Past exam of the mathematics course of the University of Cambridge / 2005 / iii / Paper 19 (Geometrisation of 3-manifolds)

We use $[a,b]=a^{-1}{b}^{-1}ab$. To obtain a Wirtinger presentation, choose an oriented regular knot diagram, cut its strands at undercrossings, and give each resulting arc a positively oriented meridian of a knot. The complement of the arcs above the projection plane has a free fundamental group on these meridians. At a crossing, pushing an understrand meridian past the overstrand conjugates it by the overstrand meridian. Thus, if $a$ is the overstrand generator and $b,c$ are the incoming and outgoing understrand generators, the relation is $c=a^{ϵ}b{a}^{-ϵ}$, where $ϵ$ is determined by the crossing orientation. Applying the Seifert-van Kampen theorem while adjoining the crossing neighborhoods gives one such relation per crossing. The remaining three-dimensional region adds no fundamental-group relation. For a connected diagram one crossing relation is redundant. This derives a group presentation from any tame knot diagram, rather than just specifying a recipe without its topological justification.

For the Borromean rings, choose three meridians $x,y,z$ and call the other three arc meridians $x^{\prime },y^{\prime },z^{\prime }$. Base paths and orientations can be chosen so the six crossing relations readEliminate the primed generators with the first row. The second row says respectively that $x$ commutes with $(z^{\prime }{)}^{-1}z=[y^{-1},z]$, that $y$ commutes with $(x^{\prime }{)}^{-1}x=[z^{-1},x]$, and that $z$ commutes with $(y^{\prime }{)}^{-1}y=[x^{-1},y]$. The last relation is redundant, just as one crossing relation is redundant in the Wirtinger presentation. Alternatively the Hall-Witt identity, after conjugating its three factors, shows that any two of these cyclic relations imply the third. Hence the Borromean link group isBoth relators have zero exponent sum in each generator. Passing from the free group to its abelianization therefore imposes no relations beyond commutation, andEquivalently the three linking-number maps to $\mathbb{Z}$ independently detect the three meridians.

Here is an explicit representation, followed by a faithfulness argument:Each determinant is one. Direct multiplication verifies the two relators. For example $[Y^{-1},Z]=\left(\begin{array}{cc}-1& 2i\\ 0& -1\end{array}\right)$, which commutes with $X$. Merely checking those relators would not prove faithfulness, so we identify the image's complete group presentation.

Write $B=PS{L}_2(\mathbb{Z}[i])$, the Gaussian Bianchi group, and use its four generators $T,U,L,A$ from the allowed presentation. In $B$ the three proposed generators areThe last equality is in $PS{L}_2$, so an overall minus sign in matrix multiplication is immaterial. Reduce matrices modulo $4$ in $\mathbb{Z}[i]$, and then quotient by $\{\pm I\}$. The finite image $Q$ has order $1536$: the residue field is ${\mathbb{F}}_2$, $S{L}_2({\mathbb{F}}_2)$ has order six, and each of the three subsequent local-ring layers has a kernel of order $2^3$, before the final quotient by $\{\pm I\}$. Elementary elimination over this local ring shows that the images of $T,U,A$ generate it. The subgroup $\bar{H}=⟨\bar{X},\bar{Y},\bar{Z}⟩$ has order $64$, as multiplication in this finite ring verifies. Let $K$ be its inverse image in $B$; then $[B:K]=24$.

