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The twisted Spin Dirac index is the integral of A-hat times ch(E)

Statement

Assume full AC. Let $M$ be a compact, boundaryless, oriented, Riemannian manifold of even dimension $2 m$ with $m \geq 0$ , and let $E \to M$ be any finite-rank complex Hermitian vector bundle with a unitary connection. For $m \geq 1$ , assume a specified Spin lift and use the ordinary chiral Spinor bundle. For $m = 0$ , no Spin lift is input: at a point with orientation sign $ε_{x}$ , use the reference line from The A-hat form and relative Chern character, in even degree for $ε_{x} = 1$ and odd degree for $ε_{x} = - 1$ . Thus $D_{S \otimes E}^{+} = 0 : ⨁_{ε_{x} = 1} E_{x} \to ⨁_{ε_{x} = - 1} E_{x}$ . Then the chiral operator is Fredholm and ${ind}_{C} (D_{S \otimes E}^{+}) = \int_{M} {[\hat{A} (T M, \nabla^{L C}) tr \exp ⁣ (- \frac{F^{E}}{2 π i})]}_{2 m} .$ The rank-zero twist gives index and integral both equal to zero. The zero-dimensional convention does not assert an unstabilized Spin $(0)$ lift or assign it an $H^{1}$ torsor. This is a complex Fredholm index, not a real KO index.

Facts & Assumptions

Given: Full AC and either a positive-dimensional closed even-dimensional Spin manifold with its specified spin structure and a finite-rank Hermitian twist $E$ , or a closed zero-dimensional manifold with the finite-rank Hermitian twist and convention in the Statement.

[F1]

The spinor bundle, chirality, and spin connection are defined in Even-dimensional spinor bundles, chirality, and the spin connection; their global associated-bundle and connection properties are verified in Spinor and geometric Clifford bundles and connections are well defined.

[F2]

For the product Clifford module $V = S \hat{\otimes} E$ , the relative curvature is $F^{V / S} = F^{E}$ and the relative Chern character is the ordinary trace form (Spinor and geometric Clifford bundles and connections are well defined, The A-hat form and relative Chern character).

[F3]

The closed even-dimensional geometric Clifford-module index theorem gives $ind (D_{V}^{+}) = \int_{M} [\hat{A} (T M) ch (V / S)]_{2 m}$ (Atiyah–Singer index formula for a closed even-dimensional geometric Clifford module).

[A1]

The even and full complex algebras used in the spinor decomposition are supplied by Even subalgebras of real and complex Clifford algebras and Real and complex Clifford algebra classification with explicit signature. Full AC remains assumed through the geometric index theorem.

[F4]

The zero-dimensional reference line has grading $ε_{x} I$ ; with $E$ even its multiplicity space is $E$ , and the relative character is $rank E_{x}$ (The A-hat form and relative Chern character). Integration is $\int_{M} f = \sum_{x} ε_{x} f (x)$ (Integral of a compactly supported top form).

Proof

Given: The bundle $V = S \hat{\otimes} E$ with its product Clifford action and product connection.

1.1F3F4givenconstructalgebra

If $m = 0$ , compactness makes $M$ a finite discrete set. By [F4], the chiral operator is the zero map $⨁_{ε_{x} = 1} E_{x} \to ⨁_{ε_{x} = - 1} E_{x}$ , so it is Fredholm with index $\sum_{x} ε_{x} {rank}_{C} E_{x}$ . All curvature two-forms vanish; ${\hat{A}}_{0} = 1$ and the ordinary Chern character is the rank. The signed integral in [F4] is therefore this same sum, and [F3] agrees because the multiplicity bundle is $E$ in even degree. No Spin $(0)$ lift or torsor classification is asserted. The argument includes both point signs, disconnected mixtures, rank-zero twists and the empty manifold.

1.2A1F1F3given

Suppose $m \geq 1$ . The tensor-product connection is unitary and Clifford-compatible, and the product grading is the spinor chirality grading tensored with the even grading on $E$ . Hence [F3] applies to $V$ and gives a Fredholm chiral operator with index equal to the integral of $\hat{A}$ times its relative Chern character.

2.1A1F2F3step 1.2algebra∎

By [F2], the relative curvature for this product is exactly $F^{E}$ , and the relative supertrace is the ordinary trace because $E$ is ungraded. Substituting this equality in [F3] gives the displayed formula. If $E$ has rank zero, the module, both $L^{2}$ spaces, and the trace form are zero. The result uses the exact suppliers in [A1]; it remains a complex index.

Depends on

Used by

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

Dependency tree · two levels

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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)