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For a Riemannian metric $g$, write $s(t)=\sqrt{g_{\gamma (t)}(\dot{\gamma }(t),\dot{\gamma }(t))}$. Use the energy of a curve normalization compatible with the requested inequality:The Cauchy-Schwarz inequality here has the direct variance formEquality holds exactly when $s$ is constant almost everywhere, hence everywhere for a smooth curve. With the alternative energy of a curve convention containing a factor $1/2$, the same assertion reads $L^2\le 2E$.

In the upper half-plane write $\gamma =u+iv$, $v>0$, and use $d{s}^2=(d{u}^2+d{v}^2)/v^2$. Let the endpoints be $P=ip$, $Q=iq$, $p,q>0$. ThenThe first equality requires $\dot{u}=0$ throughout. Endpoint values force $u=0$. The second equality requires the derivative of $\log v$ to have one sign, so $v$ is monotone. Conversely every smooth monotone positive $v$ with these endpoints attains this bound. Consequently the absolute length minimizers are precisely the vertical curves with monotone height, including a constant curve when $P=Q$, and the distance is $|\log (q/p)|$.

For curves stationary for the energy of a curve, fixed-endpoint variations give the Euler–Lagrange equations for $F=({\dot{u}}^2+{\dot{v}}^2)/v^2$:Thus $\dot{u}=c{v}^2$ for a constant $c$. Since $u(1)-u(0)=0$, integration gives $c{\int }_0^1{v}^{2}dt=0$, hence $c=0$. The second equation then becomes $(\dot{v}/v{)}^{\prime }=0$. ThereforeIt has constant hyperbolic speed, is one of the absolute length minimizers, and also has minimum energy of a curve $E=[\log (q/p){]}^2$.

Use the Levi-Civita connection and the Riemann curvature tensor conventionfor orthonormal $T,E$. Thus the round sphere has positive sectional curvature. Work on a fixed compact parameter interval $[a,b]$ inside the given open interval; this keeps all differentiations and endpoint evaluations well-defined. The base geodesic is regular and has constant speed, so a constant rescaling of $t$ makes it unit-speed. Put $T={\partial }_t\varphi$, $J={\partial }_s\varphi$, $D_t={\nabla }_T$, $D_s={\nabla }_J$ and evaluate at $s=0$. The torsion-free connection gives $D_{s}T=D_{t}J$ because the two parameter derivatives commute. The geodesic equation is $D_{t}T=0$.

Differentiating the length of a curve once and twice, using metric compatibility, givesandLet $A=D_{s}J$. The curvature commutator gives $D_s{D}_{t}J=D_{t}A+\mathcal{R}(J,T)J$, and the skew-adjoint curvature symmetry gives $⟨\mathcal{R}(J,T)J,T⟩=-⟨\mathcal{R}(J,T)T,J⟩$. Integration by parts therefore proves the moving-endpoint second variation of Riemannian arc length:Equivalently, with $W=J-⟨J,T⟩T$, the integral is $I(W,W)$, the Riemannian index form. The endpoints of this geodesic variation generally move, so the displayed boundary term must be retained. For a base speed $c$ instead of one, the boundary and integral have an overall factor $1/c$, and the subtracted square inside the integral is $⟨D_{t}J,T{⟩}^2/c^2$. Returning to the original parameter, every ruling has constant speed, so $L_{[a,b]}(s)=(b-a)|{\partial }_t\varphi (0,s)|$ and the full open-interval length is $2\delta |{\partial }_t\varphi (0,s)|$. ConsequentlyThe preceding curvature formula, with the base-speed factor $1/c$ when necessary, therefore evaluates the original full length derivative using any compact subinterval. No extension to the missing endpoints is required.

