Past exam of the mathematics course of the University of Cambridge / 2017 / iii / Paper 133 (Geometric Group Theory)

The Milnor–Švarc lemma applies to an isometric properly discontinuous group action that is also a cocompact group action on a nonempty proper metric space that is a geodesic metric space. It concludes that the group is finitely generated and, for a basepoint $x$ and any finite generating set of a group $S$, the orbit mapHere $d_S$ is the word metric, and a quasi-isometry means a map satisfying two-sided coarse distance bounds with a uniformly coarsely dense image. A free action is not required: finite stabilizers are compatible with proper discontinuity. The properness and cocompactness hypotheses ensure that only finitely many group elements move a chosen basepoint by the bounded amount used to generate the group. Replacing a finite generating set by another does not change the conclusion, by equivalence of finite word metrics.

The surviving conclusion is a quasi-isometry from a word metric for a possibly infinite generating set of a group. Finite generation cannot be asserted. Choose $x\in X$ and $R\ge 0$ so that $Gx$ is $R$-dense; the cobounded group action hypothesis provides such a choice. PutThis set is symmetric because $d_X(x,g^{-1}x)=d_X(gx,x)$. The metric geodesic subdivision argument in the next part givesIn particular $S$ generates $G$. Apply the same estimates to $g^{-1}h$ to obtain the two-sided bounds for the orbit map; its image is $R$-dense. Thus a cobounded isometric action on a metric geodesic space admits an orbit quasi-isometry for a suitable, possibly infinite, generating set. This is the cobounded orbit map lemma.

Infinite stabilizers cause no obstruction to these bounds: all their nonidentity elements belong to $S$ and have length one, although their orbit displacement is zero. The additive constant permits precisely this collapse. For example, every group acts coboundedly on a point; its complete Cayley graph for $S=G\setminus \{1\}$ is bounded, but the group need not be finitely generated.

Retain the metric geodesic subdivision and orbit-density parts of the usual Milnor–Švarc lemma proof. Drop the step using proper discontinuity and compactness to prove that the displacement-bounded generating set is finite.

More explicitly, let $L=d_X(x,gx)$ and $m=\max \{1,\lceil L\rceil \}$. Subdivide a metric geodesic from $x$ to $gx$ into $m$ segments of length at most one. Choose orbit points $g_{i}x$ within distance $R$ of the subdivision points, choosing the endpoints exactly: $g_0=1$, $g_m=g$. Thenso every nonidentity increment $g_{i-1}^{-1}{g}_i$ lies in $S$. Their telescoping product is $g$, and henceThis also covers $L=0$, when the one increment may be a nontrivial stabilizer element. Conversely, for any word $g=s_1\cdots s_k$ in $S$, the triangle inequality and invariance under the isometric action giveTaking the shortest word gives the other inequality. Thus the proof works unchanged except that the resulting generating set may be infinite. Neither local compactness nor a finite stabilizer is used in this weaker argument.

Each factor is a free abelian group of rank two. Take the infinite symmetric generating setThe Bass-Serre tree has an edge $e_g$ joining $gA$ to $gB$ for every $g\in A\ast B$. Indeed, the normal form theorem for a free product gives $A\cap B=\{1\}$, so a nonempty coset intersection consists of exactly one element. Let $m_g$ be the midpoint of $e_g$ and send the Cayley vertex $g$ to $m_g$.

Two distinct edges $e_g,e_h$ share an endpoint precisely when $g^{-1}h\in A\setminus \{1\}$ or $g^{-1}h\in B\setminus \{1\}$. This is exactly adjacency in the Cayley graph for $S$. A path in that graph therefore gives a path between the midpoints of the same length in the tree. Conversely, the unique tree segment between two edge midpoints passes through a succession of adjacent edges and gives a Cayley path of that length. Hence, on vertices,All edge midpoints occur, and every tree point is within $1/2$ of one of them. Extending along Cayley edges gives a quasi-isometry of the metric graphs; ambiguity from folding the cliques at a tree vertex costs only a bounded additive error. This is the free product with factor generating set construction.

For finite generating sets the answer changes. Use $S_0=\{a,a^{\prime },b,b^{\prime }{\}}^{\pm 1}$. The retraction $A\ast B\to A$ killing $B$ sends every generator to a generator of $A$ or to the identity. Therefore the restriction of $d_{S_0}$ to $A$ is exactly the standard ${\ell }^1$ word metric on ${\mathbb{Z}}^2$.

