= Solution
Take the <orientation-preserving affine group of the real line>, written as
$$
G=\left\{\begin{pmatrix}a&b\\0&1\end{pmatrix}:a>0,\ b\in\mathbb R\right\},\qquad
(a,b)(a',b')=(aa',b+ab').
$$
It 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 $h\mapsto ghg^{-1}$ 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,
$$
T=\begin{pmatrix}0&1\\0&0\end{pmatrix},\qquad g=\begin{pmatrix}2&0\\0&1\end{pmatrix}.
$$
Direct multiplication gives $\operatorname{Ad}_gT=gTg^{-1}=2T$. Metric invariance would then imply $\|T\|^2=\|2T\|^2=4\|T\|^2$, impossible for a nonzero vector in a positive <inner product>. Therefore \b[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.
Back to article page
