Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 24 / 1

Put $P=|p|$ and $\zeta =e^{2\pi i/p}$. The lens space is the quotientThe cyclic group has order $P$. Its action is free: if $g^j$ fixes a point with $z_1\ne 0$, then $p\mid j$; if $z_1=0$, then $z_2\ne 0$ and $p\mid qj$, which again implies $p\mid j$ by coprimality. A finite free smooth action gives a closed orientable three-manifold, and the quotient projection is a covering space.

Since $S^3$ is simply connected, this is the universal covering and its deck group is the cyclic group of order $P$. HenceFor $P=1$ the quotient is $S^3$ and the group is trivial. With this unoriented quotient convention, replacing $p$ by $-p$ reverses the chosen generator and leaves the quotient unchanged; oriented surgery conventions may record the signs differently.

Take two solid tori $V_1,V_2$, each with a meridian-longitude basis $({\mu }_i,{\lambda }_i)$ on its boundary. Use opposite boundary-basis orientations on the two sides, so that a determinant-one coefficient matrix describes an orientation-reversing gluing map. For $A\in SL(2,\mathbb{Z})$ choose a torus homeomorphism $f:\partial V_2\to \partial V_1$ inducingand define $M_A=V_1{\cup }_f{V}_2$. The linear map on ${\mathbb{R}}^2/{\mathbb{Z}}^2$ realizes the matrix. Torus homeomorphisms with the same action on first homology are isotopic, and an isotopy extends over a boundary collar, so the resulting three-manifold depends only on $A$. Specifying opposite boundary conventions is necessary: gluing two identically oriented boundary bases by a determinant-one map would not describe the usual oriented gluing.

We now identify this gluing directly in the quotient from part (a). Split $S^3$ into the two invariant solid tori $W_1=\{|z_2|\le 1/\sqrt{2}\}$ and $W_2=\{|z_1|\le 1/\sqrt{2}\}$. On $W_1$ write $z_1=\sqrt{1-|w{|}^2}{e}^{2\pi ix}$, $z_2=w$. The quotient coordinatesidentify $W_1/⟨g⟩$ with $S^1\times D^2$. They are invariant under $g$; choosing a $p$th root of $u$ reconstructs $(x,w)$, and different choices are exactly the same cyclic orbit. Thus this is a genuine solid-torus coordinate model.

On the common boundary, measure arguments in turns as $(x,y)$. Its quotient lattice isThe meridian and longitude of the first quotient torus can be chosen as ${\mu }_1=(0,1)$ and ${\lambda }_1=(1/p,q/p)$. They form a lattice basis, since $(1,0)=p{\lambda }_1-q{\mu }_1$. The meridian of the second torus is precisely that $(1,0)$ loop. Consequently

The determinant condition is $qr+ps=-1$. Put $k=-r$, so $kq\equiv 1(\mathrm{mod}p)$. On $W_2$, the analogous quotient coordinates use $e^{2\pi ipy}$ and $e^{-2\pi iky}{z}_1$, and a longitude is $(k/p,1/p)$. Choose its negative for the second boundary-basis convention. Thensince $ps+qr=-1$. Thus the two quotient tori are glued by exactly $A$. This proves the genus-one gluing model of a lens space andDifferent choices of $r,s$ satisfying the determinant condition differ by $(s,r)\mapsto (s,r)+j(-q,p)$. This changes $f$ by a meridional Dehn twist on $V_2$, which extends over $V_2$: in solid-torus coordinates it is $(z,e^{i\theta })\mapsto (e^{ij\theta }z,e^{i\theta })$. Therefore there is no hidden dependence on that choice of longitude.

Use the quotient lens spaces and set $\zeta =e^{2\pi i/7}$. Let $g_j(z_1,z_2)=(\zeta z_1,{\zeta }^j{z}_2)$. The mapis a homeomorphism and satisfiesbecause $3\cdot 2\equiv -1(\mathrm{mod}7)$. Since $2$ is invertible modulo $7$, $g_3^2$ generates the entire target deck group. Hence $F$ maps cyclic orbits bijectively to cyclic orbits, as does its inverse, and descends to a homeomorphism:This is an explicit map, so no classification theorem for lens spaces is required. It reverses orientation: complex conjugation in one coordinate reverses the ambient real orientation, while swapping the two complex coordinates preserves it.