The negative sign identifies hyperbolic geometry. The numerical value depends on the length unit used.

Area becomes \(4(\pi/6)=2\pi/3\) and curvature becomes \(-1/4\). Their product stays \(-\pi/6\), so the angle sum stays \(5\pi/6\). Doubling lengths does not double the angle defect. An isometry preserves lengths within one geometry; this rescaling compares metrics and is not an isometry.

For negative curvature rewrite the ratio as \([\sinh(hR/2)/(hR/2)]^2\); for positive curvature use \([\sin(hR/2)/(hR/2)]^2\). Both tend to one. At zero curvature the ratio is identically one.

The ratios are \(\sinh(hR)/(hR)\), one, and \(\sin(hR)/(hR)\). The first and last tend to one by their derivatives at zero.

Translation by the displacement between two points maps the first to the second. But the only translation fixing a point is identity, so it cannot turn one tangent direction into a different direction. Euclidean distance is nevertheless invariant under translations.

For any starting and target pair, choose in each factor a transformation taking its starting coordinate to its target coordinate. The pair of transformations lies in the product group and carries the entire starting pair to the target pair.

Both terms are nonnegative and their sum vanishes exactly when both coordinate pairs agree. Symmetry follows term by term. Add the two factor triangle inequalities to obtain the product triangle inequality. Applying a product transformation leaves each term unchanged, establishing invariance.

This is the plane’s translation geometry. It is homogeneous and admits invariant metrics, but the specified group has no rotations about a point and is not isotropic.

Approach. Normalize the figure using an isometry, then apply the definition.

Setting \(k=1\) gives matrices with \(|a|^2+|b|^2=1\), antipodal relation \(z_1\bar z_2=-1\), line element \(2|dz|/(1+|z|^2)\) and its spherical area element. Setting \(k=-1\) gives \(|a|^2-|b|^2=1\), reflection relation \(z_1\bar z_2=1\), and line element \(2|dz|/(1-|z|^2)\) on the open unit disk. The ideal boundary is excluded from the hyperbolic space; including it would make the metric singular.

Approach. Normalize the figure using an isometry, then apply the definition.

For \(k=0\), \(|a|=1\) and \(T(z)=(az+b)/\bar a=a^2z+ab\), exactly rotations followed by translations. The limiting geodesic condition says the cline passes through infinity, so the geodesics are Euclidean lines. The line element is \(2|dz|\) and the area element \(4\,dx\,dy\), so coordinates have a constant factor of two in length relative to ordinary Euclidean coordinates. Infinity is not a finite-distance point of this plane.

Approach. Normalize the figure using an isometry, then apply the definition.

On \(S_1\times S_2\), \((T_1,T_2)\) acts by \((x,y)\mapsto(T_1x,T_2y)\). Each such map is a bijection, with inverse \((T_1^{-1},T_2^{-1})\). The identity is \((I_1,I_2)\), and composition is componentwise: \((T_1,T_2)(U_1,U_2)=(T_1U_1,T_2U_2)\). Associativity follows in each coordinate.

Approach. Normalize the figure using an isometry, then apply the definition.

For manifold models, local coordinates from the first factor and the second factor combine independently, producing \(k+h\) coordinates. Thus the product has dimension \(k+h\). This uses the manifold meaning of dimension, not a claim for arbitrary sets without a dimension theory.

Approach. Normalize the figure using an isometry, then apply the definition.

Use \(M(a,b)=\begin{pmatrix}a&b\\-k\bar b&\bar a\end{pmatrix}\) with \(|a|^2+k|b|^2=1\). Multiplication gives the same form with \(A=ac-kb\bar d\) and \(B=ad+b\bar c\); determinant multiplicativity gives \(|A|^2+k|B|^2=1\). The inverse is \(M(\bar a,-b)\), and \(M(1,0)\) is the identity.

For finite nonpole points, direct expansion gives \[1+kT(z_1)\overline{T(z_2)}=\frac{1+kz_1\bar z_2}{(-k\bar bz_1+\bar a)(-kb\bar z_2+a)}.\] Thus the identification is preserved; limits handle poles. For \(k<0\), the identity \[1+k|Tz|^2=\frac{1+k|z|^2}{|-k\bar bz+\bar a|^2}\] also shows preservation of the metric disk. Here the group is \(G_k\); \(A_k\) denotes the geometry, not literally the group.

