Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 302 / 1

The integers are closed under addition, addition is associative, zero is an identity, and $-n$ is the additive inverse of $n$. Thus $(\mathbb{Z},+)$ is an abelian group.

The nonzero integers are closed under multiplication and contain the identity, but most elements have no inverse in the set: for example, the multiplicative inverse of $2$ is $1/2\notin {\mathbb{Z}}^{\ast }$. Hence $({\mathbb{Z}}^{\ast },\times )$ is not a group.

Composition is associative, andagain has nonzero slope. The identity is $(1,0)$ andThus these maps form the real affine group of the line, a nonabelian group.

In coordinates the Real Heisenberg group law isCommuting this with every $(q^{\prime },r^{\prime },s^{\prime })$ requires $q{r}^{\prime }-q^{\prime }r=0$ for all $q^{\prime },r^{\prime }$, hence $q=r=0$. Therefore

The Lie algebra consists of strictly upper-triangular matricesThe only nonzero basis bracket is $[Q,R]=S$. Thus the independent nonzero structure constant of a Lie algebra values are $f_{QR}{}^S=1$ and $f_{RQ}{}^S=-1$.

Since $X^3=0$,Every group element has the unique preimage $t=s-qr/2$, so the Exponential map of a Lie group is bijective. It is a diffeomorphism from ${\mathbb{R}}^3$ onto $G$, proving that $G$ is connected and simply connected.

The line $\mathbb{R}S$ is a nonzero central ideal, so the Heisenberg Lie algebra is not simple. It is two-step nilpotent and hence solvable; a nonzero solvable Lie algebra is not semisimple. It is therefore neither simple nor semisimple.

Strictly, the normalized functions form the unit sphere rather than a vector space; let $V=L^2(\mathbb{R})$ and restrict to normalized states when interpreting wavefunctions. The group law givesTranslation preserves Lebesgue measure and both exponential factors have unit modulus, so $D(g)$ preserves the inner product and is a unitary representation. This is the Schrödinger representation of the Heisenberg group: $q$ translates position, $r$ translates momentum, and $s$ contributes the physically irrelevant overall phase.