In this grid, take vertices $(0,0),(2N,0),(0,2N)$. Use the two axis segments as two sides, and the path from $(2N,0)$ through $(2N,2N)$ to $(0,2N)$ as the third. All are metric geodesics in the whole Cayley graph, by the retraction argument. The corner $(2N,2N)$ is at distance $2N$ from either of the other sides. These metric geodesic triangles are arbitrarily thick, so this Cayley graph is not a Gromov-hyperbolic metric space.

A tree is $0$-hyperbolic. Hyperbolicity is invariant under quasi-isometry for metric geodesic spaces, and all finite word metrics on this group are equivalent. Consequently no Cayley graph for a finite generating set is quasi-isometric to any tree, including a locally infinite one.

A subset is a quasiconvex subset with constant $C\ge 0$ if every metric geodesic segment with endpoints in it lies within distance $C$ of it. Explicitly,for every choice of the metric geodesic $[p,q]$. The distance is computed in the ambient geodesic metric space, not in the intrinsic path metric of $N$. One only asks the metric geodesic to stay close to $N$, rather than requiring it to lie in $N$; the latter makes $N$ a convex subset of a geodesic metric space.

The usual convention takes $C\ge 0$. If the printed allowance $C\in \mathbb{R}$ is interpreted literally using the distance inequality, a negative $C$ cannot work for a nonempty $N$, since the constant metric geodesic at a point of $N$ has distance zero. For the empty subset the endpoint condition is vacuous.

The Greendlinger lemma says that a nonempty freely reduced null-homotopic combinatorial loop in a complex satisfying the metric small cancellation condition $C^{\prime }(1/6)$ contains a consecutive segment of a cell boundary longer than half that boundary. In a reduced disc diagram without boundary spurs this is supplied by a boundary disc diagram shell, or by the single-cell case. The long segment can be replaced by the strictly shorter complementary segment.

The useful precise diagram version also allows degenerate diagrams. A reduced disc diagram is either a single vertex, a single closed cell, a disc diagram ladder, or has at least three disc diagram spurs or shells whose inner paths consist of at most three pieces. This is the Greendlinger ladder theorem. For such a shell, writing the perimeter as $QP$, where $Q$ is the exterior arc and $P$ is the union of its interior pieces, the strict $C^{\prime }(1/6)$ inequalities giveA ladder is a chain of cells and possibly connecting edges, with the two boundary paths running along its opposite sides. In a nontrivial ladder with at least two blocks its two ends are shells or spurs. The single-cell case also has the long exterior arc conclusion when viewed against an appropriate boundary decomposition.

A tree diagram with no two-cells should not be described as containing a shell: it has a spur unless it is a point. Retaining the spur and ladder alternatives is essential for the quasiconvexity argument below.

The universal cover $\tilde{X}$ is simply connected. Hence the closed combinatorial path $\gamma {\sigma }^{-1}$ is null-homotopic in its two-dimensional CW complex. The van Kampen lemma supplies a finite disc diagram mapping to $\tilde{X}$ with precisely that boundary path.

Choose such a diagram with the smallest possible number of two-cells, and then with the smallest number of edges. If two adjacent cells formed a cancellable pair, removing that pair would preserve the boundary path and reduce the area, a contradiction. ThusThis does not assume that $\tilde{Y}$ is simply connected. Only the loop's null-homotopy in $\tilde{X}$ is used. If the two boundary paths agree, a diagram with no two-cells is allowed; if they share initial or final segments, the diagram may have tree portions or cut vertices. A disc diagram need not be an embedded topological disk.

Let $L$ be the maximum perimeter of a two-cell of $X$, taking $L=0$ if there are none. Compactness of the combinatorial complex makes $L$ finite. We show that the vertex-endpoint metric geodesic $\gamma$ stays within $L/2$ of ${\tilde{Y}}^{(1)}$.

Fix a shortest path $\sigma$ in ${\tilde{Y}}^{(1)}$ and choose the reduced diagram of minimum area, then minimum edge count, as in the preceding part. There is no spur in the interior of either boundary side, since both sides are reduced metric geodesic paths in their respective graphs.

