# When and why can the spin connection term of the Dirac Operator be omitted?

The Dirac Operator $$D$$ is defined by $$$$\tag{1} D=i\gamma^a\nabla_a=i\gamma^a\nabla_{e_a}=i\underbrace{\gamma^a{e_a}^\mu}_{=\gamma^\mu}\nabla_{\partial_\mu}=i\gamma^\mu\nabla_\mu=i\gamma^\mu(\partial_\mu+\omega_\mu+A_\mu)$$$$ and $$$$\omega_a=-\frac{1}{4}\omega_{abc}\gamma^{bc}\psi={e_a}^\mu\omega_\mu.$$$$ However, in his derivation of the Atiyah Singer Index theorem$$^1$$, Fujikawa (chapter $$5.5$$ of Path Integrals and Quantum Anomalies) assumes $$$$D=i\gamma^\mu\nabla_\mu=i\gamma^\mu(\partial_\mu+A_\mu).$$$$ One might think that Fujikawa's $$D$$ is simply another operator, but Nakahara - who derives the same equation in section $$13.2.1$$ (Fujikawa's method) - says that the spin connection "plays no role" under certain assumptions:

We compactify the space in such a way that the geometry (the spin connection) plays no role. For example, this can be achieved by compactifying $$\mathbf{R}^4$$ to $$S^4=\mathbf{R}^4\cup\{\infty\}$$, for which the Dirac genus $$\hat{A}(TM)$$ is trivial.

I know that $$S^n\cong\mathbf{R}^n\cup\{\infty\}$$, but I don't see why this implies that the spin connection "plays no role".

$$^1$$ By the "Atiyah Singer Index theorem" I mean the equation $$$$\mathrm{ind}\,D_+=-\frac{1}{8\pi^2}\int_M\mathrm{tr}(F_{ij}F_{kl})\epsilon^{ijkl}\cdot\omega,$$$$ where $$M$$ is a $$4$$-dimensional Riemannian manifold with euclidean signature and $$\omega$$ is the volume form. (I am aware of the fact that $$D\colon\Gamma(M,S\otimes E)\to\Gamma(M,S\otimes E)$$ is a Fredholm operator and that $$(1)$$ is only valid after the choice of a local gauge - here we assume that $$E$$ is the the associated vector bundle induced by the adjoint representation.)

If you compacify to a torus then, in Cartesian coordinates, the spin connection vanishes and so is irrelevent. If you compactify to a sphere, as Fujikawa suggests, it is far less obvious that $$\hat A(TM)$$ is not needed. That it plays no role is because $$\hat A(TM)$$ is a genus and so cobordism invariant. This means that the curvature contribution to the index is zero when $$M$$ is a boundary: $$M=\partial N$$, and the spin connection can be extended through $$N$$. This is true in the case of a sphere. I do not know a simple proof of the cobordism property though.
• To prove the ASIT, it doesn't suffice to prove $\mathrm{ind}\,D_+=-\frac{1}{8\pi^2}\int_M\mathrm{tr}(F_{ij}F_{kl})\epsilon^{ijkl}\cdot\omega$ like I did, but one needs to show that RHS is equal to the topological index and that's where the A hat genus joins the party. Considering the operator $i\gamma^\mu(\partial_\mu+A_\mu)$ is useful to prove $\mathrm{ind}\,D_+=-\frac{1}{8\pi^2}\int_M\mathrm{tr}(F_{ij}F_{kl})\epsilon^{ijkl}\cdot\omega$, but $i\gamma^\mu(\partial_\mu+A_\mu)$ is not the Dirac operator. (A confirmation would be appreciated.) Apr 7, 2021 at 19:10
• Yes. There are curvature terms coming from $\hat A$ in in the expression for the index general spin manifolds. Even when $\hat A$ makes not contribution you will need the spin connection to makes ense of the Dirac operator. Apr 7, 2021 at 19:29