For completeness the following gives the coset computation needed to check the subgroup presentation explicitly. Cosets are right cosets $Kr$; columns record right multiplication by $T,U,L,A$. Representatives are words in the allowed generators.Apply the Reidemeister–Schreier theorem: for a transition $j{⟶}^{g}k$, use generator $r_{j}g{r}_k^{-1}$, omitting the 23 tree transitions. Rewrite each of the eight allowed ambient relators starting at each of the 24 rows. There are 73 remaining Schreier generators. Adjoin the names $x=T$, $y=A{U}^{-2}{A}^{-1}$, $z=A{T}^{-2}{A}^{-1}L{U}^{-1}$ and eliminate the Schreier generators using relations in which they occur once. The remaining relations, with an uppercase letter denoting the inverse of the corresponding lowercase one, are the wordsThis table and the rewriting rule determine every relation without an appeal to a guessed presentation. In this notation cyclically reduced versions of the two desired relators are $r=ZYzXZyzYxy$ and $q$. Elementary substitutions of relator segments, with cyclic reduction at each step, giveusing $r,q$; conversely $r⟶p⟶1$ using $p,q$. These substitutions show equality of the two normal closures. For example the first substitution replaces $xyX$ in the cyclic rotation of $p$ by $zxZyzXZ$, as allowed by $q$, and then the initial $ZYz$ is replaced by $YXyZYzx$ using $r$. Thus $K$ has exactly the presentation of $G$, with $x,y,z$ represented by $X,Y,Z$. In particular these three matrices generate $K$, and

To identify the actual link complement as a hyperbolic three-manifold, rather than only on a manifold with an isomorphic group, cut along the projection-plane regions, making the usual small crossing detours above and below. The octahedral projection graph has eight triangular regions. The upper and lower pieces are ideal octahedra; corresponding faces are matched with one-third-turn corner permutations, with opposite directions on adjacent checkerboard faces. Reversing these cuts reconstructs the link complement. Give both pieces the regular ideal hyperbolic space octahedron metric. All triangular face maps are isometries and each edge cycle has four dihedral angles $\pi /2$, so the metric has no edge singularity. Equal horospherical truncations match in Euclidean squares and assemble into three flat tori, giving complete torus cusps. The compact truncated remainder and these cusps show completeness and finite volume. Thus the Borromean complement is a complete finite-volume hyperbolic three-manifold. This also supplies a geometric discrete faithful holonomy, independently of the subgroup calculation.

Start with the prime decomposition of a closed orientable three-manifold. Its fundamental group is a free product of the prime-summand groups. A nontrivial abelian group cannot be a free product of two nontrivial groups. The same is true of a nontrivial finitely generated nilpotent group: it has nontrivial center, whereas such a free product has trivial center. Thus all nontrivial fundamental group comes from a single prime summand; simply connected summands do not affect the argument.

A prime but nonirreducible orientable summand is $S^2\times S^1$, giving $\Gamma \cong \mathbb{Z}$. Otherwise the infinite-group prime summand is an irreducible three-manifold. The sphere theorem for three-manifolds gives ${\pi }_2=0$. Its simply connected universal cover is noncompact, so its top ordinary homology vanishes; its first and second homology also vanish by simple connectivity and the Hurewicz theorem. It is therefore acyclic and contractible by the Whitehead theorem. The summand is aspherical, and its closed orientable three-dimensional classifying space gives integral cohomological dimension of a group equal to three: the cellular resolution gives the upper bound and top Poincare duality gives the lower bound.

If $\Gamma$ is torsion-free abelian, finite generation makes it ${\mathbb{Z}}^r$. Its integral cohomological dimension of a group is $r$, as one sees from the standard $r$-torus. The aspherical case therefore has $r=3$. We obtainBoth occur, on $S^2\times S^1$ and the three-torus.

For the infinite torsion-free nilpotent case, the same reduction gives either $\mathbb{Z}$ or an aspherical group of cohomological dimension three. By the Malcev completion of a torsion-free nilpotent group, the latter is a lattice in a simply connected three-dimensional nilpotent Lie group. A nonabelian three-dimensional nilpotent Lie algebra has abelianization of dimension at least two: a nilpotent algebra generated modulo its derived algebra by one element would be generated by that element and thus be abelian. Its derived algebra is therefore one-dimensional, and nilpotence makes that line central. Hence it has a basis with a single nonzero bracket $[Y,T]=X$; the other possibility is an abelian algebra. In the nonabelian case the lattice's center is an isolated infinite cyclic subgroup, generated by $x$, and the quotient by it is ${\mathbb{Z}}^2$. Choose lifts $y,t$ of a basis of that quotient. Their commutator is $x^b$ for a nonzero integer $b$, and normal forms giveThe subgroup $A=⟨x,y⟩\cong {\mathbb{Z}}^2$ is normal and ${\Gamma }_b/A\cong \mathbb{Z}$. In its ordered basis $x,y$, conjugation by $t$ acts byThis is the unipotent torus-bundle group. Indeed, its normal forms multiply according toProjection to the $n,k$ coordinates, followed by inspection of the central $m$ coordinate, proves torsion-freeness. All commutators lie in $⟨x^b⟩$, which is central, proving nilpotence. The mapping torus of the torus automorphism with this matrix realizes every integer $b$ as a closed orientable three-manifold; $b=0$ gives ${\mathbb{Z}}^3$, while $b\ne 0$ gives the Nil cases. Thus the complete list of groups under the nilpotent hypotheses is $\mathbb{Z}$ and the groups ${\Gamma }_b$.

