# Dirac operator on Kähler manifold

Reference: John Morgan's book on Seiberg-Witten theory. (pg 110)

I was working out the computational details of formulation of Dirac operator on Kähler manifold. If we choose the $$\mathrm{Spin}^{\mathbb{C}}$$-structure as the natural one determined by the underlying almost complex structure, and we choose the Chern connection on the anticanonical line bundle $$K^{-1}$$. We can then compute the Dirac operator as $$D=\sqrt{2}(\bar\partial+\bar\partial^*): \bigwedge ^{0,2}\oplus\bigwedge^0 \rightarrow \bigwedge^{0,1}$$.

Now consider the complex spin bundle obtained by the above tensored with a complex line bundle $$L_0$$, which is the complex spin bundle associated with another $$\mathrm{Spin}^{\mathbb{C}}$$-structure. Choose a unitary connection on the determinant line bundle $$L$$ as $$A$$. This is equivalent of choosing a unitary connection $$A_0$$ on $$L_0$$ satisfies $$A_0^2=A_{K} \otimes A$$. Here $$A _K$$ is the Chern connection on canonical line bundle.

We then have the Dirac operator $$D_A: \bigwedge ^{0,2}(L_0)\oplus\bigwedge^0(L_0) \rightarrow \bigwedge^{0,1}(L_0)$$.

The author went on claims that $$D_A=\sqrt{2}(\bar\partial_{A_0}+ \bar \partial_{A_0}^*)$$, this is the operator obtained by "coupling" $$\sqrt{2}(\bar\partial+\bar\partial^*)$$ with $$\nabla_{A_0}$$.

I don't understand what does "coupling" mean here. Also, I didn't sucessfully conclude what $$D_A$$ looks like, here is my attempt (with $$s$$ a section of original spin bundle and $$t$$ a section of $$L_0$$):

$$D_A (s\otimes t)=e_i \cdot \nabla_{e_i}^A(s\otimes t)= e_i \cdot(\nabla_{e_i}s\otimes t+s\otimes\nabla_{e_i}^{A_0}t)= D(s)\otimes t+e_i \cdot (s\otimes\nabla_{e_i}^{A_0}t)$$

But then how do we deal with the rest part?

You are considering a Kähler manifold of real dimension 4. The bundle $$(\Lambda^{0,2}\oplus\Lambda^0)\oplus\Lambda^{0,1}$$ is a $$\mathbb Z_2$$-graded Clifford module with compatible hermitian metric and connection (induced by the Levi-Civita connection), i.e. there is an action of the Clifford bundle satisfying the Leibniz rule with respect to the connection. Clifford modules are not unique, but you can tensor them with any hermitian bundle with compatible connection to obtain a new Clifford module with compatible connection. Every Clifford module with compatible connection admits a natural Dirac operator $$D=c\circ\nabla$$ where $$c$$ means contraction (i.e. applying the Clifford multiplication with (co)tangent vectors). If you start with a spin$$^\mathbb C$$ bundle (a special type of Clifford module, like the bundles in your question) and tensor it with a complex line bundle (with hermitian connection) you get a new spin$$^\mathbb C$$ bundle. The new Dirac operator is the old Dirac operator coupled with the connection. The basic properties (index, kernel, etc.) change by coupling with non-trivial connections.
Concerning your last question: if $$s$$ is a section of $$\Lambda^0$$ then $$\sum e_i\cdot s\otimes \nabla^A_{e_i} t$$ is just $$\sqrt{2}s\otimes\bar\partial^At$$, i.e.,it is by the Leibniz rule the difference between $$\sqrt{2}\bar\partial\otimes \text{id}$$ on $$\Lambda^0\otimes L_0$$ and $$\sqrt{2}\partial^A$$ on $$\Lambda^0\otimes L_0$$. Analogously, you can show the statement for $$s$$ being a section of $$\Lambda^{0,2}$$ using the adjoint operators.