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A smooth real function $\lambda$ on the unit tangent bundle determines this unit-speed smooth flow. Its base velocity is $v$, so its generator never vanishes. In the canonical frame its divergence in Liouville volume of a surface geodesic flow is $V\lambda$. The special case where $\lambda$ depends only on the base is a magnetic flow; when $\lambda (x,v)=⟨E(x),iv⟩$ it is a Gaussian thermostat.

A connected complete Riemannian manifold is disconnected at infinity when the complement of some compact set has at least two unbounded connected components. Equivalently it has at least two ends. A line in a Riemannian manifold is a unit-speed geodesic $\ell :\mathbb{R}\to M$ satisfying $d(\ell (s),\ell (t))=|s-t|$ for every $s,t$.

Let $K$ be the compact separating set, enlarged inside a fixed compact metric ball if needed for bounds. Choose $x_j,y_j$ in two different unbounded components of $M\setminus K$, with both distances from $K$ tending to infinity. By the Hopf-Rinow theorem, join them by unit-speed minimizing geodesics. Every such segment meets $K$, since a path avoiding it cannot join its different components. Reparametrize a segment so that one crossing point is at time zero. Its parameter interval is $[-a_j,b_j]$, where $a_j,b_j\to \infty$ by the distances from the endpoints to $K$.

The unit tangent bundle over $K$ is compact, so a subsequence of the initial position/velocity pairs converges. Continuous dependence of the geodesic equation on its initial data, and completeness, give a limiting geodesic defined on all of $\mathbb{R}$. For any fixed $s<t$, the approximating segments are defined on $[s,t]$ for large $j$ and satisfy $d({\gamma }_j(s),{\gamma }_j(t))=t-s$. Continuity of the distance passes this equality to the limit. Thus the limiting geodesic is a line, proving a line from disconnection at infinity.

Pointwise positive Ricci curvature does not imply compactness or a finite diameter. The complete positively curved paraboloid $z=x^2+y^2$ has induced metricand Gaussian curvatureFor example, this follows from the graph formula $K=\det D^{2}z/(1+|Dz{|}^2{)}^2$. The metric dominates the Euclidean plane metric, so it is complete by upward stability of Riemannian completeness. Its diameter is infinite since its distance dominates Euclidean distance. On a surface the Ricci tensor is $Kg$, so it is strictly positive everywhere, but has no uniform positive lower bound.

There is no positive-Ricci metric on $S^2\times S^1$. If one existed, compactness of the unit tangent bundle would give ${\mathrm{Ric}}_{\mathrm{tr}}\ge cg$ for some $c>0$. The lifted metric on the universal cover is complete by Q2 and satisfies the same positive lower bound. Part (a) makes this cover compact. A covering fiber is closed and discrete and therefore finite, implying a finite fundamental group. Butwhich is infinite. This is the finite fundamental group from a uniform positive Ricci bound obstruction; it rules out every possible metric, not just the standard product metric.

Choose a complex projective line in each Complex projective plane summand. Make the connected sum at small disks on these lines, so the lines themselves join through the neck to form a smoothly embedded sphere $S\cong S^2$. Its class is the sum $h_1+h_2$ of the two line classes. The intersection form is diagonal with entries one, henceThe oriented normal bundle of $S$ is a rank-two real bundle with Euler number two, because its Euler number equals the self-intersection number. Oriented plane bundles over $S^2$ are classified by that integer, so this normal bundle is isomorphic to $T{S}^2$. The boundary of a closed tubular neighborhood $N$ is consequently the unit tangent bundle of $S^2$.

A unit tangent pair $(x,v)$ determines the oriented orthonormal frame $(x,v,x\times v)$ in ${\mathbb{R}}^3$. Thus $UT{S}^2\cong SO(3)\cong {\mathbb{RP}}^3$, the latter identification coming from the double covering by unit quaternions. Therefore $\partial N$ is the required smoothly embedded ${\mathbb{RP}}^3$. Its complement is disconnected: the nonempty interior of $N$ and the nonempty exterior $X\setminus N$ are disjoint open subsets of $X\setminus \partial N$ that exhaust it. The neighborhood can be chosen small enough that its exterior is nonempty. This is the construction of separating projective three-space in a positive connected sum.

