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Fix $p$ on a Riemannian surface. For sufficiently small $r>0$, the geodesic circles of radius $r$ centered at $p$ and at every sufficiently nearby distinct $q$ meet transversally at exactly two points. In a strongly convex normal neighbourhood, write $q={\exp }_p(\epsilon v)$ and $x(\theta )={\exp }_p(re(\theta ))$. First variation givesThe quotient of the bracketed function by $\epsilon$ extends smoothly and has two simple zeros at zero. The implicit function theorem, uniformly over unit $v$, preserves these two zeros and excludes others. The nonzero angular derivatives prove transversality. The distinct-centre hypothesis is essential: coincident circles are not transverse.

For a fixed closed orientable Riemannian surface, the smooth metrics with a specified scalar Laplace-Beltrami operator spectrum form a compact set modulo diffeomorphisms, in the smooth topology. This is a compactness assertion, not uniqueness or finiteness of the number of Riemannian isometry classes. It restricts geometric degeneration within an isospectral set and uses more information than the leading heat invariants alone. The theorem was established in the 1988 compactness paper.

A surface metric has a cone singularity of total angle $\Theta >0$ if a punctured neighbourhood has the metric obtained by joining the radial boundaries of a sector of angle $\Theta$. For constant Gaussian curvature $-1$ away from the vertex, its polar expression is $d{r}^2+(\Theta /2\pi {)}^2{\sinh }^{2}rd{\theta }^2$, where $\theta$ has period $2\pi$. A circle of small radius $r$ has length asymptotic to $\Theta r$. A smooth Riemannian surface instead has circumference asymptotic to $2\pi r$, so the metric extends smoothly at the vertex exactly when $\Theta =2\pi$.

Two Riemannian metrics lie in the same conformal class when one is a positive smooth scalar multiple of the other. They measure the same angles, while lengths may change from point to point. On an oriented Riemannian surface, rotation of each tangent plane through $\pi /2$ determines a complex structure which depends only on this conformal class.

For a finite isometric group action on a closed Riemannian surface, choose a disk in the free locus whose translates are disjoint. Modify the Riemannian metric there, and identically on its translates, by small supported changes with three distinguished high Gaussian curvature maxima. Give their heights distinct values occurring nowhere else, arrange them as a tight noncollinear cluster, and keep separate clusters far apart. Such markers can be made by local conformal bump functions controlling curvature and its derivatives. Every quotient isometry must send an entire labelled cluster to another. After identifying the marked disks with the original one, an isometry fixes its three noncollinear points and is therefore the identity locally. If $p_i:M\to U_i\backslash M$ are free finite quotients, any quotient isometry $F$ consequently makes $F\circ p_1=p_2\circ t$ on an open disk for some element $t$ of the original group. Two local isometries agreeing on an open set agree on the connected surface, by propagation along geodesics. The equality therefore holds globally and implies $t{U}_1{t}^{-1}\subseteq U_2$, hence equality for equal subgroup orders. Nonconjugate equal-order subgroups thus have nonisometric quotients for this marked invariant metric; constant curvature is not required.

For a compact smooth Riemannian surface with smooth boundary and a pure Dirichlet boundary condition or pure Neumann boundary condition, the constant coefficient in the heat trace isThe equality is the Gauss-Bonnet theorem. Thus such isospectral manifolds have the same Euler characteristic. A simply connected compact connected bordered surface is a disk with $\chi =1$, whereas a nonorientable surface with $c\ge 1$ crosscaps and $b\ge 1$ boundary components has $\chi =2-c-b\le 0$. These cannot be isospectral in this setting. Corners and mixed boundary conditions contribute additional heat coefficients, so the smooth pure-boundary-condition hypothesis is essential.

At unit speed on a Riemannian surface, the right-hand side is $⟨E,iv⟩iv$. This removes the velocity-parallel component of the forcing, so differentiating $|\dot{\gamma }{|}^2$ gives zero. The dynamics need not preserve the ordinary Liouville volume of a surface geodesic flow.

For $r$ smaller than the normal-coordinate radius, the geodesic circle on a Riemannian surface consists of points at distance $r$ from $p$. The exponential map restricted to the radius-$r$ circle in $T_{p}S$ gives its smooth embedded parameterization. Outside the injectivity range a distance sphere need not be smooth.

