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A line twist on the two-sphere has nonzero index and curvature term

Statement

Assume full AC.

Let $S^{2}$ have its unit round metric and complex orientation. Its tangent bundle admits a Spin lift; fix one. For each integer $k$ , there is a Hermitian line bundle $L_{k}$ with unitary curvature $F^{L_{k}} = - 2 π i k ω, ω = \frac{{area}_{S^{2}}}{4 π}, \int_{S^{2}} ω = 1.$ The twisted chiral Spin Dirac index is $k$ . In an oriented orthonormal frame the twisting term in the Lichnerowicz formula is $- \frac{k}{2} Γ$ , where $Γ = i c (e^{1}) c (e^{2})$ is spinor chirality. Thus for $k \neq 0$ the twist contributes a nonzero curvature endomorphism.

Facts & Assumptions

Given: Full AC, the round two-sphere, its Spin structure, an integer $k$ , and the line bundle with the displayed local connection.

[F1]

$T S^{2} ≅ T {C P}^{1}$ has $c_{1} = 2 x$ , so $w_{2} = 0$ and a Spin lift exists (The tangent bundle of complex projective space and its Pontryagin classes, Mod-two reduction of Chern classes, Spin structures exist exactly when w2 vanishes and then form an H1 torsor).

[F2]

The normalized Chern character is $tr \exp (- F / (2 π i))$ , and the A-hat form has no degree-two term (The A-hat form and relative Chern character, Chern, Pontryagin, and Euler characteristic forms).

[F3]

The curvature form of a unitary connection represents the real image of the topological first Chern class (Characteristic forms represent topological characteristic classes over the reals).

[F4]

The spin Dirac index formula applies to any finite-rank Hermitian twist, and its Lichnerowicz curvature term is $\sum_{i < j} c (e^{i}) c (e^{j}) F_{i j}^{E}$ (The twisted Spin Dirac index is the integral of A-hat times ch(E), Lichnerowicz formula for a geometric Dirac operator).

[F5]

For $S^{2} ≅ {C P}^{1}$ with its complex orientation, $x = c_{1} (γ^{*})$ generates $H^{2} (S^{2}; Z)$ and $⟨ x, [S^{2}] ⟩ = 1$ (The tangent bundle of complex projective space and its Pontryagin classes).

[A1]

Full AC is inherited through the Spin-lift criterion, Chern–Weil, and the spin index formula (The Axiom of Choice).

Proof

Given: Spherical coordinates $(θ, ϕ)$ away from the poles, with area form $\sin θ d θ \land d ϕ$ .

1.1A1F1givenconstructalgebra

By [F1], $T S^{2}$ admits a Spin lift; fix one. The unit sphere has area $4 π$ , so $ω = (4 π)^{- 1} \sin θ d θ \land d ϕ$ . On north and south charts define unitary connection forms $A_{N} = - \frac{i k}{2} (1 - \cos θ) d ϕ$ and $A_{S} = \frac{i k}{2} (1 + \cos θ) d ϕ$ . They extend smoothly over their respective poles, and $A_{S} - A_{N} = i k d ϕ = g^{- 1} d g$ for the unitary transition $g (ϕ) = e^{i k ϕ}$ . Hence they define a Hermitian line bundle connection; differentiation gives $F = d A_{N} = d A_{S} = - (i k / 2) \sin θ d θ \land d ϕ = - 2 π i k ω$ .

2.1A1F3F4F5step 1.1algebra

By [F3]–[F5], the real Chern class of this connection integrates to $k$ , so its integral first Chern class is $c_{1} (L_{k}) = k x$ ; the integral group is torsion-free by [F5], so its image in real cohomology determines it. The line-bundle curvature term in [F4] is $c (e^{1}) c (e^{2}) F_{12} = c (e^{1}) c (e^{2}) (- i k / 2) = - \frac{k}{2} Γ$ , because $Γ = i c (e^{1}) c (e^{2})$ . This is nonzero when $k \neq 0$ .

3.1A1F2F4step 1.1step 2.1∎

In dimension two $\hat{A}$ has only its degree-zero term $1$ , while $tr \exp (- F / (2 π i)) = 1 + k ω$ because $ω^{2} = 0$ . Applying [F4] yields ${ind}_{C} (D_{S \otimes L_{k}}^{+}) = \int_{S^{2}} k ω = k$ . For $k = 0$ both the curvature term and index vanish. The construction is smooth at both poles by the displayed vanishing factors, the orientation is the complex orientation, and the formula uses the same actual spin-module interface as [F4].

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Sources

Liviu I. Nicolaescu, Notes on the Atiyah–Singer Index Theorem, revision 15 November 2013 (standard reference, not scraped)
Konstantin Wernli, Lecture Notes on Spin Geometry (2019) (standard reference, not scraped)