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Use the round unit sphere with its complex orientation and unique spin structure. Its chiral spinor bundles are $S^{+}=\mathcal{O}(-1)$ and $S^{-}={\Lambda }^{0,1}\otimes \mathcal{O}(-1)$, the latter a smooth line bundle of degree $+1$. The Dirac operator isHere Clifford multiplication has $c(v{)}^2=-|v{|}^2$, and $D$ acts self-adjointly on the L2 spinors with first-order Sobolev domain. The geometric spin connection is invariant under the lifted $SU(2)$ action, hence $D$ is equivariant.

We justify the spectral properties needed for its phase. The Peter-Weyl theorem and Frobenius reciprocity for compact groups give the decomposition of sections of an isotropy-weight-$q$ line bundle: $E_n$ occurs once exactly when $n\ge |q|$ and $n\equiv q(\mathrm{mod}2)$, because its torus weights are $n,n-2,\dots ,-n$. The spinor isotropy weights are $+1,-1$, so each chirality contains $E_1,E_3,E_5,\dots$, each once. The horizontal connection Laplacian is the total Casimir minus the vertical generator square, ${\nabla }^{\ast }\nabla =C-1/4$ on these weight-one bundles. Expanding the Dirac square in a normal orthonormal frame gives the diagonal connection-Laplacian terms and the off-diagonal curvature contraction. The Clifford relations contract spin curvature to $\mathrm{Scal}/4$, the Lichnerowicz formula for a twisted Dirac operator with no twisting bundle. Since the scalar curvature here is two,Thus there is no kernel, the eigenvalues of $D$ are $\pm 1,\pm 2,\dots$ with finite multiplicities, and the resolvent is compact by spectral truncation. These round-sphere spectral facts agree with the explicit calculation in Abrikosov's paper.

Define the chiral phase of the round-sphere Dirac operatorIt is unitary because both chiral kernels vanish; equivariance follows from that of $D$. The full phase is $F=D|D{|}^{-1}=\left(\begin{array}{cc}0& V^{\ast }\\ V& 0\end{array}\right)$.

For a smooth scalar $f$, the Leibniz rule gives $[D,M_f]=c(df)$, a bounded bundle endomorphism. The spectral integral for the sign operator gives, in the strong topology,Write $R_{\pm }=(D\mp it{)}^{-1}$. Since $2D(D^2+t^2{)}^{-1}=R_{+}+R_{-}$ and $[R_{\pm },M_f]=-R_{\pm }[D,M_f]R_{\pm }$, the commutator integral is norm convergent: its integrands are compact and bounded by $\Vert [D,M_f]\Vert /(1+t^2)$. A norm limit of their integrals is compact. Thus $[F,M_f]$ is compact, and its off-diagonal blocks say $V{M}_f^{+}-M_f^{-}V$ is compact. Uniform approximation of continuous $f$ by smooth functions extends this to all $f\in C(S^2)$, using the bound $2\Vert f-g{\Vert }_{\infty }$ for the change of commutator.

The chiral phase is an equivariant unitary which intertwines the two multiplication representations modulo compacts. This specifies the grading implicit in the subsequent index question; treating the full self-adjoint phase as an endomorphism would yield a different, zero, compressed index.