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A cobounded group action by isometries on a nonempty geodesic metric space admits an orbit map that is a quasi-isometry for a possibly infinite generating set of a group. If $Gx$ is $R$-dense, take $S=\{g\ne 1:d(x,gx)\le 2R+1\}$. Subdivide a metric geodesic into segments of length at most one and approximate subdivision vertices by orbit points, keeping the endpoints exact. The resulting group increments generate and giveThe first estimate includes zero displacement: a nontrivial stabilizer element has length one. No finiteness assertion about $S$ survives without properness.

A subset $Y$ of a geodesic metric space is convex when every metric geodesic whose endpoints lie in $Y$ is contained in $Y$.

A spur is an exposed degree-one vertex and its incident edge in a disc diagram. The exterior boundary runs out along that edge and immediately back. Thus a spur in the interior of a boundary side contradicts that side being a reduced combinatorial path or a metric geodesic. A diagram with no two-cells may be a tree and have spurs rather than shells.

An energy has $\lambda$-displacement convexity when the displayed inequality holds along every displacement interpolation between endpoints in its domain. The case $\lambda =0$ is displacement convexity; $\lambda >0$ gives uniform convexity along these metric geodesics. This is distinct from convexity along ordinary linear mixtures of probability measures. On subintervals, the inequality is equivalent to convexity of $t\mapsto E({\mu }_t)-\lambda W_2({\mu }_0,{\mu }_1{)}^2{t}^2/2$; if the energy is twice differentiable, it is equivalent to the corresponding second derivative lower bound. Without that regularity the lower bound is interpreted using distributions.

A metric space in which every two points are joined by a metric geodesic segment. Equivalently, they admit a continuous path whose metric path length equals their distance. In this general setting no Riemannian metric or covariant derivative is required.

For geodesic metric spaces, a quasi-isometry preserves the property of being a Gromov-hyperbolic metric space. Images of metric geodesic sides are uniform quasigeodesics. The Morse lemma for quasi-geodesics keeps them uniformly close to metric geodesic sides in the hyperbolic target. Thinness there and the coarse lower distance bound pull a uniform triangle-thinness constant back to the source. Coarse surjectivity supplies a quasi-inverse for the reverse implication. Consequently a metric geodesic space quasi-isometric to a tree is hyperbolic, with no local finiteness assumption on the tree.

If $g$ has infinite order in a hyperbolic group, the map $n\mapsto g^n$ is a quasi-isometric embedding of the integers. Equivalently $|g^n|\ge c|n|-C$ for some $c>0$. This is the homogeneous quasigeodesic theorem: local-to-global control of metric geodesics in a hyperbolic Cayley graph gives a quasigeodesic orbit for a suitable conjugate of a positive power; undoing conjugation and passing between consecutive powers changes only uniform constants. It implies $\tau (g)>0$. The theorem and its hypotheses are given in Löh, Theorem 6.5.9. It is not true for arbitrary finitely generated groups, as the Baumslag-Solitar group relation $ab{a}^{-1}=b^3$ shows.

A metric geodesic triangle consists of three points of a geodesic metric space and a chosen metric geodesic segment between each pair. Multiple choices are allowed, and coincident vertices give degenerate triangles. A Gromov-hyperbolic metric space has a uniform bound on the distance from each side to the other two sides for every such choice. This definition applies to Cayley graphs and trees without tangent vectors or curvature assumptions; it does not invoke the surface angle formula for a geodesic triangle.

For a continuous path $\gamma :[a,b]\to X$ in a metric space, take the supremum over finite partitions $a=t_0<\cdots <t_n=b$ of ${\sum }_{i=1}^{n}d(\gamma (t_i),\gamma (t_{i-1}))$. This definition allows $+\infty$ and needs no derivative. A metric geodesic parametrized on $[a,b]$ has length $b-a$. In a Riemannian manifold, it agrees with the usual arc length integral for a smooth curve.

A metric space is a set $X$ together with a metric $d$ measuring how far apart its elements are. The elements can be positions, functions, or other objects: what matters is that the metric obeys the same basic rules. This lets one discuss convergence in a metric space, continuous maps, and Cauchy sequences without first assigning Cartesian coordinates. The useful abstraction is that many arguments require only comparisons of distances, rather than the particular objects being compared.

Even the same set can support several informative metrics. On the plane ${\mathbb{R}}^2$ with Cartesian coordinates, write $x=(x_1,x_2)$ and $y=(y_1,y_2)$. Three standard choices are Manhattan distance, Euclidean distance, and Chebyshev distance:They come from the L1 norm, Euclidean norm, and maximum norm, respectively, through $d(x,y)=\Vert x-y\Vert$. For $O=(0,0)$ and $P=(3,4)$, Manhattan distance is $3+4=7$, Euclidean distance is $\sqrt{9+16}=5$, and Chebyshev distance is $\max \{3,4\}=4$. All three answers are correct for their respective metrics.