A perpendicular Jacobi field on a Riemannian surface has scalar equation $\ddot{y}+Ky=0$. For strictly negative curvature on a closed manifold, the displayed derivative is uniformly positive definite, so the quadratic-form criterion for an Anosov flow makes the geodesic flow Anosov. When $K=-1$, the stable and unstable lines in the canonical frame are respectively $\mathbb{R}(H-V)$ and $\mathbb{R}(H+V)$, with factors $e^{-t}$ and $e^t$.

For a smooth Anosov flow generated by a surface thermostat on a closed oriented Riemannian surface, a smooth base function plus a smooth one-form is a smooth flow coboundary precisely when the base function is zero and the one-form is an exact differential form. This is the nontrivial degree-zero/one kernel theorem. It does not require negative curvature or preservation of Liouville volume. The converse follows directly from $F(f\circ \pi )=df(v)$. A primary proof of the kernel theorem is arxiv.org/abs/math/0601039.

Write ${\theta }_i$ for the internal angle at $v_i$, using the numbering in part (b), and use the hyperbolic plane with Gaussian curvature $-1$. The necessary and sufficient conditions are equal lengths for each pair of equally labelled polygon sides, and total angle $2\pi$ at each vertex equivalence class. Explicitly, the two angle conditions areNecessity follows because the identification must preserve arc length along each seam, and a smooth interior point of a Riemannian surface has total angle $2\pi$. For sufficiency, equally long hyperbolic geodesic segments can be identified by isometries. Across their interiors, two half-disc neighbourhoods join to a hyperbolic disc. At an identified vertex, the incident sectors form a cyclic disc with total angle $2\pi$, so no cone singularity of a surface metric remains. These local isometries therefore extend the Riemannian metric smoothly everywhere. This is the smooth hyperbolic polygon gluing criterion.

The ten internal angles sum to $4\pi$. The area of a hyperbolic geodesic polygon consequently givesEquivalently, the Gauss-Bonnet theorem gives $-\mathrm{Area}(X)=2\pi \chi (X)=-4\pi$.

Use the following permitted kernel result, stated in the form needed here. For an Anosov flow generated by a smooth surface thermostat on a closed oriented Riemannian surface, if a smooth function $u$ on $SM$ satisfieswhere $h$ is a smooth base function and $\eta$ is a smooth one-form, then $h=0$ and $\eta$ is an exact differential form. Conversely these two conditions give a smooth solution by taking a base primitive. This is the kernel of the thermostat X-ray transform in degrees zero and one. In particular it applies to Gaussian thermostats without an additional curvature condition; it is stronger than the magnetic or geodesic special case.

Put $\theta =E^{♭}$, so ${\theta }_x(v)=⟨E(x),v⟩$. Suppose the thermostat preserves a smooth volume form. Write it as $\rho \mu$ with smooth $\rho >0$, choosing its sign separately on each component if necessary. By the invariant volume criterion for a smooth flow and the divergence computation in part (b),Thus $u=\log \rho$ is a smooth solution of $Fu={\theta }_x(v)$. Apply the stated kernel theorem with $h=0$ to obtain $\theta =df$ for a smooth base function $f$. The definition of the gradient then gives $E={\mathrm{grad}}_{g}f$.

Conversely, suppose $E={\mathrm{grad}}_{g}f$. Since $d\pi (F)=v$, the vertical part has no effect on a base function andThe divergence of a Gaussian thermostat and the volume-change formula therefore giveConsequentlyOnly the necessary direction uses the Anosov kernel theorem; the displayed invariant density works for any potential Gaussian thermostat. In particular its exponent is positive, as required to cancel the negative divergence found in part (b).

Let $p=|a|$, $q=|b|$ and $r=|ab|$ denote element orders. Remove three points from an oriented sphere. Its fundamental group has peripheral generators ${\gamma }_a,{\gamma }_b,{\gamma }_c$ with ${\gamma }_a{\gamma }_b{\gamma }_c=1$. Send them to $a,b,(ab{)}^{-1}$. Since $a,b$ generate $T$, the kernel defines a connected regular covering of the punctured sphere with degree $|T|$ and deck transformation group $T$.

Complete the lifted punctures by disks. Above the three removed points the monodromy cycles have lengths $p,q,r$, respectively, and the disk projection has the local form $z\mapsto z^p$, $z^q$ or $z^r$. This gives a closed connected oriented topological surface $M$ and an orientation-preserving $T$-action. At the three kinds of filled-in points the stabiliser subgroups are conjugates of $⟨a⟩$, $⟨b⟩$ and $⟨ab⟩$, respectively. Every other point has trivial stabiliser subgroup. Orders equal to one simply correspond to unbranched filled points.

