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Every pair of coordinate basis fields commutes:Therefore a frame containing two fields with nonzero Lie bracket of vector fields cannot be induced by a coordinate system.

For a local basis $e_a$ of the tangent bundle, an affine connection has components defined by ${\nabla }_{e_c}{e}_b={\Gamma }^a{}_{bc}{e}_a$. If $V=V^b{e}_b$, then $({\nabla }_{e_c}V{)}^a=e_c(V^a)+{\Gamma }^a{}_{bc}{V}^b$. The connection components depend on the chosen basis and have an inhomogeneous transformation law; they are not themselves a tensor field. In a coordinate basis, a torsion-free connection has symmetry in the two lower indices.

On Euclidean space, every Killing vector field is a constant translation generator plus a constant infinitesimal rotation generator:Here $A$ is a constant vector and $B$ is an antisymmetric matrix. The second covariant derivative of a Killing vector vanishes for the Euclidean metric, so the components are affine functions of Cartesian coordinates; the Killing equation makes the linear coefficient antisymmetric.

In the Euclidean plane with metric tensor $g=d{x}^2+d{y}^2$, two useful examples areThe notation ${\partial }_x$ denotes a coordinate basis vector; $dx$ denotes a differential one-form. The corresponding flow of a vector field maps areFor translation, all arrows have the same direction and length. For rotation, arrows are tangent to circles, have length $\sqrt{x^2+y^2}$, and vanish at the origin. The rotation flow keeps $x^2+y^2$ fixed, and its nonstationary integral curves of a vector field are circles. Both flows preserve the Euclidean metric. See Carroll's original notes, section 5, equations (5.45)-(5.46).

The circular integral curves of a vector field are symmetry orbits, whereas geodesics of the Euclidean plane are straight lines. A Killing vector field does not require its own orbits to be geodesics. The geodesic conserved quantity from a Killing vector pairs the field with the velocity of a separate geodesic; it does not require the velocity to equal the Killing vector field.

The Killing equation tests whether a candidate vector field generates a local isometry, without needing to solve for its flow of a vector field. For the Levi-Civita connection of a metric tensor $g$, it isParentheses mean averaging over the two index orders. The equation says the symmetric part of the covariant derivative vanishes, not that the entire covariant derivative vanishes. The Lie derivative of the metric tensor gives the coordinate formThe last two terms account for how tangent vectors move with the field; checking only $K^{\rho }{\partial }_{\rho }{g}_{\mu \nu }$ misses this. See Carroll, section 5, equations (5.33) and (5.41)-(5.42).

For the Euclidean Killing vector field $K=-y{\partial }_x+x{\partial }_y$, the Euclidean metric in Cartesian coordinates makes covariant derivatives ordinary partial derivatives. Its diagonal equations give ${\partial }_x{K}_x={\partial }_y{K}_y=0$, and its mixed equation gives ${\partial }_x{K}_y+{\partial }_y{K}_x=1-1=0$. Thus a varying vector field can preserve a constant metric tensor. In contrast, $K=x{\partial }_x+y{\partial }_y$ gives $({\mathcal{L}}_{K}g{)}_{xx}=2$ and fails the Killing equation, despite the constant metric components.

If $K={\partial }_z$ is a coordinate basis vector, its components are constant and the coordinate form reduces to ${\partial }_z{g}_{\mu \nu }=0$. Hence independence of the metric components from $z$ supplies a Killing vector field. This shortcut applies to that coordinate generator, not to an arbitrary varying vector field.

On the punctured Euclidean plane, the orthonormal polar frame isIts bracket is $[\hat{r},\hat{\theta }]=-\hat{\theta }/r$, so it is not a coordinate basis even though $({\partial }_r,{\partial }_{\theta })$ is one.

The covariant derivative differentiates a vector field while accounting for how its coordinate basis changes between neighboring points. For the Levi-Civita connection,The Christoffel symbol term makes this a tensor with one contravariant and one covariant index; ordinary partial derivatives of vector components do not generally have that transformation law. Equivalently, one compares the neighboring vector with the original after parallel transport.

For a curve $x^{\gamma }(\lambda )$ with tangent $U^{\gamma }=d{x}^{\gamma }/d\lambda$, contraction gives the covariant derivative along a curveIt is the rate of change measured relative to a parallel transport frame along the curve. It vanishes precisely when $v$ undergoes parallel transport along that curve. If $v=U$, it is the covariant acceleration, and its vanishing is the geodesic equation with an affine parameter.

For a contravariant vector field $A^{\lambda }$, the covariant derivative isThe Levi-Civita connection is the unique torsion-free connection satisfying metric compatibility. Its Christoffel symbols in a coordinate basis areThere is generally no symmetry interchanging the upper index with a lower one. The Christoffel symbols themselves are not the components of a tensor.

