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A nondegenerate convex regular spherical $n$-gon contained in an open hemisphere has equal interior angles $\alpha$ in the displayed interval. The spherical excess formula gives area $n\alpha -(n-2)\pi >0$. Conversely, put equally spaced vertices at colatitude $t\in (0,\pi /2)$ around a pole. Dividing each central triangle in half gives $\cot (\alpha /2)=\cos t\tan (\pi /n)$ by the spherical law of cosines. As $t$ increases, the angle ranges continuously from the Euclidean limiting angle to $\pi$.

For the positively oriented boundary of a spherical triangle on the unit sphere, parallel transport is rotation by its angle excess modulo $2\pi$. The tangent is parallel along each geodesic side and turns by $\pi -\alpha$ at a corner of angle $\alpha$. The three corner turns sum to $3\pi -\alpha -\beta -\gamma$, and combining them with parallel transport returns the initial tangent. Hence the transport angle is the negative of that sum modulo $2\pi$, giving the formula. The spherical excess formula identifies the angle with enclosed area; reversing traversal changes its sign.

For a spherical triangle on a sphere of radius $R$, bounded by shorter great circle arcs and with interior angles $\alpha ,\beta ,\gamma$, the spherical excess formula, the triangular form of the Gauss-Bonnet theorem, isHere is an area proof. On the unit sphere, a spherical lune of angle $\alpha$ has area $2\alpha$: rotating around its vertices shows it occupies a fraction $\alpha /(2\pi )$ of the sphere when one uses the full angular width, or directly integrates ${\int }_0^{\alpha }{\int }_0^{\pi }\sin \theta d\theta d\phi$. The three side great circles of the triangle divide the sphere into eight regions, including the triangle $T$ and its antipodal triangle $-T$. Choose, at each vertex, the lune containing $T$, and also its antipodal lune. Each of $T$ and $-T$ is covered three times by these six lunes; each other region is covered once. Thus, with $A$ the area of $T$,This proves $A=\alpha +\beta +\gamma -\pi$; scaling lengths by $R$ scales area by $R^2$.

Radially project each face of the regular dodecahedron from its centre onto the unit sphere. Each edge and the origin lie in a plane, so its image is a shorter great circle arc. The twelve faces give congruent convex regular spherical polygons partitioning the sphere. At each projected vertex three congruent face angles fill the tangent plane. Hence each spherical pentagon has angle $2\pi /3$ at each of its five vertices. Splitting a pentagon into three spherical triangles and summing the spherical excess formula givesThis also agrees with $4\pi /12$, as required by the spherical partition.

Work first on the unit sphere. A spherical lune of angle $\alpha$ has area $2\alpha$: its fraction of sphere area is $\alpha /(2\pi )$. The three great circles extending the sides of a nondegenerate spherical triangle divide the sphere into four antipodal pairs of triangular regions. Call the desired region $T$, of area $A$, and its three adjacent regions $T_1,T_2,T_3$, of areas $A_1,A_2,A_3$. These four regions contain one region from each antipodal pair, so $A+A_1+A_2+A_3=2\pi$. The lune at each vertex consists of $T$ and one of these adjacent regions. Therefore, if the three interior angles are $\alpha ,\beta ,\gamma$,which proves the spherical excess formula, the spherical-triangle form of the Gauss-Bonnet theorem:For a sphere of radius $s$, multiply this area by $s^2$.

Triangulate a convex spherical $n$-gon by diagonals from a vertex. There are $n-2$ spherical triangles, and their angles add to the polygon's interior angle sum. Thus its unit-sphere area isFor a nondegenerate convex regular polygon in an open hemisphere, area is positive and each angle is less than $\pi$, giving $(n-2)\pi /n<\alpha <\pi$. Every angle in this interval occurs. Indeed place $n$ vertices at equal longitude spacings and colatitude $t\in (0,\pi /2)$ about a pole, and join consecutive vertices by shorter great circle arcs. The polygon is convex and regular. Split a central isosceles spherical triangle along the perpendicular to its edge. The resulting right spherical triangle has angles $\pi /n$, $\alpha /2$, $\pi /2$, and hypotenuse $t$. The angular spherical law of cosines yieldsso $\cot (\alpha /2)=\cos t\tan (\pi /n)$. As $t$ increases from $0$ to $\pi /2$, $\alpha$ increases continuously through exactlyThe endpoints are degenerate limits, not strict convex polygons.