Use the metric on $M$ induced by its given Euclidean embedding. For two smooth maps define the C1 metric on a smooth mapping space bywhere the differential has domain $T_{x}M$ with the induced inner product. Compactness makes both suprema finite. The norm properties give symmetry and the triangle inequality, and the first term makes distance zero imply equality of maps. Thus convergence in $d_1$ is uniform convergence of both maps and differentials.

Because $i$ is an immersion, $D{i}_x$ is injective at every point. The continuous function $|D{i}_{x}v|$ on the compact unit tangent bundle has a positive minimum $a$. ThereforeFor all sufficiently large $n$, ${\sup }_x\Vert D{f}_{n,x}-D{i}_x\Vert <a/2$, and thenConsequently every sufficiently late $f_n$ is an immersion. If $M$ has dimension zero it is a finite set, the immersion assertion is automatic, and eventual injectivity follows directly from separation of its finitely many image points; we henceforth treat positive dimension.

For injectivity, choose finitely many coordinate maps ${\varphi }_b$ on convex balls, with smaller coordinate neighborhoods covering $M$ and closures lying inside the larger balls. Write $I_b=i\circ {\varphi }_b$ and $F_{n,b}=f_n\circ {\varphi }_b$. On the finitely many relevant compact coordinate closures, the differentials of the $I_b$ have a common positive lower bound $s$, and convergence in $d_1$ givesfor all late $n$. The assumed uniform second-derivative bound gives a common bound $L$ for $D^2{F}_{n,b}$ on those larger coordinate neighborhoods. If the bound is expressed using an intrinsic Hessian, the coordinate chain rule also involves first derivatives and derivatives of the fixed charts; their bounds are uniform by C1 convergence and compactness. Thus $L$ can be larger than the printed constant $C$, but is independent of $n$.

For $u,v$ whose segment lies in one such coordinate ball, Taylor formula with integral remainder givesChoose $\delta >0$ with $L\delta \le s/2$. Then, whenever $0<|u-v|<\delta$,The Lebesgue number lemma applied to the finite smaller-chart cover, together with uniform continuity of its coordinate inverses on the relevant compact sets, gives $\rho >0$ such that every pair $x,y\in M$ with $0<|x-y|<\rho$ lies in one such chart with coordinate distance below $\delta$. The displayed estimate therefore excludes all sufficiently close pairs of equal images, uniformly in late $n$.

The remaining pairs form the compact setIf it is nonempty, injectivity of the smooth embedding $i$ gives a positive minimum $b$ of $|i(x)-i(y)|$ on it. Uniform convergence makes ${\sup }_x|f_n(x)-i(x)|<b/3$ for all sufficiently large $n$, and hence on $K_{\rho }$,If $K_{\rho }$ is empty there is nothing left to check. Combining the near-pair and distant-pair bounds makes $f_n$ injective for every sufficiently late $n$. It is already an immersion, and a continuous injection from compact $M$ into a Hausdorff space is a homeomorphism onto its image. By compact injective immersion is an embedding,This proves the requested stability of compact smooth embeddings using the stated second-derivative assumption. That assumption supplies a convenient Taylor constant; the stronger C1 openness result can replace it by the uniform continuity of the limiting differential on finitely many small coordinate balls.