Here is the center argument, with its literal qualification. If a nilpotent group with abelian quotient $\Gamma /A$ were nonabelian, the last nonzero term ${\gamma }_c\Gamma$ of its lower central series would satisfyConsequently $Z(\Gamma )\cap A=1$ forces the group to be abelian. For the specific nontrivial rank-two $A$ in the printed family the hypothesis cannot actually occur: when $b\ne 0$, $x\in Z({\Gamma }_b)\cap A$; when $b=0$, the intersection is all of $A$. The implication is valid, but should not be used to invent an example satisfying this impossible extra condition.

A torsion-free finite extension of a lattice with quotient $C_4$ is realized by the quarter-turn flat three-manifold. On ${\mathbb{R}}^3$ letLet $c$ be unit translation in the third coordinate. Then $sa{s}^{-1}=b$, $sb{s}^{-1}=a^{-1}$, $sc{s}^{-1}=c$, and $s^4=c$. The lattice $\Lambda =⟨a,b,c⟩\cong {\mathbb{Z}}^3$ is normal and the quotient is $C_4$. A representative $a^m{b}^n{c}^k{s}^j$, $0\le j<4$, with $j\ne 0$ has third-coordinate displacement $k+j/4\ne 0$, so has no fixed point; for $j=0$ a nonidentity element is a nonzero translation. This proves torsion-freeness. The action is properly discontinuous, cocompact and orientation preserving, so it gives a closed orientable flat three-manifold with precisely the required extension.

For faithful linear holonomy on the maximal translation lattice, the holonomy groups of closed orientable flat three-manifolds areThe cyclic examples replace the quarter-turn by a planar lattice rotation of order $m=2,3,4,6$ and a translation of $1/m$ along its axis. The square and triangular plane lattices realize these orders; the trace condition $2\cos (2\pi /m)\in \mathbb{Z}$ explains the crystallographic restriction. The remaining case is obtained fromTheir squares translate by one in the first and second directions, and $(\alpha \beta {)}^2$ translates by minus one in the third. Their linear parts generate $C_2\times C_2$. The three nontranslation cosets have half-integral displacement along their respective fixed axes, so every element acts without a fixed point. These constructions realize every member of the orientable three-dimensional Euclidean holonomy classification.

The maximality/faithfulness restriction is essential. Taken literally as a question about arbitrary finite group quotients, the wording has more answers: $\mathbb{Z}$ has every finite cyclic quotient, and ${\mathbb{Z}}^3$ has, for example, quotient $C_5$. An exact criterion for the literal extension question is as follows. Choose an action $Q\to \mathrm{Aut}(A)$ and an extension class $e\in H^2(Q,A)$, where $A=\mathbb{Z}$ or ${\mathbb{Z}}^3$. For every subgroup $⟨q⟩$ of prime order $p$, requireTo prove this, lift $q$ to $g$. Its $p$th power is in $A^q$, and replacing $g$ by $ag$ changes that power by the displayed norm of $a$. Thus a lift of order $p$ exists exactly when this restriction is zero. Any torsion element has a prime-order power, so the condition is both necessary and sufficient. For an oriented flat realization in rank three the linear action must also preserve orientation. This distinguishes a complete extension criterion from the finite faithful-holonomy list.