An ordinary partial derivative of the components of a vector field includes the apparent change caused by the varying coordinate basis. The connection term corrects for that change, so ${\nabla }_{\nu }{A}^{\lambda }$ obeys the tensor component transformation law. It gives coordinate-independent ways to express parallel transport, geodesics and stress-energy conservation in curved spacetime.

The metric tensor in a basis ${\mathbf{e}}_a$ is $g_{ab}={\mathbf{e}}_a\cdot {\mathbf{e}}_b$. Differentiating this inner product with the product rule and using the assumed basis-derivative relation yieldsThus the coefficients satisfy metric compatibility. For a coordinate basis with zero torsion tensor, symmetrizing this relation in $a,c$, adding the corresponding two equations and subtracting the third givesThey are therefore the Christoffel symbols of the Levi-Civita connection. Metric compatibility by itself does not establish zero torsion tensor; that also uses the coordinate basis and its commuting coordinate derivatives.

For an embedded curved surface, the stated ambient basis-derivative relation needs a qualification: the Gauss formula is ${\partial }_c{\mathbf{e}}_a={\Gamma }_{ac}^b{\mathbf{e}}_b+h_{ac}\mathbf{n}$, where $\mathbf{n}$ is a unit normal and $h_{ac}$ is the second fundamental form. The normal term has zero inner product with each tangent basis vector, so the displayed identity for ${\partial }_c{g}_{ab}$ still holds. For a concrete counterexample to dropping that term, on a two-sphere one has ${\partial }_{\theta }{\mathbf{e}}_{\theta }=-R\mathbf{n}$, whereas every ${\Gamma }_{\theta \theta }^b$ is zero. Thus the intrinsic connection describes the tangential projection of the ambient derivative.

In standard spherical coordinates, parametrize the two-sphere byThe coordinate basis transforms from the fixed Cartesian basis according to ${\mathbf{e}}_a=(\partial x^i/\partial q^a){\mathbf{e}}_i$. ThereforeTheir inner products are ${\mathbf{e}}_{\theta }^2=R^2$, ${\mathbf{e}}_{\theta }\cdot {\mathbf{e}}_{\varphi }=0$ and ${\mathbf{e}}_{\varphi }^2=R^2{\sin }^2\theta$. Thus the induced metric isThe source's intermediate display places an extra $R^2$ outside entries already labelled $g_{\theta \theta },\dots$. Interpreting those entries as the actual metric tensor components would double-count the factor: for example, $R=2$ gives $g_{\theta \theta }=4$ directly from ${\mathbf{e}}_{\theta }^2$, not $16$. Its final matrix is the correct induced metric; the entries of its intermediate matrix must instead denote dimensionless coefficients.

The coordinate basis vector $K={\partial }_v$ has constant components $K^a={\delta }_v^a$. Its norm isso it is a nonzero null vector everywhere. The Lie derivative of the metric tensor isThus $\boxed{K={\partial }_v\text{ is a null Killing vector field}.}$ In fact it is a parallel vector field: $K_b=g_{bv}={\delta }_b^u$ and the Christoffel symbols found below give ${\nabla }_a{K}_b=-{\Gamma }^u{}_{ab}=0$. This stronger property is characteristic of this plane-fronted gravitational wave.

The geodesic deviation equation compares neighboring geodesics by covariant derivatives. Replacing those derivatives by ordinary derivatives of coordinate components requires a parallel transport frame along the reference geodesic. There is also an affine parameter normalization to specify: the argument of $M$ is $u$, and it becomes $\tau$ only after choosing $u=\tau$.

For the intended local tidal calculation, take the reference timelike geodesicHere $g(V,V)=-1$, so $\tau$ is proper time. Both $F$ and its first partial derivatives vanish on $\gamma$. All Christoffel symbols therefore vanish along this curve, and its coordinate basis is parallel transport along it. With the Riemann curvature tensor from part (b),The Jacobi field equation thus becomes, in this adapted frame,A purely transverse initial Jacobi field, with ${\sigma }^u={\sigma }^v={\dot{\sigma }}^u={\dot{\sigma }}^v=0$ initially, remains transverse. This transverse tidal equation of a Brinkmann plane wave describes the relative tidal acceleration of the freely falling bodies.

In vacuum write the symmetric matrix asIts instantaneous eigenvalues are $\pm \sqrt{A^2+B^2}$. The tidal acceleration stretches along one principal axis and squeezes along the perpendicular axis, with the axes allowed to vary with $u$. This describes accelerations; it does not imply monotonically growing and shrinking separations for arbitrary initial velocities. For the simple constant profile $A={\omega }^2$, $B=0$,Thus the geodesic deviation can be defocusing in one direction and oscillatory in the other. For a varying profile, diagonalizing at each instant does not decouple the ordinary differential equations without also including derivatives of the rotating eigenvectors.

