I'm trying to understand Atiyah--Singer by looking at the usual starting point of $CP^1$ and the Dirac--Dolbeault operator. If I've reduced everything down correctly, then in this case the theorem gives $$ \text{Index}(\overline{\partial} + \overline{\partial}^*) = \frac{1}{2}\int_{CP^1} \text{ch}_1(T(CP^1)), $$ where ch$_1(T(CP^1))$ is the first Chern class of the tangent bundle $CP^1$. I would like a direct explanation of why this is true.
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asked Jun 11, 2011 at 19:03
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$\begingroup$ For the RHS, see en.wikipedia.org/wiki/… $\endgroup$– Steve HuntsmanCommented Jun 11, 2011 at 20:44
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$\begingroup$ I don't know helpful my cryptic comment will be, but you can compute both sides explicitly in this case. The right side is $1$ because $ch_1=c_1=2$ (times fundamental class). So is the left: use the Dolbeault resolution and the fact the holomorphic functions on $CP^1$ are constant... $\endgroup$– Donu ArapuraCommented Jun 11, 2011 at 21:21
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1$\begingroup$ @ Steve, Donu: Thanks for your answers, but I'm not really looking to show that both sides are equal (as you show, this is easy) I'm looking for someone who understands the Atiyah--Singer proof to show what it reduces down to in this simple case. $\endgroup$– Jean DelinezCommented Jun 11, 2011 at 22:02
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$\begingroup$ Yes, I see. I have to admit, I've only really looked as special cases, e.g Hirzebruch-Riemann-Roch. In any case, my (limited) understanding is there are two steps. The first (and probably the deepest) is to see that there is a universal formula in Chern classes of the manifold and symbol. Then to discover the explicit form by computing enough examples. Hopefully, someone else gave give you more insight. $\endgroup$– Donu ArapuraCommented Jun 11, 2011 at 22:57
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5$\begingroup$ I'm a little confused about what you want - the Atiyah-Singer index theorem for the Dolbeault operator on a Kahler manifold is precisely the Hirzebruch-Riemann-Roch theorem. So for $CP^1$ it looks like you're asking about the classical Riemann-Roch theorem. The proof of Atiyah-Singer doesn't really tell you anything new about complex geometry (in fact, if you follow the global proof then the first step is to embed $CP^1$ into Euclidean space...), and the proof that Atiyah-Singer implies Riemann-Roch just involves playing around with Clifford algebras and characteristic classes. $\endgroup$– Paul SiegelCommented Jun 12, 2011 at 0:44
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