Differentiating $D_{t}T=0$ in $s$ yields the Jacobi field equationThe same ruling curves are geodesics of the surface with its induced metric, because the tangential part of their ambient acceleration is zero. Write its second fundamental form as $II$ and its induced Levi-Civita connection as $D^N$. In particular $II(T,T)=0$. The scalar $f=⟨J,T⟩$ satisfies $f^{\prime \prime }=0$, by the Jacobi field equation, so $W=J-fT$ is a Jacobi field both intrinsically and in the ambient manifold. In a regular surface chart $h=|W|>0$, and $E=W/h$ is the perpendicular unit tangent. Along a two-dimensional geodesic, $D_t^{N}E=0$: its derivative is orthogonal to both $E$ and the parallel field $T$. The intrinsic Jacobi field equation consequently givesFor the ambient derivative, the Gauss formula gives $D_{t}W=h^{\prime }E+hII(T,E)$. Applying the same norm differentiation used above to $h=|W|$ givesComparing these identities proves the curvature comparison for a geodesically ruled surface:This is also the Gauss equation in a curved ambient manifold with $II(T,T)=0$. It applies locally to each regular patch of the surface.

For strict inequality, take the Euclidean embedded submanifold parametrized by $\varphi (t,s)=(t,s,ts)$. Each $t$-curve is a straight-line geodesic. Its first fundamental form has coefficients $E_0=1+s^2$, $F_0=st$, $G_0=1+t^2$; the normal $(-s,-t,1)/\sqrt{1+t^2+s^2}$ gives second fundamental form coefficients $e=0$, $f_0=1/\sqrt{1+t^2+s^2}$, $g_0=0$. HenceThus equality can fail everywhere, although every ruling is an ambient geodesic.

The length differentiation requires a nonconstant base ruling; a regular parametrized surface patch guarantees this. If “image of a smooth map” is read literally without a rank assumption, a degenerate member can fail even to have a first length derivative. For example, in the flat cylinder $\mathbb{R}\times S^1$, the map $\varphi (t,s)=(st,e^{is})$ for $|t|<\delta$, $|s|<4\pi$ has geodesic rulings and a two-dimensional open image: every nonzero-$s$ point has full-rank parametrization, and the sole zero-$s$ image point also has a full-rank preimage at $s=2\pi ,t=0$. Yet $L({\gamma }_s)=2\delta |s|$ is not differentiable at zero. The variation formula therefore applies to nonconstant members, as usual. The curvature inequality still holds on the whole image: regular values are dense by Sard theorem, and both curvatures extend continuously to the remaining points.

For $p,q\in M$, define the Riemannian distance byA connected smooth manifold is locally path connected, so any two points are joined by a finite concatenation of coordinate paths. Thus the infimum is finite. It is nonnegative, $d_g(p,p)=0$, and reversing a curve proves symmetry. Concatenating two curves whose lengths approach their respective infima proves the triangle inequality.

For positivity and the topology, choose a coordinate ball about $p$ whose closure is contained in a chart. On its closure there are constants $0<c<C$ withA curve staying in the ball has length at least $c$ times its coordinate displacement. A curve leaving it must first reach its boundary, costing at least $c$ times the positive coordinate distance from $p$ to that boundary. Therefore a distinct $q$ has $d_g(p,q)>0$, whether it lies inside or outside the ball. For $q$ sufficiently close to $p$, the coordinate line segment gives $d_g(p,q)\le C|x(q)-x(p)|$, while the preceding lower bounds prevent a short curve from taking a shortcut outside the chart. These inequalities prove that the Riemannian distance induces the manifold topology. In particular it defines a genuine metric space.

A Riemannian manifold is geodesically complete if every maximal affinely parametrized geodesic exists for all real parameter values. We prove both implications of the Hopf-Rinow theorem rather than assuming global distance minimizers.

First assume completeness as a metric space. Let a constant-speed geodesic have a finite maximal right endpoint $b$. The bound $d_g(\gamma (t),\gamma (s))\le c|t-s|$ makes $\gamma (t)$ a Cauchy sequence as $t\uparrow b$, hence gives a limit $p\in M$. The distance topology just established puts the final portion inside a relatively compact coordinate ball about $p$. The positive lower bound on the Riemannian metric bounds all coordinate velocity components. The Christoffel symbols are bounded on the closure of that ball, and the geodesic equationthen bounds the coordinate acceleration. The velocity has a limit $v$ as $t\uparrow b$. Local existence and uniqueness for the geodesic equation with data $(p,v)$ continue $\gamma$ through $b$, contradicting maximality. The same argument applies at a finite left endpoint. Hence metric completeness implies geodesic completeness.

