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A contravariant vector has coordinate components obeying the displayed transformation law. A tangent vector to a parametrized curve provides the basic example: the chain rule transforms $d{x}^a/ds$ in precisely this way. Pairing it with a covector produces a coordinate-independent scalar. Conversely, if an array pairs invariantly with every contravariant vector, coefficient comparison forces the covector transformation law for that array.

The cotangent space at $p$ is the dual vector space $T_p^{\ast }M=(T_{p}M{)}^{\ast }$. Its elements are covectors on the tangent space. In a manifold chart $x$, the basis $d{x}^i{|}_p$ is dual to ${\partial }_i{|}_p$. Under a coordinate change to $y$, the coefficients of $\alpha =a_{i}d{x}^i=b_{j}d{y}^j$ obey $b_j={\sum }_i{a}_i\partial x^i/\partial y^j$. The disjoint union of these spaces is the cotangent bundle.

For a source-free plane electromagnetic wave in vacuum, let $n^{\mu }=(1,\mathbf{s})$ be its null propagation Lorentz four-vector. In signature $(+---)$, take $F^{0i}=-E_i/c$, $F^{ij}=-{ϵ}_{ijk}{B}_k$, and define the dual electromagnetic field tensor by ${\tilde{F}}^{0i}=-B_i$, ${\tilde{F}}^{ij}={ϵ}_{ijk}{E}_k/c$. Lower both indices with the Minkowski metric. Direct contraction givesBoth contractions vanish exactly when the transverse plane electromagnetic wave Maxwell equations hold. Being zero covectors, they remain zero in every inertial frame. Thus the transformed spatial wavevector is perpendicular to both transformed fields, even though the separate three-dimensional dot products are not themselves Lorentz scalars for arbitrary fields.

One contraction alone is insufficient without the other Maxwell equations. In particular, adding a magnetic field parallel to $\mathbf{s}$ leaves $n^{\mu }{F}_{\mu \nu }$ unchanged but violates $\mathbf{s}\cdot \mathbf{B}=0$. Conversely the two scalar perpendicularity conditions alone do not impose the required relation $\mathbf{B}=\mathbf{s}\times \mathbf{E}/c$ between field strengths and directions.

A covector is an element of the dual space, hence a linear functional on a vector space. In finite-dimensional coordinates, it acts as a row multiplied by a coordinate vector column. The row $\left(\begin{array}{cc}2& -1\end{array}\right)$ acts on $\left(\begin{array}{c}3\\ 1\end{array}\right)$ to give five. The terminology emphasizes its role as a scalar-valued measurement and its behavior under a change of basis.

For example, the total derivative of $f(x,y)=x^2+y^2$ at $(1,2)$ sends a small displacement $h$ to the linear change $df(h)=2h_x+4h_y$. This covector describes the first-order response. The gradient $\nabla f=(2,4)$ represents the same rule through the chosen Euclidean inner product. In differential geometry, the cotangent space collects these scalar-valued measurements of tangent vectors.

Differentiate the advection equation componentwise: $D_t({\partial }_i\theta )=-({\partial }_i{u}_j){\partial }_j\theta$. Along a Lagrangian trajectory, the scalar gradient therefore evolves by the negative transpose of the velocity gradient tensor. If $F$ is the deformation gradient, this says $\mathbf{k}(t)=F(t{)}^{-T}\mathbf{k}(0)$, as expected for a covector. Fluid material vectors and scalar gradients transform oppositely under stretching.

Let the columns of an invertible matrix $B$ be a new basis expressed in the old basis. The same vector has old coordinate vector $x$ and new coordinate vector $x^{\prime }=B^{-1}x$. If a covector has old row $\ell$, its new row is ${\ell }^{\prime }=\ell B$, sinceThus the scalar dual pairing stays unchanged. If covector coefficients are stored as a column $c={\ell }^T$, their new column is $c^{\prime }=B^{T}c$; this is the inverse transpose of the transformation $x\mapsto B^{-1}x$ on vector coordinates. These conventions express the covariance and contravariance of vectors. ^[ref].

For the basis $b_1=(1,0)$, $b_2=(1,1)$,The linear functional $\phi (x,y)=2x-y$ has rows $\ell =\left(\begin{array}{cc}2& -1\end{array}\right)$ and ${\ell }^{\prime }=\left(\begin{array}{cc}2& 1\end{array}\right)$. Both calculations give five: $2\cdot 3-1\cdot 1=2\cdot 2+1\cdot 1$. Changing the basis changes the descriptions, not the measured value.