Use the Levi-Civita connection to represent a tangent vector to the unit tangent bundle at $(x,v)$ by the initial data $(J(0),D_{t}J(0))$ of the Jacobi field generated by a variation of the unit-speed geodesic. The unit-speed constraint gives $⟨D_{t}J(0),v⟩=0$. Along the geodesic, writeHere $a$ is constant, while the perpendicular Jacobi field equation isThe flow direction is represented by $(a,y,\dot{y})=(1,0,0)$. Consequently $(y,z)=(y,\dot{y})$ are coordinates on the transverse quotient. Define its nondegenerate quadratic form by $Q(y,z)=yz$. Its signature of a quadratic form is $(1,1)$, andSince the Gaussian curvature is strictly negative and the surface is a closed manifold, there is $k_0>0$ with $K\le -k_0$. HenceThe quadratic-form criterion for an Anosov flow now proves the geodesic flow is Anosov. This is the geodesic hyperbolicity from negative Gaussian curvature argument. If the surface is not oriented, the local choice of perpendicular unit vector changes both $y$ and $z$ in sign; their product is unchanged, so the quadratic form still exists globally.

When $K=-1$, the perpendicular solutions are $y=A{e}^t+B{e}^{-t}$. The pure contracting data have $(y,z)=(B,-B)$ and the pure expanding data have $(y,z)=(A,A)$. In the canonical coframe of a surface unit tangent bundle, $H$ represents $(iv,0)$ and $V$ represents $(0,iv)$, soEquivalently the Anosov stable bundle consists of pairs $(\eta ,-\eta )$ with $\eta \perp v$, and the Anosov unstable bundle consists of $(\eta ,\eta )$. The derivative multiplies these data by $e^{-t}$ and $e^t$, respectively, using parallel transport along the geodesic to identify the perpendicular directions. Together with $\mathbb{R}X$ they give the required invariant splitting.

A simple Riemannian manifold has strictly convex boundary, no conjugate points, and a unique connecting geodesic depending smoothly on its endpoints. In particular its unit-speed geodesics are nontrapping. Write $\nu$ for the inward unit normal, and identify inward and outward parts of its boundary unit tangent bundle byThe signs are a convention; here $+$ means inward. If $(x,v)$ is strictly inward, let $\tau (x,v)>0$ be its first exit time. The geodesic scattering relation isThus it records both the exit point and the outward unit velocity, not just the exit point. On strictly outward vectors extend $\alpha$ by following the same geodesic flow backwards to its previous entry point, retaining the velocity orientation. On tangential vectors set $\alpha (x,v)=(x,v)$. The extended scattering relation is an involution on the full boundary unit bundle, since its forward and backward constructions undo one another. On a simple Riemannian manifold, the travel time equals the boundary distance function between the two endpoints; travel time is often recorded separately as lens data.

Let $\pi :SM\to M$ be the unit tangent bundle projection. Write $X$ for the geodesic flow generator and $V$ for the vertical vector field of a surface unit tangent bundle, positively rotating $v$ toward $iv$. The acceleration prescribed in the question changes only this velocity angle, so the surface thermostat generator isIt never vanishes: $d\pi (F)=v\ne 0$. Also $F$ and $V$ are linearly independent, since $d\pi (V)=0$.

The vertical Maslov cycle of a surface thermostat isThere is exactly one such plane in each fibre, so ${\Lambda }_V$ is the image of a smooth section. A two-plane containing $F$ corresponds to a line in the rank-two quotient $T(SM)/\mathbb{R}F$; thus $\Lambda (SM)$ is a projective bundle with fibre ${\mathbb{RP}}^1$. Its total dimension is four and the cycle has dimension three, making it a smooth hypersurface. In this two-dimensional transverse space the Maslov cycle means the projective line meeting the distinguished vertical line, which here is just that distinguished line itself in each fibre.

For a nonsingular smooth flow generated by $F$ on a closed manifold, the Anosov flow condition is a continuous $D{\varphi }_t$-invariant splittingand constants $C,\lambda >0$ such that, for all $t\ge 0$,The norm can come from any auxiliary Riemannian metric; compactness makes the resulting definitions equivalent. The centre direction is the flow direction, with $F$ nowhere zero.

