Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

share|improve this question
2  
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. –  Paul Siegel May 12 '12 at 19:41
add comment

1 Answer

up vote 3 down vote accepted

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).

share|improve this answer
    
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. –  Satoshi Nawata May 15 '12 at 4:54
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.