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A Lorentzian inner product is a nondegenerate symmetric bilinear form with one negative and the remaining positive eigenvalues in the mostly-plus convention. Two nonzero future causal vectors have nonpositive inner product; if either is timelike, it is strictly negative.

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 four-vector is a collection of four components $A^{\mu }=(A^0,\mathbf{A})$ that transforms by the same Lorentz transformation as $(ct,\mathbf{x})$. Use the metric signature $(-,+,+,+)$. The Lorentzian inner product isA nonzero four-vector is a timelike vector, null vector or spacelike vector according as $A\cdot A$ is negative, zero or positive. Reversing the metric signature reverses the signs used to name these three classes, without changing their geometric meaning.

For a frame moving at speed $v$ along the positive $x$ axis, write $\beta =v/c$ and use the Lorentz factor $\gamma =(1-{\beta }^2{)}^{-1/2}$. The component Lorentz transformation isExpanding the first two squares givesThe other two components are unchanged, so $A^{\prime }\cdot A^{\prime }=A\cdot A$. Consequently a timelike vector remains timelike under a Lorentz transformation.

For a nonzero null vector, $|\mathbf{A}|=|A^0|$ and $A^0\ne 0$. Set $a=A^0$ and $\hat{\mathbf{n}}=\mathbf{A}/a$. Then $|\hat{\mathbf{n}}|=1$ and $A=a(1,\hat{\mathbf{n}})$. If the zero four-vector is included among null vectors, take $a=0$ and any unit vector.

For two future-pointing null vectors, write $A=a(1,\hat{\mathbf{n}})$ and $B=b(1,\hat{\mathbf{m}})$ with $a,b>0$. The sum of future-pointing null vectors satisfiesbecause the ordinary inner product of two unit vectors is at most one. The sum is null if their spatial directions coincide, and timelike otherwise. Its positive time component makes the sum nonzero and future-pointing.

Work in the Boyer-Lindquist coordinates exterior $r>r_{+}=M+\sqrt{M^2-a^2}$, away from the axes. There $\Delta >0$, $\Sigma >0$, andInverting the $t,\varphi$ block of the Kerr metric givesThus $W^a=-g^{ab}(dt{)}_b=-{\nabla }^{a}t$ has $g(W,W)=g^{tt}<0$, and the smooth nonvanishing timelike vector field $W$ defines a time orientation. Choose it future-directed, matching increasing $t$ near infinity. For every nonzero future causal vector $Z$, the Lorentzian inner product satisfies $g(W,Z)<0$, henceSo $t$ increases strictly along every regular future causal curve in this exterior. Notice that ${\partial }_t$ itself can be spacelike in the Kerr ergoregion; the timelike object used here is $-\nabla t$. Direct substitution also gives $D=\Delta {\sin }^2\theta$.

With metric signature $(-,+,+,+)$, a nonzero vector $v$ is spacelike when its Lorentzian inner product with itself satisfies $g(v,v)>0$. In Minkowski spacetime, this means its spatial components have larger squared Euclidean length than its time component. The classification is preserved by Lorentz transformations. A timelike vector has negative squared Lorentzian inner product, and a null vector has zero squared Lorentzian inner product.

In Minkowski spacetime with metric signature $(-,+,+,+)$, write two nonzero future-pointing null vectors as $A=a(1,\mathbf{n})$ and $B=b(1,\mathbf{m})$, where $a,b>0$ and $\mathbf{n},\mathbf{m}$ are unit vectors. Their Lorentzian inner product givesThus their sum is a future-pointing causal vector. It is a null vector exactly when the spatial directions coincide, and a timelike vector otherwise. The positivity of the time components is essential: adding past- and future-pointing null vectors can instead produce a spacelike vector.