Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 133 (Geometric Group Theory)

For a finite group presentation $⟨S\mid R⟩$ and a word $w\in F(S)$ representing the identity, the area of a null-homotopic word isThe Dehn function of the presentation isEquivalently, area is the least number of two-cells in a van Kampen diagram for $w$, and the Dehn function is the worst such area among null words of length at most $n$.

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The normal form theorem for an amalgamated free product says that, after choosing left coset representatives for $C$ in $A$ and $B$, each element of $A{\ast }_{C}B$ has a unique normal form consisting of an initial element of $C$ followed by an alternating word in nontrivial representatives from the two factors. In particular, every nonempty reduced alternating word whose syllables lie outside $C$ is nonidentity.

For the free product $A\ast B$, the amalgamated subgroup is trivial. Henceis nontrivial whenever, after omitting a possibly empty initial or final syllable, every displayed $A$-syllable and $B$-syllable is nonidentity. It is then a nonempty reduced normal form.

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The infinite dihedral group is $D_{\infty }\cong C_2\ast C_2$, with factors $A=⟨a⟩$ and $B=⟨b⟩$. Its Bass-Serre tree has vertex setand one edge indexed by each $g\in D_{\infty }$, joining $gA$ to $gB$. Since both factors have order two, every vertex has degree two. The connected tree is therefore a bi-infinite line.

The action is cocompact, and its vertex stabilizers are the finite conjugates of $A$ and $B$, so it is proper. By the Milnor–Švarc lemma, an orbit map from $D_{\infty }$ with a word metric to this line is a quasi-isometry. A simplicial bi-infinite line is quasi-isometric to $\mathbb{R}$, hence so is $D_{\infty }$.

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Map both $a$ and $b$ to the nonidentity element of $C_2$. Both relators $a^2,b^2$ map to the identity, so this gives a homomorphism $D_{\infty }\to C_2$. A word of length $n$ maps to the parity class of $n$; consequently a null word has even length.

Now let $w$ be a null word of positive even length. Interpreting $a^{-1}=a$ and $b^{-1}=b$, the free-product normal form theorem says that a nonempty alternating word cannot be trivial. Thus $w$ has two adjacent equal letters. Delete this $a^2$ or $b^2$, using one conjugate of a defining relator, and apply induction to the resulting null word of length $n-2$. This gives

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Part d gives $\delta (2n)\le n$. For the reverse inequality, consider $w=a^{2n}$. Under the homomorphism $F(a,b)\to \mathbb{Z}$ with $a\mapsto 1$ and $b\mapsto 0$, every conjugate of $a^{\pm 2}$ has image $\pm 2$ and every conjugate of $b^{\pm 2}$ has image zero. Any expression of $a^{2n}$ as a product of conjugates of relators therefore uses at least $n$ factors. HenceThe opposite inequality follows by applying $a^2$ exactly $n$ times, so the Dehn function satisfies $\delta (2n)=n$.

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Let $x,y\in \mathrm{Fix}(\varphi )$. A tree has a unique geodesic segment $[x,y]$. The isometry $\varphi$ sends this segment to the segment $[\varphi x,\varphi y]=[x,y]$. An isometry of a segment that fixes both endpoints fixes every point of it, soThus the fixed-point set of an elliptic isometry of a tree is a convex subtree, in particular it is path-connected.

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Let $A=\mathrm{Axis}(\varphi )$ and let $p$ be the nearest-point projection of $x$ to $A$. If the translation length is $\tau >0$, the unique path from $x$ to $\varphi x$ is the concatenationThe first and last pieces both have length $d(x,A)$, while the middle one has length $\tau$. ThereforeThe displacement is minimized exactly when $x\in A$, proving that the axis is the minimum set of the displacement function.

On $A$, the power ${\varphi }^r$ translates through $r\tau$ when $r>0$ and in the opposite direction through $|r|\tau$ when $r<0$. The same formula applied to ${\varphi }^r$ shows that its minimum set is $A$. Hence

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Suppose that $a$ acted as a hyperbolic isometry of a tree, with translation length $\tau >0$. Nonzero powers have the same axis andTranslation length is invariant under conjugacy, whereas the defining relation in the Baumslag-Solitar group says that $a^m$ and $a^n$ are conjugate. Hence $m\tau =n\tau$, contradicting $m\ne n$. Therefore $a$ acts elliptically in every combinatorial tree action.

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Suppose $G=A\ast B$ were a nontrivial free product. Its Bass-Serre tree action has trivial edge stabilizers and no global fixed vertex. Part c makes $a$ elliptic. Since $a$ has infinite order, the fixed set of every nonzero power $a^r$ is a single vertex: it is nonempty, while fixing two vertices would fix the intervening edge and put the infinite-order element $a^r$ in a trivial edge stabilizer.

