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For the standard catenoid, the two principal curvatures are opposite, so its mean curvature vanishes. Its Gaussian curvature is $K=-{\mathrm{sech}}^{4}u<0$, which also shows that no neighbourhood is locally isometric to a plane, cylinder, or regular part of a cone, all of which have zero Gaussian curvature.

A local isometry from a nonempty complete Riemannian manifold to a connected Riemannian manifold is a surjective covering map. Lift radial geodesics in a normal ball using completeness; the exponential map at each inverse-image centre maps a tangent ball diffeomorphically onto a sheet. Uniqueness of the geodesic initial-value problem makes these sheets disjoint and exhaustive. Surjectivity follows by lifting piecewise geodesic paths. Conversely a Riemannian covering of a complete target is complete, by lifting complete target geodesics and using the Hopf-Rinow theorem. If the target is disconnected, the first assertion gives a covering only of the union of components met by the map.

The first fundamental form of a smooth surface $S\subseteq {\mathbb{R}}^3$ is the restriction of the ambient inner product to each tangent space $T_{p}S$. For a regular parametric surface $X(u,v)$, it is

Here $X_u^2$ means $X_u\cdot X_u$, not a componentwise square. The coefficients form the Gram matrix of the two coordinate tangent vectors. Regularity makes this a positive-definite matrix, so $E>0$ and $EG-F^2>0$.

To calculate the length of a curve $\gamma (t)=X(u(t),v(t))$, apply the chain rule and integrate its speed:The same inner product gives the angle $\alpha$ between nonzero tangent vectors $V,W$ through $\cos \alpha =I(V,W)/\sqrt{I(V,V)I(W,W)}$, and gives the area element $dA=\sqrt{EG-F^2}dudv$ as a positive density. One compact expression therefore supplies rulers, protractors, and surface area measurements.

For the lateral surface of a circular cylinder, take the circumferential arc length $u$ rather than the angle:Direct inner products give $E=G=1,F=0$, hence $I=d{u}^2+d{v}^2$. This is exactly the Euclidean metric on the unrolled plane. For instance, a path with $u=3t,v=4t$, $0\le t\le 1$, has length $5$ on either surface. Every corresponding path retains its length, not merely this example, so sufficiently small neighborhoods retain their intrinsic distances. The local isometry preserves angles and surface areas too. Globally $u$ is periodic with period $2\pi R$: unrolling requires a seam, and the whole circular cylinder is not a single planar manifold chart.

On a connected regular radial interval, trivial holonomy around circular fibres forces the continuous function $f^{\prime }$ to be integer-valued, hence constant. The resulting Riemannian metrics have $f(r)=nr+b$, with $n\in \mathbb{Z}$ and $f>0$ on the interval. Their Gaussian curvature is $-f^{\prime \prime }/f=0$. For $n\ne 0$, $\rho =f/|n|$ and $\theta =|n|\varphi$ give a local isometry to the Euclidean plane; for $n=0$, $r,b\varphi$ give a flat cylindrical chart. The polar developing map is injective on the regular domain when $|n|=1$, and is a multiple covering when $|n|>1$. Trivial holonomy therefore does not imply a global Euclidean chart. The slope is a discrete parameter, not an unrestricted real one.

Rolling a sheet into a circular cylinder bends it without changing its internal lengths and angles. Fitting that same sheet smoothly onto a sphere requires distortion of its first fundamental form. Gaussian curvature distinguishes these situations: it measures the intrinsic curvature of a smooth surface, even though its simplest definition uses how the surface bends in Euclidean space.

The Gaussian curvature at a point is the determinant of the shape operator, or the product of the two principal curvatures:Here $I=Ed{u}^2+2Fdudv+Gd{v}^2$ is the first fundamental form and $II=ed{u}^2+2fdudv+gd{v}^2$ is the second fundamental form, with $S=-dN$. Regularity makes $EG-F^2>0$. Reversing the unit normal negates $e,f,g$ and both principal curvatures, leaving Gaussian curvature unchanged. ^[ref]

The sign describes bending at the point: positive Gaussian curvature means the principal curvatures have the same nonzero sign, as at an elliptic point; negative Gaussian curvature means opposite signs, as on a saddle; zero Gaussian curvature means at least one principal curvature vanishes. Zero Gaussian curvature at a single point does not imply that a neighborhood is a plane. ^[ref]

Four worked models make the signs concrete:

The patches use a common length unit, with $R=1$ for the sphere and circular cylinder. The marked red and blue curves have principal directions at the black point; the arrow is the chosen unit normal. For physical lengths, principal curvature and mean curvature have units of inverse length, while Gaussian curvature has units of inverse squared length. Rescaling a surface by $\lambda >0$ divides its principal curvatures by $\lambda$ and its Gaussian curvature by ${\lambda }^2$. Writing the saddle as $h=(x^2-y^2)/(2L)$ restores an explicit length scale: $K(0)=-1/L^2$.

The Theorema Egregium says that Gaussian curvature depends only on the first fundamental form and its first two coordinate derivatives. A local isometry therefore preserves Gaussian curvature, even if the shape operator and the principal curvatures change. ^[ref]

Here is the outline, rather than a full proof. The Gauss formula splits second derivatives of a parametric surface into tangential Christoffel symbols and the second fundamental form. The Christoffel symbols are determined by the first fundamental form and its first derivatives. Commuting the next partial derivatives gives the Gauss equation, which identifies $\det S$ with the intrinsic sectional curvature computed from the Riemann curvature tensor. Thus the apparently extrinsic product $k_1{k}_2$ is determined by the Riemannian metric. ^[ref]

One can see the distinction directly by rolling a plane strip with coordinates $(u,v)$ into the circular cylinder $X(u,v)=(R\cos (u/R),R\sin (u/R),v)$. Both have $I=d{u}^2+d{v}^2$, so the map is a local isometry, and is one-to-one on an open strip of width less than $2\pi R$. Extending $u$ over all real numbers wraps around periodically; it does not give a global isometry between the entire plane and the entire circular cylinder. A nonempty open patch of a sphere cannot be flattened by a local isometry into a plane, because its Gaussian curvature $1/R^2$ cannot become zero. The obstruction concerns preservation of the first fundamental form, rather than a particular drawing or choice of seam.

An isometric immersion is an immersion $f:M\to N$ between Riemannian manifolds whose derivative preserves the inner products of tangent vectors; equivalently, $f^{\ast }{g}_N=g_M$. Its derivative is injective and the target may have larger dimension. A local isometry additionally requires a local diffeomorphism, so its tangent maps are linear isomorphisms. For example, a sphere with its induced Riemannian metric is isometrically immersed in Euclidean space, although its positive intrinsic sectional curvature differs from that of the ambient space.

Two local isometries from a connected Riemannian manifold that have the same value and differential at one point agree everywhere. They preserve geodesics, so their equality follows on a normal neighborhood from $f({\exp }_{p}v)={\exp }_{f(p)}(d{f}_{p}v)$. The set of points where both value and differential agree is consequently open, and continuity makes it closed. Connectedness completes the argument.

The helicoid $H(s,v)=(\sinh s\cos v,\sinh s\sin v,v)$ and catenoid $C(s,v)=(\cosh s\cos v,\cosh s\sin v,s)$ have the same first fundamental form ${\cosh }^{2}s(d{s}^2+d{v}^2)$. Therefore matching equal parameter pairs gives a local isometry. The result is local: a full turn changes the height on the helicoid while the angular coordinate on the catenoid repeats. Equal intrinsic lengths do not require the same embedding into Euclidean space.

A smooth Riemannian metric with no nonidentity local isometry between open subsets. Equivalently distinct nonempty open subsets never have isometric induced metrics. This strong local rigidity is also called bumpy in some spectral-geometry treatments; it is distinct from the closed-geodesic nondegeneracy convention for bumpy metrics.

Parametrize a nonconstant hyperbolic geodesic by unit speed. Its energy is $({\dot{x}}^2+{\dot{y}}^2)/(2y^2)$, and the Euler-Lagrange equation in the cyclic coordinate $x$ gives $\dot{x}/y^2=c$. Unit speed gives ${\dot{x}}^2+{\dot{y}}^2=y^2$. If $c=0$ the curve is vertical. Otherwise, on either side of a possible highest point,The two branches join smoothly at the top. Conversely these curves with their unit-speed parametrizations satisfy the energy equations. Thus the unparametrized complete geodesics are vertical half-lines and upper semicircles centered on the real axis.

Let the tractrix point have positive height $r(\theta )$. Its tangent segment to the axis has length one and makes angle $\theta$, so $r=\sin \theta$. Since its slope is $\tan \theta$, $h^{\prime }(\theta )={\cos }^2\theta /\sin \theta$ for its axial coordinate. Rotating $X(\theta ,v)=(h(\theta ),\sin \theta \cos v,\sin \theta \sin v)$ gives orthogonal coordinate vectors with squared lengths ${\cot }^2\theta$ and ${\sin }^2\theta$. Hence the stated first fundamental form follows.

Now put $x=v$, $y=\csc \theta$. Direct substitution givesOn the specified intervals this coordinate map is a diffeomorphism onto the stated rectangle, so is an isometry there and a local isometry on the tractoid.

Set $b=\csc \beta$, $a=\csc \alpha$. In the unwrapped coordinate, connect $(x_0,b)$ to $(x_0+2\pi ,b)$ along the semicircle centered at $(x_0+\pi ,0)$ of radius $\sqrt{b^2+{\pi }^2}$. Its height lies between $b$ and that radius. The hypothesis $a^2-b^2>{\pi }^2$ puts the whole arc in the closed height strip $b\le y\le a$. Projecting the angular coordinate modulo $2\pi$ therefore gives a geodesic segment in the closed tractoid annulus which winds once and returns to its starting point. Taking $x_0$ to represent any chosen angular position proves the assertion for every point of its $\theta =\beta$ boundary. The returned tangent need not equal the starting tangent, so this proves the claimed returning segment without asserting a periodically smooth closed geodesic.

Two regular coordinate patches are locally isometric under a coordinate correspondence exactly when it preserves their first fundamental forms. More explicitly, a diffeomorphism $\Phi :U\to \tilde{U}$ of parameter domains induces a local isometry if and only ifIn common coordinates this means equality of the three coefficient functions $E,F,G$. The criterion follows because the first fundamental form gives the inner product of every pair of tangent vectors. Equality therefore preserves lengths of all curves; conversely a smooth isometry preserves tangent-vector lengths and their inner products by polarization.

Using the two first fundamental forms calculated below, chooseThe surface of revolution is then a catenoid, with metric ${\cosh }^{2}s(d{s}^2+d{v}^2)$. On the helicoid, take $u=\sinh s/a$, assuming $a\ne 0$ so that its parametrization is regular. Substitution givesThus the coordinate mapis a local isometry on sufficiently small patches. The constant $a$ only changes the radial parametrization of the given helicoid, whose pitch is fixed at one. For $a=0$ the displayed helicoid parametrization is not a regular surface.

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 $-\mathrm{Area}(X)=2\pi \chi (X)=-4\pi$.

Consider the subsetIt is nonempty by the given initial data. A local isometry preserves the Levi-Civita connection and therefore affinely parametrized geodesics. On a sufficiently small normal neighborhood of $x\in A$, this givesThe right sides agree, so $f_1=f_2$ on that neighborhood and their differentials agree there as well. Consequently $A$ is open.

It is also closed. If $x_j\in A$ converges to $x$, continuity first gives $f_1(x)=f_2(x)$. Choose coordinate charts around $x$ and this common image. The coordinate matrices of the two differentials depend continuously on the base point; their equalities at $x_j$ therefore pass to $x$. Thus $x\in A$. Since $M$ is connected, the nonempty subset $A$ that is both open and closed equals $M$. Hence $f_1=f_2$ everywhere. This proves local isometries are determined by first-order data without requiring completeness.

Let $\pi :\tilde{M}\to M$ be the universal cover and equip it with the pullback Riemannian metric $\tilde{g}={\pi }^{\ast }g$. The covering map is a local isometry, so $\tilde{M}$ has constant sectional curvature one.

We must check completeness of the lifted Riemannian metric. A geodesic on $\tilde{M}$ projects to a geodesic on $M$. By geodesic completeness of $M$, the projected geodesic extends for all real time. Its unique lift with a specified initial point extends the original lifted geodesic, because $\pi$ is a local isometry. Hence $\tilde{M}$ is geodesically complete, and the Hopf-Rinow theorem also gives metric completeness.

The allowed constant-curvature form of Cartan-Ambrose-Hicks theorem says that a linear tangent-space isometry between complete simply connected Riemannian manifolds of the same constant sectional curvature extends to a global isometry. The unit round sphere $S^n$ is complete, simply connected for $n\ge 2$, and has sectional curvature one. Comparing it with $\tilde{M}$ therefore yieldsThis is the covering step in the classification of complete positive constant-curvature manifolds. As usual for this assertion, $n\ge 2$ is understood: in dimension one there are no sectional two-planes, and the universal cover of the circle is the line, not the circle.

Take unit-curvature Real projective space ${\mathbb{RP}}^n=S^n/\{\pm 1\}$ with $n\ge 2$, and any point $p=[v]$. It is complete with positive sectional curvature. A unit-speed geodesic through $p$ lifts toSince the two representatives of an endpoint are antipodal, Riemannian distance in the quotient is the smaller of the two spherical distances:This equals $t$ for $0\le t\le \pi /2$ and is strictly smaller just afterwards. Thus the Riemannian cut point occurs at time $\pi /2$ in every direction.

Along a curvature-one geodesic, every normal Jacobi field with initial value zero is $J(t)=\sin t{P}_{t}a$, where $P_t$ is parallel transport and $a\perp \dot{\gamma }(0)$. Its first further zero is $t=\pi$. The quotient map is a local isometry, so the Jacobi field equation and conjugacy times are unchanged. HenceThis is cut and conjugate times in round real projective space.

A connected Riemannian manifold is complete when its Riemannian distance is a complete metric space, meaning that every Cauchy sequence converges in the manifold. The Hopf-Rinow theorem says that for a connected finite-dimensional Riemannian manifold, this is equivalent to geodesic completeness, and also to compactness of every closed bounded subset. Under these equivalent conditions any two points are joined by a minimizing geodesic, and the Riemannian exponential map at every point is defined on its entire tangent space.

We use the usual connected, boundaryless-manifold convention for the global statements in Q2–Q5. Connectedness is essential, including for the diameter and group-growth conclusions. Without connectedness, the completeness and covering assertions apply to the target components met by the map; an unrelated incomplete component cannot be controlled by a local isometry into another component.

A local isometry preserves the Levi-Civita connection and hence geodesics, as well as the lengths of lifted curves. We first show that a complete source permits finite-length paths to lift all the way to their endpoints. Begin lifting a piecewise smooth path $c:[0,1]\to N$ from any chosen point above $c(0)$, using local inverse maps. If its lift is defined only up to a maximal $b\le 1$, thenCompleteness gives a limit point in $M$. A local inverse at that point continues the lift, including its endpoint. Thus such a finite obstruction is impossible. Connectedness of $N$ and piecewise smooth paths from $f(p)$ to any target point prove that $f$ is onto; no prior completeness of $N$ is used in this step.

Now lift the initial position and velocity of any geodesic in $N$. The resulting geodesic in $M$ exists for all real time by Hopf-Rinow theorem. Its image extends the original target geodesic by uniqueness of the geodesic initial-value problem. Hence $N$ is geodesically complete, and therefore complete.

The converse is false for a general local isometry, even an onto one. Give the interval $(0,4\pi )$ its ordinary metric and map it to the unit circle by $t\mapsto e^{it}$. This is an onto local isometry to a complete circle, but the sequence $t_j=1/j$ is Cauchy in the interval and has no limit there. It is not a covering map: the identity point of the circle has only one preimage, whereas nearby points have two.

The converse holds for a Riemannian covering. If $f$ is also a covering map and $N$ is complete, every target geodesic defined on $\mathbb{R}$ lifts globally with any specified initial point. A local isometry makes that lift a geodesic. Its initial velocity can be any prescribed source velocity because $df$ is an isomorphism. Thus all source geodesics extend for all time, and $M$ is complete by the Hopf-Rinow theorem.

Let $p:N\to M=U\backslash N$ be a finite normal Riemannian covering, so $U$ acts freely by deck isometries. With $x,y$ any lifts of $\bar{x},\bar{y}$, the heat kernel on a finite isometric quotient satisfiesThere is no factor $1/|U|$ in this kernel formula. Simultaneous invariance of $H_N$ under isometries proves independence of the lifts, by reindexing $U$. Each term solves the lifted heat equation. If $F_U$ is a fundamental domain and $f$ a function on $M$, the integral of the right side against $f(\bar{y})$ over $F_U$ equalsbecause the translates $u{F}_U$ tile $N$ up to sets of measure zero. This converges to $f(p(x))$ as $t\downarrow 0$, proving the initial condition. Uniqueness of the heat equation solution on the closed manifold gives the formula. For a noncompact finite cover, use the minimal heat kernels, or compatible Friedrichs heat realizations. The normalized pullback $f\mapsto |U{|}^{-1/2}f\circ p$ is unitary onto the deck-invariant $L^2$ subspace and preserves the Dirichlet energy, so it intertwines the heat semigroups and yields the same kernel formula. No uniqueness of unrestricted noncompact heat-equation solutions is needed.

Set $I_t(g)={\int }_N{H}_N(t,x,gx)dV(x)$ for an isometry $g$. Diagonal integration and the covering degree giveUnlike the kernel formula, the trace formula has an averaging factor: integration of a quotient function over $N$ is $|U|$ times integration over $U\backslash N$.

Suppose now $U\le T$, with $T$ any larger group of isometries of the closed manifold $N$. Change variables $x=hy$ and use simultaneous heat-kernel invariance to obtainThus $I_t$ is a class function, and the Sunada orbital heat-trace formula can be grouped by conjugacy classes $C$ of $T$:Only classes meeting the finite subgroup $U$ contribute, so this class sum is finite even if the larger isometry group $T$ is infinite.

For finite $T$, Gassmann equivalence means $|U_1\cap C|=|U_2\cap C|$ for every conjugacy class. Summing these counts gives $|U_1|=|U_2|$, so the displayed formula gives identical heat traces. Their spectra, with multiplicities, agree: as $t\to \infty$, the constant limit identifies the multiplicity of zero; after subtracting it, the slowest-decaying exponential identifies the least positive eigenvalue and its multiplicity. Repeating determines every subsequent eigenvalue. This proves the Sunada theorem in the stated covering situation.

There is an important convention when the universal cover $N$ has an infinite deck group $T$. For compact quotient manifolds, the subgroups must have finite index; the appropriate generalized Gassmann condition is equality of the characters of the finite coset permutation representations, not equality of possibly infinite cardinalities of conjugacy-class intersections. Let $K$ be the intersection of the two subgroup cores. Each core is the kernel of the corresponding finite coset action, so $K$ is normal and of finite index in $T$, and $K\subseteq U_1\cap U_2$. The manifold $X=K\backslash N$ is then a compact finite normal cover of $M_0$. The finite group $G=T/K$ acts on $X$, and $H_i=U_i/K$ have equal coset permutation characters in $G$. For a finite group, the fixed-coset count isEquality at the identity gives equal subgroup orders, and then equality for all $g$ gives almost-conjugacy. Applying the proved finite theorem to $H_i\backslash X=U_i\backslash N$ establishes isospectrality also in this finite-index interpretation. It avoids taking the divergent heat trace of a noncompact universal cover.

To realize a finitely presented group $T=⟨a_1,\dots ,a_r\mid R_1,\dots ,R_s⟩$, begin with the closed smooth four-manifold $B={\#}_{j=1}^r(S^1\times S^3)$, whose fundamental group is free on the $a_j$. Represent the relator words by disjoint smoothly embedded circles; general position permits this in dimension four, including separating self-intersections of representatives. Their oriented normal rank-three bundles are trivial, so perform framed surgeries replacing $S^1\times D^3$ by $D^2\times S^2$. Removing the circles' tubular neighborhoods leaves the fundamental group unchanged, by codimension-three general position. The Seifert-van Kampen theorem shows that each replacement kills exactly the normal closure of its relator. The resulting closed smooth $M_0$ has ${\pi }_1(M_0)\cong T$. This gives a concrete closed-manifold realization of a finitely presented fundamental group. Equip it with any smooth Riemannian metric and lift that metric to the universal and intermediate covers; all covering maps become local isometries.