For the spherical triangle reflection group, positive area requires $1/p+1/q+1/r>1$. Each angle is less than $\pi$, so the integer denominators are at least two. Arrange $p\le q\le r$. If $p\ge 3$ the sum is at most one, hence $p=2$. If $q\ge 4$ the sum is again at most one; thus either $q=2$ and any $r\ge 2$ works, or $q=3$ and $r=3,4,5$. These triangles exist, for example as the reflection chambers of a regular dihedral configuration, tetrahedron, cube, and dodecahedron, respectively.

To count group elements using the assumed tessellation, first label the original triangle's three vertices by their types and propagate these labels through side reflections. Around a type-$p$ vertex there are $2p$ sectors; the associated alternating reflections make a full turn and return the labels unchanged. The same holds at the other types. Every closed path in the dual tessellation is generated by these circuits around vertices, since the sphere is simply connected. Thus the labels are globally consistent, including when two angle denominators coincide. Every generated isometry preserves these labels. An element stabilizing a triangle therefore fixes its three vertices and is the identity. Consequently group elements correspond bijectively to triangles, and area givesThe possibilities, including permutations of the entries, are

Here is an explicit dodecahedron construction for the last case, avoiding an assumption about which vertices of the tessellation to join. Put $\phi =(1+\sqrt{5})/2$, and take the convex hull of the twenty pointswith independent signs. They all have squared radius three. The supporting plane $\phi y+z={\phi }^2$ contains the five verticesin cyclic order. Consecutive distances are $2/\phi$, and nonconsecutive distances are $2$, so this is a regular pentagon. Changing the two signs in its normal $(0,\phi ,1)$ and cyclically permuting coordinates gives twelve supporting planes, all with a congruent pentagon. Each listed vertex lies on three of these faces; the sixty face-edge incidences pair into thirty edges. The faces form a closed convex polyhedral surface with $20-30+12=2$, hence exhaust the boundary of the convex hull. This constructs a regular dodecahedron.

Choose the spherical directions of a face center $F=(0,\phi ,1)$, the midpoint $M=(0,\phi ,0)$ of its edge between $(\pm {\phi }^{-1},\phi ,0)$, and that edge's endpoint $V=({\phi }^{-1},\phi ,0)$. Normalize these to the unit sphere. The three side planes of triangle $FMV$ are $x=0$, $z=0$, and $\mathbf{n}\cdot \mathbf{x}=0$ with $\mathbf{n}=(-\phi ,{\phi }^{-1},-1)$. Their reflections are coordinate sign changes in the first two cases andin the third, since $\Vert \mathbf{n}{\Vert }^2=4$. Substitution in the twenty listed vertices, using ${\phi }^2=\phi +1$, shows that all three reflections permute the vertex set. The side-plane angles give interior angles $\pi /2$ at $M$, $\pi /3$ at $V$, and $\pi /5$ at $F$ because $|n_z|/\Vert \mathbf{n}\Vert =1/2$ and $|n_x|/\Vert \mathbf{n}\Vert =\phi /2=\cos (\pi /5)$. Thus this triangle is congruent to $\Delta$, and its side-generated group is $G$, conjugated by the aligning isometry.

It already has $120$ elements. Conversely, a symmetry of the dodecahedron is determined by the image of one face, one vertex on that face, and the orientation along its boundary: there are at most $12\cdot 5\cdot 2=120$ choices. An isometry fixing these choices fixes that face pointwise and must preserve the polyhedron's interior side, hence is the identity. Therefore the complete symmetry group of the constructed dodecahedron is $G$, including orientation-reversing symmetries.

Work on the unit sphere with geodesic edges. A spherical lune of angle $\alpha$ has area ${\int }_0^{\alpha }{\int }_0^{\pi }\sin \theta d\theta d\phi =2\alpha$. The three great circles through the sides of a spherical triangle partition the sphere into four antipodal pairs of triangles. If the original triangle has area $A$ and the other pair representatives have areas $B,C,D$, then $A+B+C+D=2\pi$. The three lunes at its vertices have areas $A+B,A+C,A+D$, soThis proves the spherical excess formula $\boxed{A=\alpha +\beta +\gamma -\pi }$. Triangulate a convex spherical $k$-gon from one vertex into $k-2$ triangles; summing areas and vertex angles givesFor a decomposition of the whole sphere, angles around every vertex sum to $2\pi$ and each edge belongs to two faces. Summing the polygon area formula over all faces therefore givesNow $F=m+n+p$, $2E=4m+5n+6p$, and $3V=2E$, since three edges meet at each vertex. Substituting into Euler's formula yields $2m+n=12$. Nonnegative integer solutions therefore give at most seven pairs:At least three distinct pairs are realized: a dodecahedron gives $(0,12)$, a pentagonal prism gives $(5,2)$, and a cube gives $(6,0)$. Center each convex polyhedron around an interior origin and radially project its faces onto the sphere; edges become great-circle arcs and the resulting spherical faces are convex, preserving the incidence counts. This supplies the required actual decompositions.