Approach. Normalize the figure using an isometry, then apply the definition.

The relation gives \(|z_1||z_2|=1/|k|=R^2\), where \(R=1/\sqrt{|k|}\). If one modulus is less than \(R\), the other is greater; if one equals \(R\), so does the other. For \(k>0\) the boundary representatives are antipodal, and the closed disk with opposite boundary points identified is the single elliptic model.

For \(k<0\) a boundary point is paired with itself, but the length density is singular there. Exclude the boundary and choose the inside representative from each inside/outside pair to get the open disk model. The scan's displayed \(R=\sqrt{|k|}\) is inconsistent with its identification equation; the reciprocal square root is required.

Approach. Normalize the figure using an isometry, then apply the definition.

Set \(w=\sqrt{k}\,z\). This sends the representative disk to the unit disk and the identification to antipodal identification. Its geodesics become great circles, so any two projective lines meet. Lengths are \(1/\sqrt{k}\) times unit-model lengths, and areas \(1/k\) times unit-model areas. Angles are unchanged. Hence the unit elliptic relation gives \(\alpha+\beta+\gamma=\pi+kA>\pi\) for a nondegenerate triangle.

Approach. Normalize the figure using an isometry, then apply the definition.

Write \(\kappa=\sqrt{-k}\) and \(w=\kappa z\). The identification becomes reflection in the unit circle; geodesics become diameters and circle arcs orthogonal to it. In original coordinates they are diameters and clines orthogonal to \(|z|=1/\kappa\). The metric disk is open. Area scales by \(1/\kappa^2\), so the unit hyperbolic defect formula gives \(\pi-(\alpha+\beta+\gamma)=\kappa^2A=-kA>0\).

Approach. Normalize the figure using an isometry, then apply the definition.

For \(k>0\), rescale coordinates by \(\sqrt{k}\) to the unit sphere model. This preserves angles and multiplies areas by \(k\), so its angular-excess theorem gives the result. For \(k<0\), rescale by \(\sqrt{-k}\); the unit hyperbolic formula gives \((-k)A=\pi-\sum\alpha\), equivalent to the same equation. For \(k=0\), Euclidean triangle angles sum to \(\pi\), so both sides are zero. This case says nothing about Euclidean area; division by \(k\) would be invalid.

Approach. Normalize the figure using an isometry, then apply the definition.

Put \(q=|(z_2-z_1)/(1+k\bar z_1z_2)|\). For \(k<0\), writing \(\kappa=\sqrt{-k}\) gives \[d=\frac{2}{\kappa}\operatorname{artanh}(\kappa q),\quad A=\frac{2\pi}{\kappa^2}(\cosh(\kappa R)-1),\quad C=\frac{2\pi}{\kappa}\sinh(\kappa R).\] For \(k=0\), \(d=2|z_2-z_1|\), \(A=\pi R^2\), and \(C=2\pi R\).

For \(k>0\), writing \(\kappa=\sqrt{k}\) gives spherical distance \(D=2\arctan(\kappa q)/\kappa\) and projective distance \(d=\min(D,\pi/\kappa-D)\). Also \[A=\frac{2\pi}{k}(1-\cos(\kappa R)),\qquad C=\frac{2\pi}{\kappa}\sin(\kappa R).\] The projective circumference formula assumes \(0<R<\pi/(2\kappa)\); at the endpoint the boundary line has half the limiting length. Spherical cap formulas use \(0\le R\le\pi/\kappa\) without that boundary identification.

Derivation: move the center to zero and integrate \(2\,dr/(1+kr^2)\) for radius, \(4r\,dr\,d\theta/(1+kr^2)^2\) for area, and \(2r\,d\theta/(1+kr^2)\) for circumference; substitute the radius relation in each case.

Approach. Normalize the figure using an isometry, then apply the definition.

