Past exam of the mathematics course of the University of Cambridge / 2004 / iii / Paper 19 / 5 / c / Solution

Take the orientation-preserving affine group of the real line, written asIt is a connected Lie group, since its underlying manifold is $(0,\infty )\times \mathbb{R}$. We use the permitted alternative to part (a), a direct obstruction from the Adjoint representation of a Lie group.

If a bi-invariant Riemannian metric existed, conjugation $h\mapsto gh{g}^{-1}$ would be an isometry, being a composition of a left and a right translation. Its differential at the identity would preserve the positive inner product on the Lie algebra. Let $T$ be the nonzero infinitesimal translation,Direct multiplication gives ${\mathrm{Ad}}_{g}T=gT{g}^{-1}=2T$. Metric invariance would then imply $\Vert T{\Vert }^2=\Vert 2T{\Vert }^2=4\Vert T{\Vert }^2$, impossible for a nonzero vector in a positive inner product. Therefore this connected affine Lie group has no bi-invariant Riemannian metric. This is the adjoint dilation obstruction to a bi-invariant Riemannian metric; allowing an indefinite metric would be a different question.