A vector field along a curve $\gamma :[a,b]\to M$ is a smooth section of the pullback tangent bundle ${\gamma }^{\ast }TM$: it assigns $V(t)\in T_{\gamma (t)}M$ smoothly. It need not be the restriction of one ambient vector field, since its values at different times with the same image point may differ.

Given a Koszul connection, use a local frame $e_j$ of $TM$ and write $V(t)={\sum }_j{v}^j(t)e_j(\gamma (t))$. Its covariant derivative along a curve isThe first term differentiates the time-dependent coefficients, and the second uses the connection on the local frame. Under a change of frame, differentiating the change matrix supplies exactly the extra term in the connection transformation, so the expression is independent of the frame. This definition also works when $\dot{\gamma }(t)=0$. In coordinate components it isA field is parallel when $D_{t}V=0$.

Thus a parallel field solves the linear system $\dot{v}=-A(t)v$, with smooth coefficient matrix $A(t)$. Given its value at one time, existence and uniqueness for a linear ordinary differential equation determine it on each coordinate segment. To obtain it on all of $[a,b]$, cover the compact parameter interval by finitely many frame neighborhoods and subdivide so that each closed subinterval lies in one such neighborhood. The coefficients are bounded on each subinterval, and the standard estimate $\Vert v(t)\Vert \le \Vert v(t_0)\Vert \exp (\int \Vert A(s)\Vert ds)$ prevents finite-time blowup. The solutions match uniquely at the subdivision points. This proves existence and uniqueness of the requested global parallel field.

Define parallel translation by ${\tau }_t(v)=V_v(t)$, where $V_v(a)=v$ and $D_t{V}_v=0$. Linearity of the equation and uniqueness make ${\tau }_t$ linear. Solving the same equation backwards gives its inverse; equivalently, a solution zero at time $t$ must be identically zero. Therefore

A metric connection for a Riemannian metric satisfies $\nabla g=0$, orPulling this identity along the curve givesFor parallel fields the right side vanishes, provingFor a piecewise smooth path compose the parallel transport maps on its smooth segments. On a closed path their domain and codomain are the same tangent space, so the resulting map is orthogonal. Its determinant need not be positive on a nonorientable surface.

An explicit reflection example is the reflection holonomy along a closed projective geodesic. Give ${\mathbb{RP}}^2$ the round metric induced from the sphere by the antipodal quotient. The Levi-Civita connection descends because the antipodal map is an isometry, so the quotient map is a local isometry carrying geodesics and parallel fields to their counterparts. The great-circle arcprojects to a smooth closed geodesic: $c(t+\pi )=-c(t)$, including all derivatives under the quotient identification. Along it the fieldsare parallel. The first is the velocity of a great-circle geodesic. The second is tangent, unit length, and orthogonal to the first. Metric compatibility makes its covariant derivative orthogonal to both frame vectors: differentiate its constant length and its zero inner product with the parallel first vector. Its derivative therefore vanishes in the two-dimensional tangent plane. At $t=\pi$, $E_1(\pi )=-E_1(0)$ and $E_2(\pi )=E_2(0)$. Identifying the endpoint with the start through the derivative of the antipodal map, namely $-I$, fixes the returned $E_1$ and negates the returned $E_2$. HenceThis is a reflection fixing the geodesic direction.

Finally orient the unit sphere outward and traverse a spherical triangle with its interior on the left. Let its angles be $\alpha ,\beta ,\gamma$. Its parallel transport preserves both the metric and orientation, so is a rotation by some angle $\delta$ in the initial tangent plane. On each geodesic side the unit tangent is parallel. At a vertex with angle $\alpha$, the actual forward tangent turns left by $\pi -\alpha$ relative to the transported incoming tangent. Tracking the tangent around all three sides and making those three corner turns returns it to its initial direction.

