The Heisenberg Lie group and algebra

Matrix multiplication gives $h(x,y,z)h(x^{\mathrm{\prime }},y^{\mathrm{\prime }},z^{\mathrm{\prime }})=h(x+x^{\mathrm{\prime }},y+y^{\mathrm{\prime }},z+z^{\mathrm{\prime }}+x{y}^{\mathrm{\prime }})$ and $h(x,y,z{)}^{-1}=h(-x,-y,-z+xy)$. Thus ${\mathbb{R}}^3$ with these polynomial formulas is a Lie group embedded in ${\mathrm{GL}}_3$.

Differentiation at the identity gives the span of $E_{01},E_{12},E_{02}$. From the product law in step 1.1, the corresponding left-invariant fields are $X^L={\mathrm{\partial }}_x$, $Y^L={\mathrm{\partial }}_y+x{\mathrm{\partial }}_z$, and $Z^L={\mathrm{\partial }}_z$. Their commutators are $[X^L,Y^L]=Z^L$ and $[X^L,Z^L]=[Y^L,Z^L]=0$. Hence [F1] gives $[X,Y]=Z$ and $Z$ central.

The group is nonempty and three-dimensional; its Lie algebra is two-step nilpotent but the bracket is degenerate because $Z$ is central. No metric, interval, endpoint, or biconditional occurs. ${\mathrm{A}\mathrm{C}}_{\omega }$ is propagated only through the current tangent-bracket supplier, and finite coordinates add no choice.