Qualification of the coordinate claim. The displayed equation for ${\sigma }^v$ is not true for arbitrary coordinate displacements along arbitrary reference geodesics. For synchronized neighboring curves with $u=\tau$ and ${\sigma }^u=0$, the exact geodesic equations areTheir variation givesThe second expression need not vanish. A concrete vacuum counterexample is $M=\mathrm{diag}(1,-1)$ and the one-parameter family of timelike geodesicsEach has $g(V,V)=-1$. Differentiating with respect to $A$ produces a Jacobi fieldThis even has $g(V,\sigma )=0$, so it is a physical orthogonal separation. The covariant derivative form of geodesic deviation remains correct; the unrestricted assertion about ordinary coordinate derivatives needs the adapted reference and frame used above. If instead $\dot{u}=c$ is a constant affine parameter normalization, the transverse tidal coefficient is $c^{2}M(u)$.

A covariant derivative on the tangent bundle is an affine connection: it assigns a vector field ${\nabla }_{X}Y$ to vector fields $X,Y$, is linear over smooth functions in $X$, is real-linear in $Y$, and obeys the Leibniz rule ${\nabla }_X(aY)=X(a)Y+a{\nabla }_{X}Y$. Thus it differentiates a vector field while taking account of how its local basis changes. It extends to tensor fields by the Leibniz rule, compatibility with tensor contraction, and ${\nabla }_{X}a=X(a)$ for a scalar.

In a local basis $e_{\mu }$, the connection components are defined by ${\nabla }_{e_{\rho }}{e}_{\nu }={\Gamma }^{\mu }{}_{\nu \rho }{e}_{\mu }$. In a coordinate basis, $e_{\rho }={\partial }_{\rho }$, soA covariant derivative of a tensor field has tensorial transformation properties, but the connection components themselves do not: changing the coordinate basis introduces second derivatives of the coordinate change. They describe the connection in a chosen basis, rather than components of a tensor.

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.

Write the covector field as $\omega ={\omega }_{\mu }d{x}^{\mu }$. From the contraction and product rule properties,Take $Y={\partial }_{\mu }$. The Lie bracket of vector fields is $[X,{\partial }_{\mu }]=-({\partial }_{\mu }{X}^{\nu }){\partial }_{\nu }$, soEquivalently, the covariant-factor contribution comes from ${\mathcal{L}}_{X}d{x}^{\mu }=d{X}^{\mu }=({\partial }_{\nu }{X}^{\mu })d{x}^{\nu }$. The formula holds in any coordinate basis and is the one-form case of the coordinate tensor Lie derivative.

The unit fields are $\hat{r}={\partial }_r$ and $\hat{\theta }=r^{-1}{\partial }_{\theta }$. Relation (3) and $[{\partial }_r,{\partial }_{\theta }]=0$ giveA coordinate basis must commute, soThis is the noncommuting orthonormal polar frame.

In a coordinate basis, parallel transport isAlong the circular orbit, $dr=d\theta =0$ and $d\varphi /dt=\Omega$. Dividing by $dt/d\tau$ gives

The retarded and advanced null coordinates invert toThusand the Minkowski metric becomes

The coordinate basis transforms by the chain rule:or equivalentlyIn the $(z,t)$ diagram, ${\partial }_t$ points vertically upward and ${\partial }_z$ horizontally right. The vector ${\partial }_u$ points along the future-left null ray and ${\partial }_v$ along the future-right null ray; their factors of one half affect length in the coordinate drawing but not direction.

Using these derivative relations, the Minkowski wave operator in null coordinates is

Differentiate the position vector with respect to the three coordinates to obtain the coordinate basis vectors:The chain rule gives $d\mathbf{x}={\mathbf{h}}_{u}du+{\mathbf{h}}_{v}dv+{\mathbf{h}}_{\varphi }d\varphi$. The scale factors of orthogonal coordinates are the corresponding norms:Direct dot products give ${\mathbf{h}}_u\cdot {\mathbf{h}}_v=uv-uv=0$, ${\mathbf{h}}_u\cdot {\mathbf{h}}_{\varphi }=0$, and ${\mathbf{h}}_v\cdot {\mathbf{h}}_{\varphi }=0$. Thus the parabolic coordinates are orthogonal wherever the coordinate vectors are nondegenerate. Their degeneracy on $uv=0$ is the usual axis degeneracy of the azimuthal coordinate.

The Jacobian determinant is the oriented scalar triple product of the three coordinate basis vectors:Here ${\mathbf{h}}_u\times {\mathbf{h}}_v=(u^2+v^2)(-\sin \varphi ,\cos \varphi ,0)$, soThe sign is nonnegative on the stated range, and the absolute Jacobian determinant also equals $h_u{h}_v{h}_{\varphi }$ as expected for orthogonal coordinates.

The three coordinate basis vectors are the partial derivatives of the position vector with respect to the three coordinates. Direct differentiation givesThe multivariable chain rule then gives $d\mathbf{x}={\mathbf{h}}_{\xi }d\xi +{\mathbf{h}}_{\theta }d\theta +{\mathbf{h}}_{\varphi }d\varphi$. These are coordinate tangent vectors, not necessarily unit vectors.