Rotations of oriented two-dimensional tangent planes commute with orientation-preserving parallel transport. Consequently the total angle in this tracking procedure is the transport angle plus the three corner turns:This determines parallel transport around a spherical triangle asBy the spherical excess formula, this is rotation by the enclosed area on the unit sphere. Reversing traversal negates the angle. For an octant triangle all three angles are $\pi /2$, giving a positive $\pi /2$ rotation.

On the unit sphere, the spherical excess formula gives the area of a spherical triangle asFor a sphere of radius $R$, the right side is multiplied by $R^2$.

Join an interior point of a convex spherical $n$-gon to its vertices by geodesics. The resulting $n$ spherical triangles contain all polygon angles and angles summing to $2\pi$ at the interior point. Adding their spherical excesses givesFor a regular polygon with interior angle $\alpha$, positivity of area requires $n\alpha >(n-2)\pi$, while strict convexity requires $\alpha <\pi$. Both limiting values can be approached by regular polygons, so

Spherical geometry studies lengths and angles measured along the surface of a sphere, rather than through its interior. It is a useful model whenever travel is constrained to a round surface. Its straightest paths are great circle arcs: they are geodesics, even though they appear curved in the surrounding Euclidean space. Two distinct great circles meet at an antipodal pair, so the parallel-line picture of Euclidean geometry does not apply.

For a sphere of radius $R>0$, write two positions as $Ru,Rv$, with $u,v$ unit vectors. Their shorter surface route has great-circle distancewhere the dot product gives the central angle in radians. The straight chord through the surrounding Euclidean space instead has length $2R\sin (d/(2R))$. For a quarter-circle separation these are $\pi R/2\approx 1.571R$ and $\sqrt{2}R\approx 1.414R$: specifying which distance we mean has a measurable effect. On the unit sphere, $R=1$ and great-circle distance equals the central angle.

Consider the spherical triangle with vertices $(R,0,0)$, $(0,R,0)$ and $(0,0,R)$, joined by the shorter great circle arcs. Each side has length $\pi R/2$. At each vertex the two outgoing tangent vectors point along the other coordinate axes and are orthogonal, so all three interior angles are $\pi /2$. This is the positive coordinate octant of the sphere, not a flat triangle drawn inside it.

For a convex minor spherical triangle with interior angles $\alpha ,\beta ,\gamma$ in radians, the spherical excess formula statesOur octant therefore has area $R^2(3\pi /2-\pi )=\pi R^2/2$. Independently, eight congruent octants partition the sphere, giving $4\pi R^2/8$ again. Thus angles alone determine area when $R$ is fixed, unlike Euclidean geometry, where similar triangles can have arbitrarily different areas. The Gauss-Bonnet theorem explains the connection: the sphere's constant Gaussian curvature $1/R^2$ times the area equals this excess. For spherical triangles whose area is small compared with $R^2$, the excess is small, recovering the familiar flat picture.

A spherical lune is the region between two great semicircles joining antipodal points. On a unit sphere a lune of opening angle $\alpha$ has area $2\alpha$, obtained by integrating area over longitude width $\alpha$. Lune area counting yields the spherical excess formula.

A spherical triangle is bounded by three great circle arcs. For the usual convex minor spherical triangle, the vertices are neither antipodal nor on one great circle, and its angular sides and interior angles lie in $(0,\pi )$. An interior angle is measured between outgoing tangent vectors to the two sides at a vertex, rather than between straight chords inside the sphere. The spherical law of sines and spherical law of cosines determine its side-angle relations; the spherical excess formula determines its area.

Reflections in the sides of a spherical triangle of angles $\pi /p,\pi /q,\pi /r$ generate a finite group when its images tessellate the sphere. The triangle area is $\pi (1/p+1/q+1/r-1)$ by the spherical excess formula. Label its three vertex types and continue those labels across each reflected side; they are consistent around every vertex and, because the sphere is simply connected, globally. The generated group preserves these labels, so a chamber stabilizer fixes all three vertices and is trivial. Dividing sphere area by chamber area gives its order. Up to permutation the possibilities are $(2,2,n)$ for $n\ge 2$, and $(2,3,3),(2,3,4),(2,3,5)$, of orders $4n,24,48,120$.