Question about the index of two elliptic operators over a 4-dimensional Riemannian manifold

Question

I've already asked this question in: https://math.stackexchange.com/questions/4899825/question-about-the-index-of-two-elliptic-operators-over-a-4-dimensional-riemanni, and I've been suggested to ask it on overflow.

Let $X$ be a compact Riemannian 4-manifold, $P$ a principal $G$-bundle over $X$, and $\mathfrak{g}$ be its adjoint bundle. Let $\omega$ be a self-dual conneciton on $P$ (i.e. its curvature $\Omega \in \Omega^2(\mathfrak{g})$ is self-dual). It induces a covariant derivative $d^\omega:\Omega^0(\mathfrak{g})\to \Omega^1(\mathfrak{g})$, which extends to $d^\omega:\Omega^k(\mathfrak{g})\to \Omega^{k+1}(\mathfrak{g})$ (https://en.wikipedia.org/wiki/Connection_(vector_bundle)#Exterior_covariant_derivative_and_vector-valued_forms). Since $\omega$ is self-dual, the sequence $$ 0\to \Omega^0(\mathfrak{g})\xrightarrow{d^\omega} \Omega^1(\mathfrak{g})\xrightarrow{d^\omega_-}\Omega^2_-(\mathfrak{g})\to 0$$ is a complex, where $d^\omega_-$ is the composition $\Omega^1(\mathfrak{g})\xrightarrow{d^\omega}\Omega^2(\mathfrak{g})\xrightarrow{\text{proj.}_-}\Omega^2_-(\mathfrak{g})$ and $\Omega^2_-(\mathfrak{g})$ is the space of anti-self-dual $\mathfrak{g}$-valued 2-forms.

This complex is elliptic, and considering the adjoint $d^*:\Omega^1(\mathfrak{g})\to\Omega^0(\mathfrak{g})$ (https://en.wikipedia.org/wiki/Hodge_star_operator#Codifferential) of $d^\omega$, the index of the complex is equal to the index of the elliptic operator $$ d^* \oplus d^\omega :\Omega^1(\mathfrak{g})\to \Omega^0(\mathfrak{g}) \oplus \Omega^2_-(\mathfrak{g})$$

Now my question is the following: Why does this operator has the same index as the Dirac operator $$D:\Gamma(V_+\otimes V_-\otimes \mathfrak{g})\to \Gamma(V_-\otimes V_-\otimes \mathfrak{g})?$$ This is asserted in top of p.445 of Atiyah's paper https://www.math.miami.edu/~cscaduto/teaching/782-fall-2023/notes/AHS.pdf, and it is written that: these two elliptic operators have the same symbol and factor through the same connection. But I can't understand what this mean and that why it implies that two have the same index. (Here, $V_\pm$ are the bundles of self-dual/anti-self-dual spinors, defined in p.428 of the paper.)

Answer 1

Consider the full bundle of exterior forms. The Hodge star operator $\ast$ acts on it with eigenvalues $\pm1$. The $-1$-eigenbundle is isomorphic to $V^{\pm1}\otimes V^-$. The map $\frac{1-\ast}2$ is a projection onto this subbundle. Restricted to $\Lambda^0T^*M\oplus\Lambda^1T^*M\oplus\Lambda^-T^*M$, it becomes an isomorphism of vector bundles. All this still holds if everything is twisted by the adjoint bundle.

Now one can check that the operator $d^*\oplus d^\omega$ under the isomorphism above becomes an elliptic operator that can be deformed into $D$ within the class of elliptic operators. Therefore both have the same index.

Answer 2

It's something along the lines of spinors can be identified with the space of homomorphisms preserving some structure between two 2 dimensional complex vector spaces. If you have a sequence $0\to \mathbb{C}\to U\to \Lambda^2U\to 0$ where $U$ is a 2 dim complex vector space and the maps are both $\delta u$, multiplication by some fixed $u\in U$, you can trivialize the third term and use a metric to take an adjoint of the second map (corresponds to your $d*$) and then for each $u$, you have a map $\delta u \oplus \delta u* :\mathbb{C}\oplus \Lambda^2 U\to U$. This identifies $U$ with the space of maps $\mathbb{C}\oplus \Lambda^2 U\to U$, so it identifies $U$ with a space of spinors. If you consider these as bundles, you get an identification of $d+d*$ to $D$ in some way which I forget the details of. The details and a significantly better explanation can be found in Donaldson's geometry of four manifolds.