Conversely assume geodesic completeness. Fix $p,q$, put $r=d_g(p,q)>0$, and choose $0<ϵ<r$ so small that a closed radius-$2ϵ$ normal ball about $p$ lies inside a convex normal neighborhood. By the Gauss lemma, its radial geodesics minimize; a curve leaving the larger ball cannot improve on a radius-$ϵ$ radial segment. Thus the radius-$ϵ$ geodesic sphere $S_{ϵ}(p)$ is compact and consists of points at distance exactly $ϵ$ from $p$.

The distance splitting through a small geodesic sphere has an elementary proof here. Every path from $p$ to $q$ must cross $S_{ϵ}(p)$, and its initial portion has length at least $ϵ$. HenceThe triangle inequality proves the reverse bound. Continuity of $d_g(\cdot ,q)$ and compactness give a point $x$ achieving the minimum, so $d_g(x,q)=r-ϵ$. Let $\gamma (t)={\exp }_p(tv)$ be the complete unit-speed radial geodesic through $x$ at time $ϵ$.

Consider the closed subsetIt is nonempty since $ϵ\in A$, and has a largest member $t_{\ast }$. Suppose $t_{\ast }<r$. At $x_{\ast }=\gamma (t_{\ast })$, choose a small normal sphere of radius $\eta <r-t_{\ast }$ and repeat the splitting argument. It supplies a radial endpoint $y$ withThe triangle inequality gives $d_g(p,y)\ge t_{\ast }+\eta$, while the broken path following $\gamma$ to $x_{\ast }$ and then the short radial segment to $y$ has exactly that length. It is therefore minimizing. Its two velocities at the joining point must agree: if unit incoming and outgoing velocities are $u,w$, moving the join in direction $w-u$ changes the sum of lengths by $-|u-w{|}^2$ to first order, using the first variation of geodesic energy or the equivalent unit-speed length formula. A corner would strictly shorten it. Uniqueness of the geodesic equation consequently gives $y=\gamma (t_{\ast }+\eta )$, contradicting the maximality of $t_{\ast }$. Thus $t_{\ast }=r$, $\gamma (r)=q$, and $\gamma {|}_{[0,r]}$ is a minimizing geodesic.

This radial continuation proof of Hopf-Rinow also proves compactness of closed balls. For every $R\ge 0$,The inclusion from right to left follows from the length of a radial geodesic; the reverse inclusion uses the minimizing geodesic just constructed. The Riemannian exponential map is defined on all of $T_{p}M$ by geodesic completeness, and the closed tangent ball is compact in a finite-dimensional vector space. Its continuous image is compact. Every Cauchy sequence is bounded, hence has a convergent subsequence in such a ball, and the Cauchy property makes the whole sequence converge. Therefore geodesic completeness implies metric completeness, completing both directions.

For the hyperbolic plane, use the upper half-plane model with $g=(d{x}^2+d{y}^2)/y^2$, $y>0$. Its geodesic equation isEvery nonconstant unit-speed solution is either a vertical curve $x=a$, $y=b{e}^{\pm t}$ or, after translating and possibly reversing $t$,Substitution verifies both equations and $g(\dot{\gamma },\dot{\gamma })=1$. These curves cover every unit initial tangent: a nonvertical tangent at $(x,y)$ determines the circle center $a=x+y{y}^{\prime }/x^{\prime }$ and radius $R=\sqrt{(x-a{)}^2+y^2}$, and its orientation determines the sign of the parameter. Constant solutions and constant-speed rescalings cover arbitrary initial velocities. In every case $y(t)>0$ for all finite $t$, and the solution is defined for every real $t$. Thus the real hyperbolic plane is geodesically complete, directly, and the proved Hopf-Rinow theorem also makes its distance complete.

Use the curvature convention $\mathcal{R}$ of Solution 1, and put $T=\dot{\gamma }$. On continuous piecewise smooth perpendicular vector fields with zero endpoint values, the Riemannian index form is the symmetric bilinear formA Jacobi field is a smooth field $J$ along $\gamma$ satisfying $D_t^{2}J+\mathcal{R}(J,T)T=0$. The point $\gamma (t_0)$ is a conjugate point to $\gamma (0)$ along this geodesic when there is a nonzero Jacobi field with $J(0)=J(t_0)=0$. Such a field is perpendicular to $T$, because $⟨J,T⟩$ is affine and vanishes at both endpoints.

