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The Levi-Civita connection is the unique affine connection which is both a metric connection and a torsion-free connection. Uniqueness is encoded by the Koszul formulaIt follows by adding the metric-compatibility identities with derivative fields $X,Y$, subtracting the one with derivative field $Z$, and replacing differences of covariant derivatives by Lie brackets. Nondegeneracy of $g$ determines ${\nabla }_{X}Y$ uniquely from all its pairings with $Z$. The existence and uniqueness of the Levi-Civita connection supplies existence; no existence proof is needed here.

The torsion tensor of an affine connection isThe connection is symmetric, or a torsion-free connection, exactly whenIn a coordinate frame, write ${\nabla }_{{\partial }_i}{\partial }_j={\sum }_k{\Gamma }_{ij}^k{\partial }_k$. Since coordinate fields commute, the condition is ${\Gamma }_{ij}^k={\Gamma }_{ji}^k$; this is the reason for the word symmetric.

Take $\mathbf{n}$ to be the unit normal, so $P(v)=v-⟨v,\mathbf{n}⟩\mathbf{n}$ is orthogonal projection onto $TM$. With $D$ the Euclidean Levi-Civita connection, the proposed derivative isIt is a smooth tangent field, and independence of the extensions is given. Linearity follows from linearity of $D$ and $P$. Moreover, for $f\in C^{\infty }(M)$, take local extensions of $f,X,Y$ to obtainsince $PY=Y$. Thus it is an affine connection on $M$.

For tangent fields $Y,Z$, the normal component of $DY$ pairs to zero with $Z$. Differentiating the ambient inner product along $X$ givesHence it is a metric connection for the induced Riemannian metric. The ambient torsion vanishes, soHere the restriction of the ambient bracket is the intrinsic bracket and is tangent to $M$: derivations of functions restricted to $M$ give the same bracket, or this follows in submanifold coordinates. Thus the projected derivative is also a torsion-free connection. Uniqueness now provesThis is the projected ambient connection. The projection formula requires a unit normal. If a nonunit normal is used, its normal term must instead be $⟨DY,\mathbf{n}⟩\mathbf{n}/⟨\mathbf{n},\mathbf{n}⟩$. For example, on the unit circle, taking $\mathbf{n}=2p$ and differentiating its unit tangent along itself gives $DY=-p$; the unnormalized printed expression would be $3p$, which is not tangent. Normalizing the normal removes that ambiguity.

Assume the level set is regular and non-null. The latter hypothesis is necessary: a null hypersurface has a degenerate induced metric and hence no ordinary intrinsic Levi-Civita connection determined by that metric. Normalize the normal vector and define the non-null hypersurface projection byHere $h$ denotes the induced metric restricted to tangent vectors, while its ambient extension annihilates $n$. The intrinsic covariant derivative is $D_{X}Y=P{\nabla }_{X}Y$ for tangent vector fields. Projection of the ambient torsion-free connection and metric compatibility shows that $D$ is the Levi-Civita connection of $h$. Define the extrinsic curvature by $K(X,Y)=g({\nabla }_{X}n,Y)$, or $K_{ab}=P^c{}_a{P}^d{}_b{\nabla }_c{n}_d$. It is symmetric: differentiating the defining function twice and projecting removes the derivative of its normalization, leaving a multiple of its symmetric Hessian. This sign convention makes the scalar second fundamental form $g(n,{\nabla }_{X}Y)=-K(X,Y)$.

Let $S$ be the tangent linear map characterized by $h(SX,Y)=K(X,Y)$. Differentiating $g(n,n)=\epsilon$ and $g(n,Y)=0$ gives the Gauss formulaFix the Riemann curvature tensor conventionApplying the Gauss formula twice givesSubtracting the expression with $X,Y$ interchanged, and then subtracting the bracket derivative, provesThus the Gauss–Codazzi equations for a non-null hypersurface give the intrinsic Riemann curvature tensor asAll free indices here are tangent to the hypersurface. The normal component of the same calculation isAfter antisymmetrizing and using $D_{X}Y-D_{Y}X=[X,Y]$, this yields the curved-ambient Codazzi equationThe extrinsic curvature measures how tangent planes turn in the ambient manifold; the Gauss equation says that this turning contributes to intrinsic curvature even when the ambient Riemann curvature tensor is zero.

For the quadric in Minkowski spacetime, write its position vector as $X^a$. On the hypersurface, $g(X,X)=1$ and differentiating this identity in a tangent direction gives $g(X,Y)=0$. Hence $n=X$ is a unit normal and a spacelike vector, $\epsilon =1$, and ${\nabla }_{Y}n=Y$ in flat Cartesian coordinates. Therefore $K=h$. The ambient Riemann curvature tensor vanishes, soThis is unit-radius de Sitter spacetime. Reversing the normal changes $K$ to $-K$ but does not change this curvature. An opposite overall convention for the Riemann curvature tensor would reverse both curvature signs below.

To retain a four-dimensional Lorentzian metric while reversing the curvature, take a flat ambient metric with two negative directions and the quadricNow the position vector is a unit normal with negative squared norm, so $\epsilon =-1$ while $K=h$ remains true. The Gauss equation givesThis is unit-radius Anti-de Sitter spacetime. If a positive-definite induced metric is acceptable, an even simpler alternative is to keep the original one-time ambient Minkowski metric and change the quadric level to $g(X,X)=-1$. Each sheet then gives four-dimensional hyperbolic space with the same negative curvature, but its induced signature differs from that of de Sitter spacetime.

Work on a regular patch with a smooth nonzero conformal factor. Its sign is constant there, so replace it by $|\Omega |$ if necessary and assume $\Omega >0$. Put $u=\log \Omega$, and let all partial derivatives, index raising on $u_i,{\Omega }_i$, and the operators $\Delta ,|\nabla \cdot {|}^2$ in the following formulas use the flat Cartesian metric $\delta$. The orthonormal coframe $E^i=\Omega d{x}^i$ hasThe proposed connection 1-forms reduce toLowering the first frame index with $\delta$ makes these antisymmetric, which establishes metric compatibility. Moreover,Thus Cartan's first structure equation has zero torsion. Uniqueness of the metric-compatible torsion-free connection identifies these forms as the Levi-Civita connection; this also explains why merely guessing a skew connection would not have sufficed.

To calculate the Riemann curvature tensor, use Cartan's second structure equation. Direct differentiation and multiplication giveIntroduce the Schouten tensor in Cartesian coordinate components,The preceding calculation groups into the curvature 2-formsKeep the convention $R_{ijkl}=g({\partial }_i,R({\partial }_k,{\partial }_l){\partial }_j)$, with $R(X,Y)=[{\nabla }_X,{\nabla }_Y]-{\nabla }_{[X,Y]}$, as in the first solution. Converting frame indices to coordinate indices gives the curvature of a conformally Euclidean metric:These are coordinate components. Components in the dual orthonormal frame are ${\Omega }^{-4}{R}_{ijkl}$, hence the same expression with the overall factor ${\Omega }^{-2}$. This distinction is important when contracting the Riemann curvature tensor.

Contract with $g^{ik}={\Omega }^{-2}{\delta }^{ik}$. In four dimensions the Ricci tensor becomes $R_{jl}=2S_{jl}+{\delta }_{jl}{\sum }_k{S}_{kk}$, and thereforeTaking the remaining metric trace yields the Ricci scalarThe cancellation of the gradient-squared terms in the last expression is special to dimension four.

For the Tolman wormhole, compute on $r>0$:Substitution into the Ricci tensor and Ricci scalar givesFor example, the coordinate radial eigenvalue is $-12\lambda /(r^2+\lambda {)}^2$, while each of the three tangential eigenvalues is $4\lambda /(r^2+\lambda {)}^2$. Their sum vanishes, but none vanishes when $\lambda \ne 0$. The Vacuum Einstein equations with zero cosmological constant require $R_{ij}-\frac{1}{2}R{g}_{ij}=0$ and hence $R_{ij}=0$. The metric is scalar-flat but not a vacuum solution for any nonzero $\lambda$; only $\lambda =0$ gives flat vacuum space. A nonzero cosmological constant does not rescue it: its trace would force $R=4\Lambda$, so $R=0$ would first force $\Lambda =0$.

The smooth wormhole interpretation uses $\lambda >0$; $r=0$ is another asymptotically flat end, not an included origin, as inversion $\rho =\lambda /r$ shows. For $\lambda <0$, the conformal factor vanishes at $r=\sqrt{-\lambda }$, where the metric is degenerate. The local non-vacuum calculation remains valid on each regular region away from that sphere, but such a metric is not a smooth wormhole across it.

Use the Levi-Civita connection interpretation of the curvature convention. Its Riemann curvature tensor satisfiesThe first antisymmetry comes from the derivative commutator; the second comes from compatibility with the metric tensor. The cyclic identity is the first Bianchi identity, and together these imply exchange symmetry of the two antisymmetric index pairs. Contracting gives the symmetric Ricci tensor, $\boxed{R_{ab}=R_{ba}}$. The differential Bianchi identity is ${\nabla }_{[a}{R}_{bc]de}=0$.

There is a hypothesis qualification. A metric connection need not be a torsion-free connection. For a general such connection, the tensorial derivative commutator contains $-T^c{}_{ab}{\nabla }_c$ as well as curvature; pair exchange, the stated cyclic identity and Ricci symmetry need not hold. The printed pure-curvature commutator, and the scalar identity requested later, use the torsion-free interpretation. Under that intended interpretation all the boxed symmetries follow. They are not assertions about arbitrary metric-compatible connections with torsion tensor.

For example, with a constant Minkowski metric a connection having only ${\Gamma }^1{}_{02}=1$ and ${\Gamma }^2{}_{01}=-1$ nonzero is metric compatible, since its connection matrix in direction zero is a spatial rotation. It has $T^1{}_{02}=1$. For the scalar $f=x^1$ the tensorial Hessian commutator is $[{\nabla }_0,{\nabla }_2]f=-1$, rather than zero. This explicitly demonstrates the missing zero-torsion condition used in the subsequent scalar proof.

For a torsion-free connection, the covariant Hessian of a scalar is symmetric, so $[{\nabla }_a,{\nabla }_b]f=0$. Moreover, a commutator of two covariant derivatives acts as a derivation on tensor products. Applying the Leibniz rule twice, the terms containing one derivative on each factor cancel between the two orders:Projecting with the parallel symplectic form preserves this identity for the chiral spinor curvature operator. Since its scalar projection obeys ${\Delta }_{AB}f=0$, we obtainThus the chiral spinor curvature operator is algebraic over scalar functions even though its displayed expression uses second derivatives. If torsion tensor is allowed, $[{\nabla }_a,{\nabla }_b]f=-T^c{}_{ab}{\partial }_{c}f$ need not vanish, explaining why metric compatibility alone would not prove the claim.

An affine connection $\nabla$ is a metric connection for the Riemannian metric $g$ whenEquivalently the induced covariant derivative of the metric tensor satisfies $\nabla g=0$. This is metric compatibility; it imposes no condition on the torsion tensor. A Levi-Civita connection has the additional property of being a torsion-free connection.

Use coordinate indices $\mu ,\nu$ and Lorentz-frame indices $A,B$, and take ${\eta }_{AB}=\mathrm{diag}(-1,1,1,1)$. Locally choose four independent one-forms $e^A=e^A{}_{\mu }d{x}^{\mu }$ satisfyingOne construction begins with a smooth timelike vector field in a sufficiently small neighborhood, normalizes it, and applies the Gram-Schmidt process to its spacelike orthogonal complement. The dual orthonormal coframe in spacetime gives the vierbein; equivalently its matrix factorizes the metric as $g=e^T\eta e$. The inverse matrix gives the frame $e_A=e_A{}^{\mu }{\partial }_{\mu }$ with $e_A{}^{\mu }{e}^B{}_{\mu }={\delta }_A^B$. This is a local construction: a global frame is not guaranteed by a metric, and a globally defined spinor field requires a compatible spin structure.

The metric does not determine a unique vierbein. A local Lorentz transformation changes the coframe components byThe coordinate index is unchanged in this frame transformation. Restrict to the proper orthochronous Lorentz group for its connected spin lift; orientation-reversing transformations require further structure and are not part of this local spin-group discussion.

Let constant gamma matrices obey the Clifford algebra $\{{\gamma }^A,{\gamma }^B\}=2{\eta }^{AB}$, and write ${\gamma }^{AB}=\frac{1}{2}[{\gamma }^A,{\gamma }^B]$. A spinor field transforms by the spin lift of the Lorentz transformation, not by its vector matrix:For $\Lambda =1+\alpha +O({\alpha }^2)$, with ${\alpha }_{AB}=-{\alpha }_{BA}$, the Clifford algebra identityshows explicitly thatIndeed the commutator of ${\gamma }^C$ with this infinitesimal spin matrix is ${\alpha }^C{}_B{\gamma }^B$. A finite transformation connected to the identity is obtained by exponentiating these generators along a path. Its lift has two possible signs, so spinors are not ordinary Lorentz vectors: a $2\pi$ spatial rotation acts by $-1$, and $4\pi$ by $+1$. Consistent transition lifts define the spin structure; the metric alone does not select it.

For the metric connection, take the Levi-Civita connection and determine the spin connection from the tetrad postulateIt is equivalently determined by $d{e}^A+{\omega }^A{}_B\wedge e^B=0$ and ${\omega }_{\mu AB}=-{\omega }_{\mu BA}$. Explicitly, in terms of the inverse frame,Thus both the coordinate connection and the Lorentz connection 1-form are determined by the metric and chosen frame; no independent torsion has been supplied.

An ordinary derivative fails covariance because ${\partial }_{\mu }(S\psi )=S{\partial }_{\mu }\psi +({\partial }_{\mu }S)\psi$. Seek ${\nabla }_{\mu }\psi =({\partial }_{\mu }+{\Omega }_{\mu })\psi$. Covariance requiresThe coframe's connection obeys ${\omega }^{\prime }=\Lambda \omega {\Lambda }^{-1}-d\Lambda {\Lambda }^{-1}$. Applying the spin representation of the same Lorentz algebra, whose infinitesimal generator was just determined, suppliesThis proves the transformation law rather than choosing the coefficient arbitrarily. It is the spinor covariant derivative induced by the spin connection. With curved matrices ${\gamma }^{\nu }(x)=e_A{}^{\nu }{\gamma }^A$, the same coefficient and tetrad postulate giveso differentiation is compatible with Clifford multiplication and $\{{\gamma }^{\mu }(x),{\gamma }^{\nu }(x)\}=2g^{\mu \nu }(x)$.

Finally, expand the commutator of these derivatives:The spin representation preserves the Lorentz algebra bracket. Consequently $d\Omega +\Omega \wedge \Omega =\frac{1}{4}{R}_{AB}{\gamma }^{AB}$, where $R^A{}_B=d{\omega }^A{}_B+{\omega }^A{}_C\wedge {\omega }^C{}_B$. In components the spinor curvature identity isFor the full second derivative, ${\nabla }_{\mu }({\nabla }_{\nu }\psi )$ also includes $-{\Gamma }_{\mu \nu }^{\rho }{\nabla }_{\rho }\psi$, because its first derivative has a covector index. This term cancels on antisymmetrization for the torsion-free connection. With unit-weight brackets, $X_{[\mu \nu ]}=(X_{\mu \nu }-X_{\nu \mu })/2$, the requested expression is thereforeThis convention distinguishes the antisymmetrized second derivative from the full commutator by its factor of two. In a noncoordinate orthonormal frame, an operator commutator must additionally subtract the derivative along the frame-vector commutator; using the full tensorial second derivative incorporates that correction.

Let $e(\omega )$ denote exterior multiplication, $e(\omega )\alpha =\omega \wedge \alpha$. For a $q$-form $\alpha$, evaluation of the proposed operator on vector fields $Y_0,\dots ,Y_q$ givessince ${\sum }_i{\omega }_i(Y_a)X_i=Y_a$. This expression is independent of the chosen local basis.

Expand each covariant derivative into the derivative of $\alpha (Y_0,\dots ,{\hat{Y}}_a,\dots ,Y_q)$ minus the terms obtained by differentiating each argument. For every pair $a<b$, those argument terms combine intoThe Levi-Civita connection is a torsion-free connection, so the vector in this expression is $[Y_a,Y_b]$. The result is exactly the defining evaluation formula for the exterior derivative:Therefore $\boxed{d=\sum _{i}e({\omega }_i){\nabla }_{X_i}}$. No orthonormality or compactness is needed for this identity; the dual-frame property and vanishing torsion tensor suffice.

The torsion tensor of an affine connection is $\mathcal{T}(X,Y)={\nabla }_{X}Y-{\nabla }_{Y}X-[X,Y]$, where the last term is the Lie bracket of vector fields. A torsion-free connection has $\mathcal{T}=0$. Since a coordinate basis commutes, this is equivalent there to ${\Gamma }^{\mu }{}_{\nu \rho }={\Gamma }^{\mu }{}_{\rho \nu }$; in a general basis its commutator must also be included.

The Levi-Civita connection is the unique torsion-free connection with metric compatibility, $\nabla g=0$. To derive its Christoffel symbols, lower the first index: ${\Gamma }_{\sigma \nu \rho }=g_{\sigma \mu }{\Gamma }^{\mu }{}_{\nu \rho }$. Metric compatibility givesWrite down this identity with the derivative indices $\rho ,\nu ,\sigma$, add the first two identities and subtract the third. Symmetry in the last two indices, supplied by the torsion-free connection, cancels all terms except twice ${\Gamma }_{\sigma \nu \rho }$. Raising the first index yieldsThis calculation proves uniqueness. Conversely, these Christoffel symbols are symmetric in $\nu ,\rho$ and satisfy $\nabla g=0$ by substitution, proving existence as well.

Use precisely the stated convention for the Riemann curvature tensor, and let $D={\nabla }_T$. The geodesic equation says ${\nabla }_{T}T=0$ because proper time is an affine parameter. The commuting variation and the torsion-free connection giveThus the geodesic deviation equation isHere $(R(X,Y)Z{)}^a=R^a{}_{bcd}{Z}^b{X}^c{Y}^d$. Declaring this index convention fixes the sign: writing $R(S,T)T$ instead would require a minus sign. A deviation vector satisfying this equation is a Jacobi field.

A Killing vector field $\xi$ generates a local one-parameter family of isometries of the metric tensor. Equivalently, its tensor Lie derivative of the metric vanishes:The latter is the Killing equation; the equivalence uses metric compatibility and the torsion-free connection.

A Levi-Civita connection on a Riemannian manifold is a covariant derivative $D$ that is a torsion-free connection, $D_{X}Y-D_{Y}X=[X,Y]$, and a metric connection, $Xg(Y,Z)=g(D_{X}Y,Z)+g(Y,D_{X}Z)$. A covariant derivative is $C^{\infty }(M)$-linear in $X$, real linear in $Y$, and obeys $D_X(fY)=X(f)Y+f{D}_{X}Y$.

Combining metric compatibility for the cyclic triples and eliminating reversed derivatives with the torsion identity gives the Koszul formulaSince $g$ is nondegenerate, this determines $D_{X}Y$ uniquely. For existence, use the right-hand side to define $D_{X}Y$: expansion of the Lie bracket of vector fields shows it is $C^{\infty }(M)$-linear in $Z$, hence defines a smooth one-form, which the musical isomorphism converts to a smooth vector field. The same expansion shows $D_{fX}Y=f{D}_{X}Y$ and $D_X(fY)=X(f)Y+f{D}_{X}Y$. Subtracting the formula with $X,Y$ exchanged gives torsion zero; adding its versions for $Y,Z$ exchanged gives metric compatibility. Thus it is a Levi-Civita connection. Equivalently its coefficients in a manifold chart are the Christoffel symbolsThe intrinsically defined Koszul formula ensures these local expressions fit together. This proves the fundamental theorem of Riemannian geometry.

The product Riemannian metric on $M\times N$ isThe two factor tangent spaces are orthogonal, and the sum is positive definite. Lift $X$ from $M$ and $Y$ from $N$. We check $D_{X}Y$ against lifted local frame fields from both factors, which together span each product tangent space.

If $Z$ is lifted from $M$, then $g(Y,Z)=g(X,Y)=0$, while $g(Z,X)$ depends only on $m$, so $Yg(Z,X)=0$. Also $[Y,Z]=[X,Y]=0$ and $[Z,X]$ is horizontal, hence orthogonal to $Y$. Every term in the Koszul formula is zero. If $W$ is lifted from $N$, $g(W,X)=g(X,Y)=0$ and $g(Y,W)$ depends only on $n$, so $Xg(Y,W)=0$. Now $[W,X]=[X,Y]=0$ and $[Y,W]$ is vertical, hence orthogonal to $X$. Again every term is zero. Thus $D_{X}Y$ is orthogonal to both spanning frame families, and positive definiteness yieldsThis calculation uses the lifts' independence of the other factor coordinates; a vector field with varying coefficients in those coordinates can have a nonzero mixed derivative.

Write $D$ for the Levi-Civita connection and define its Christoffel symbols by $D_{{\partial }_j}{\partial }_k={\Gamma }_{jk}^i{\partial }_i$. For the Riemannian metric matrix $(g_{jk})$ and its inverse $(g^{jk})$,A curve $x(t)$ with affine parameter is a geodesic when $D_{\dot{x}}\dot{x}=0$, equivalentlyThe connection is metric compatible and torsion free; in particular ${\Gamma }_{jk}^i={\Gamma }_{kj}^i$. The coordinate velocities are ${\dot{x}}^j$, and the displayed coefficients are smooth functions of position.

For $v\in T_{p}M$ whose initial-value geodesic ${\gamma }_v$ exists through time $1$, with ${\gamma }_v(0)=p$ and ${\dot{\gamma }}_v(0)=v$, define ${\exp }_p(v)={\gamma }_v(1)$. The domain is open and contains a neighborhood of zero. Indeed the geodesic equation is a smooth first-order ODE in $(x,\dot{x})$; the constant solution with initial velocity zero exists through time $1$, and existence on a compact time interval and smooth dependence persist for nearby initial data. Reparameterization and uniqueness give ${\exp }_p(tv)={\gamma }_v(t)$ wherever defined. Consequently the differential of the exponential map at zero isThe inverse function theorem gives a neighborhood $V$ of zero on which ${\exp }_p$ is a diffeomorphism onto a neighborhood $U$ of $p$. Shrink $V$ to a ball in $T_{p}M$ and choose an orthonormal basis there. Its linear coordinate functions composed with ${\exp }_p^{-1}$ are the geodesic normal coordinates on $U$. This establishes well-defined local coordinates without assuming geodesic completeness.

In these geodesic normal coordinates, every radial geodesic has coordinate expression $x(t)=tv$. Its equation at $t=0$ reads ${\Gamma }_{jk}^i(p)v^j{v}^k=0$ for every $v$. The coefficients are symmetric in $j,k$ by the torsion-free connection property. Evaluating at the coordinate unit vectors and their pairwise sums, or using the polarization identity, therefore givesThis is the fact that Christoffel symbols vanish at the center of normal coordinates. Also $g_{ij}(p)={\delta }_{ij}$, since $(d{\exp }_p{)}_0$ is the identity in an orthonormal basis.

For sufficiently small $r>0$, the geodesic sphere is $S_r(p)={\exp }_p\{v:|v{|}_{g_p}=r\}$. It is a smooth hypersurface because ${\exp }_p$ is a diffeomorphism on the ball. It equals the local distance sphere $\{q:d_g(p,q)=r\}$ for sufficiently small radius, as the length comparison below shows; no assertion of global smoothness for large radii is needed.

The full Gauss lemma isfor $v$ in a star-shaped domain of ${\exp }_p$ and $w\in T_{p}M$. To prove it, set $F(t,s)={\exp }_p(t(v+sw))$, $T={\partial }_{t}F$ and $S={\partial }_{s}F$, for $s$ near zero. Every $t$-curve is a geodesic and has squared speed $|v+sw{|}_{g_p}^2$. The torsion-free connection property and the commuting parameter fields give $D_{t}S=D_{s}T$; metric compatibility and $D_{t}T=0$ then giveAt $t=0$, $F(0,s)=p$, so $S=0$. Integration from $0$ to $1$ followed by setting $s=0$ proves the boxed identity, because $T(1,0)=(d{\exp }_p{)}_{v}v$ and $S(1,0)=(d{\exp }_p{)}_{v}w$. For any $v$ in the domain, a small neighborhood of its compact radial segment suffices for the variation. The zero vector is covered directly.

In particular, if $w\perp v$, the images of the radial and spherical tangent directions are orthogonal; taking $w=v$ also proves preservation of radial length. Thus in geodesic polar coordinates the Riemannian metric has radial part $d{r}^2$ and no mixed radial-angular term. Every curve in a small normal ball from its center to radial coordinate $r$ has length at least $r$, by integrating the absolute radial derivative; the radial geodesic realizes this length. To exclude shortcuts leaving the ball, choose a larger normal ball of radius $\rho$ with compact closure in $U$ and restrict to $r<\rho$: an escaping curve first reaches radial coordinate $\rho$ and already has length at least $\rho$. This proves the asserted small-radius distance interpretation and orthogonality of radial geodesics to geodesic spheres.

Use the standard meaning of static spacetime: a stationary spacetime chart whose time-space components also vanish. The printed static definition needs to inherit the timelike requirement $g_{00}<0$; without it, the requested timelike conclusion does not follow.

The preceding argument shows that $V={\partial }_0$ is a timelike Killing vector field. Its metric-dual one-form is $v=V_{\mu }d{x}^{\mu }=g_{00}d{x}^0$. Its exterior derivative satisfiesFor a torsion-free connection, antisymmetrizing the covariant derivative of a one-form gives its exterior derivative. Consequently the Killing vector has hypersurface orthogonality:Geometrically, the hypersurfaces $x^0=\mathrm{constant}$ are orthogonal to $V$.