Framework And Geometry — Examples

The component functions of $A$ in the global chart are the constants $A_{ab}=\mathrm{diag}(-1,1,1,1{)}_{ab}$, hence $C^{\mathrm{\infty }}$; by [L1] the family $A$ is therefore a smooth covariant $2$-tensor field, and the component matrix is symmetric, invertible with $\det =-1$, and has three positive and one negative diagonal entries, so each $A_p$ is symmetric, nondegenerate and has inertia $(3,1,0)$, i.e., is a Lorentz form of signature $(-,+,+,+)$ by [L2]. Hence $(M,A)$ is a smooth Lorentzian four-manifold.

The vector field $N={\mathrm{\partial }}_0$ has constant components $(1,0,0,0)$ in the global chart, so it is smooth by [L4], and $A_p(N_p,N_p)=-1<0$ at every $p$; by [L3] this makes $(M,A,N)$ a time-oriented smooth Lorentzian four-manifold.

For a nonzero $v={\sum }_a{v}^a{\mathrm{\partial }}_a{\mathrm{\mid }}_p$, $A_p(v,v)=-(v^0{)}^2+\mid v_{\mathrm{v}\mathrm{e}\mathrm{c}}{\mid }^2$, so $v$ is causal exactly when $\mid v^0\mid \ge \mid v_{\mathrm{v}\mathrm{e}\mathrm{c}}\mid$ (equality being the null case), that is, when $v^0\ge \mid v_{\mathrm{v}\mathrm{e}\mathrm{c}}\mid$ or $v^0\le -\mid v_{\mathrm{v}\mathrm{e}\mathrm{c}}\mid$; among these, by [L3] and step 2.1, $v$ is future-directed exactly when $A_p(v,N_p)=-v^0<0$, that is, when $v^0>0$. Hence the future-directed causal vectors are exactly the nonzero $v$ with $v^0\ge \mid v_{\mathrm{v}\mathrm{e}\mathrm{c}}\mid$, which is the flat case of the description proved in [L3] for the field $N$.

Evaluating the description of step 3.1: $v_{(1)}=(1,0,0,0)$ has $v_{(1)}^0=1>0=\mid v_{(1),\mathrm{v}\mathrm{e}\mathrm{c}}\mid$, so $v_{(1)}$ is future timelike; $v_{(2)}=(1,1,0,0)$ has $A_p(v_{(2)},v_{(2)})=-1+1=0$ with $v_{(2)}\mathrm{\ne }0$ and $v_{(2)}^0=1=\mid v_{(2),\mathrm{v}\mathrm{e}\mathrm{c}}\mid$, so $v_{(2)}$ is future null; $v_{(3)}=(0,1,0,0)$ has $A_p(v_{(3)},v_{(3)})=1>0$, so $v_{(3)}$ is spacelike.

$\mathrm{\Phi }$ is smooth on the open set $W$ as a tuple of polynomials and trigonometric functions, and its differential has the matrix with rows obtained by differentiating the four coordinate functions; a direct computation of the $4\times 4$ determinant gives $\det D\mathrm{\Phi }(u)=r^2\sin \vartheta$, which is nonzero on $W$ because $r>0$ and $0<\vartheta <\pi$.

$\mathrm{\Phi }$ is injective on $W$: if $\mathrm{\Phi }(u)=\mathrm{\Phi }(u^{\mathrm{\prime }})$, then $(u^1{)}^2=(\mathrm{\Phi }(u{)}^1{)}^2+(\mathrm{\Phi }(u{)}^2{)}^2+(\mathrm{\Phi }(u{)}^3{)}^2=(u^{\mathrm{\prime }1}{)}^2$, so $u^1=u^{\mathrm{\prime }1}>0$; then $u^1\cos u^2=\mathrm{\Phi }(u{)}^3=u^1\cos u^{\mathrm{\prime }2}$ gives $\cos u^2=\cos u^{\mathrm{\prime }2}$ with $u^2,u^{\mathrm{\prime }2}\in (0,\pi )$, hence $u^2=u^{\mathrm{\prime }2}$; then $\sin u^2>0$ and the pair of first two spatial components gives $(\cos u^3,\sin u^3)=(\cos u^{\mathrm{\prime }3},\sin u^{\mathrm{\prime }3})$ with $u^3,u^{\mathrm{\prime }3}\in (0,2\pi )$, hence $u^3=u^{\mathrm{\prime }3}$; finally $u^0=u^{\mathrm{\prime }0}$ from the first component.