For coordinates $x^{\mu }=(ct,x,y,z)$, the four-gradient of a scalar field is its covector of coordinate derivatives. Under a constant Lorentz transformation, ${\partial }_{\mu }^{\prime }=({\Lambda }^{-1}{)}^{\nu }{}_{\mu }{\partial }_{\nu }$. In metric signature $(+,-,-,-)$, raising its index gives ${\partial }^{\mu }=(c^{-1}{\partial }_t,-\nabla )$. Contracting it with a four-vector field produces the Lorentz scalar four-divergence. In particular ${\partial }_{\mu }{J}^{\mu }=0$ is charge continuity in every inertial frame.

Generalized geometry studies structures on the generalized tangent bundle $TM\oplus T^{\ast }M$, combining tangent vectors and covectors on a smooth manifold $M$. It has a natural split-signature pairingNo Riemannian metric is needed to define this pairing. The factor one-half is a convention. A Courant bracket combines the Lie bracket of vector fields with derivatives of one-forms, adding differential information to the bundle.

For a two-form $B$, the transformation $X+\xi \mapsto X+\xi +{\iota }_{X}B$ preserves the pairing: the two new cross terms cancel because $B$ is alternating. When $dB=0$, it also preserves the standard Courant bracket. These transformations enlarge the ordinary diffeomorphism symmetries. The framework places complex structures and symplectic forms within the common language of generalized complex geometry.^[ref]

The generalized tangent bundle of a smooth manifold $M$ is the direct sum of vector bundles $\mathbb{T}M=TM\oplus T^{\ast }M$. A fibre contains a tangent vector and a covector together. In real dimension $n$, it has rank $2n$ and a canonical symmetric bilinear formof signature $(n,n)$. No Riemannian metric is needed.^[ref]

Both $TM$ and $T^{\ast }M$ are isotropic: the pairing of any two elements of either summand is zero. A two-form $B$ gives the transformation $X+\xi \mapsto X+\xi +{\iota }_{X}B$. The new pairing terms sum to $B(X,Y)+B(Y,X)=0$, so the transformation preserves the pairing. These extra symmetries explain the utility of the bundle in generalized geometry, where complex structures and symplectic forms can be treated in one framework.

A smooth hypersurface in an $n$-dimensional smooth manifold is an embedded submanifold of codimension one. Locally it can be written as $\{\varphi =0\}$ withat every point of that zero set. The regular level set theorem gives this local description, and its tangent space is $\ker d{\varphi }_p$.

The covector $d\varphi$ is defined without a metric. If the ambient manifold has a Riemannian metric, its metric dual gradient $\nabla \varphi$ is a normal vector, since $g(\nabla \varphi ,v)=d\varphi (v)=0$ for every tangent $v$. A pseudo-Riemannian metric gives the analogous normal, which can be null. A global scalar defining function need not exist, as the normal-bundle obstruction to being a regular level set shows.

Author's graduate lecture notes, arXiv:gr-qc/9712019, 1997. Section 2 develops tangent vectors, covectors and the failure of ordinary tensor derivatives under nonlinear coordinate changes; section 3 constructs covariant differentiation, parallel transport and curvature. Metric and curvature signs must be translated when combining conventions.

The wave phase labels a physical stage of an oscillation at a spacetime event. Treating it as a Lorentz scalar makes its gradient $K_{\mu }={\partial }_{\mu }\Phi$ a covector; raising the index gives the four-wavevector $K^{\mu }=(\omega /c,\mathbf{k})$. A boost along $x$ gives ${\omega }^{\prime }=\gamma (\omega -v{k}_x)$ and $k_x^{\prime }=\gamma (k_x-v\omega /c^2)$. Thus angular frequency and wavevector transform together rather than independently.

The six smallest complex irreducible representations of SU(3) have highest weights $(0,0),(1,0),(0,1),(2,0),(0,2),(1,1)$ and dimensions $1,3,3,6,6,8$. Their Young diagram row lengths are $(0,0),(1,0),(1,1),(2,0),(2,2),(2,1)$. They are realized by scalars, defining vectors, covectors (equivalently antisymmetric tensors of degree two), upper and lower symmetric tensors of degree two, and mixed tensors $T^i{}_j$ with zero trace, respectively. The Weyl dimension formula is $\dim (p,q)=(p+1)(q+1)(p+q+2)/2$. For $p+q\ge 3$ this is at least ten, proving that the list exhausts the six smallest representations.