The punctured sphere has Euler characteristic $-1$, so its cover has Euler characteristic $-|T|$. Filling the ends adds $|T|/p+|T|/q+|T|/r$ disks. This proves the formula for the branched surface action from two generators:

For a subgroup $U$, its quotient map is always regular as a branched quotient. It is an ordinary unbranched normal covering map exactly when $U$ acts freely, that is,Necessity follows from the listed stabiliser subgroups. Conversely, these conditions eliminate every possible nontrivial point stabiliser. A free finite action gives a covering space, and its deck transformations are $U$, acting transitively on each fibre; hence the cover is normal. In this caseThe condition $U⊴T$ concerns regularity of the different map $U\backslash M\to T\backslash M$; it is not needed for $M\to U\backslash M$ when the action of $U$ is free.

For the specified order-$96$ group, $p=q=3$. Set $c=ab$ and $z=c^3$. Since $z$ has order two, $c^6=1$ and $c^3\ne 1$, so $|c|$ is either two or six. The order-two alternative would make the displayed formula give $\chi (M)=96(1/3+1/3+1/2-1)=16$. That is impossible for a connected closed oriented topological surface, whose Euler characteristic is $2-2g\le 2$. Thus $|c|=6$, and

Each $U_i$ has order eight. By Lagrange's theorem, it contains no element of order three or six, so it meets all conjugates of $⟨a⟩$ and $⟨b⟩$ trivially. In $⟨c⟩$, the only nonidentity element whose order divides eight is $z=c^3$. Since $z$ is central, it is also the unique such element in every conjugate of $⟨c⟩$, and it is excluded from both $U_i$ by hypothesis. Both subgroup actions are therefore free. The quotients are closed oriented surfaces with

The branched-cover construction supplies a smooth structure, with the stabilisers acting by rotations in the disk charts. Average a smooth Riemannian metric over $T$ to make the whole action isometric. The Gassmann equivalence of $U_1,U_2$ then gives isospectral manifolds: for every upstairs eigenspace with character ${\chi }_{\lambda }$, its downstairs multiplicities are $|U_i{|}^{-1}{\sum }_{u\in U_i}{\chi }_{\lambda }(u)$, and these are equal because the two subgroups have equal intersection sizes with every conjugacy class. This proves the spectral assertion for any such invariant metric.

To guarantee nonisometry, choose the invariant metric with curvature markers distinguishing finite-cover quotients. Here are the geometric details, since nonconjugacy alone would not guarantee nonisometry for an arbitrary metric. Choose a small disk in the free locus of the $T$-action, with all its translates disjoint. In that disk choose three noncollinear points in a geodesically convex neighbourhood. Supported conformal smooth bump functions can give them three distinct high values as strict Gaussian curvature maxima, occurring nowhere else in the unmodified metric. Arrange these three maxima as one tight cluster, much closer to each other than to any translated cluster, and copy the same modification to every translate. Curvature and its first few derivatives can be adjusted by the second and higher derivatives of the conformal factor; narrow bumps give high distinguished curvature with small change in metric distances. Further small supported perturbations remove accidental maxima of these exact heights. Thus the resulting metric remains smooth and $T$-invariant, and its marked clusters are intrinsic and uniquely labelled by their three curvature values.

Any proposed quotient isometry $F:U_1\backslash M\to U_2\backslash M$ must carry a whole labelled cluster to another. Lift its restriction near that cluster to the corresponding two marked disks in $M$, and identify these disks using the appropriate group element $t\in T$. The resulting local isometry fixes all three marked points. Its differential at the first point fixes the two independent directions of the unique geodesics to the other two; hence it is the identity, and the local isometry itself is the identity near that point. If $p_i:M\to U_i\backslash M$ are the quotient maps, we therefore haveon an open set. Both sides are local isometries on the connected complete Riemannian surface $M$. A local isometry is determined along each geodesic by its value and differential at the initial point, so the equality propagates to all of $M$.

For $u\in U_1$, this equality implies $p_2\circ tu=p_2\circ t$. Evaluating at a free point of the $T$-action shows $tu{t}^{-1}\in U_2$; equivalently these maps differ by a deck transformation of $p_2$. Thus $t{U}_1{t}^{-1}\subseteq U_2$, and equality of subgroup orders gives equality. This contradicts their assumed nonconjugacy. For this marked invariant metric the two quotients are nonisometric, isospectral Riemannian surfaces of genus two. No constant-curvature requirement is imposed or used.

