Past exam of the mathematics course of the University of Cambridge / 2012 / iii / Paper 55 / 1

Use the Lie bracket convention $[X,Y]=XY-YX$. If $E_{ij}$ denotes a matrix unit, then $E_{ij}{E}_{kl}={\delta }_{jk}{E}_{il}$. Here this givesThus, writing $[X_j,X_k]=f^i{}_{jk}{X}_i$, the nonzero structure constants of a Lie algebra areAll other entries vanish. This is the Lie algebra of the translation-dilation group of the plane: its two-dimensional translation ideal is abelian, and $X_0$ acts on that ideal as the identity. In group coordinates the multiplication and inverse areThese formulas also make the semidirect product structure explicit.

The Maurer-Cartan form is $g^{-1}dg={\lambda }^j{X}_j$. Direct matrix multiplication givesThese are left-invariant differential forms: for a constant group element $a$, $(ag{)}^{-1}d(ag)=g^{-1}dg$. Equivalently, left translation sends $(\rho ,x)$ to $(a\rho ,b+ax)$, so all three coordinate differentials and the denominator acquire the same factor $a>0$.

The left-invariant coframe of the translation-dilation group is pointwise linearly independent because $\rho >0$. Consequentlyis positive definite and unchanged under every left translation. It therefore defines the required left-invariant metric, with precisely the coordinate expression obtained from this sum. The notation $d{\rho }^2$ in that metric means $d\rho \otimes d\rho$, not the differential of ${\rho }^2$.

Taking the exterior derivative of the left-invariant differential forms yieldsFor example, $d({\rho }^{-1}d{x}^1)=-{\rho }^{-2}d\rho \wedge d{x}^1$. With the antisymmetric structure constants of a Lie algebra found above, the two terms with indices $(0,1)$ and $(1,0)$ combine asThe same computation holds for superscript $2$, and all superscript-$0$ terms vanish. Hence the Maurer-Cartan equation isThe factor $1/2$ avoids counting each antisymmetric pair twice. If coefficients were not required to be antisymmetric in $j,k$, their symmetric parts would be invisible to the wedge product of differential forms; the standard structure constants of a Lie algebra fix this ambiguity.

Take the frame dual to the left-invariant coframe of the translation-dilation group. It isIndeed ${\lambda }^i(L_j)={\delta }_j^i$. These are left-invariant vector fields, either because they are dual to invariant forms or because $L_j(g)=(d{L}_g{)}_e{X}_j$. For instance, the curve $g\exp (t{X}_0)$ differentiates to $\rho {\partial }_{\rho }$, while $g\exp (t{X}_1)$ differentiates to $\rho {\partial }_{x^1}$.

The coordinate Lie brackets satisfyFor the first identity only the derivative of the coefficient $\rho$ contributes. The linear map $\boxed{X_j⟼L_j}$ is a Lie algebra isomorphism: it preserves all basis brackets, is injective by independence of the frame, and is onto the three-dimensional space of left-invariant vector fields.