For \(k>0\) it is an ellipse, with semiaxes \(1\) and \(1/\sqrt{k}\) (a circle when \(k=1\)). For \(k<0\) it is a hyperbola opening along the \(x\)-axis. For \(k=0\) it is the pair of parallel lines \(x=\pm1\), a degenerate conic; it is not an ordinary parabola. The source calls this a degenerate parabola, referring to its degenerate parabolic conic type.

Approach. Normalize the figure using an isometry, then apply the definition.

Move the first triangle's included vertex to the second's, then rotate its first side ray onto the corresponding ray. If the oriented included angles agree, the second side ray also coincides. On each chosen ray the point at a specified distance is unique, provided spherical sides are minor arcs and avoid the antipodal ambiguity. The two equal side lengths therefore place both remaining vertices at the corresponding points, so the triangles coincide.

If the triangles have opposite orientation, first reflect one before aligning. Thus the usual unoriented SAS statement uses the full isometry group, including reflections. With the book's orientation-preserving Möbius subgroup alone, the corresponding oriented-angle condition is necessary.

Approach. Normalize the figure using an isometry, then apply the definition.

Euclidean case: the cosine rule, or its standard side-angle comparison proof, gives \(a>b\) exactly when \(A>B\).

Hyperbolic case, in unit-curvature lengths: subtracting the two cosine-law expressions yields \[\cos A-\cos B=\frac{\sinh(a-b)[\cosh c-\cosh(a+b)]}{\sinh a\sinh b\sinh c}.\] The denominator is positive and \(c<a+b\), so the bracket is negative. Since cosine decreases on \((0,\pi)\), \(A-B\) has the sign of \(a-b\).

Spherical case: for minor sides of a convex spherical triangle, subtraction similarly gives \[\cos A-\cos B=\frac{\sin(a-b)[\cos(a+b)-\cos c]}{\sin a\sin b\sin c}.\] Here \(0<a,b,c<\pi\) and \(|a-b|<c<\min(a+b,2\pi-a-b)\). The bracket is negative, while \(\sin(a-b)\) has the sign of \(a-b\). Thus the same conclusion holds. Rescaling proves other nonzero curvatures. A projective triangle must be interpreted through a consistent convex spherical lift; arbitrary long arcs are not covered.

Approach. Normalize the figure using an isometry, then apply the definition.

The broken path along two sides connects the endpoints of the third. A shortest geodesic has no greater length. Equality can occur only when the broken path itself is a minimizing geodesic with no corner: at the middle vertex, noncollinear tangent directions can be rounded locally to shorten the path. A nondegenerate triangle has a genuine corner, so the inequality is strict. This applies in the Euclidean and hyperbolic planes and to shortest arcs in the spherical/projective models.

Approach. Normalize the figure using an isometry, then apply the definition.

For \(k>0\), \(A=4\pi\sin^2(\sqrt{k}r/2)/k\), so the ratio is \([\sin u/u]^2\) with \(u=\sqrt{k}r/2\), which tends to one. For \(k<0\), replace sine by hyperbolic sine, and \(\sinh u/u\to1\). For \(k=0\) the ratio is identically one. More precisely \(A=\pi r^2-k\pi r^4/12+O(r^6)\).

Approach. Normalize the figure using an isometry, then apply the definition.

For \(k>0\), the ratio is \(\sin(\sqrt{k}r)/(\sqrt{k}r)\); for \(k<0\), it is \(\sinh(\sqrt{-k}r)/(\sqrt{-k}r)\); for \(k=0\), it equals one. Each tends to one. The expansion is \(C=2\pi r-k\pi r^3/3+O(r^5)\).

Approach. Normalize the figure using an isometry, then apply the definition.

Given \((a,b)\) and \((c,d)\), homogeneity supplies \(T_1\) with \(T_1a=c\) and \(T_2\) with \(T_2b=d\). Their product transformation sends \((a,b)\) to \((c,d)\). Hence the product action is transitive.

Approach. Normalize the figure using an isometry, then apply the definition.

If the factor metrics are \(d_1,d_2\), define \[d((a,b),(c,e))=d_1(a,c)+d_2(b,e).\] It is nonnegative, vanishes exactly when both coordinates agree, and is symmetric. Adding the two factor triangle inequalities gives the product triangle inequality. Componentwise transformations preserve each summand, so the metric is invariant. Other choices, such as \(\sqrt{d_1^2+d_2^2}\), are possible; a product action alone does not select a unique metric.

