Commutation of left- and right-invariant vector fields #
This file proves that left- and right-invariant differentiation commute. The proof uses Clairaut symmetry for a smooth scalar function evaluated on two multiplied exponential lines, then uses the tangent-vector/point-derivation equivalence to identify the vector-field bracket.
This supplies a prerequisite for Deliverable A, Layer 1 of
TauCetiRoadmap/RepresentationTheory/LieGroups/README.md.
Main results #
contDiff_comp_mulInvariantExp_mul_mulInvariantExp: a vector-valued function on two exponential lines multiplied around a fixed group element is smooth in both parameters.spatialFDeriv_mulInvariantExp_mul_mulInvariantExpandtimeFDeriv_mulInvariantExp_mul_mulInvariantExp: identify its two partial derivatives with left- and right-invariant differentiation.mvfderiv_mulRightInvariantVectorField_mulInvariantVectorField_commute: left- and right-invariant scalar differentiation commute at every group point.mlieBracket_mulRightInvariantVectorField_mulInvariantVectorField: the corresponding vector-field bracket vanishes everywhere.mlieBracket_mulInvariantVectorField_mulRightInvariantVectorField: the same vanishing result with the bracket arguments reversed.
References #
- Lie groups and the Lie algebra correspondence roadmap, Deliverable A, Layer 1, "The infinitesimal adjoint".
A smooth vector-valued function evaluated on two exponential lines multiplied around a fixed group element is smooth in both parameters.
Differentiating the second exponential parameter gives left-invariant differentiation by
Y.
Differentiating the first exponential parameter gives right-invariant differentiation by
X.
Left- and right-invariant scalar differentiation commute at every group point.
Right- and left-invariant vector fields commute everywhere.
Left- and right-invariant vector fields commute everywhere, with the bracket arguments in the opposite order.