Past exam of the mathematics course of the University of Cambridge / 2018 / ii / Paper 1 / 21H (Algebraic Topology)

Write $v(x,y,z)$ for the displayed upper-unitriangular matrix. Matrix multiplication givesThus $V$ is the real Heisenberg group and $\Gamma$ is its Integer Heisenberg group. The subgroup $\Gamma$ is discrete. Choose an identity neighbourhood $U\subset V$ so small thatThen the translates $\gamma U$ are pairwise disjoint, and the quotient map restricts on each one to a homeomorphism onto the same open subset of $N$. Hence every point has an evenly covered neighbourhood andThis is the general quotient of a topological group by a discrete subgroup construction.

The mapis continuous and satisfies $g\circ f=f\circ g={\mathrm{id}}_T$. Therefore

For $u=e^{2\pi iy}$ and $w=e^{2\pi iz}$, defineChanging $y$ or $z$ by an integer changes $v(-t,y,z)$ by left multiplication by an element of $\Gamma$, so $\Phi$ is independent of the chosen lifts. It also respects the mapping-torus seam becausewhich is exactly the identification

Every $\Gamma$-orbit has a representative with its first coordinate in $[-1,0]$, so $\Phi$ is surjective. If two such representatives lie in one orbit, the integer shift of their first coordinates is zero unless they are opposite endpoints; in the first case their torus coordinates agree, and in the endpoint case they are related by the displayed seam. Thus $\Phi$ is injective. It is continuous, $M$ is compact, and $N$ is Hausdorff, so a continuous-bijection theorem givesThis realizes the Heisenberg nilmanifold as a torus mapping torus.

The space $V$ is homeomorphic to ${\mathbb{R}}^3$, hence contractible and simply connected. Part (a) therefore identifies $V\to N$ as the universal covering map. Its deck transformation group consists of the left translations by $\Gamma$, so covering-space theory givesCombining this with the homeomorphism from part (c),