Fix a bounded tile in Euclidean space $\Omega$ with labelled flat boundary faces. For each face label $s$, encode an assembly of $n$ copies by an involutive real symmetric signed permutation matrix $M_s$: a paired interface contributes a two-by-two swap, an unglued face with a Neumann boundary condition contributes $+1$ on the diagonal, and an unglued face with a Dirichlet boundary condition contributes $-1$. Use the same face-identification convention on each copy. If a second assembly has matrices $N_s$ and there is an invertible constant matrix $C$ satisfyingthen the map $u=(u_1,\dots ,u_n)\mapsto Cu$ is an isomorphism between corresponding Laplace-Beltrami operator eigenspaces. This is the transplantation theorem.

To prove it, restrict an eigenfunction to the tile copies and identify each copy with $\Omega$. Along a face, let $a$ be its vector of boundary values and $b$ its vector of outward normal derivatives. At a glued interface the values agree and the outward derivatives are opposite. At an unglued face with a Neumann boundary condition the derivative vanishes, and at a face with a Dirichlet boundary condition the value vanishes. All three cases are encoded byThe intertwining identities give $N_s(Ca)=Ca$ and $N_s(Cb)=-Cb$, so the transplanted function satisfies the second assembly's interface and boundary conditions. On each tile its Laplace-Beltrami operator is the same constant linear combination of the original Laplacians, hence it has the same eigenvalue. Across an interface, matching values and opposite normal derivatives make it a weak solution; elliptic regularity, or the allowed reflection principle, makes it a genuine smooth solution there. At corners the operator is understood by its finite-energy, Friedrichs extension; the same matching argument applies to its quadratic form domain. Applying $C^{-1}$ reverses the construction. Thus dimensions of all eigenspaces agree, including multiplicities, and the two specified boundary-value spectra coincide.

The requested pair exists with explicitly specified mixed boundary conditions and polygonal boundary. Take four copies $Q_0,Q_1,Q_2,Q_3$ of an octagonal regular polygon in Euclidean space. Label four alternate sides $a,b,c,d$ in cyclic order; the four intervening sides always have a Neumann boundary condition. The labelled sides are disjoint even at their endpoints. Parameterize every side in the positive boundary direction of the reference octagon. When pairing same-labelled sides, identify equal parameters. This produces a locally flat surface with boundary: a neighbourhood of an interior point of an interface consists of two half-disks glued along their diameters. All tile vertices remain boundary points, with boundary angles $3\pi /4$ or $3\pi /2$; there are no interior cone points.

Here are the complete gluing and boundary conditions. An entry $ij$ pairs the corresponding sides of $Q_i,Q_j$. The indicated unpaired sides have a Dirichlet boundary condition; every other unpaired side has a Neumann boundary condition.Both assemblies are compact and connected. The Laplace-Beltrami operator is the nonnegative operator $P=-\Delta$ associated with the finite-energy quadratic form $\int |\nabla u{|}^2$, with zero Sobolev trace on the specified Dirichlet arcs. Thus the mixed boundary condition is defined without requiring classical differentiability at a corner or at a Dirichlet–Neumann junction.

The four signed permutation matrices for $D$ areFor $B$ they areUse the Hadamard matrixThe intertwining relations can be checked on all columns at once. Writing the columns of $C$ as $h_0,h_1,h_2,h_3$, direct application of the displayed $N_s$ givesFor the four intervening faces both assemblies have the identity face matrix, so these intertwining relations hold as well. Also $C^{T}C=C^2=4I$. This four-tile mixed-boundary transplantation between a disk and a nonorientable surface gives a bijection of every pair of eigenspaces, preserving eigenvalues and multiplicities.

There is also a direct argument on the quadratic form domains that handles all boundary corners. The orthogonal matrix $U=C/2$ acts on tile restrictions, preserves their L2 inner product and the sum of their Dirichlet energies, and commutes with taking a Sobolev trace. The condition $M_{s}u=u$ on each face encodes equality of the two traces at a paired interface and vanishing trace at an unpaired Dirichlet face. Intertwining takes these conditions exactly to $N_s(Uu)=Uu$. Consequently $U$ maps the whole form domain onto the other form domain and preserves its quadratic form. The associated self-adjoint operators are unitarily equivalent, proving isospectrality for the stated mixed boundary conditions without any assumption of smoothness at a Dirichlet–Neumann junction.