One construction is a suspension flow. The hyperbolic toral automorphism of ${\mathbb{T}}^2$ defined byhas eigenvalues ${\lambda }_{\pm }=(3\pm \sqrt{5})/2$, with $0<{\lambda }_{-}<1<{\lambda }_{+}$. Form the closed three-dimensional mapping torus $({\mathbb{T}}^2\times [0,1])/((x,1)\sim (Ax,0))$ and move in the second coordinate at unit speed. Each crossing of the section applies $A$. The two one-dimensional eigenspaces give invariant transverse bundles, and after $k$ crossings their derivatives multiply by ${\lambda }_{-}^k$ and ${\lambda }_{+}^k$. The difference between $k$ and elapsed time is bounded, so any smooth Riemannian metric gives the required uniform exponential estimates. The vertical direction supplies the centre line.

A different construction is the geodesic flow on the unit tangent bundle of a closed oriented hyperbolic surface. This bundle is a closed three-manifold. In curvature $-1$, a normal Jacobi field along a unit-speed geodesic satisfies $j^{\prime \prime }-j=0$. The initial data $(j,j^{\prime })=(a,-a)$ give $j(t)=a{e}^{-t}$, while $(a,a)$ give $j(t)=a{e}^t$. The linearized geodesic flow transports these initial data as $(j(t),j^{\prime }(t))$, so they give one stable and one unstable direction with precisely those exponential rates. The remaining tangent direction is the geodesic vector field. Thus this also constructs an Anosov flow, using geometric rather than return-map hyperbolicity.

The geodesic flow of a closed negatively curved surface is a topologically transitive Anosov flow. In particular, its Jacobi-field hyperbolicity and the usual negative-curvature transitivity theorem allow the smooth Livsic theorem to be applied. Set $b(x,v)={\beta }_x(v,v)$. The given closed-geodesic integrals are exactly its periodic-orbit obstructions on $SM$. Therefore

We make the differential operators and signs explicit. On the unit tangent bundle, let $V$ generate positive fibre rotation, $X$ be the geodesic vector field and $H=[V,X]$. For the canonical coframe of a surface unit tangent bundle, writeand let $\omega$ be the Levi-Civita connection form with $\omega (V)=1$. This uses $\theta$ for a one-form, keeping $\beta$ for the given tensor. The dual frame is $(X,H,V)$, and the structure equations areThese follow by writing the two horizontal forms as a rotated local orthonormal coframe and $\omega =d\vartheta +{\omega }_0$, with ${\omega }_0$ its connection 1-form. Equivalently, differentiating the sine and cosine components of $X$ twice in the fibre angle givesThe curvature equation also gives $[X,H]=KV$. The Liouville volume of a surface geodesic flow will be taken positive in the base/positive-fibre orientation, namely $\mu =\alpha \wedge \theta \wedge \omega$.

In an oriented orthonormal frame at $x$, put $v=(\cos \vartheta ,\sin \vartheta )$. The symmetric tensor evaluation isSince $V={\partial }_{\vartheta }$, this proves $(V^3+4V)b=0$.

The Lie bracket of vector fields identities give an operator identity that supplies the required equation. First,Using $VH=HV-X$ and $XV=VX-H$, differentiate this expression once more to obtainFor $\phi =V^{2}u+u$ and $Xu=b$, the first term vanishes by the explicit fibre calculation. Hence

The Riemannian distance on a connected manifold iswhere the infimum is over piecewise smooth curves joining $p$ to $q$. A connected manifold is path connected, so this infimum is finite. The metric is positive definite and induces the manifold topology; locally this follows from comparison with the Euclidean metric in a coordinate chart.

Geodesic completeness means that every geodesic with prescribed initial point and velocity extends to all parameter values in $\mathbb{R}$. The Hopf-Rinow theorem says that, for a connected finite-dimensional Riemannian manifold without boundary, this is equivalent to completeness of the distance metric and equivalent to compactness of every closed bounded set. It is also equivalent to the exponential map at one, and hence every, point being defined on the whole tangent space. Under these equivalent conditions, any two points are joined by a distance-minimizing geodesic.

