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Curvature of an extending Bott connection lies in the transverse differential ideal

Statement

Assume Countable Choice ${A C}_{ω}$ . Let $F$ be a codimension- $q$ regular foliation of a smooth manifold $M$ with normal bundle $ν = T M / T F$ , let $\nabla^{B}$ be the Bott partial connection, and let $\nabla$ be any connection on $ν$ extending it: $\nabla_{X} = \nabla_{X}^{B}$ for every $X \in Γ (T F)$ . Let $I \subseteq Ω^{*} (M)$ be the differential ideal generated by the $1$ -forms that vanish on $T F$ , i.e. the ideal locally generated by a coframe of the annihilator bundle of $T F$ . Then, in a local frame of $ν$ that is parallel along the leaves for $\nabla^{B}$ , the curvature matrix entries of $\nabla$ lie in $I$ ; consequently every coefficient of the curvature two-form $R^{\nabla} \in Ω^{2} (M; E n d ν)$ lies in $I$ , and $I^{q + 1} = 0$ .

Facts & Assumptions

Given: Assume ${A C}_{ω}$ . A codimension- $q$ regular foliation $F$ with tangent distribution $T F$ and normal bundle $ν = T M / T F$ , the Bott partial connection $\nabla^{B}$ on $ν$ , and a connection $\nabla$ on $ν$ with $\nabla_{X} = \nabla_{X}^{B}$ for every $X \in Γ (T F)$ .

[F1]

The Bott partial connection is well defined, is $C^{\infty}$ -linear in the vector-field variable, satisfies the Leibniz rule, and has vanishing curvature $R^{B} (X, Y) s = 0$ for leaf-tangent fields $X, Y$ . (The Bott partial connection is well defined and flat along leaves).

[F3]

In a local frame with connection matrix $ω$ and curvature matrix $Ω$ , the structure equation $Ω = d ω + ω \land ω$ holds. (Curvature two-form structure equation).

[F4]

For homogeneous smooth forms $α, β$ of degrees $p, q$ one has $d (α \land β) = d α \land β + (- 1)^{p} α \land d β$ . (The exterior derivative is a graded derivation).

Proof

technique · direct

1.1givenF1F4

In a foliation chart $(x_{1}, \dots, x_{n - q}, y_{1}, \dots, y_{q})$ , the annihilator of $T F$ is spanned by $d y_{1}, \dots, d y_{q}$ . Thus the ideal $I$ consists locally of sums $\sum_{j} d y_{j} \land α_{j}$ , and is closed under $d$ , because $d (d y_{j}) = 0$ and $d (d y_{j} \land α_{j}) = - d y_{j} \land d α_{j}$ . The classes $e_{j} = π (\partial_{y_{j}})$ form a local frame of $ν$ parallel for the Bott connection: if $X = \sum_{i} X^{i} \partial_{x_{i}}$ then $[X, \partial_{y_{j}}] = - \sum_{i} (\partial_{y_{j}} X^{i}) \partial_{x_{i}}$ is leaf-tangent.

2.1givenstep 1.1

Use the extending connection supplied in the statement. Its connection entries $θ^{a}_{b}$ in the frame of step 1.1 vanish on every leaf direction, since $\nabla_{X} e_{b} = \nabla_{X}^{B} e_{b} = 0$ there. Hence $θ^{a}_{b} \in I$ .

3.1F3step 1.1step 2.1

The structure equation gives $Ω^{a}_{b} = d θ^{a}_{b} + \sum_{c} θ^{a}_{c} \land θ^{c}_{b}$ . Both summands lie in $I$ , by its differential-ideal property and step 2.1. A frame change conjugates the curvature matrix by smooth function matrices, so every curvature entry in every frame lies in $I$ .

4.1step 1.1step 3.1∎

Every product of $q + 1$ local elements of $I$ contains $q + 1$ factors drawn from the $q$ forms $d y_{j}$ ; alternating multiplication forces a repeated factor and gives zero. Thus $I^{q + 1} = 0$ , including $q = 0$ , where $I = 0$ . This proves the assertions without asserting existence of an extension or choosing a global cover.

Depends on

Used by

Bott vanishing for real Pontryagin monomials of a foliation Theorem

Dependency tree · two levels

76 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Raoul Bott, Lectures on Characteristic Classes and Foliations (Lecture Notes in Mathematics 279; complete scan of the 178-page volume) (standard reference, not scraped)