Let us consider the Dirac complex
\begin{equation}
D\_{\rm Dirac}:S^+\to S^-
\end{equation}
where $S^{\pm}$ are the chiral-spinor bundles on $\mathbb{R}^4$.
Using the fact that the bundle $S^+$ is given by $\Omega^{0,0} \oplus \Omega^{0,2}$ twisted by $K^{1/2}$ while $S^-$ is given by $\Omega^{0,1}$ twisted by $K^{1/2}$ where $K$ is the canonical bundle, the equivariant index of the Dirac complex with respect to $T=U(1)\_1\times U(1)\_2$ action $(z\_1,z\_2)\mapsto (t\_1 z\_1,t\_2,z\_2)$ can be computed by  

\begin{eqnarray}
{\rm ind} D\_{\rm Dirac}&=& \frac{ t\_1^{1/2} t\_2^{1/2} + t\_1^{-1/2} t\_2
 ^{-1/2} 
 - ( t\_1^{1/2} t\_2^{-1/2} + t\_1^{-1/2} t\_2^{1/2})}
{ (1-t\_1)(1 -t\_1^{-1})(1-t\_2)(1-t\_2^{-1})} \cr
&=& \frac { t\_1^{1/2} t\_2^{1/2}}{ (1 -t\_1)(1-t\_2)}
\end{eqnarray}
I would like to know the reason why the spinor bundles are equivalent to the Dolbeault complex twisted by the square root $K^{1/2}$ of the canonical bundle. Why is the index of the Dirac operator equal to the one of the twisted Dolbeault operator?


This question comes from the computation of one-loop determinant done by Pestun. (see p.35-36 in the [paper](http://arxiv.org/abs/0712.2824) and p.34 in the [paper](http://arxiv.org/abs/1105.2568)) It was shown in those papers that, to compute one-loop determinant
\begin{equation}
\frac{\det\_{{\rm Coker} D} T}{\det\_{{\rm Ker} D} T} \ ,
\end{equation}
one can use the Atiyah-Singer index theorem for transversally elliptic operators.



The equivariant index of Dirac operator
To see why the spinor bundle is the bundle $\Omega^{0,\* }\otimes K^{1/2}$, you need to understand the relation between the spinor representation $S$ of $Spin(2n)$ and the exterior algebra representations $\Lambda^\* (\mathbf C^n)$ of $U(n)$.


If you choose an orthogonal complex structure $J$ on $\mathbf R^{2n}$, this picks out a subgroup $U(n)\subset SO(2n)$, with double-cover $\widetilde{U(n)}\subset Spin(2n)$. The notion of a "square root of the top exterior power" $(\Lambda^n(\mathbf C^n))^{1/2}$ makes sense as a $\widetilde{U(n)}$ representation. One finds (by computing characters, or from your favorite construction of the spinor representation) that, restricted to $\widetilde{U(n)}$, the spinor representation $S$ of $Spin(2n)$ is $\Lambda^\* (\mathbf C^n)\otimes (\Lambda^n(\mathbf C^n))^{-1/2}$ 


There's more about this in some of my class notes here


<http://www.math.columbia.edu/~woit/LieGroups-2012/spinors.pdf>


I learned this first from some beautiful lectures of Atiyah which explicitly discuss the Dirac/Dolbeault operator relation, see


"Classical groups and classical differential operators on manifolds" CIME, Varenna (1975).


