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Germs of orientation-preserving diffeomorphisms of the line at zero are torsion-free

Statement

Every finite subgroup of the group Diff⁡0+(R,0) of germs at 0 of orientation-preserving local diffeomorphisms of R fixing 0 is trivial: if h is such a germ and hn=id in the group of germs for some n≥1, then h=id. Equivalently, the group of germs of orientation-preserving local diffeomorphisms fixing a point of a one-dimensional manifold is torsion-free.

Facts & Assumptions

Given: A germ h∈Diff⁡0+(R,0) and a positive integer n with hn=id in the group of germs.

[F1]

A germ of local diffeomorphisms of R at 0 fixing 0 is an equivalence class of local diffeomorphisms f:U→V with 0∈U, f(0)=0, two representatives being equivalent when they agree on a neighbourhood of 0; the product is represented by the composite and the group structure is as in Germs of local diffeomorphisms at a point form a group (Germs of local diffeomorphisms at a point).

[F2]

Composition of representatives induces a well-defined associative operation with identity and inverses on Diff⁡0(R), so it is a group, and an element is the identity germ exactly when one (equivalently every) representative equals the identity on a neighbourhood of 0 (Germs of local diffeomorphisms at a point form a group).

[F3]

A local diffeomorphism of R is a C∞ map with C∞ local inverse; an orientation-preserving one fixing 0 has positive derivative at 0 and is therefore strictly increasing on a neighbourhood of 0 (Diffeomorphisms and local diffeomorphisms of manifolds).

Proof

technique · direct
1.1F1F2F3

(A representative with controlled iterates.) By [F1, F2] choose a representative h:I→R defined on an open interval I containing 0, with h(0)=0 and h orientation-preserving; by [F3] h has positive derivative at 0 and is strictly increasing on a neighbourhood of 0. Since hn is the identity germ, some neighbourhood of 0 is mapped identically by hn [F2]. First shrink I so h is strictly increasing throughout I. Continuity at the fixed point then gives an open interval J∋0 so that hn=id on J and every iterate hk(J), 0≤k≤n, lies in I [F1, F2, F3].

2.1F2F3step 1.1

(No displacement.) Suppose h is not the identity germ. Then, by [F2], for every neighbourhood of 0 there is a point t of that neighbourhood with h(t)≠t. Choose such a point t∈J. If h(t)>t, then strict increase of h on I gives hk+1(t)>hk(t) for every k<n, hence hn(t)>t, contradicting hn(t)=t; if h(t)<t, the same monotonicity gives hk+1(t)<hk(t) and hn(t)<t, again a contradiction. Hence no such t exists and h agrees with the identity on a neighbourhood of 0, that is, h=id as a germ.

3.1step 2.1

(Finite subgroups.) Let H≤Diff⁡0+(R,0) be a finite subgroup and let h∈H. The cyclic subgroup generated by h is contained in H, hence finite, so hk=id for some k≥1; step 2.1 applied with that k gives h=id. Therefore every element of H is the identity germ and H is the trivial subgroup: the group of germs is torsion-free.

4.1step 3.1∎

For a one-dimensional manifold T and x∈T, choose a chart at x; a germ of an orientation-preserving local diffeomorphism of T at x is represented in this chart by a germ of an orientation-preserving local diffeomorphism of R at 0, and composition and the identity are preserved by the chart change. Hence the same argument shows that the group of germs at x is torsion-free.

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