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Holonomy respects path concatenation and reversal

Statement

Assume Countable Choice ACω (The Axiom of Countable Choice (ACω)). Let a be a leafwise path from x to y, let b be a leafwise path from y to z, and let T,S,R be local transversals at x,y,z (Local transversals to a regular foliation). Then, with a∗b the concatenation (traverse a, then b), ha∗b(R,T)=hb(R,S)∘ha(S,T). Also ha−1(T,S)=(ha(S,T))−1, where a−1 is the reversed leafwise path.

Facts & Assumptions

Given: Leafwise paths a from x to y and b from y to z in a regular foliation F of M, local transversals T at x, S at y, R at z, and the concatenation a∗b and reversal a−1 of leafwise paths.

[F1]

The holonomy germ ha(T′,T):(T,x)→(T′,y) of a leafwise path is well defined, independent of the chart chain, the subdivision and the auxiliary transversals, and in a single foliation chart it is the germ matching points of the transversals with equal transverse coordinates (The holonomy germ is independent of the foliation chart chain, Plaques of a flat chart, Regular foliation atlases).

[F2]

Concatenation and reversal of leafwise paths are leafwise paths: the concatenation traverses a on the first half and b on the second, the reversal traverses a backwards; both lie in the common leaf (Leafwise paths and leafwise homotopy relative to endpoints).

[F3]

Germs of local diffeomorphisms at a point form a group under composition, so germs have inverses and composites of germs are germs; two germs are equal when representatives agree near the source point (Germs of local diffeomorphisms at a point form a group, Germs of local diffeomorphisms at a point).

Proof

technique · direct
1.1F1F2F3

Single-chart computation. Suppose that both a and b have images in a single foliation chart U with coordinates (x,y); then so does a∗b, and x,y,z lie in one plaque of U because a leafwise path segment in a chart stays in a plaque. By [F1] each of the three transports matches transverse coordinates in U: ha(S,T) sends u∈T to the point of S with the same y-coordinate, hb(R,S) sends that point to the point of R with the same y-coordinate, and ha∗b(R,T) sends u directly to the point of R with the same y-coordinate. The composite therefore agrees with the direct transport on a neighbourhood of x, so their germs are equal by [F3]. Similarly, traversing a backwards exchanges source and target and inverts the coordinate matching, so ha−1(T,S) is the inverse germ of ha(S,T).

1.2F1F2

General chain computation. Choose a chart chain for a with endpoint transversals T,S and a chart chain for b with endpoint transversals S,R. Concatenating the two chains and the two subdivisions at the middle time gives a chart chain for a∗b with endpoint transversals T,R and the intermediate transversal S at the middle point. By definition of the holonomy germ as the composite of the chart-wise transports, the germ obtained from this concatenated chain is exactly hb(R,S)∘ha(S,T); by chain independence [F1] it equals the intrinsic germ ha∗b(R,T).

2.1F1F3step 1.1

Reversal. Choose a chart chain for a; reading the same charts and subdivision backwards gives a chart chain for a−1 with the endpoint transversals exchanged. In each chart the reversed transport is the inverse of the forward transport by step 1.1, and by the group law for germs [F3] the composite of the inverses is the inverse of the composite, so ha−1(T,S)=(ha(S,T))−1 by [F1].

3.1step 1.2step 2.1∎

Conclusion. Steps 1.2 and 2.1 give ha∗b(R,T)=hb(R,S)∘ha(S,T) and ha−1(T,S)=(ha(S,T))−1 for arbitrary leafwise paths a,b and endpoint transversals.

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