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

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.