First correct the group-ring condition in the PDF. For an infinite group, a left-invariant element of $\mathbb{Z}\Gamma$ would have constant coefficients on the whole group; finite support then forces it to be zero. Hence $H^0(\Gamma ,\mathbb{Z}\Gamma )=0$. For a Poincare duality group of positive dimension, the correct condition isIt must be accompanied by the finiteness condition: the trivial module has a finite-length resolution by finitely generated projective modules. The module $D$ has a right group action, possibly an orientation-sign action. The degree-zero exception only applies to the trivial $P{D}^0$ group, not simultaneously to positive-dimensional groups.

Here is how the duality follows. Put $R=\mathbb{Z}\Gamma$ and take a length-$n$ projective resolution $P_{\ast }\to \mathbb{Z}$. Dualize to $P^{\ast }={\mathrm{Hom}}_R(P_{\ast },R)$. Its cohomology is zero except in degree $n$, where it is $D$. Reversing degrees therefore makes $Q_j=P_{n-j}^{\ast }$ a projective right-$R$ resolution of $D$. Finite generation and projectivity give, for every left $R$-module $V$, an isomorphism of cochain complexesTaking cohomology and reversing the grading yieldsSince $D$ has underlying abelian group $\mathbb{Z}$, its action is a sign character $\omega :\Gamma \to \{\pm 1\}$. Twisting the resolution by this character identifies the right side with group homology having coefficient module $D{\otimes }_{\mathbb{Z}}V$, with diagonal left action $\gamma (d\otimes v)=\omega (\gamma )d\otimes \gamma v$. ThusFor an orientable Poincare duality group, $D$ is trivial and this is untwisted Poincare duality. The natural isomorphism can equivalently be realized by cap product with the group fundamental class. The reversed finite projective complex explains why the top group-ring condition produces duality for every coefficient module, rather than just matching integral Betti numbers.

Now let $H$ have finite index in $\Gamma$. There is an isomorphism of left $\Gamma$-modulesOne explicit map sends $r\in \mathbb{Z}\Gamma$ to the $H$-linear function $g\mapsto {\mathrm{pr}}_H(gr)$, where ${\mathrm{pr}}_H$ retains only terms in $H$. Finite index ensures that all such functions arise in this way. Shapiro's lemma gives the isomorphismas abelian groups. If $\Gamma$ is a Poincare duality group, restricting its finite projective resolution to $H$ preserves projectivity and finite generation: the group ring is finite free over $\mathbb{Z}H$. This proves that $H$ is $P{D}^n$.

For the converse one needs the usual ambient torsion-free hypothesis. The standard finite-index resolution lemmas say that type $F{P}_{\infty }$ is preserved under finite-index passage, and that a torsion-free group of finite virtual cohomological dimension has the same integral cohomological dimension as a finite-index subgroup. Thus if $H$ is $P{D}^n$ and $\Gamma$ is torsion-free, $\Gamma$ has a finite resolution by finitely generated projective modules of length $n$. The preceding explicit coinduction and Shapiro calculation transfers exactly the required group-ring cohomology, and proves the finite-index invariance of Poincare duality for torsion-free groups.

Without that hypothesis the claimed converse is false. Take $\Gamma =\mathbb{Z}\times C_2$ and $H=\mathbb{Z}\times 1$. Then $H$ is index two and is $P{D}^1$, while $\Gamma$ contains torsion. Its integral cohomological dimension of a group is infinite: restriction of any putative finite projective resolution to $C_2$ would contradict the nonzero cohomology $H^{2j}(C_2,\mathbb{Z})=\mathbb{Z}/2$ for every $j\ge 1$. Also, orientability need not ascend: the torus subgroup of the Klein-bottle group has trivial orientation action, whereas the full group has a nontrivial one. The invariant theorem concerns the possibly twisted notion of $P{D}^n$.