A homogeneous Riemannian manifold has an isometry group acting transitively on its points. Fix $o\in M$. Local compactness and the local metric-topology comparison give an $r>0$ such that $\overline{B}(o,r)$ is compact: choose a relatively compact coordinate neighborhood of $o$, then a small closed metric ball contained in it. Homogeneity makes $\overline{B}(p,r)$ isometric to this fixed compact ball for every $p\in M$.

Let $(p_j)$ be a Cauchy sequence. Its tail lies inside $\overline{B}(p_N,r)$ for some $N$. Compactness gives a convergent subsequence, and the Cauchy property forces the whole sequence to converge to the same limit. Thus the distance metric is complete. Hopf-Rinow now gives every homogeneous Riemannian manifold is geodesically complete. This is the completeness of homogeneous Riemannian manifolds; the compact ball used here is local and does not presuppose global completeness.

A two-point homogeneous Riemannian manifold has isometries acting transitively on ordered pairs at each fixed distance. We prove the unit tangent transitivity characterizes two-point homogeneity equivalence. First suppose two-point homogeneity holds. Applying it to pairs with identical endpoints gives ordinary homogeneity. Given $p,q$ and unit tangent vectors $u,v$, choose $a>0$ sufficiently small that $au$ and $av$ lie in normal neighbourhoods at $p$ and $q$. The Gauss lemma and local minimizing property giveThere is therefore an isometry $f$ taking these ordered pairs to one another. Isometries preserve the Levi-Civita connection and geodesics, soBoth vectors on the right lie in the injectivity neighborhood of ${\exp }_q$, because $d{f}_p$ preserves length. It follows that $d{f}_{p}u=v$. Thus isometries act transitively on the unit tangent bundle.

Conversely, suppose the stated transitivity on unit tangent vectors holds. In positive dimension it implies point homogeneity, and hence completeness by the preceding argument. Consider two ordered pairs at the same distance $a>0$. Hopf-Rinow supplies unit-speed minimizing geodesics ${\gamma }_1,{\gamma }_2:[0,a]\to M$ joining their respective endpoints. Choose an isometry taking ${\gamma }_1(0)$ to ${\gamma }_2(0)$ and ${\dot{\gamma }}_1(0)$ to ${\dot{\gamma }}_2(0)$. Then $f\circ {\gamma }_1$ and ${\gamma }_2$ solve the same geodesic initial-value problem, so they agree throughout $[0,a]$. In particular $f$ takes both endpoints of the first pair to those of the second. For $a=0$, point homogeneity suffices. If the connected manifold has dimension zero, it is a single point and both properties are immediate. Hence two-point homogeneity is equivalent to transitivity on unit tangent vectors. Completeness is the step allowing the local initial-direction condition to control pairs at arbitrary distance.

For unit vectors, the defining inequality for $Y$ is equivalent to $v\cdot w\ge 0$. WriteThen $t=(1-\Vert u{\Vert }^2{)}^{1/2}$, sois a homeomorphism $D(T{S}^{2n})\to Y$. Replacing $v$ by $-v$ gives the same description of $Y^{\prime }$. Their intersection isthe unit tangent bundle, equivalently a Stiefel manifold. Since $e(T{S}^{2n})$ evaluates to $\chi (S^{2n})=2$, its Gysin sequence gives the cohomology of the unit tangent bundle of an even-dimensional sphere:Every product of positive-degree classes vanishes.

Let $G$ consist of the eight signed permutations of the two factors:It is the group of signed permutation matrices in dimension two, hence isomorphic to the dihedral group $D_8$, and every element preserves the equation $v\cdot w=0$. If $a,b$ generate $H^{2n}(X;\mathbb{Z})$ from the two sphere factors, these maps act by the corresponding signed permutation matrix, because the antipodal map of the even-dimensional sphere has degree $-1$. The eight actions are distinct, so only the identity can be homotopic to $1_X$.

