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 and p.34 in the paper) 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.

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    $\begingroup$ I'm not sure if these remarks will answer your specific question, but I think the Dirac operator on the Dolbeault complex (regarded as a Dirac complex) and the Dolbeault operator have the same index because the Dirac operator and the Dolbeault operator have the same principal symbol (they differ by a 0th order operator) and the index depends only on the K-theory class of the symbol. The same sort of argument ought to apply in your twisted setting. $\endgroup$ May 12, 2012 at 19:41

1 Answer 1


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


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).

  • $\begingroup$ Thank you very much for explaining me and telling me the reference. It is very much appreciated. I also found the paper by Atiyah very beautiful. $\endgroup$ May 15, 2012 at 4:54

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