A nonzero covector $\ell$ is null when its squared dual Lorentzian inner product is zero: $g^{\alpha \beta }{\ell }_{\alpha }{\ell }_{\beta }=0$. Raising its index gives a null vector. Null covectors describe normals to characteristic light cones and are the test directions in the classical null condition for wave equations.

A differential 1-form is a smooth section of the cotangent bundle; it assigns a covector to every point of a smooth manifold. In planar coordinates, $\alpha =P(x,y)dx+Q(x,y)dy$ evaluates a tangent vector $v$ as $P(x,y)v_x+Q(x,y)v_y$.

Along a parametric curve $\gamma (t)=(x(t),y(t))$, its line integral isReversing the direction reverses the integral. A force expressed as a one-form therefore accumulates work along a directed path. In Euclidean coordinates its components can be displayed as arrows, but identifying a one-form with a vector field uses an inner product; the covector field itself does not require a Riemannian metric.

The paper uses $T_{\mu \nu }{}^{\lambda }={\Gamma }_{\nu \mu }{}^{\lambda }-{\Gamma }_{\mu \nu }{}^{\lambda }$, the negative of a common torsion tensor convention. Keep that sign throughout. Also, ${\nabla }_{\mu }{\nabla }_{\nu }\phi$ denotes the covariant Hessian, so the second derivative acts on the lower index $\nu$ of the covector ${\nabla }_{\nu }\phi$; it is not just a composition of two directional derivatives on a scalar.

Since the covariant derivative of a scalar field is its ordinary derivative,Commuting the ordinary partial derivatives therefore givesFor a vector field, write $D_{\nu }{X}^{\lambda }={\partial }_{\nu }{X}^{\lambda }+{\Gamma }_{\nu \rho }{}^{\lambda }{X}^{\rho }$. This is a tensor with one covariant and one contravariant index. HenceSubtracting the expression with $\mu ,\nu$ exchanged, the first two terms combine toThe coefficients of first derivatives of $X$ in those two terms cancel. The last terms give $({\Gamma }_{\nu \mu }{}^{\sigma }-{\Gamma }_{\mu \nu }{}^{\sigma })D_{\sigma }{X}^{\lambda }$. Identifying the curvature coefficient yieldsHere the bracket denotes antisymmetrization of the tensorial second covariant derivative, consistently with the preceding calculation.

To check tensoriality directly, antisymmetrize the supplied transformation law of the connection coefficients in the two lower indices. The inhomogeneous term vanishes because mixed coordinate second derivatives are symmetric. The remaining transformation law iswhich is precisely that of a $(1,2)$ tensor. Intrinsically, with the paper's sign,The derivative terms involving the components of $X,Y$ cancel against their Lie bracket, leaving $({\Gamma }_{\nu \mu }{}^{\lambda }-{\Gamma }_{\mu \nu }{}^{\lambda })X^{\mu }{Y}^{\nu }$. This also shows that the expression is linear over smooth functions in each argument.

Finally take the torsion-free connection and the commuting parameter fields $Y,Z$ along the geodesic family. Vanishing torsion gives ${\nabla }_{Y}Z={\nabla }_{Z}Y$, while affine geodesic parametrization gives ${\nabla }_{Y}Y=0$. The curvature operator in the specified convention isThereforeIn components this is the geodesic deviation equationThe connecting field represents infinitesimal separation within the family, so this is a linear equation for relative free-fall acceleration, rather than an exact coordinate difference for widely separated worldlines.

A Killing vector field $K$ generates local isometries: its flow of a vector field ${\Phi }_s$ satisfies ${\Phi }_s^{\ast }g=g$. Equivalently, the Lie derivative of the metric tensor vanishes. For the Levi-Civita connection, metric compatibility converts the coordinate expression into the Killing equation:The Levi-Civita connection is an implicit extra assumption in this question. The printed assumption of vanishing torsion tensor alone does not imply metric compatibility. For example, on Minkowski spacetime, take the affine connection whose only nonzero Christoffel symbol is ${\Gamma }^0{}_{00}=1$. The vector field $K={\partial }_0$ preserves the Minkowski metric, but $K_0=-1$ gives ${\nabla }_0{K}_0=1$, contradicting the Killing equation. The following calculation, and the covariant formulas in part (b), use the Levi-Civita connection; this is the role of metric compatibility in Killing identities.