Let $z\in \mathrm{\Phi }(W)$ and take a single $u\in W$ with $\mathrm{\Phi }(u)=z$. Step 1.1 and [L2] supply open neighbourhoods $u\in O\subseteq W$ and $z\in V=\mathrm{\Phi }(O)$ on which $\mathrm{\Phi }{\mathrm{\mid }}_O$ has a smooth inverse. In particular $z\in V\subseteq \mathrm{\Phi }(W)$; since every image point has such an open neighbourhood, $\mathrm{\Phi }(W)$ is open. By step 1.2, the inverse on this one $V$ equals the unique set-theoretic inverse $y={\mathrm{\Phi }}^{-1}$ restricted to $V$. Consequently $y$ is smooth near each image point, hence smooth on $\mathrm{\Phi }(W)$, without selecting neighbourhoods simultaneously. It and its smooth inverse $\mathrm{\Phi }$ form a homeomorphism with smooth transitions to the identity chart, so [L3] gives the asserted smooth chart.

Apply the transformation law [L1] with the chart $y$ and $x(u):=\mathrm{\Phi }(u)$, so that $\frac{\mathrm{\partial }{x}^c}{\mathrm{\partial }{y}^a}(u)=D\mathrm{\Phi }(u{)}^c{}_a$. The first column of $D\mathrm{\Phi }$ is $(1,0,0,0)$, so $A_{a0}^{\mathrm{\prime }}=A_{0a}^{\mathrm{\prime }}=-1$ for $a=0$ and $0$ for $a\mathrm{\ne }0$. The remaining columns are the partial derivatives of $(r\sin \vartheta \cos \phi ,r\sin \vartheta \sin \phi ,r\cos \vartheta )$ with respect to $r,\vartheta ,\phi$, whose squared Euclidean norms are $1$, $r^2$ and $r^2{\sin }^2\vartheta$, and whose pairwise Euclidean inner products are $0$; since $A_{cd}=\mathrm{diag}(-1,1,1,1)$ and the spatial indices carry the positive signs, the transformation law gives $A_{11}^{\mathrm{\prime }}=1$, $A_{22}^{\mathrm{\prime }}=r^2$, $A_{33}^{\mathrm{\prime }}=r^2{\sin }^2\vartheta$ and $A_{ab}^{\mathrm{\prime }}=0$ for the remaining off-diagonal pairs.

The component functions of step 3.1 are smooth on $W$, so by [L1] the transformed family is a smooth covariant $2$-tensor field on $\mathrm{\Phi }(W)$; its component matrix has determinant $-r^4{\sin }^2\vartheta \mathrm{\ne }0$ on $W$ and diagonal entries of both signs, with exactly one negative entry, so the form has inertia $(3,1,0)$ and the chart expression defines a Lorentzian metric on this chart domain. This shows that a chart change without a change of the field produces non-constant components and that the Lorentzian character is preserved by the transformation law of [L1].

The operator $\mathrm{\nabla }$ is well defined on smooth fields because the component functions $Y^b$ of [L5] are smooth so $X(Y^b)$ is a smooth function. It is additive and $\mathbb{R}$-linear in $Y$ because the components do and $X$ is $\mathbb{R}$-linear; it is function-linear in $X$ because $(fX)(Y^b)=fX(Y^b)$; and ${\mathrm{\nabla }}_X(fY)={\sum }_{b}X(f{Y}^b){\mathrm{\partial }}_b={\sum }_b(X(f)Y^b+fX(Y^b)){\mathrm{\partial }}_b=X(f)Y+f{\mathrm{\nabla }}_{X}Y$. Hence $\mathrm{\nabla }$ is an affine connection by [L1].

For the coordinate fields, $Y={\mathrm{\partial }}_b$ has constant components $Y^c={\delta }^c{}_b$, so $X(Y^c)=0$ and the definition of step 1.1 gives ${\mathrm{\nabla }}_{{\mathrm{\partial }}_a}{\mathrm{\partial }}_b=0$. By [L2] the Christoffel symbols of $\mathrm{\nabla }$ in the global chart are ${\mathrm{\Gamma }}^c{}_{ab}=0$.

