Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 302 / 1

A Lie group is a smooth manifold $G$ equipped with smooth maps $m:G\times G\to G$, $m(g,h)=gh$, and $\iota :G\to G$, $\iota (g)=g^{-1}$, for which multiplication is associative, there is an identity $e$, and every $g$ has the inverse $g^{-1}$. In equations,Thus the group axioms and the smooth structure are compatible.

In a coordinate chart centred at the identity, smooth multiplication has a local group lawThe identity gives $F(0,x)=x$ and $F(y,0)=y$, while associativity gives the functional equation $F(u,F(y,x))=F(F(u,y),x)$. Smooth inversion determines coordinates $\bar{x}$ with $F(\bar{x},x)=F(x,\bar{x})=0$. Changes of local coordinates alter the symmetric part of $B^k{}_{ij}$, whereas its antisymmetric part supplies the coordinate-independent Lie bracket on the tangent space at the identity.

Write $g(x)=I+x^i{T}_i+O(x^2)$ in a matrix representation and $[T_i,T_j]=c^k{}_{ij}{T}_k$. Direct multiplication givessoThe cancellation of the constant and linear terms explains why the group commutator first detects the Lie bracket at quadratic order.

The matrix is invertible precisely when $a\ne 0$, andThe largest group is thereforethe real affine group $\mathbb{R}⋊{\mathbb{R}}^{\ast }$ under $(a,b)(a^{\prime },b^{\prime })=(a{a}^{\prime },b+a{b}^{\prime })$.

As a smooth manifold, $G\cong {\mathbb{R}}^{\ast }\times \mathbb{R}$. It has two connected components, distinguished by the sign of $a$: each of $a>0$ and $a<0$ is connected, and a continuous path in $G$ cannot cross $a=0$.

The defining two-dimensional group representation is reducible, because the line $\mathbb{R}{e}_1$ is an invariant subspace:For $b\ne 0$, the complementary line $\mathbb{R}{e}_2$ is not invariant, so this representation need not split as a direct sum of one-dimensional representations.

The map $(a,b)\mapsto \mathrm{sgn}(a)$ is a surjective group homomorphism with kernel $H_0$, so $H_0$ is a normal subgroup and $G/H_0\cong C_2$. The subgroup $H_1$ is not normal: for a translation $t=(1,c)$,which leaves $H_1$ when $a\ne 1$ and $c\ne 0$. Finally, $(a,b)\mapsto a$ is a surjective homomorphism onto ${\mathbb{R}}^{\ast }$ with kernel $H_2$, so $H_2◃G$ and the quotient group $G/H_2\cong {\mathbb{R}}^{\ast }$.