Every element of $G$ acts trivially on $H^{\ast }(Z;\mathbb{Z})$. The degree-$2n$ group is $\mathbb{Z}/2$, so its automorphism is forced to be the identity. On the top class, regard $Z$ as the Stiefel manifold of two-frames. The signed permutations are the right action of $O(2)$. The determinant-one component is connected, while a determinant-minus-one element is homotopic within that component to $(v,w)\mapsto (v,-w)$, the antipodal map on the $S^{2n-1}$ fibers. That map has degree $+1$, so the top class is also fixed.

The subgroup preserving $Y$ iswhere $A(v,w)=(-v,w)$, $B(v,w)=(v,-w)$, and $S(v,w)=(w,v)$. It consists exactly of the elements with ${\epsilon }_1{\epsilon }_2=1$. Under the deformation retraction $Y\simeq S^{2n}$ onto the diagonal, $1$ and $S$ act as the identity, while $AB$ and $ABS$ act as the antipodal map. They therefore act on $H^{2n}(Y;\mathbb{Z})$ by $+1,+1,-1,-1$, respectively.

The Thom isomorphism theorem identifies $H^{\ast }(Y,Z)$ with a degree-$2n$ shift of $H^{\ast }(S^{2n})$. On the Thom class in degree $2n$, the same four elements act by $+1,+1,-1,-1$: the factor swap reverses each normal vector, which preserves the orientation because the normal rank $2n$ is even, while the simultaneous antipodal map reverses the oriented tangent fiber. On the degree-$4n$ relative class every element acts by $+1$, since the base and fiber signs for the last two elements cancel.

When $n=1$, $Z=V_2({\mathbb{R}}^3)\cong SO(3)$. Every signed permutation of $(v,w)$ extends, after sending the third frame vector $v\times w$ to the required sign, to right multiplication by an element of $SO(3)$. Since $SO(3)$ is path-connected, every right translation is homotopic to the identity. Therefore

Tube the projective lines in the two summands of ${\mathbb{CP}}^2\#{\mathbb{CP}}^2$ through the connecting neck. Their connected sum is an embedded two-sphere in the class $h_1+h_2$, with self-intersection two. Its oriented normal disk bundle has Euler number two, so its boundary is the unit tangent bundle of $S^2$, namely $SO(3)\cong {\mathbb{RP}}^3$. The boundary separates the interior of the tubular neighborhood from its nonempty exterior.

// Destination: geometry-and-topology.bigb

The real Stiefel manifold $V_k({\mathbb{R}}^n)$ consists of ordered orthonormal $k$-frames in ${\mathbb{R}}^n$. In particular, $V_2({\mathbb{R}}^{m+1})$ is naturally the unit tangent bundle of $S^m$.

The transform integrates a base function and a velocity-evaluated one-form along the closed trajectories of a surface thermostat. Every exact differential form of degree one integrates to zero, because $F(f\circ \pi )=df(v)$. The relevant kernel theorem concerns these functions of velocity degree zero and one, rather than arbitrary functions on the unit tangent bundle.

A connected Riemannian manifold is two-point homogeneous if its isometry group is transitive on ordered pairs at each fixed Riemannian distance. The equal-distance condition is necessary because isometries preserve distance. Round spheres and Euclidean spaces are examples. The unit tangent transitivity characterizes two-point homogeneity lemma shows that this pair condition is equivalent to transitivity on the unit tangent bundle.

For a connected Riemannian manifold, two-point homogeneity is equivalent to the isometry group being transitive on the unit tangent bundle. One direction follows by taking short equal-length radial geodesic segments: the Gauss lemma identifies their distance, and injectivity of the exponential map identifies the initial directions after the endpoints are matched. Conversely, unit tangent transitivity implies point homogeneity, hence completeness. The Hopf-Rinow theorem supplies minimizing geodesics for arbitrary equal-distance pairs. Matching their initial unit tangent vectors and using uniqueness of the geodesic equation matches their other endpoints.

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 $[X,H]=KV$.