The metric components $A_{ab}=\mathrm{diag}(-1,1,1,1{)}_{ab}$ are constant, so ${\mathrm{\partial }}_a{A}_{bc}=0$; the compatibility identity of [L3] reads $0=0+0$ and holds, and the symmetry identity ${\mathrm{\Gamma }}^c{}_{ab}={\mathrm{\Gamma }}^c{}_{ba}$ reads $0=0$ and holds. By [L3] the connection is torsion-free and $A$-compatible, hence metric-selected by [L4].

Define ${\mathrm{\nabla }}_{X}Y:={\sum }_{a,b,c}{X}^a{Y}^b{\mathrm{\Gamma }}^c{}_{ab}{\mathrm{\partial }}_c+{\sum }_{b}X(Y^b){\mathrm{\partial }}_b$ for smooth fields $X={\sum }_a{X}^a{\mathrm{\partial }}_a$, $Y={\sum }_b{Y}^b{\mathrm{\partial }}_b$ in the global chart, with the given coefficients. Then ${\mathrm{\nabla }}_{fX}Y=f{\mathrm{\nabla }}_{X}Y$ because the coefficients depend bilinearly on the components of $X,Y$, ${\mathrm{\nabla }}_X(fY)={\sum }_{a,b,c}{X}^a(fY{)}^b{\mathrm{\Gamma }}^c{}_{ab}{\mathrm{\partial }}_c+{\sum }_{b}X(f{Y}^b){\mathrm{\partial }}_b=f{\mathrm{\nabla }}_{X}Y+X(f)Y$ by the product rule, and $\mathbb{R}$-bilinearity holds componentwise; hence $\mathrm{\nabla }$ is an affine connection by [L1].

By [L2] the Christoffel symbols of $\mathrm{\nabla }$ in the global chart are the given ${\mathrm{\Gamma }}^0{}_{01}={\mathrm{\Gamma }}^0{}_{10}=1$ and ${\mathrm{\Gamma }}^a{}_{bc}=0$ otherwise; in particular ${\mathrm{\Gamma }}^a{}_{bc}={\mathrm{\Gamma }}^a{}_{cb}$ for all $a,b,c$, so $\mathrm{\nabla }$ is torsion-free by [L3].

The compatibility identity of [L3] fails at every point: for $(a,b,c)=(1,0,0)$ the left-hand side is ${\mathrm{\partial }}_1{A}_{00}=0$ because $A_{00}=-1$ is constant, while the right-hand side is ${\mathrm{\Gamma }}^0{}_{10}{A}_{00}+{\mathrm{\Gamma }}^0{}_{10}{A}_{00}=1\cdot (-1)+1\cdot (-1)=-2\mathrm{\ne }0$. By the equivalence [L3], $\mathrm{\nabla }$ is not compatible with $A$; it is therefore not metric-selected by [L4].

The connection of steps 1.1-1.2 is an affine connection that is torsion-free but not $A$-compatible, so it refutes the displayed statement. Since it is not compatible it also shows that the two conditions in the definition of a metric-selected connection are independent in the sense that torsion freedom alone does not force compatibility.

The Jacobian of the coordinate change is diagonal: $\frac{\mathrm{\partial }{x}^0}{\mathrm{\partial }{u}^0}=c$, $\frac{\mathrm{\partial }{x}^i}{\mathrm{\partial }{u}^i}=1$ for $i=1,2,3$, and all off-diagonal entries vanish. Applying [L1] to the diagonal matrix $A=\mathrm{diag}(-1,1,1,1)$ gives $A_{00}^{\mathrm{\prime }}=c^2{A}_{00}=-c^2$, $A_{ij}^{\mathrm{\prime }}=A_{ij}={\delta }_{ij}$ for $i,j\ge 1$, and $A_{0i}^{\mathrm{\prime }}=A_{i0}^{\mathrm{\prime }}=0$; hence $A^{\mathrm{\prime }}=\mathrm{diag}(-c^2,1,1,1)$.

