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Consider the orientation-preserving affine group of the real line, realized asThe coordinates $(\log a,b)$ identify it smoothly with ${\mathbb{R}}^2$, so this Lie group is connected. Choose the Lie algebra element $E$ and the group element $D$:Direct matrix multiplication gives ${\mathrm{Ad}}_{D}E=DE{D}^{-1}=2E$.

If a bi-invariant Riemannian metric existed, every conjugation map $x\mapsto Dx{D}^{-1}$ would be an isometry fixing the identity. Its differential there, the Adjoint representation of a Lie group, would preserve the positive-definite inner product at the identity. Thusa contradiction since $E\ne 0$. Therefore this connected Lie group admits no bi-invariant Riemannian metric. This is the adjoint dilation obstruction to a bi-invariant Riemannian metric, an elementary alternative permitted by the question. The positive definite Riemannian metric interpretation matters; no claim about indefinite metrics is needed.

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.