Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 133 (Geometric Group Theory)

PutFor every nonzero integer $n$,LetThese are disjoint nonempty subsets of ${\mathbb{R}}^2$. If $(x,y)\in B$, thenso $u^n(B)\subseteq A$. Similarly, if $(x,y)\in A$, thenso $v^n(A)\subseteq B$.

The ping-pong lemma now identifies the subgroup generated by $u$ and $v$ withBoth generators have infinite order, so this is a free group of rank two inside the special linear group ${\mathrm{SL}}_2(\mathbb{Z})$.

Use the injective group homomorphism $F_2\hookrightarrow {\mathrm{SL}}_2(\mathbb{Z})$ from part (a). If $1\ne g\in F_2$, its image is an integral matrix $M\ne I$. Choose a prime number $p$ that does not divide one nonzero entry of $M-I$. The reduction modulo a prime in an integral matrix group homomorphismthen sends $M$ to a nonidentity element. Its target is a finite group, so the composite map separates $g$ from the identity. Hence $F_2$ is a residually finite group.

Write an element of the Integer Heisenberg group as $(x,y,z)$. Matrix multiplication givesThe subgroupis normal and isomorphic to ${\mathbb{Z}}^2$. If $t=(1,0,0)$, thenThus, on the coordinate column $(y,z{)}^T$, conjugation by $t$ is the linear map with matrixEvery element has a unique expression $(0,y,z)t^x$, so

The commutators $[t,(0,y,z)]$ fill the central subgroup of matrices $(0,0,z)$, while the quotient by this subgroup is generated freely and abelianly by the images of $(1,0,0)$ and $(0,1,0)$. Equivalently, $(A-I){\mathbb{Z}}^2$ is the second coordinate axis. Therefore the abelianization is

Let $N={\mathbb{Z}}^2$ and let $t$ generate the other factor of ${\Gamma }_B=N{⋊}_B\mathbb{Z}$. For $n\in N$,After abelianization, the image of $N$ is thereforeIf $1$ is not an eigenvalue of $B$, then $\det (B-I)\ne 0$. Hence $(B-I)N$ is a full-rank sublattice of $N$ with finite index of a subgroup $|\det (B-I)|$. The displayed quotient, and therefore the image of the ${\mathbb{Z}}^2$ factor in ${\Gamma }_B^{\mathrm{ab}}$, is finite. In fact the abelianization of a semidirect product by the integers gives

TakeIts eigenvalues areNeither is a root of unity, so $1$ is not an eigenvalue of $C^k$ for any positive $k$.

Let $L$ be a finite-index subgroup of ${\Gamma }_C$, let $N={\mathbb{Z}}^2$, and project ${\Gamma }_C$ onto its $\mathbb{Z}$ factor. Then $N^{\prime }=L\cap N$ has finite index in $N$, while the image of $L$ is $k\mathbb{Z}$ for some $k>0$. Choose $s\in L$ projecting to $k$. Every element of $L$ has a unique expression $n{s}^j$ with $n\in N^{\prime }$, and conjugation by $s$ on $N^{\prime }$ is the restriction of $C^k$. ThusBecause $1$ is not an eigenvalue of $C^k$, part (b), applied to the finite-rank lattice $N^{\prime }$, shows that the image of $N^{\prime }$ in $L^{\mathrm{ab}}$ is finite. The quotient by that finite image is generated by the image of $s$, so $L^{\mathrm{ab}}$ is a finite extension of an infinite cyclic group. It is therefore a virtually cyclic group.

For each generator $t\in T$, let $|t{|}_S$ be its length in the generating set $S$, and putA shortest $T$-word for $h_1^{-1}{h}_2$ has $d_T(h_1,h_2)$ letters. Replacing each letter by an $S$-word of length at most $C$ givesThus inclusion of any finitely generated subgroup is Lipschitz continuous for the corresponding word metrics.

Let $r:G\to H$ be a retraction. Part (a) supplies a constant $C$ such thatThe finite set $r(S)$ generates $H$. PutIf a shortest $S$-word represents $h_1^{-1}{h}_2\in H$, applying the group homomorphism $r$ gives a $T$-word for the same element of length at most $D{d}_S(h_1,h_2)$. HenceThe inclusion is therefore bilipschitz and in particular a quasi-isometric embedding. Thus every finitely generated retract subgroup is quasi-isometrically embedded.

This is the Baumslag-Solitar group $G=\mathrm{BS}(1,2)$. Its defining relation gives, by mathematical induction,In the cyclic subgroup $H=⟨a⟩$ with generator $a$,whereas in $G$, with generators $a,b$,If inclusion were a $(\lambda ,\epsilon )$-quasi-isometric embedding, its lower bound would implyfor every $n$, which is impossible because an exponential function eventually dominates every linear function. Hence $H$ is not quasi-isometrically embedded; it is an exponentially distorted subgroup.

SetThe triangle inequality makes $f$ Lipschitz continuous and hence a continuous function. Since $\gamma$ is an isometric embedding,so $f(t)\to \infty$ as $|t|\to \infty$. The function is therefore coercive, and the extreme value theorem on a sufficiently large compact interval gives a minimizing parameter.

Suppose $t_1,t_2$ both minimize $f$, put $p=\gamma (t_1)$ and $q=\gamma (t_2)$, and letThe restriction of $\gamma$ between the two parameters is a geodesic from $p$ to $q$. Let $m$ be its midpoint. In the geodesic triangle with vertices $x,p,q$, the thin geodesic triangle condition gives a point $y$ on one of the other two sides with $d(m,y)\le \delta$. By symmetry suppose $y\in [x,p]$. ThensoBut $m$ lies on $\gamma (\mathbb{R})$ and $p$ is a closest point, so $R\le d(x,m)$. Therefore $L\le 4\delta$, which is stronger than the requiredThis is the closest point on a geodesic line in a hyperbolic metric space estimate.

The defining formulas imply the matching-distance identitiesFor example,and the other two follow cyclically. Thus $p,q,r$ are the tripod points of a geodesic triangle.

By the thin geodesic triangle condition, $p$ lies within $\delta$ of some point $p^{\prime }$ on $[x,y]$ or $[y,z]$. If $p^{\prime }\in [x,y]$, then the triangle inequality givesSince $q$ is the point of $[x,y]$ at distance $d(x,p)$ from $x$, the geodesic parametrization gives $d(p^{\prime },q)\le \delta$, and hence $d(p,q)\le 2\delta$. If instead $p^{\prime }\in [y,z]$, comparison of distances from $z$ gives $d(p,r)\le 2\delta$.

Applying the same argument cyclically, each of $p,q,r$ lies within $2\delta$ of at least one of the other two. The graph on these three points whose edges join pairs at distance at most $2\delta$ therefore has no isolated vertex, so it is connected. Any two vertices are joined by at most two edges, and the triangle inequality yields