By [L2] the factor $c^2$ carries dimension ${\mathrm{m}}^2{\mathrm{s}}^{-2}$, so $A_{00}^{\mathrm{\prime }}=-c^2$ carries that dimension, while $A_{11}^{\mathrm{\prime }}=A_{22}^{\mathrm{\prime }}=A_{33}^{\mathrm{\prime }}=1$ are dimensionless; in particular the components are no longer all dimensionless. In the $u$ chart, $d{u}^0=dt$ carries seconds and $d{u}^i$ metres, so $A_{00}^{\mathrm{\prime }}(d{u}^0{)}^2$ and $A_{ij}^{\mathrm{\prime }}d{u}^{i}d{u}^j$ both carry dimension ${\mathrm{m}}^2$, and so does their sum $d{s}^2$, in agreement with [L2]; in the $x$ chart the same expression reads $A_{ab}d{x}^{a}d{x}^b$ with all coordinates in metres.

The two component matrices therefore differ by the factor $c^2$ in the time-time entry, and the dimension of a single component depends on the chart while the invariant $d{s}^2$ does not; this is exactly the content of part (b) of SI unit and coordinate conventions of the classical gravitational framework, and the example asserts no numerical value of $c$.

The component matrix $\mathrm{diag}(1,1,-1,-1)$ is constant, hence $C^{\mathrm{\infty }}$, and symmetric with determinant $1\mathrm{\ne }0$; by [L1] the family $B$ is a smooth symmetric nondegenerate covariant $2$-tensor field on $M$.

In the global chart the matrix is already diagonal with two positive entries and two negative entries and no zero entry; by [L2] the inertia of each $B_p$ is therefore $(2,2,0)$, independent of any further basis choice.

Since $(2,2,0)\mathrm{\ne }(3,1,0)$, $B_p$ is not a Lorentz form by [L3], so $(M,B)$ is not a smooth Lorentzian four-manifold; hence the refuted statement is false. The same computation with the constant matrix $\mathrm{diag}(1,1,1,1)$ gives a smooth symmetric nondegenerate field of inertia $(4,0,0)$, a Riemannian metric, which also fails the Lorentzian signature condition; both witnesses show that the signature clause in the definition of a Lorentzian metric is a genuine additional requirement beyond symmetry, smoothness and nondegeneracy.

The field $N^{\mathrm{\prime }}=-N$ has component functions the negatives of those of $N$ in every chart, hence is smooth, and $g_p(N_p^{\mathrm{\prime }},N_p^{\mathrm{\prime }})=g_p(N_p,N_p)<0$ for every $p$; by [L1] the triple $(M,g,N^{\mathrm{\prime }})$ is a time-oriented smooth Lorentzian four-manifold.

For the pair $(N,N^{\mathrm{\prime }})$ the function of [L2] is $f(p)=g_p(N_p,-N_p)=-g_p(N_p,N_p)>0$ for every $p$; so $f$ is positive on every connected set, and the positive-sign case of [L2] gives $C_p^{+}(N^{\mathrm{\prime }})=C_p^{-}(N)$ at every $p$.

By [L3] applied in the two time orientations, the future-directed causal vectors of $N^{\mathrm{\prime }}$ are $\mathrm{cl}(C_p^{+}(N^{\mathrm{\prime }}))\setminus \{0\}=\mathrm{cl}(C_p^{-}(N))\setminus \{0\}$, which are exactly the past-directed causal vectors of $N$; and symmetrically the future-directed causal vectors of $N$ are the past-directed ones of $N^{\mathrm{\prime }}$.

In the flat model, $N^{\mathrm{\prime }}=-{\mathrm{\partial }}_0$ and $g_p({\mathrm{\partial }}_0,N^{\mathrm{\prime }})=-g_p({\mathrm{\partial }}_0,{\mathrm{\partial }}_0)=1>0$ by [L4], so by [L1] the vector ${\mathrm{\partial }}_0$ is past-directed for $N^{\mathrm{\prime }}$; it is future-directed for $N$ since $g_p({\mathrm{\partial }}_0,N)=-1<0$. This is the concrete instance of step 3.1, and it shows that the future/past attribution of a fixed causal vector is relative to the adopted time orientation.