Let this vertex be $v$. The relation $b{a}^m{b}^{-1}=a^n$ givesand both sides are the singleton $\{v\}$. Thus $b$ also fixes $v$. Since $a$ and $b$ generate the Baumslag-Solitar group, the entire group fixes $v$, contradicting the Bass-Serre action of a nontrivial free product. Hence no such decomposition exists.

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Any two word metrics from finite generating sets on the same group are bilipschitz equivalent. Indeed, if $S,S^{\prime }$ are finite, let $L={\max }_{s\in S}|s{|}_{S^{\prime }}$; then $|g{|}_{S^{\prime }}\le L|g{|}_S$, and the reverse inequality follows symmetrically. Apply this once to the two finite generating sets of $G$ and once to those of $H$. Composing these bilipschitz identity maps with the inclusion changes only the multiplicative and additive constants in the quasi-isometric embedding inequalities. Thus being a quasi-isometrically embedded subgroup is independent of $S$ and $T$.

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Let $h\in H$ and letbe a geodesic in the Cayley graph ${\mathrm{Cay}}_S(G)$. By $K$-quasiconvexity choose $h_i\in H$ with $d_S(x_i,h_i)\le K$, taking $h_0=1$ and $h_N=h$. ThenThe elements $h_i^{-1}{h}_{i+1}$ telescope to $h$, so the finite setgenerates $H$.

Moreover $|h{|}_U\le |h{|}_S$, while $|h{|}_S\le (2K+1)|h{|}_U$. Thus the inclusion $(H,d_U)\to (G,d_S)$ is a quasi-isometric embedding, and $H$ is quasi-isometrically embedded.

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Take $G={\mathbb{Z}}^2$ with the standard generating set $S=\{(1,0),(0,1)\}$, and letIntrinsic distance in $H$ between $0$ and $(n,n)$ is $|n|$, while its ambient word metric distance is $2|n|$, so $H$ is quasi-isometrically embedded. However, the ambient geodesic from $(0,0)$ to $(n,n)$ that first travels to $(n,0)$ and then to $(n,n)$ contains $(n,0)$. Its distance from the diagonal subgroup $H$ is $n$. No uniform $K$ can contain every such geodesic in the $K$-neighborhood of $H$, so $H$ is not a quasiconvex subgroup.

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Choose a finite generating set $T$ of $H$. A $d_T$-geodesic between two elements of $H$ maps under the inclusion to a uniform quasigeodesic in ${\mathrm{Cay}}_S(G)$ because $H$ is quasi-isometrically embedded. Since $G$ is a hyperbolic group, the Morse lemma for quasi-geodesics gives a constant $R$ such that this quasigeodesic and the ambient geodesic with the same endpoints have Hausdorff distance at most $R$. Every vertex of the former lies in $H$, so the latter lies in the closed $R$-neighborhood of $H$. Therefore $H$ is a quasiconvex subgroup.

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For $x,y\in X$, the triangle inequality and the fact that $\varphi$ is an isometry giveThus the displacement function is $2$-Lipschitz, hence continuous.

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Choose $y_n\in X$ with $d_{\varphi }(y_n)\to 0$. By the cocompact group action, there is a compact set $K\subseteq X$ whose translates cover $X$. Choose ${\gamma }_n\in \Gamma$ such thatIsometric invariance givesFor a fixed basepoint $x_0$, compactness of $K$ gives $C={\max }_{x\in K}d(x_0,x)<\infty$, so $d(x_0,x_n)\le C$.

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The bounded sequence $(x_n)$ lies in a compact set because $X$ is a proper metric space. After taking a subsequence, $x_n\to x$. Since $d(x_n,{\varphi }_n{x}_n)\to 0$, both $x_n$ and ${\varphi }_n{x}_n$ eventually lie in one compact neighborhood $L$ of $x$. A properly discontinuous group action has only finitely many $g\in \Gamma$ with $gL\cap L\ne \varnothing$, so some conjugate $\psi$ occurs as ${\varphi }_n$ along an infinite subsequence. Continuity then givesThus $\psi$ fixes $x$. Since $\psi =\gamma \varphi {\gamma }^{-1}$ for some $\gamma$, the original element $\varphi$ fixes ${\gamma }^{-1}x$.

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Suppose a cocompact Fuchsian group contained a nonidentity parabolic isometry of the hyperbolic plane $\varphi$. After conjugating in the upper half-plane model, write $\varphi (z)=z+c$ with $c\ne 0$. The supplied estimate givesso the infimum of the displacement function is zero.

A Fuchsian action is properly discontinuous, and the action is cocompact by hypothesis. Parts b and c therefore imply that $\varphi$ fixes a point of the hyperbolic plane. A nonidentity parabolic isometry has no fixed point inside the plane, only one on its ideal boundary. This contradiction excludes nontrivial parabolic elements.

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