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Explicit Chern–Simons transgression between two connections

Statement

Let $M$ be a finite-dimensional Hausdorff second-countable smooth manifold, possibly with boundary. Let $E \to M$ be a finite-rank real or complex smooth vector bundle with a fixed $G$ -frame reduction, and let $\nabla_{0}, \nabla_{1}$ be connections compatible with that same reduction. Put $A = \nabla_{1} - \nabla_{0}$ , $\nabla_{t} = \nabla_{0} + t A$ , and let $Ω_{t}$ be the curvature of $\nabla_{t}$ . For a homogeneous degree- $k$ $G$ -invariant polynomial $P_{k}$ with symmetric polarization, $k \geq 1$ , define

T_{P_{k}} (\nabla_{0}, \nabla_{1}) = k \int_{0}^{1} P_{k} (A, Ω_{t}, \dots, Ω_{t}) d t .

This is a global $(2 k - 1)$ -form and

P_{k} (Ω_{1}, \dots, Ω_{1}) - P_{k} (Ω_{0}, \dots, Ω_{0}) = d T_{P_{k}} (\nabla_{0}, \nabla_{1}) .

For $k = 0$ , the two constant curvature evaluations agree, so their difference is zero. The result applies to the ${GL}_{r} (C)$ , ${GL}_{r} (R)$ , $U (r)$ , and $SO (2 m)$ reductions with their respective invariant polynomials.

Facts & Assumptions

Given: The smooth base, fixed $G$ -reduction, compatible endpoint connections, and invariant polynomial in the Statement.

[F1]

In a supplied $G$ -frame, curvature is $Ω = d ω + ω \land ω$ ; its invariant-polynomial evaluation patches to a global form (Evaluation of an invariant polynomial on curvature).

[F2]

The symmetric polarization is invariant under simultaneous adjoint action by $G$ (Invariant symmetric polynomials on a matrix Lie algebra).

[F3]

The difference of two connections is a global endomorphism-valued one-form, whose frame matrix is $ω_{1} - ω_{0}$ (The difference of two connections is an endomorphism valued one form).

[F7]

A complex connection is $C$ -linear, so a difference of complex connections is complex-linear on the underlying real bundle (Complex-linear and metric-compatible bundle connections).

[F4]

For homogeneous $g$ -valued forms, differentiating the invariant-polynomial extension is the signed sum obtained by applying $D^{\nabla}$ in each slot (Invariant polynomials cancel connection commutators).

[F5]

For each connection with curvature $Ω_{t}$ , the covariant exterior derivative satisfies $d^{\nabla_{t}} Ω_{t} = 0$ (Second Bianchi identity for a bundle connection).

[F6]

On a manifold with boundary, the exterior derivative is defined by locally extendible coefficients and obeys the graded Leibniz rule (The de Rham complex and pullback extend to manifolds with boundary).

Proof

Proof technique: differentiate the invariant curvature form along the affine path and integrate its exact derivative.

1.1F1F2F3F7givenalgebra

By [F3], $A$ is a global endomorphism-valued one-form; for a complex bundle [F7] ensures that the difference remains complex-linear. In every supplied $G$ -frame its matrix $a = ω_{1} - ω_{0}$ is $g$ -valued. If $g_{β α}$ is a transition matrix, then $a_{β} = g_{β α}^{- 1} a_{α} g_{β α}$ and $ω_{t, β} = g_{β α}^{- 1} ω_{t, α} g_{β α} + g_{β α}^{- 1} d g_{β α}$ ; hence $ω_{t} = ω_{0} + t a$ is a compatible connection for every $t \in [0, 1]$ . The curvature transforms by conjugation, so [F2] makes $β_{t} = P_{k} (A, Ω_{t}, \dots, Ω_{t})$ agree in all frames. Its coefficients are smooth in $(x, t)$ , hence integrating on the compact interval defines a global smooth form of degree $1 + 2 (k - 1) = 2 k - 1$ .

2.1F1F4step 1.1algebra

In a fixed $G$ -frame, [F1] gives $Ω_{t} = d ω_{t} + ω_{t} \land ω_{t}$ . Differentiating this expression in $t$ yields ${\dot{Ω}}_{t} = d a + a \land ω_{t} + ω_{t} \land a = D^{\nabla_{t}} A$ , since for a one-form $a$ the local covariant derivative is $D^{\nabla_{t}} a = d a + ω_{t} \land a + a \land ω_{t}$ .

3.1F2F4F5step 2.1algebra

Set $α_{t} = P_{k} (Ω_{t}, \dots, Ω_{t})$ . Symmetry of $P_{k}$ and the even degree of every curvature factor give ${\dot{α}}_{t} = k P_{k} ({\dot{Ω}}_{t}, Ω_{t}, \dots, Ω_{t})$ . Applying [F4] to $(A, Ω_{t}, \dots, Ω_{t})$ leaves $d β_{t} = P_{k} (D^{\nabla_{t}} A, Ω_{t}, \dots, Ω_{t})$ : every other term contains $D^{\nabla_{t}} Ω_{t} = 0$ by [F5]. Here the local $D^{\nabla_{t}}$ in [F4] is the covariant exterior derivative $d^{\nabla_{t}}$ in [F5]. Therefore ${\dot{α}}_{t} = k d β_{t}$ .

4.1F1F6step 3.1algebra

Integrating the identity from step 3.1 gives $α_{1} - α_{0} = k \int_{0}^{1} d β_{t} d t = d (k \int_{0}^{1} β_{t} d t)$ . To justify the last equality, write $β_{t} = \sum_{I} b_{I} (x, t) d x^{I}$ in a chart; each $b_{I}$ is smooth, and each coordinate derivative commutes with its integral over compact $[0, 1]$ , so the equality holds coefficient by coefficient. On boundary charts the same calculation restricts from local extensions by [F6]. For $k = 0$ both endpoint forms are the same constant and their difference is zero. No axiom of choice is used. $□$

Explicit Chern–Simons transgression between two connections

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources