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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.

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.

At a fixed point $p$ of an $n$-dimensional smooth manifold $M$ without boundary, the tangent space $T_{p}M$ is the real vector space of all tangent vectors at $p$. In manifold charts $x=(x^1,\dots ,x^n)$ and $y=(y^1,\dots ,y^n)$, a tangent vector has componentsThe coordinate basis transforms inversely, so the tangent vector is unchanged. This invertible Jacobian matrix also preserves vector addition and scalar multiplication, giving a vector space of dimension $n$ independently of coordinates. ^[ref]

For example, at $(x,y)=(1,0)$ in ${\mathbb{R}}^2$, use the alternative manifold chart $u=x$, $w=y+x^2$. The Jacobian matrix sends $(v_x,v_y)=(2,3)$ to $(v_u,v_w)=(2,7)$. Indeed, the smooth curve $(x(t),y(t))=(1+2t,3t)$ gives $w(t)=1+7t+4t^2$. Conversely, ${\partial }_u{|}_p={\partial }_x{|}_p-2{\partial }_y{|}_p$ and ${\partial }_w{|}_p={\partial }_y{|}_p$, so $2{\partial }_u{|}_p+7{\partial }_w{|}_p=2{\partial }_x{|}_p+3{\partial }_y{|}_p$. The numbers change; their action on every smooth function does not.

For an embedded submanifold $M\subseteq {\mathbb{R}}^N$, the tangent space can be identified with the vector subspaceA proof outline explains why every coordinate vector is possible: in a manifold chart $x$, lift the short straight path $x(p)+ta$ using $x^{-1}$. Its tangent vector has components $a$. For a regular local parametrization $X:U\subseteq {\mathbb{R}}^n\to M$, the same chain rule gives $T_{p}M=\mathrm{im}D{X}_q$, where $X(q)=p$. This is the tangent space from a local parametrization.

For the sphere $S_R^2=\{p\in {\mathbb{R}}^3:p\cdot p=R^2\}$, differentiating that equation gives $p\cdot v=0$. Conversely, for any nonzero $v$ with $p\cdot v=0$, the smooth curvelies on the sphere and has ${\gamma }_v^{\prime }(0)=v$; use a constant smooth curve for $v=0$. Thus $T_p{S}_R^2=\{v:p\cdot v=0\}$, not merely a subset of that plane. At $p=(0,0,R)$ this vector subspace is $v_z=0$, while the affine tangent plane drawn through $p$ is $z=R$.

This also explains a mechanical constraint without requiring a force calculation. In the ideal mathematical model of a particle confined to a stationary sphere, its allowed velocities at $p$ are exactly $T_p{S}_R^2$: the radial component must vanish, but either independent tangential direction is available. For a general stationary holonomic constraint $F(q)=0$ with independent equations, the regular level set theorem gives $T_{q}M=\ker D{F}_q$. These are instantaneous kinematic possibilities; they do not determine the actual subsequent motion. On a general smooth manifold, tangent spaces at different base points cannot be identified canonically; an affine connection supplies parallel transport when such comparisons are needed.