Past exam of the mathematics course of the University of Cambridge / 2017 / iii / Paper 133 / 1

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.