There can be no shell with its entire exterior arc $Q$ on $\gamma$: the Greendlinger lemma gives a complementary path $P$ shorter than $Q$, contradicting the ambient metric geodesic property. There can be no such shell with $Q$ entirely on $\sigma$ either. If its image cell $R$ belongs to $\tilde{Y}$, the shorter complementary path belongs to ${\tilde{Y}}^{(1)}$ and contradicts the intrinsic metric geodesic property of $\sigma$. If $R$ is outside $\tilde{Y}$, the given perimeter decomposition has an arc $I$ of length at least half the perimeter all of whose edges are outside $\tilde{Y}$. Any contiguous boundary arc whose edges lie in $\tilde{Y}$ must then be contained in the complementary arc $O$ and has length at most half the perimeter. This contradicts $|Q|>|\partial R|/2$.

Thus every shell or spur of the diagram must straddle one of the two marked boundary corners $p,q$. Distinct shell exterior arcs have disjoint interiors, so there can be at most two such exposed features. The Greendlinger ladder theorem now says that the diagram is a single cell or a ladder, apart from common edge paths and the zero-area case. Its two ends contain the marked corners; each intervening cell has one boundary arc on $\gamma$ and one on $\sigma$. Common edge paths already map into $\tilde{Y}$.

For a point $z$ of the $\gamma$ arc of a cell, choose a vertex on its $\sigma$ arc. One of the two paths around that cell boundary to the chosen vertex has length at most half the perimeter. Its image is an ambient path to ${\tilde{Y}}^{(1)}$, soThe same argument treats a single cell. In the zero-area case the reduced boundary sides coincide and already lie in ${\tilde{Y}}^{(1)}$.

Finally allow endpoints in the interiors of edges of ${\tilde{Y}}^{(1)}$. The initial and final partial edges of a graph metric geodesic lie in ${\tilde{Y}}^{(1)}$; what remains is a metric geodesic between vertices of that CW subcomplex. If both endpoints are joined without reaching a vertex, the whole segment is already in its edge. Thus the same bound applies to every graph metric geodesic, andThis is quasiconvexity from no missing shells. The intrinsic path $\sigma$ was never assumed to be an ambient metric geodesic, and the bound is independent of the endpoints and of its length.

Use the fact that infinite cyclic subgroups of hyperbolic groups are undistorted: if $z$ has infinite order in a hyperbolic group, then its stable word lengthis strictly positive. The limit exists by subadditivity and the Fekete lemma. It is invariant under conjugation, because conjugating $z^m$ changes its length by at most twice the conjugator's length. It also satisfies $\tau (z^k)=|k|\tau (z)$.

In the given Baumslag-Solitar group, $b$ has infinite order. To verify this without presuming the presentation's normal form, let $a$ act on the real line by $t\mapsto 3t$ and $b$ by $t\mapsto t+1$. These affine bijections satisfy $ab{a}^{-1}=b^3$, and every nonzero power of $b$ is a nonidentity translation. Thus the element $b$ in the presented group cannot have finite order.

If an injective group homomorphism to $G$ existed, its images $z$ of $b$ and $u$ of $a$ would satisfy $uz{u}^{-1}=z^3$, with $z$ still of infinite order. Consequentlyforcing $\tau (z)=0$, a contradiction. ThereforeEquivalently, the relation gives $a^{k}b{a}^{-k}=b^{3^k}$, so powers of $z$ with exponent $3^k$ would have length at most $2k|u|+|z|$. This contradicts the linear lower bound for an undistorted infinite cyclic subgroup. This alternative also makes the exponential compression obstruction explicit.

The two commutation relations make $c$ central, and the remaining relation is $ab=cba$. Repeatedly move $a$ past $b$, keeping the central powers in front. Induction first on $s$ and then on $r$ givesThereforeThis is the commutator identity in the integer Heisenberg group. The order of the factors fixes the positive sign of the exponent.

To check the claimed infinite order, map the generators to matricesTheir commutator is $I+E_{13}$, which commutes with both other matrices, so all defining relations hold. The image of $c^m$ is $I+m{E}_{13}$; hence $c$ has infinite order. This realizes the presentation in the Integer Heisenberg group and is sufficient to certify the cyclic subgroup, without needing to prove faithfulness of the whole representation.

