \\ &= \frac{1}{\sqrt{h(z,\bar{z})}} \frac{\partial s}{\partial \bar{z}}(z,\bar{z}) \, \sqrt{d\bar{z}} \end{align} I hope that all of this makes sense.