We now verify the actual topology using the topology of disk tiles glued along disjoint boundary arcs. Because the gluing arcs in each tile are disjoint, the resulting surface with boundary is a thickening of its gluing graph, with one vertex for each closed disc tile and one edge for each paired arc. One can construct this homotopy equivalence by taking a small central closed disc in each tile and narrow strips to its paired arcs. What remains is a collar neighbourhood of the boundary, which retracts onto this connected system of closed discs and strips. Shrinking each central closed disc and each strip then gives the gluing graph. Equivalently, begin with a spanning tree of the gluing graph: its closed disc tiles joined along boundary arcs form one closed disc, and every remaining pair of boundary arcs adds one band. Thus for $n$ tiles and $e$ paired arcs the fundamental group is a free group of rank $e-n+1$, and the Euler characteristic is $n-e$.

For $D$, the gluing graph is the pathGluing a new closed disc to a closed disc along one boundary arc again gives a closed disc; applying this three times proves that $D$ is homeomorphic to a closed disc. In particularFor $B$, the five paired arcs give the graph edges $13,12,02,01,23$. It is connected, and henceHere $F_2$ is the free group on two generators, so $B$ is not simply connected. To decide orientation of a surface, give every reference tile its counterclockwise orientation. Our same-parameter gluing preserves boundary directions, so a global orientation requires opposite orientation signs on paired tiles: the two induced boundary orientations at an interior interface must be opposite. Such signs exist exactly when the gluing graph is bipartite. In $B$ the cyclehas length three. Following it forces the sign to reverse three times and return to its starting tile with the opposite sign. This proves that $B$ is a nonorientable surface. It has nonempty boundary because every intervening octagon side remains unpaired. Thus $D$ is a simply-connected orientable surface with boundary, whereas its isospectral partner $B$ is neither simply connected nor orientable, exactly as required, for the specified mixed boundary problem.

The boundary qualification is essential. With smooth boundary and the same pure Dirichlet boundary condition or pure Neumann boundary condition on both surfaces, the following existing obstruction applies. Polygonal corners and the mixed boundary conditions in the construction above place it outside that obstruction.

With the interpretation of compact smooth Riemannian surfaces with the same pure Dirichlet boundary condition or pure Neumann boundary condition, the printed simply-connected/nonorientable pair cannot exist. The two-dimensional heat trace has the following expansion (the smooth-boundary coefficients are given in the heat-coefficient formulas, Theorems 5.1–5.2):where the boundary sign is minus for Dirichlet and plus for Neumann. The Gauss-Bonnet theorem identifies the constant term with $\chi (M)/6$. Isospectrality therefore forces equal Euler characteristics. A compact connected simply-connected surface with boundary is a disk and has $\chi =1$. A connected nonorientable surface with $c\ge 1$ crosscaps and $b\ge 1$ boundary components has $\chi =2-c-b\le 0$. Thus no such smooth pure-Dirichlet or pure-Neumann pair exists. This is the Euler characteristic from the heat trace of a bordered surface.

Piecewise-smooth boundaries introduce corner terms, and mixed boundary conditions change the heat coefficients, as the explicit construction above demonstrates. Moreover, in the usual pure-Neumann polygonal reflection-transplantation setup, the pure Neumann reflection transplantation preserves Euler characteristic argument applies even with corners. The numbers of tiles agree. For each face label the number of edge classes is $(n+\mathrm{tr}{M}_s)/2$, so these agree too. For each tile-vertex label, vertex classes are orbits of the subgroup generated by incident-face involutions. Their number is the dimension of invariant vectors in that permutation representation, which is unchanged by the intertwiner. Hence the numbers of vertices agree, proving equal $V-E+F$. Thus this standard tiled construction also cannot supply a disk on one side and a nonorientable surface on the other.

For comparison, the following existing construction gives an orientable and a nonorientable pair with the same spectrum for the pure Neumann boundary condition. Both have nontrivial fundamental groups; this is a separate example, not the construction used to meet the simply-connected requirement above. This eight-tile Neumann transplantation across orientability makes the comparison explicit. The finite seed is the affine modulo-eight example in the original orientability construction; the intertwiner and the topology checks are derived below. Take eight copies of a regular Euclidean hexagon and label three pairwise nonadjacent sides $\sigma ,t,u$. Every remaining side is a side with a Neumann boundary condition. Index the copies by $j\in \mathbb{Z}/8\mathbb{Z}$. For the first assembly pair same-labelled faces according tofor the second, usewith all arithmetic modulo eight. A fixed point means an unglued side with a Neumann boundary condition. Pair faces by the reflection convention, matching the same endpoint parameter in the reference tiles. The distinguished sides are disjoint, so no interior vertex singularities are created; the resulting surfaces have polygonal boundary corners.