The choice encodes the question being asked. Manhattan distance models travel along a rectangular street grid. Euclidean distance measures unrestricted travel along a line segment. Chebyshev distance measures the time needed when two Cartesian coordinates can change simultaneously, each at speed at most one. Thus choosing a metric is part of modelling, rather than a consequence of knowing the endpoints. These examples have the same metric topology, but different metric path lengths and different collections of metric geodesics.

A historical reference for the general study of metric spaces is Maurice Fréchet's 1906 paper Sur quelques points du calcul fonctionnel, published in volume 22 of the Rendiconti del Circolo Matematico di Palermo. ^[ref]

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.

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.

Put $D=d_2(\mu ,\nu )$ and $z_t(x,y)=(1-t)x+ty$. The marginal distributions of the optimal transport plan $\Pi$ give ${\mu }_0=\mu$ and ${\mu }_1=\nu$. Alsoso ${\mu }_t$ has a finite second absolute moment. For $0\le s<t\le 1$, the pushforward measure $(z_s,z_t{)}_{\#}\Pi$ is a coupling of ${\mu }_s,{\mu }_t$. ThereforeApply the triangle inequality for the second Wasserstein distance along ${\mu }_0,{\mu }_s,{\mu }_t,{\mu }_1$:All inequalities are equalities, including when $D=0$. Thus this displacement interpolation is a constant speed metric geodesic:

Define the barycenter of a measure by $m_{\mu }=\int xd\mu (x)$. Under $\Pi$, the mean displacement is $\bar{q}=m_{\nu }-m_{\mu }$, and optimality gives $\int |q{|}^{2}d\Pi =D^2$. Expanding the square under the product measure yields the variance identity for independent copies:In the classical regime of part (c), the lower bound on the Hessian matrix of the interaction potential therefore givesFor equal barycenters, this becomes

For the full stated class of probability measures, prove the displacement convexity inequality directly. Pointwise strong convexity of $W$ givesIntegrating, including the $1/2$ in the interaction energy, and using the displayed variance identity for independent copies givesAll terms are finite by quadratic growth and finite second absolute moments. Equal barycenters give the claimed $\lambda$-displacement convexity. Also $m_{{\mu }_t}=(1-t)m_{\mu }+t{m}_{\nu }$, so this metric geodesic remains in the prescribed fixed-mean class. Applying the same argument to subintervals proves the second derivative bound in the sense of distributions even when a classical derivative has not been justified. The mean restriction is natural: translating all particles together leaves their pairwise differences and hence the interaction energy unchanged.

Let $X$ be a finite combinatorial 2-complex satisfying $C^{\prime }(1/6)$, and let $Y$ be a connected CW subcomplex of its universal cover with no missing shells. If the largest cell perimeter is $L$, then $Y^{(1)}$ is $(L/2)$-quasiconvex in the universal cover's unit-edge 1-skeleton. Bound an ambient metric geodesic and an intrinsic metric geodesic of $Y^{(1)}$ by a reduced disc diagram. A shell on the ambient side would shorten a metric geodesic. A shell on the intrinsic side belongs to $Y$ and would shorten its intrinsic metric geodesic. Spurs are likewise excluded from side interiors. The Greendlinger ladder theorem forces a ladder between the two marked corners. Each cell gives a path of length at most half its perimeter from either side to the other.

A subset $Y$ of a geodesic metric space is $K$-quasiconvex when every metric geodesic between points of $Y$ lies in the closed $K$-neighborhood of $Y$.

A metric satisfies $d(x,z)\le d(x,y)+d(y,z)$: visiting an intermediate point cannot beat the direct endpoint distance. A norm satisfies the corresponding inequality $\Vert u+v\Vert \le \Vert u\Vert +\Vert v\Vert$.

Take $O=(0,0)$, $Q=(3,0)$, and $P=(3,4)$. For Euclidean distance, the triangle inequality reads $5\le 3+4=7$. The direct line segment has metric path length $5$, while the route through $Q$ has metric path length $7$. For Manhattan distance, the same comparison becomes $7=3+4$: both routes are shortest. The line segment and the bent route both become metric geodesics when progress is measured by metric path length. For Chebyshev distance, the direct line segment has metric path length $4$, while the same bent route still costs $7$.

Repeated application of the triangle inequality bounds the endpoint distance by the sum along any finite chain of intermediate points. This explains why metric path length is always at least the distance between endpoints. Equality reveals a shortest route, but its shape and uniqueness depend on the chosen metric.