Finally, a closed aspherical surface has a two-dimensional contractible universal cover and exactly this compact-support duality, so its surface group is $P{D}^2$. Conversely the algebraic condition forces the same local duality dimension and, in the orientable case, a nonsingular skew pairing on first cohomology, the familiar algebra of a closed orientable surface. Infinite-index subgroups behave like noncompact surfaces rather than closed ones. These are strong reasons to expect a surface group; in fact the $P{D}^2$ classification theorem confirms that every integral $P{D}^2$ group is the fundamental group of a closed aspherical surface. This includes the torus and, with orientation twist, the Klein bottle; the sphere and projective plane do not give $P{D}^2$ groups.

The Loop theorem converts a nonzero kernel of ${\pi }_1(F)\to {\pi }_1(M)$, for a boundary surface $F$, into a properly embedded compression disk. It is stronger than simply producing a singular nullhomotopy, and does not assert that every original self-intersecting loop is the boundary of the final disk.

The proof outline uses a strengthened version: keep the boundary class outside a prescribed normal subgroup $N◃{\pi }_1(F)$. Begin with a singular disk in general position. Local cut-and-paste removes branch points; when a disk splits, at least one boundary stays outside $N$, since otherwise their product would lie in $N$. Take a regular neighborhood of its image and lift the disk through connected double covers of successive neighborhoods. Each lift decreases a finite lexicographic singularity complexity, so the tower terminates. At the top no double cover remains; mod-two duality forces its boundary components to be spheres. A boundary curve of the disk neighborhood lying outside the pulled-back $N$ then bounds an embedded disk on such a sphere, pushed inward. Project down one stage at a time. Exchange disk sectors along double curves, retaining a boundary outside $N$, to remove the projection's double intersections. Repeating yields an embedded disk downstairs. Taking $N=1$ proves the Loop theorem; controlling a prescribed simple boundary gives Dehn's lemma. The key points are termination of the covering tower and preservation of an essential boundary during surgery, not an unjustified removal of all intersections by general position alone.

Two classification applications are particularly useful. First, a two-sided embedded surface is incompressible precisely when its inclusion is fundamental-group injective, with the usual exclusion of inessential sphere components. A compression disk immediately supplies a nontrivial kernel element. In the other direction, cut along the surface and use the Seifert-van Kampen theorem: if both boundary-copy maps were injective, the amalgamated-product or HNN normal-form theorem would make the original inclusion injective. Hence a nontrivial kernel appears on one cut side, and the Loop theorem produces a compression disk. This is what makes algebraically detected essential surfaces usable in three-manifold splitting and incompressible hierarchies.

Second, consider a compact orientable irreducible three-manifold with one torus boundary. If that boundary is not fundamental-group injective, the Loop theorem supplies a compression disk. An essential simple curve on a torus is nonseparating, so compression changes the torus into a sphere. Pushing this sphere into the manifold, irreducibility gives a ball on the side away from the original torus boundary; the region between the sphere and the torus is the compression neighborhood. Undoing compression therefore attaches one one-handle to a ball. This proves the solid-torus recognition by a compression disk criterion:As a concrete consequence, a knot with knot group $\mathbb{Z}$ is the unknot. Its exterior is irreducible: a sphere disjoint from a connected knot bounds a ball on the side not containing the knot. The map from its boundary group ${\mathbb{Z}}^2$ to $\mathbb{Z}$ cannot be injective, so its exterior is a solid torus. The boundary of a meridian disk of this exterior lies in the kernel of the linking-number map; it is therefore the primitive preferred longitude. Joining that longitude to the knot through its tubular neighborhood makes an embedded spanning disk for the knot. This proves the cyclic-knot-group unknot criterion, rather than merely identifying a familiar consequence by name.

A smooth manifold alone has no distinguished notion of an isometry; one must first choose a Riemannian metric $g$. We therefore interpret the essay as concerning a smooth connected Riemannian manifold $(M,g)$. Its isometry group consists of diffeomorphisms $f$ with $f^{\ast }g=g$. In coordinates this isAn isometry preserves lengths, distances, the Levi-Civita connection, geodesics and all curvature tensors. The Myers-Steenrod theorem ensures that the group is a finite-dimensional Lie group acting smoothly, and that even distance-preserving bijections have the required smoothness.