Let $R{e}_j=e_{j+1}$ on the eight-dimensional tile-index space and setIt commutes with the permutation matrices for $t$ and $\sigma$, because its coefficients are invariant under reversing cyclic order. For $u_1,u_2$, write $C_{ij}=c_{i-j}$, where $c_0=c_4=1$, $c_1=c_7=2$, and all other $c_r$ vanish. The required relation $C{M}_{u_1}=M_{u_2}C$ is exactly $c_r=c_{3r+4}$, verified from these four nonzero entries. Thus all three intertwining relations hold.

The cyclic Fourier vectors diagonalize $C$. Its eigenvalues arenamely $6,2\sqrt{2},2,-2\sqrt{2},-2,-2\sqrt{2},2,2\sqrt{2}$. None is zero, so $C$ is invertible and the two assemblies are isospectral manifolds for the Neumann boundary condition.

We prove their topology as well. Their gluing graphs are connected: the first contains the spanning path $0,1,3,6,2,7,5,4$, and the second contains the spanning cycle $0,1,7,2,6,3,5,4,0$. Each has eight tile vertices and ten paired-face edges. Since tiles are disks and gluing arcs are disjoint, each surface retracts up to homotopy to this multigraph; thus its Euler characteristic is $8-10=-2$ and its fundamental group is a free group of rank three.

An orientation assigns a sign to each reference tile, and a reflection gluing requires opposite signs at the ends of every graph edge. The second graph admits this assignment: take sign $+$ on $\{0,5,6,7\}$ and sign $-$ on $\{1,2,3,4\}$. Every paired face joins opposite signs, proving orientability. In the first graph the cyclehas odd length. Transporting orientation around it reverses the initial sign, proving nonorientability. Hence the first surface is also not simply connected. The second surface is orientable but likewise not simply connected; the disk required in the question is supplied instead by the four-tile mixed-boundary construction above.

The four-tile construction resolves the printed simply-connected requirement under explicit mixed boundary conditions. The eight-tile comparison only concerns orientation of a surface under a pure Neumann boundary condition.

For a smooth curve $\gamma (t)$ on a Riemannian surface, its energy of a curve isChoose a local parameterization $\mathbf{X}(u,v)$ and write $\gamma (t)=\mathbf{X}(u(t),v(t))$. With coefficients of the first fundamental formthe Lagrangian isThe Euler-Lagrange equations are equivalentlyAfter multiplying by the inverse metric these become the geodesic equationwhere the Christoffel symbols are determined by $E,F,G$.

If a straight line segment lies in the surface, parameterize it byThen $\ddot{\gamma }=0$, so its acceleration has zero tangential component. The two displayed equations hold directly, and the segment is a geodesic.

For the one-sheeted hyperboloidputTwo distinct ruling lines through $P$ areIndeed, the direction $d_{\pm }$ satisfiesso substitution shows that every point of the line lies in $H$. These are geodesics by the straight-line argument.

A third geodesic is the meridian through $P$. Choose $s_0$ with $\sinh s_0=z_0$; thenIts acceleration is ${\gamma }_m$, which is normal to $H$, so it is a geodesic. If $z_0\ne 0$, these give the required three distinct subsets.

If $z_0=0$, there is also the equatorial circle. Writing $P=(\cos {\theta }_0,\sin {\theta }_0,0)$,Its acceleration is normal to $H$ along $z=0$, so it is a fourth geodesic distinct from the meridian and the two rulings.

Finally, write $\rho =\sqrt{x^2+y^2}=\sqrt{1+z^2}$. Clairaut's relation gives the conserved quantityChoose initial data in $z>0$ withSince $|\sin \psi |\le 1$, every point of the resulting geodesic satisfies $\rho \ge c$, and thereforeContinuity keeps the geodesic in the component $z>0$, and the stated completeness assumption defines it for every real time. The continuum of choices of $c$ supplies infinitely many such geodesics.

On an oriented Riemannian surface, the vertical vector field generates positive rotation in each fibre of its unit tangent bundle: $Vf(x,v)={{\partial }_{t}f(x,\cos tv+\sin tiv)|}_{t=0}$. Hence $V(v)=iv$ and $V(iv)=-v$. In the canonical frame, $[V,X]=H$, $[V,H]=-X$ and $[X,H]=KV$.