Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 133 / 1

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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