The conjugate-point criterion for the Riemannian index form, stated without proof, isAt a first conjugate endpoint the form is nonnegative but has a nonzero nullspace, consisting of endpoint-vanishing Jacobi fields. A conjugate point strictly inside gives a negative direction. This explicitly distinguishes the non-strict and strict senses of positivity in the question; in terminology where “positive definite” already means strict positivity, the second line is the positive-definiteness condition.

Choose $n-1$ perpendicular orthonormal fields $E_1,\dots ,E_{n-1}$ by parallel transport along $\gamma$, and put $V_i(t)=\sin (\pi t/L)E_i(t)$. They are admissible andThe tangent has unit length. The assumed Ricci curvature bound and the two elementary sine-square and cosine-square integrals give the sine index-form bound for positive Ricci curvature:Since $n\ge 2$, at least one of the admissible $V_i$ therefore has $I(V_i,V_i)<0$.

If the Riemannian manifold is complete, the Hopf-Rinow theorem supplies a minimizing geodesic between any two points. Its Riemannian index form is nonnegative: any smooth fixed-endpoint variation has nonnegative second variation of geodesic energy, since a length minimizer of constant speed also minimizes energy on a fixed parameter interval by the Cauchy-Schwarz inequality. The same conclusion for piecewise smooth fields follows by smooth approximation, or by piecewise variations. The negative direction above excludes a minimizing segment longer than $\pi /\sqrt{\kappa }$. Taking the supremum over pairs yields the diameter conclusion of the Bonnet-Myers theorem:

The completeness assumption is essential. An explicit incomplete positively curved strip with infinite diameter isThe map $(u,v)\mapsto (\cos u\cos v,\cos u\sin v,\sin u)$ is a local isometry to the unit sphere, so $K=1$ and, in dimension two, $\mathrm{Ric}=g$. Thus the required Ricci curvature bound holds with $n=2$, $\kappa =1$. The unit-speed meridian $t\mapsto (t,0)$ reaches the missing boundary $u=\pi /4$ in finite time, so the strip is geodesically incomplete. On the other hand every curve joining $(0,0)$ to $(0,A)$ hassince $\cos u\ge 1/\sqrt{2}$ throughout the strip. Therefore its intrinsic diameter is infinite, in particular larger than $\pi$. This is the unwrapped strip with $v\in \mathbb{R}$, not a strip with longitude identified modulo $2\pi$; that distinction prevents spherical shortcuts from invalidating the example.

The equality with part (i) follows from Green's theorem. For the coefficients $P=-1$ and $Q=x^2/(2y)$,They are smooth on a neighborhood of this region, since $y>0$. Thus for the positively oriented boundary,The term $-dx$ itself has zero line integral around the closed curve, although it contributes to the four individual edge integrals.

Divide the velocity by the speed calculated above. The unit tangent vector isIts derivative isBecause $ds/dt=\sqrt{2}(1+t^2)>0$, the principal normal vector has the direction of $T^{\prime }(t)$, givingThis is a unit vector, and $N\cdot e_z=0$ for every real $t$. The curvature of a space curve is $|T^{\prime }|/(ds/dt)=1/(1+t^2{)}^2>0$, so the principal normal vector is defined everywhere on this curve. The same perpendicularity follows from constant tangent component forces perpendicular principal normals: $T\cdot e_z=1/\sqrt{2}$ is constant, and differentiating this identity forces the normal component along $e_z$ to vanish.

Differentiate the parametrized curve to obtain ${\mathbf{x}}^{\prime }(t)=(1-t^2,2t,1+t^2)$. Its speed obeysSince $1+t^2>0$, the arc length is

For a parametrized curve on the surface, the energy of a curve iswhere the first fundamental form coefficients are evaluated at $(u(t),v(t))$. A fixed-endpoint variation of a curve has coordinate variation fields $\eta ,\xi$ vanishing at $a,b$. Differentiating under the integral and integrating their derivative terms by parts givesThe fields can be chosen independently with compact support in $(a,b)$. Stationarity and the fundamental lemma of the calculus of variations therefore force both coefficients to vanish. These Euler-Lagrange equations are the geodesic equations:

For an arc-length parametrization, the unit tangent vector is $t=r^{\prime }(s)$, with $|t|=1$. Its derivative is perpendicular to $t$. The curvature of a space curve is $\kappa =|t^{\prime }|$; where $\kappa >0$, its principal normal vector is $n=t^{\prime }/\kappa$, so $t^{\prime }=\kappa n$. The binormal vector $b=t\times n$ completes the positively oriented Frenet frame.