To ensure nonisometry, one must choose nonconjugate subgroups. If $U_1=U_2$, or they are conjugate, the quotients are isometric for every lifted metric, so Gassmann equivalence alone cannot imply nonisometry. For nonconjugate subgroups, choose the base metric to have no nonidentity local isometry between open sets, using the Sunada local isometry lemma in dimension at least two. If $F:M_1\to M_2$ were an isometry, a local inverse of the first covering followed by $F$ and the second covering would be a local isometry of $M_0$, hence the identity. Thus the covering maps satisfy $p_{2}F=p_1$ everywhere. Classification of connected covers then forces $U_1,U_2$ to be conjugate in ${\pi }_1(M_0)=T$, a contradiction. Therefore nonconjugate almost-conjugate subgroups and a locally rigid base metric give isospectral nonisometric quotients.

A Riemannian isometry $F:(M,g)\to (N,h)$ is a diffeomorphism satisfying $F^{\ast }h=g$. A local isometry is a local diffeomorphism with the same pullback identity; equivalently $d{F}_p$ is an isometric linear isomorphism for every $p$. An isometric immersion into a higher-dimensional manifold does not satisfy this local-diffeomorphism requirement.

Let ${\varphi }_t$ be the local flow of the generating vector field $K$. The identity ${\varphi }_t^{\ast }g=g$ differentiates to ${\mathcal{L}}_{K}g=0$, where $\mathcal{L}$ is the Lie derivative of a tensor field. For arbitrary vector fields $X,Y$, metric compatibility and the torsion-free connection giveThis proves the Killing equation. Conversely, a smooth Killing field has a local flow, andSince ${\varphi }_0$ is the identity, ${\varphi }_t^{\ast }g=g$ wherever the flow is defined. Its inverse is ${\varphi }_{-t}$ locally, so the local flow consists of local isometries. Neither direction assumes that $K$ is complete.

Put $H(X,Y)={\nabla }_X({\nabla }_{Y}K)-{\nabla }_{{\nabla }_{X}Y}K$ and $B(X,Y,Z)=⟨H(X,Y),Z⟩$. These expressions are tensorial in $X,Y,Z$. Take the covariant derivative in direction $X$ of the Killing equation, subtracting the terms from differentiating $Y$ and $Z$. It follows thatWith the curvature convention $\mathcal{R}$ of Solution 1, the torsion-free connection also givesDefine $C(X,Y,Z)=⟨\mathcal{R}(X,K)Y,Z⟩$. It is skew in $Y,Z$ by skew-adjointness of the Riemann curvature tensor. The first Bianchi identity givesso $C$ has the same antisymmetrization in $X,Y$ as $B$. Consequently $D=B-C$ is symmetric in its first two slots and skew in its last two. These symmetries force it to vanish:We have proved the second covariant derivative of a Killing vector identity, including its sign:The printed identity uses the opposite curvature-sign convention in Killing derivative identities. If $R=-\mathcal{R}$, namely $R(X,Y)Z={\nabla }_Y{\nabla }_{X}Z-{\nabla }_X{\nabla }_{Y}Z+{\nabla }_{[X,Y]}Z$, the same result reads ${\nabla }_{X,Y}^{2}K=R(K,X)Y$, exactly as printed. The convention must change with the sign; under $\mathcal{R}$, writing $\mathcal{R}(K,X)Y$ would be incorrect.

Finally let $H_0=K-\tilde{K}$ have zero value and zero first covariant derivative at $p$. Along any geodesic issuing from $p$, the identity just proved saysThus $H_0$ is a Jacobi field with zero initial value and derivative. Uniqueness for this linear ordinary differential equation implies $H_0=0$ on each such geodesic, hence on a convex normal neighborhood of $p$. Consider the set of points where both $H_0$ and $\nabla H_0$ vanish. It is closed by continuity and open by this same local argument; on the resulting neighborhood $H_0$ vanishes identically, so its derivative does too. It is nonempty, and $M$ is connected, therefore it is all of $M$. This proves $K=\tilde{K}$ everywhere and the general statement that a Killing field is determined by its value and first covariant derivative.