Clifford Action for Kahler Manifolds I'm working with Kahler manifolds at the moment and looking at their spin$^c$ structure. I really don't know much about spin$^c$ structures in  general and don't have enough time to learn it all at present, so I was hoping someone might be able to show me a shortcut in this case.
As far as I understand, for the usual spin$^c$ structure on an $N$-dimensional Kahler manifold $M$, the spinor bundle $S$ is given by the the direct sum of all the anti-holomorphic, that is $S = \bigoplus_{i=1}^N \Omega^{(0,i)}(M)$. Let 
$$
\nabla^s:S \to S \otimes \Omega^{(1,1)}, ~~~~~~~~~ s \mapsto \sum s_i \otimes \omega_i
$$ 
be the spin$^c$ connection. 
Is there a simple direct algebraic description of the Clifford action 
$$
c:S \otimes \Omega^{(1,1)} \to S
$$
in this case? For example, an initial stupid guess might be, for $\omega_i = \omega_i^h + \omega_i^{ah}$, with $\omega_i^{h} \in \Omega^{(1,0)}$ and  $\omega_i^{ah} \in \Omega^{(0,1)}$, we would have
$$
c(\sum_i s_i \otimes \omega_i) = \sum_i s_i\omega^{ah}_i.
$$
I am quite sure this is complete rubbish, but it illustrates the kind of result-from-heaven I'm hoping exists
 A: You ask for a short-cut, and I'm going to interpret this as asking "How can I see that on a Kaehler manifold $X$ the forms $\Omega^{0,\bullet}_X$ are a complex Clifford module, without first going through all that stuff about spin groups and their representations?" 
Warning: My scalar factors are probably wrong.
I'm going to start with the Hodge Laplacian 
$$\Delta = 2(\overline{\partial}\;\overline{\partial}^\ast+\overline{\partial}^\ast\overline{\partial})$$
acting on $S:=\Omega^{0,\bullet}_X$. 
What makes $\Delta$ a Laplacian is that it's a second order differential operator whose symbol $\sigma \colon T^\ast X \to \mathrm{End}(S)$ is the quadratic map $\lambda \mapsto - \| \lambda\|^2 \mathrm{Id}_S$. (The symbol is just the coefficient matrix of the leading (i.e. second) derivative terms in $\Delta$, written invariantly.) The symbol of $\overline{\partial}$ is $\lambda\mapsto \lambda^{0,1}\wedge \cdot$, that of $\overline{\partial}^\ast$ is $\lambda\mapsto \iota(g(\lambda))$ where $g\colon T^\ast X \to TX$ is the metric tensor, and from this you can check the symbol of $\Delta$. 
$\Delta$ has an evident square root, $D=\sqrt{2}(\overline{\partial}+\overline{\partial}^\ast)$, and this is a Dirac operator: a first order differential operator whose square is a Laplacian. The (complexified) symbol $\rho\colon T^*X\otimes \mathbb{C}\to \mathrm{End}(S)$ of a Dirac operator $D\colon \Gamma(S)\to \Gamma(S)$ makes $S$ into a Clifford module. The natural Clifford multiplication on $(0,\bullet)$-forms is the symbol of $D$, that is
$$ \rho(\lambda) =  \sqrt{2}(\iota(g(\lambda))+\lambda^{0,1}\wedge \cdot ) \in \mathrm{End}(S). $$
In fact, this Clifford module is a $Spin^c$-structure, which means that pointwise on $X$ it's a isomorphic to a standard Clifford module.