For the stated curvature sign convention, the Ricci identity on a covector has a minus sign:Write $A_{\mu \nu \rho }={\nabla }_{\mu }{\nabla }_{\nu }{K}_{\rho }$. Differentiating the Killing equation gives $A_{\mu \nu \rho }=-A_{\mu \rho \nu }$. Alternately using this relation and the Ricci identity givesThe first Bianchi identity and antisymmetry of the last two indices of the Riemann curvature tensor reduce the three curvature terms to $2R^{\sigma }{}_{\mu \nu \rho }{K}_{\sigma }$. Consequently the second covariant derivative of a Killing vector is

On the extension in part (i), the hypersurface $\mathcal{N}$ has nonzero normal covector $n_a={\partial }_a\lambda$. Its squared norm isThis vanishes on $\mathcal{N}$, which is therefore a null hypersurface. Raising the index givesSince ${\partial }_t={\partial }_v$ at fixed $\lambda$, the original Killing vector field becomes the smooth nonzero field $K={\partial }_v$. The transformed metric tensor is independent of $v$, so $K$ remains a Killing vector field, andThus $K$ is both tangent and normal to $\mathcal{N}$. This is exactly a Killing horizon:The Killing vector field itself does not vanish there; a vanishing squared norm means nullness, not a zero vector.

A regular hypersurface $f=0$, with $df\ne 0$, is a null hypersurface if its normal covector has zero squared norm:Its normal vector is then also tangent to the hypersurface; the induced metric tensor is degenerate.

For the Kerr metric the normal to a constant-$r$ surface has squared normFor $0<a<M$ the two roots are positive and distinct. To verify that they are genuine regular null hypersurfaces, rather than trusting singular Boyer-Lindquist coordinates, introduce ingoing Kerr coordinatesSubstitution givesThere are no ${\Delta }^{-1}$ terms. At either root $\Sigma >0$, and the determinant is $-{\Sigma }^2{\sin }^2\theta$ away from the angular coordinate poles. Thus the metric tensor extends smoothly across each root. The raised normal on either surface isThis is a nonzero null Killing vector field normal to the surface. Both surfaces are Killing horizons, with horizon angular velocities ${\Omega }_{\pm }$.

The event-horizon terminology in the paper needs qualification. In the usual asymptotically flat Kerr black hole extension, relative to a chosen exterior's future null infinity, the future branch of $r=r_{+}$ is the event horizon, the boundary $\partial J^{-}({\mathcal{I}}^{+})$. In the intervening black-hole region $\Delta <0$, the coordinate $r$ has timelike gradient and future-directed curves move toward smaller $r$; they cannot return across the outer boundary to that exterior. The inner surface $r=r_{-}$ is a Cauchy horizon, not a second event horizon for the same asymptotic end. Nullness is a local property; identifying an event horizon additionally requires this global causal structure.^[ref]

The allowed endpoint $a=0$ provides an especially concrete counterexample to the unrestricted description of both roots. The Kerr metric then becomes Schwarzschild spacetime, $r_{+}=2M$ is its regular event horizon, but $r_{-}=0$ is a curvature singularity:It is not a regular inner null hypersurface. Thus the two-horizon calculation above assumes $0<a<M$.

Continue with the active sign convention for the linearized coordinate gauge transformation. Choose local functions $A,B,C,F$ of $u$ by integration so thatFor example, for any fixed $u_0$ one may takeA generator, specified as a covector with components in the null coordinates $(u,v,x,y)$, isIndeed, with $\delta h_{AB}={\partial }_A{\chi }_B+{\partial }_B{\chi }_A$, the mixed components cancel:All components without a $u$ index vanish as well, andSince $n^{♭}=du$, only the $uu$ component needs changing. ThusYes, $\mathring{h}$ satisfies both equations (3). Its trace is zero, ${\partial }_v\psi =0$ gives zero divergence, and the harmonic quadratic plane-wave profile givesThe two surviving functions encode the two gravitational wave polarizations; constant and linear transverse terms have been removed by the explicit linearized coordinate gauge transformation.