The equivariant index of Dirac operator - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T10:29:34Z http://mathoverflow.net/feeds/question/96793 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/96793/the-equivariant-index-of-dirac-operator The equivariant index of Dirac operator Satoshi Nawata 2012-05-12T19:12:07Z 2012-05-14T19:20:48Z <p>Let us consider the Dirac complex $$D_{\rm Dirac}:S^+\to S^-$$ 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<br> \begin{eqnarray} {\rm ind} D_{\rm Dirac}&amp;=&amp; \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 &amp;=&amp; \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?</p> <p>This question comes from the computation of one-loop determinant done by Pestun. (see p.35-36 in the <a href="http://arxiv.org/abs/0712.2824" rel="nofollow">paper</a> and p.34 in the <a href="http://arxiv.org/abs/1105.2568" rel="nofollow">paper</a>) It was shown in those papers that, to compute one-loop determinant $$\frac{\det_{{\rm Coker} D} T}{\det_{{\rm Ker} D} T} \ ,$$ one can use the Atiyah-Singer index theorem for transversally elliptic operators.</p> http://mathoverflow.net/questions/96793/the-equivariant-index-of-dirac-operator/96939#96939 Answer by Peter Woit for The equivariant index of Dirac operator Peter Woit 2012-05-14T19:20:48Z 2012-05-14T19:20:48Z <p>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 $U(n)$.</p> <p>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 $\Lambda^* (\mathbf C^n)\otimes (\Lambda^n(\mathbf C^n))^{-1/2}$ </p> <p>There's more about this in some of my class notes here</p> <p><a href="http://www.math.columbia.edu/~woit/LieGroups-2012/spinors.pdf" rel="nofollow">http://www.math.columbia.edu/~woit/LieGroups-2012/spinors.pdf</a></p> <p>I learned this first from some beautiful lectures of Atiyah which explicitly discuss the Dirac/Dolbeault operator relation, see</p> <p>"Classical groups and classical differential operators on manifolds" CIME, Varenna (1975).</p>