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In these notes on geometric quantization by Nair, on page 24, the Bohr-Sommerfeld rule in quantum mechanics is interpreted in terms of the Atiyah-Singer index theorem.

To be precise, the polarization condition $\bar{\partial}_V\Psi=0$ determines the Hilbert space of states associated with a symplectic manifold $M$, with prequantum line bundle $V$ which has curvature $\Omega$, the symplectic 2-form. Nair states that the number of normalizable solutions to the polarization condition is given by the Atiyah-Singer index theorem

$$ \operatorname{index}\left(\bar{\partial}_V\right)=\int_M \operatorname{td}(M) \wedge \operatorname{ch}(V). $$

My first question is, why is $\operatorname{index}\left(\bar{\partial}_V\right)=\operatorname{ker}\left(\bar{\partial}_V\right)$?

Nair goes on to elucidate the example of $\mathbb{CP}^1$, where $$ \operatorname{index}\left(\bar{\partial}_V\right)=\int_M \frac{\Omega}{2 \pi}+\int_M \frac{R}{4 \pi}=n+1 $$

My second question is, how does this generalize to other symplectic spaces, such as $\mathbb{CP}^2$? From what I understand, the Bohr-Sommerfeld rule says that the number of states in the Hilbert space should be linear in $n$, but it seems to me that the index theorem computation for $\mathbb{CP}^2$ will give a dependence that is roughly $an^2+bn+c$, where $a,b,c$ are constant factors, which I think follows the Hirzebruch-Riemann-Roch theorem for surfaces $$ \operatorname{index}\left(\bar{\partial}_V\right)=\frac{1}{2} c_1(V)^2+\frac{1}{2} c_1(V) c_1(M)+\frac{1}{12}\left(c_1(M)^2+c_2(M)\right) $$

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    $\begingroup$ DIM<->Volume. From the old times of quantum mechanics it was clear that DIMENSION of the Hilbert space corresponds to the symplectic VOLUME. ( Of course, god-fathers did not thought about finite-dimensional Hilbert spaces - and so one may seem that - they have no way to express that, but they did. One way was to use coherent states , and even in Landau Livshits it is written that hdpdx is a minimal volume to "fit" a quantum state ). $\endgroup$
    – Alexander Chervov
    Commented Feb 23 at 10:03
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    $\begingroup$ "Sommerfled correction". (Kind of Sommerfled correcttion to Bohr and modern understanding). Dimension = volume works pretty fine in the FLAT space. When the space is not flat - like CP^n or whatever - we need a correcting term. It comes from the curvature of the space itself. And that is why the term Td(M) appears. (It will be zero on flat spaces like tori). So the modern look dimension = Trace (Indentity) = \int Td(M) Ch(V). Where Hilbert space - are section of bundle "V", where c1(V) = \omega. $\endgroup$
    – Alexander Chervov
    Commented Feb 23 at 10:09
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    $\begingroup$ So more general look is to take Trace(AnyOperator) - not only identity operator. (For identity operator you get dimension). And so what is expected: Tr(\hat H) = \int H Ch(V) Td(M) . In some situations it is covered by the Fedosov index theorem, but there should be more general ways. There is no actual operator in Fedosov's approach. So the story is far from being finished. $\endgroup$
    – Alexander Chervov
    Commented Feb 23 at 10:13
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    $\begingroup$ Concerning dependence on "n". If I understand correctly - you scale symplectic form by "n": \omega -> n \omega , so clearly the symplectic volume on 2d-dimensional manifold will be scaled by n^d , for CP^2 : n^2 , and that is what you formula gives. So no contradction - everything is fine. You see n^2 term - its origin is also clearl ch(V) will contain \omega^d $\endgroup$
    – Alexander Chervov
    Commented Feb 23 at 12:24

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I can answer the first question.

The Atiyah-Singer theorem gives \begin{equation} \mathrm{Ind}(\overline{\partial}_V)=\int_M\mathrm{Td}(M)\wedge\mathrm{Ch}(V), \end{equation} and in general, we have \begin{equation} \mathrm{Ind}(\overline{\partial}_V)=\sum_{i=0}^{\dim_\mathbb{C}M}\dim_{\mathbb{C}}H^{(0,i)}(M,V), \end{equation} where $\dim_\mathbb{C}$ means the complex dimension.

For the case you are considering, we claim that \begin{equation} H^{(0,i)}(M,V)=0 \end{equation} for any $i\geqslant 1$, so \begin{equation} \mathrm{Ind}(\overline{\partial}_V)=\dim_{\mathbb{C}}H^{(0,0)}(M,V)=\dim_{\mathbb{C}}\ker(\overline{\partial}_V). \end{equation} as required.

To see this, we apply the Kodaira vanishing theorem: let $K_M$ be the canonical line bundle on a Kähler manifold $M$, if $V$ is a line bundle such that $V\otimes K_M^*$ is positive, then \begin{equation} H^{(0,i)}(M,V)=0 \end{equation} for any $i\geqslant 1$. In particular, here we have \begin{equation} (M,V,K_M^*)=(\mathbb{CP}^1,\mathcal{O}(n),\mathcal{O}(2)), \end{equation} and $V\otimes K_M^*=\mathcal{O}(n+2)$ is positive.

Refer to Proposition 3.72 of Berline-Getzler-Vergne's "Heat Kernels and Dirac Operators", and Proposition 2.4.3 of Huybrechts' "Complex Geometry".

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