The crucial rigidity is determination by one first jet. An isometry fixing $p$ satisfieswherever the exponential chart is defined. If also $d{f}_p=I$, it is the identity on a neighborhood. Continuing this equality through overlapping exponential neighborhoods proves it on connected $M$. Therefore the isotropy group at $p$ embeds in $O(T_{p}M,g_p)$. Its dimension is at most $n(n-1)/2$, while an orbit has dimension at most $n$. ConsequentlyOn a compact manifold the group is compact. Indeed distance preservation makes a family of isometries equicontinuous; compactness and the same argument for their inverses give subsequential limits which are inverse distance-preserving bijections. Smoothness and the Lie-group structure then follow from Myers-Steenrod theorem.

Infinitesimal isometries are Killing vector fields. Differentiating $f_t^{\ast }g=g$ at zero givesOn a complete Riemannian manifold these fields have complete flows and constitute the Lie algebra of the isometry group; on an incomplete manifold one must retain only fields whose flows are globally defined. A Killing vector field is fixed by its value and its skew derivative at one point, consistent with the same dimension bound. Along a geodesic, its restriction is a Jacobi field, so local infinitesimal data propagate rigidly.

In dimension three there is no independent Weyl curvature tensor: the Riemann curvature tensor is determined by the Ricci tensor. Large isotropy therefore strongly constrains the geometry. If the full dimension six is attained, the group is transitive with three-dimensional isotropy. That isotropy rotates all tangent two-planes transitively, forcing equal sectional curvatures at each point; transitivity makes the value constant. The simply connected complete models are the round $S^3$, Euclidean ${\mathbb{R}}^3$ and hyperbolic ${\mathbb{H}}^3$. Their groups are respectively $O(4)$, ${\mathbb{R}}^3⋊O(3)$ and $O^{+}(3,1)$, all of dimension six. The orientation-preserving hyperbolic group is $PS{L}_2(\mathbb{C})$.

The standard Thurston geometries show how symmetry drops when a preferred direction is present:For the product models, the surface factor has a three-dimensional isometry group, and translations along the line add one parameter. For Nil and the universal covering of $S{L}_2(\mathbb{R})$ with their standard geometric metrics, a three-dimensional transitive group is enlarged by a circle of isotropy. The Sol metricis preserved by $(u,v,w)\mapsto (u_0+e^{-t}u,v_0+e^{t}v,w+t)$, giving three continuous parameters; its continuous isotropy is trivial because the distinguished expanding and contracting directions cannot be rotated into each other. Additional discrete symmetries may still occur. These dimensions refer to the model metrics, not arbitrary left-invariant perturbations.

A quotient generally has fewer global isometries than its local model. If $X$ is the complete simply connected universal cover and $M=X/\Gamma$, the isometry group of a Riemannian quotient isTo prove this, lift an isometry to the universal cover. Conjugating a deck transformation by the lift again gives a deck transformation, so the lift normalizes $\Gamma$. Conversely a normalizing isometry descends, and two lifts differ by a deck transformation. A flat three-torus retains its three-dimensional translation group, with additional discrete lattice symmetries. A round spherical quotient retains only the normalizer of its finite deck group; the full six-dimensional group does not automatically descend.

For a closed hyperbolic three-manifold, there are no nonzero Killing vector fields. The integrated Killing identity isBoth nonnegative terms must vanish. The compact isometry group is therefore zero-dimensional and hence finite. The Mostow rigidity theorem additionally says that isomorphisms between the fundamental groups of complete finite-volume hyperbolic three-manifolds are induced by unique hyperbolic isometries; this fixes the complete hyperbolic metric up to isometry and contrasts with the moduli of flat tori. An arbitrary smooth three-manifold need not admit a single homogeneous metric: prime decomposition and torus splitting may lead to several geometric pieces. Thus local homogeneous symmetry, global quotient symmetry and the chosen metric must be distinguished when discussing isometries in dimension three.