Approach. Normalize the figure using an isometry, then apply the definition.

On \(\mathbb R\), translations \(T_ax=x+a\) are bijections, \(T_aT_b=T_{a+b}\), \(T_0=I\), and \(T_a^{-1}=T_{-a}\). On the unit circle, rotations \(R_\theta z=e^{i\theta}z\) are bijections, \(R_\theta R_\phi=R_{\theta+\phi}\), \(R_0=I\), and \(R_\theta^{-1}=R_{-\theta}\). Each therefore supplies a transformation group on its stated set.

Approach. Normalize the figure using an isometry, then apply the definition.

Its set is \(\mathbb R^2\) and its transformations are \((x,y)\mapsto(x+a,y+b)\), so it is the translation geometry introduced earlier. It is homogeneous and admits the Euclidean metric or the sum metric \(|\Delta x|+|\Delta y|\). It is not isotropic under this specified group: the only translation fixing a point is the identity, so there are no nontrivial rotations about that point. Having a Euclidean metric does not automatically enlarge the given transformation group.

Approach. Normalize the figure using an isometry, then apply the definition.

Represent a point by \((x,e^{i\theta})\) and embed it as \((\cos\theta,\sin\theta,x)\) on the infinite unit cylinder. Product transformations are axial translations and rotations around the axis. They are transitive, so the geometry is homogeneous. An invariant metric is \(|x-y|+\delta(\theta,\phi)\), where \(\delta\) is the shorter circular angle; the usual flat-cylinder product metric is another choice. It is not isotropic under the product group, since a transformation fixing a point must be the identity.

Approach. Normalize the figure using an isometry, then apply the definition.

Choose the flat Riemannian product metric and unroll the cylinder to the plane. Projections of ordinary lines are its geodesics: axial lines, horizontal closed circles and helices. Two cylinder points admit connecting geodesics with different winding numbers, so the first postulate's uniqueness fails globally. Geodesics extend indefinitely (some close), and right angles are locally congruent. Arbitrarily large metric circles are not ordinary Euclidean circles after wrapping, so the global circle postulate also needs qualification.

A simple contractible geodesic triangle bounding a disk lifts to a Euclidean triangle and has angle sum \(\pi\). Three arbitrary wrapped arcs may not bound a triangular disk at all, so no unconditional triangle-angle assertion should be made for them. The sum metric from the earlier exercise has different geodesics and no canonical Euclidean angle notion; specifying the metric is necessary.

Approach. Normalize the figure using an isometry, then apply the definition.

The product consists of two independent angles modulo \(2\pi\). Angle translations act transitively, so it is homogeneous. The sum of the two circular distances is an invariant metric; the flat Riemannian product metric is also invariant. The specified product group is not isotropic because its point stabilizers contain only the identity.

Topologically this is a torus, just like a doughnut surface. Geometrically its flat product metric has zero curvature everywhere. The usual torus embedded in three-dimensional Euclidean space has positive curvature outside, negative curvature inside, and zero curvature along its top and bottom circles. That induced metric is not the flat product metric and is not homogeneous. A drawing of a torus therefore does not determine its intrinsic geometry.

Initially \(k=-0.3/6=-0.05\). The new numerical area is \(100\cdot6=600\), so the new curvature is \(-0.3/600=-0.0005\). The product remains \(-0.0005\cdot600=-0.3\). The sign is unchanged, but curvature's numerical value is divided by the square of the scale factor.

Adding the same (a,b) to both points leaves both coordinate differences unchanged, proving translation invariance. The distance from (0,0) to (1,0) is 1; rotating by \(\pi/4\) sends the second point to \((1/\sqrt2,1/\sqrt2)\), whose distance is \(\sqrt2\). Thus that rotation is not an isometry for this metric. The unit circle is \(|x|+|y|=1\), a diamond with vertices (1,0),(0,1),(-1,0),(0,-1). The word circle means a distance locus relative to the chosen metric.