Past exam of the mathematics course of the University of Cambridge 2002 iii Paper 74 1 Solution Created 2026-10-03 Updated 2026-10-14
The subspace spanned by is the center of a Lie algebra. Each of their tensor squares is therefore invariant. On tensor products the infinitesimal Adjoint representation of a Lie algebra acts by a derivation, . Write and let . The only potentially nonzero checks areThe remaining generators are central, so they act trivially. This proves every claimed tensor invariant.
These tensors are contravariant. To obtain an ordinary metric tensor, take an invertible linear combination and invert it. For arbitrary real and , setIn the ordered basis its matrix isSince , , andLet be the dual left-invariant coframe. The resulting neutral invariant metrics on the free two-step Lie algebra on three generators areHere . Infinitesimal invariance of implies that its inverse is an adjoint-invariant bilinear form. Its left-translated metric is right-invariant on a connected , giving the desired four-parameter family of bi-invariant metrics. A change of variables removes the central block from the quadratic form, leaving three off-diagonal two-dimensional blocks. Each has one positive and one negative eigenvalue. Thus the metric signature is ; these are bi-invariant pseudo-Riemannian metrics, not positive-definite ones.
Connectedness, conventionally understood here, matters for the passage from infinitesimal to full group invariance. If is disconnected, its other components must preserve separately. For example the Lie algebra automorphism reversing and fixing preserves all brackets but sends to . In the semidirect product of the simply connected group with this order-two automorphism, none of these nondegenerate tensors is invariant under that component. Thus the Lie-algebra data alone give the stated family on the identity component, rather than an unconditional extension to every disconnected realization.