With the ambient word metric from $\{a,b,c{\}}^{\pm 1}$, the displayed commutator givesFor any finite generating set $T$ of $⟨c⟩$, write its elements as powers of $c$ and set $B=\max \{|j|:c^j\in T\}$, with inverses included. Since $T$ generates an infinite cyclic group, $B\ge 1$. A word of length $m$ in $T$ has exponent at most $Bm$ in absolute value, soA quasi-isometric embedding of the inclusion would require $|c^{k^2}{|}_H\ge {\lambda }^{-1}|c^{k^2}{|}_T-\epsilon$. Combining these bounds would give $4k\ge k^2/(B\lambda )-\epsilon$ for every $k$, which is impossible. Thus the inclusion of the central infinite cyclic subgroup is not a quasi-isometric embedding. Changing the ambient finite generating set only changes multiplicative constants, by equivalence of finite word metrics. This is the quadratic distortion of the Heisenberg center.

A Dehn presentation is a finite group presentation with the following strict shortening property: every nonempty freely reduced word representing the identity contains a consecutive segment $u$ of a relator $r$ satisfyingRelators are taken as cyclically reduced words with a symmetrized relator set: their inverse words and cyclic permutations are included. This convention allows the matched segment to start anywhere on either orientation of a relator. The symmetrized set remains finite and has the same maximum relator length.

If $r=uv$, the relation replaces $u$ by $v^{-1}$, of length $|r|-|u|<|u|$. The Dehn algorithm alternates this replacement with free reduction. Every step shortens the word; it terminates at the empty word exactly for words representing the identity. The strict inequality matters: a half-perimeter match alone need not shorten anything.

Finiteness is part of the convention used here and is needed for the maximum and counting argument in the last part. With an unrestricted infinite relator set, merely imposing the shortening property would not by itself justify that conclusion.

Minimality of the conjugacy representative implies that $w$ is freely reduced; otherwise free reduction produces a shorter representative of the same element. It is also a cyclically reduced word. If $w=sv{s}^{-1}$ as a freely reduced word with its first and last letters inverse, the shorter word $v$ represents a conjugate, contradicting minimality. This is the cyclic reduction of a shortest conjugacy representative.

The word is nonempty because the element has order greater than one. Since it is cyclically reduced, no cancellation occurs between successive copies in $w^n$, and $w^n$ is a nonempty freely reduced word. It represents the identity because conjugation preserves the order. Apply the defining property of the Dehn presentation to obtainThe relator belongs to the finite symmetrized set fixed in the definition. No assumption that the group itself is torsion-free is made.

Put $m=|w|$ and $\ell =|r|$. In fact minimality yields the stronger bound $m\le \lfloor \ell /2\rfloor$. Suppose instead thatBecause $|u|>\ell /2$, the first $k$ letters of the matched segment $u$ form a relator segment $v$ of length $k$. They also form a segment of the periodic word $w^n$. Since $k\le m$, that segment is contained in one cyclic permutation $w^{\prime }$ of $w$, starting at the same position in the period. A cyclic permutation represents a conjugate and has length $m$.

Rotate the symmetrized relator to write it as $vs$. Replacing the prefix $v$ of $w^{\prime }$ by $s^{-1}$ preserves the represented element and gives a word of length at mostFree reduction can only shorten further. This contradicts the choice of $w$. ThusThe periodic-segment argument treats matches that cross a boundary between copies of $w$; it would be incorrect simply to assume that the entire long segment $u$ lies in the original written copy of $w$.

Assume first that the finite relator set $R$ is nonempty and putFor every nonidentity finite-order element choose a shortest conjugacy representative. The preceding parts give a word representing its conjugacy class of length at most $M$. The identity class has the empty representative.

There are only finitely many such words because the generating alphabet is finite. If it has $q=2|A|$ formal letters and $N=\lfloor M\rfloor$, a sufficient upper bound for the number of words isDistinct classes cannot require more representatives than there are words; different words may of course represent the same class. ConsequentlyThis is the torsion conjugacy bound for a Dehn presentation. It does not claim that there are only finitely many finite-order elements.

If $R$ is empty, the maximum in the PDF is undefined. Handle this case separately: the group is a free group on the finite alphabet, and a nonempty cyclically reduced word has no freely trivial positive power. Thus there is no nonidentity torsion and only the identity conjugacy class. If the alphabet is empty, the group is trivial and the same conclusion holds.