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Send each orientation-preserving isometry to its isotopy class. This is a group homomorphismFor $g\ge 2$, a hyperbolic isometry isotopic to the identity is the identity. One proof lifts it to the hyperbolic plane: after composing with a deck transformation, its lift commutes with the surface fundamental group. It consequently fixes the endpoints at infinity of every hyperbolic deck transformation. Those endpoints are dense in the circle at infinity, so the lift fixes that circle pointwise and is the identity. The homomorphism is therefore injective, proving the injection of a finite hyperbolic isometry group into a mapping class group.

Both low-genus analogues fail. Every orientation-preserving homeomorphism of the unit sphere is isotopic to the identity, but a round sphere has nontrivial finite rotation groups. On a flat torus, translation by a nonzero torsion point is a finite-order orientation-preserving isometry isotopic to the identity.

Let $\tilde{\alpha }$ and $\tilde{\beta }$ be lifts to the compactified hyperbolic plane $\overline{{\mathbb{H}}^2}$. Suppose they have two common points. Choose two consecutive common points along one lift. If both lie in ${\mathbb{H}}^2$, the intervening subarcs contain an innermost embedded disc. The covering projection is injective on its interior; otherwise a nontrivial deck translate would produce a still smaller such disc. Its projection is a bigon between $\alpha$ and $\beta$, contrary to the hypothesis.

The same innermost-disc argument works when one corner is on the circle at infinity, after deleting a sufficiently small horoball about that corner. The only possible obstruction would identify the ideal corner by both a hyperbolic deck transformation associated with an essential return and a parabolic deck transformation stabilizing the relevant puncture. A hyperbolic and a parabolic element of the surface group cannot have that fixed point in common, so the truncated disc again projects to a bigon. If both common points are ideal, truncate at both ends and apply the same argument. Therefore any pair of lifts intersects in at most one point of $\overline{{\mathbb{H}}^2}$.

Contracting the Riemann curvature tensor gives the nonzero components of the Ricci tensor:The Ricci scalar is the contraction with the inverse metric. ThereforeIt is independent of both coordinates, so the hyperbolic plane has constant negative scalar curvature.

The relations $ak{a}^{-1}k=1$ and $bk{b}^{-1}k=1$ say that conjugation by either $a$ or $b$ sends $k$ to $k^{-1}$. Thus $K=⟨k⟩$ is a normal subgroup of order at most $3$. Quotienting by it givesthe orientation-preserving hyperbolic triangle group $\Delta (4,4,4)$. This group acts properly discontinuously and cocompactly by isometries on the hyperbolic plane. The Milnor–Švarc lemma therefore makes $\Delta (4,4,4)$ quasi-isometric to $H^2$.

The quotient map $\Gamma \to \Gamma /K$ has finite kernel, so part (b), equivalently the finite-kernel quotient quasi-isometry, makes $\Gamma$ quasi-isometric to $\Delta (4,4,4)$. By transitivity of quasi-isometry,

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.

The hyperbolic plane ${\mathbb{H}}^2$ is a geodesic Gromov-hyperbolic metric space for some universal constant ${\delta }_0$. It is not a quasi-tree. Indeed, for every $C>0$, choose two points on opposite sides of a large closed metric ball centred at the midpoint of their joining geodesic. The complement of that ball in ${\mathbb{H}}^2$ is path connected, so the endpoints can be joined by a continuous path that stays more than $C$ from the midpoint. Thus ${\mathbb{H}}^2$ fails the bottleneck property, whereas part (b) shows that every quasi-tree satisfies it.

The metric is the left-invariant metricIts right-invariant vector fields areTheir flows act by left translations, which preserve a left-invariant metric. Directly,so both are Killing vector fields and generate one-parameter isometry groups.

There is an additional Killing field. Put $y=e^a>0$ and $x=b$; thenthe hyperbolic plane of constant curvature $-1$. Its isometry algebra is three-dimensional, whereas the space of right-invariant fields here is two-dimensional. For example, the third independent Killing field can be writtenwhich is not right invariant. Hence the answer is yes.

A geodesic hexagon in the hyperbolic plane with all six interior angles $\pi /2$. Any three positive alternating side lengths determine it up to isometry. If those sides are ${\ell }_i/2$, the opposite seam satisfies $\cosh s_i=(\cosh ({\ell }_i/2)+\cosh ({\ell }_j/2)\cosh ({\ell }_k/2))/(\sinh ({\ell }_j/2)\sinh ({\ell }_k/2))$. Two congruent such hexagons glued along the other three sides form a pair of pants with boundary lengths ${\ell }_1,{\ell }_2,{\ell }_3$.

The displayed coordinates identify determinant-one real symmetric matrices of size two with the two-sheeted hyperboloid. The special linear congruence action on symmetric matrices is transitive on each sheet, with stabilizer $SO(2)$. In particular the positive sheet is the homogeneous space $SL(2,\mathbb{R})/SO(2)$, a model of the hyperbolic plane.

This metric is the hyperbolic plane with curvature scale $a$. Its nonzero connection coefficients are