Differentiate $b\cdot b=1$ to get $b^{\prime }\cdot b=0$. AlsoThus $b^{\prime }$ has only an $n$ component. Define the torsion of a space curve by $\tau =-b^{\prime }\cdot n$, giving $b^{\prime }=-\tau n$.

Next, $n^{\prime }\cdot n=0$, and differentiating the other inner products gives $n^{\prime }\cdot t=-\kappa$ and $n^{\prime }\cdot b=\tau$. Hence the Frenet-Serret formulas areThese statements require enough smoothness, for example a $C^3$ curve with positive curvature on the interval under discussion. If $\kappa =0$, the unit tangent vector is still defined, but the standard principal normal vector, binormal vector and torsion of a space curve need not be; their existence cannot be inferred just from smoothness of the curve.

Let $F=|\mathbf{F}|>0$ and choose an orientation of a regular curve following the vector field. Its unit tangent vector is $t=\sigma \mathbf{F}/F$, where $\sigma =\pm 1$ is constant along a connected regular segment. Arc length differentiation along that curve is $d/ds=(\sigma \mathbf{F}/F)\cdot \nabla$. ThereforeTaking the cross product with $t$ removes the term parallel to $\mathbf{F}$:Where the Frenet frame exists, $t\times t^{\prime }=\kappa (t\times n)=\kappa b$. ThusThe printed sign corresponds to orientation along or against the field. Taking magnitudes gives the orientation-independent curvature of an integral curve of a vector field, $\kappa =|\mathbf{F}\times (\mathbf{F}\cdot \nabla )\mathbf{F}|/F^3$. This scalar formula also applies at zero curvature, where $b$ itself may be undefined.

Set $F(z,w)=w^2-z^n+1$. Its complex gradient is $(-n{z}^{n-1},2w)$, never zero on $F=0$: simultaneous zero would give $z=w=0$, which is not on the curve. The complex implicit function theorem therefore gives a one-dimensional complex manifold, hence a Riemann surface. Where $w\ne 0$ use $z$ as coordinate, and where $z\ne 0$ use $w$; the transition maps and both projections are holomorphic.

For $\pi$, a ramification point of a holomorphic map has $w=0$ and $z=\zeta$ with ${\zeta }^n=1$. In the coordinate $w$, $z-\zeta =w^2/(n{\zeta }^{n-1})+O(w^4)$, giving ramification index $2$. Thus there are $n$ such points, and their branch points are the $n$th roots of unity.

For $\tau$, a ramification point of a holomorphic map has $z=0,w=\pm i$. In the coordinate $z$, $w-w_0=z^n/(2w_0)+O(z^{2n})$, giving ramification index $n$ at both points and branch points $\pm i$.

Write $n=2m$. At infinity set $u=1/z$, $v=w{u}^m$; then $v^2=1-u^n$. At $u=0$ there are two smooth points, $v=\pm 1$, and $u$ is a local coordinate. The extended $\pi$ has a simple pole at each, so neither is ramified. The degree-two Riemann-Hurwitz formula for the map to the Riemann sphere gives $2g-2=2(-2)+n$, henceThe list for $\tau$ above concerns the original affine Riemann surface; on the compactification its poles at the two added points have order $m$ and are additionally ramified if $m>1$.

Work over the usual algebraically closed ground field of characteristic zero. Away from $x=0$, a line of slope $t=y/x$ through the singular point meets the curve with $t^2=x-1$. ThusSubstitution verifies the defining equation. On $x\ne 0$ and $t^2+1\ne 0$, the two maps are inverse morphisms, so this is a birational map to ${\mathbb{A}}^1$. The two parameters $t=\pm i$ both map to the node $(0,0)$, explaining why the globally defined parametrization is not a global isomorphism.