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Two-dimensional base identities for the warped spherical metric: pure-trace Ricci, the vacuum reduction and the extra Killing field

Statement

Let $V \subseteq R^{2}$ be a connected nonempty open set, let $(q_{A B})$ be a smooth symmetric matrix of functions on $V$ with $D_{q} : = \det (q_{A B}) < 0$ everywhere, and let $r : V \to (0, \infty)$ be smooth; let $Γ^{A}_{B C}$ , ${Hess}_{A B} r$ , $Δ_{q} r$ , $z$ , $R^{A}_{B C D}$ , $R_{A B C D}$ , $Ric (q)_{A B}$ and $S_{q}$ be the functions defined in the statement of The warped product of a two-dimensional Lorentzian base with a round sphere: Christoffel symbols and Ricci tensor, so that all index sums run over ${0, 1}$ . Assume in addition that $z \neq 0$ everywhere on $V$ and that the warped vacuum equations

$Ric (q)_{A B} = \frac{2}{r} {Hess}_{A B} r, 1 - z - r Δ_{q} r = 0 on V$

hold, where the left-hand equation is the vanishing of the components ${Ric}_{A B}$ and the right-hand equation the vanishing of the components ${Ric}_{i j}$ of the warped metric of The warped product of a two-dimensional Lorentzian base with a round sphere: Christoffel symbols and Ricci tensor clause (c) (the mixed components vanish there identically). Define the base volume form by

$ε_{A B} : = \sqrt{- D_{q}} η_{A B}, η_{01} : = 1, η_{10} : = - 1, η_{00} : = η_{11} : = 0,$

and the volume-form field by

$K_{A} : = \sum_{B} ε_{A B} \partial^{B} r, \partial^{B} r : = \sum_{C} q^{B C} \partial_{C} r, K^{A} : = \sum_{B} q^{A B} K_{B} .$

For a smooth base vector field $ξ = ξ^{A} \partial_{A}$ , define a Killing field of $q$ to mean $L_{ξ} q = 0$ , with Lie derivative defined by $L_{ξ} q = (d / d t) ∣_{t = 0} Φ_{t}^{*} q$ for the local flow $Φ_{t}$ (The Lie derivative of a tensor field). Its coordinate components are $(L_{ξ} q)_{A B} = ξ^{C} \partial_{C} q_{A B} + q_{C B} \partial_{A} ξ^{C} + q_{A C} \partial_{B} ξ^{C}$ by The coordinate formula for the Lie derivative of a covariant tensor. Here $\nabla_{A} ξ_{B} : = \partial_{A} (q_{B C} ξ^{C}) - Γ^{C}_{A B} ξ_{C}$ and the base gradient is $\nabla r = (q^{A B} \partial_{B} r) \partial_{A}$ .

Then:

(a) Pure trace in two dimensions. $Ric (q)_{A B} = \frac{1}{2} S_{q} q_{A B}$ and $S_{q} = \sum_{A, B} q^{A B} Ric (q)_{A B}$ on all of $V$ .

(b) Vacuum reduction of the Hessian. ${Hess}_{A B} r = \frac{1}{2} Δ_{q} r q_{A B}$ , $Δ_{q} r = \frac{1 - z}{r}$ and $S_{q} = \frac{2 Δ_{q} r}{r} = \frac{2 (1 - z)}{r^{2}}$ on $V$ .

(c) Constancy of the mass parameter. $\partial_{C} z = \frac{1 - z}{r} \partial_{C} r$ for $C = 0, 1$ , hence $\partial_{C} (r (1 - z)) = 0$ ; consequently there is a constant $s \in R$ with

$r (1 - z) = s, that is, z = 1 - \frac{s}{r} on V .$

(d) The extra Killing field. $K (r) = 0$ , $\sum_{A} K^{A} \partial_{A} r = 0$ , $\sum_{A, B} K^{A} q_{A B} K^{B} = - z$ , the field is nonzero at every point, $\nabla_{A} K_{B} + \nabla_{B} K_{A} = 0$ for the base connection $Γ^{A}_{B C}$ (so $K$ is a Killing field of $q$ in the local definition above), and the fields $K$ and $\nabla r$ are pointwise linearly independent: the equalities $\sum_{A} K^{A} \partial_{A} r = 0$ , $\sum_{A, B} K^{A} q_{A B} K^{B} = - z$ and $\sum_{A, B} \partial_{A} r q^{A B} \partial_{B} r = z \neq 0$ show that neither field vanishes and that the two are not proportional.

(e) Explicit components of the Killing field. With $λ : = \sqrt{- D_{q}} > 0$ one has $K^{0} = - \frac{\partial_{1} r}{λ}$ and $K^{1} = \frac{\partial_{0} r}{λ}$ .

(f) Scope. The statements are identities of the displayed functions and use only the two warped vacuum equations, the strict positivity of $r$ and the hypothesis $z \neq 0$ ; no field equation of the four-dimensional metric beyond those two components, no global hypothesis on $V$ and no sign or physical reading of $s$ is assumed. The constant $s$ is determined on the connected set $V$ ; that it is the same constant for a larger region requires that region to be connected as well.

Facts & Assumptions

Given: A connected open set $V \subseteq R^{2}$ , a smooth symmetric matrix $q_{A B}$ with $\det q < 0$ on $V$ , a smooth $r : V \to (0, \infty)$ with $z \neq 0$ everywhere, the auxiliary functions of The warped product of a two-dimensional Lorentzian base with a round sphere: Christoffel symbols and Ricci tensor, and the two warped vacuum equations.

[L1]

The functions $Γ^{A}_{B C}$ , ${Hess}_{A B} r$ , $Δ_{q} r$ , $z$ , $R^{A}_{B C D}$ , $R_{A B C D}$ , $Ric (q)_{A B}$ and $S_{q}$ are the explicit functions of The warped product of a two-dimensional Lorentzian base with a round sphere: Christoffel symbols and Ricci tensor, in particular $R^{A}_{B C D} = \partial_{C} Γ^{A}_{D B} - \partial_{D} Γ^{A}_{C B} + \sum_{E} (Γ^{A}_{C E} Γ^{E}_{D B} - Γ^{A}_{D E} Γ^{E}_{C B})$ , $R_{A B C D} = \sum_{E} q_{A E} R^{E}_{B C D}$ , $Ric (q)_{B D} = \sum_{A, C} q^{A C} R_{A B C D}$ and $S_{q} = \sum_{B, D} q^{B D} Ric (q)_{B D}$ ; the compatibility identity $\partial_{C} q_{A B} = \sum_{D} (Γ^{D}_{C A} q_{D B} + Γ^{D}_{C B} q_{A D})$ and the trace identity $Δ_{q} r = \sum_{A} \partial_{A} (q^{A B} \partial_{B} r) + \sum_{B} (\sum_{A} Γ^{A}_{A B}) q^{B D} \partial_{D} r$ hold.

[L2]

For an invertible $2 \times 2$ matrix $q$ with inverse $q^{- 1}$ , Cramer's rule gives $q^{00} = q_{11} / D_{q}$ , $q^{11} = q_{00} / D_{q}$ and $q^{01} = q^{10} = - q_{01} / D_{q}$ , and $q^{00} q^{11} - (q^{01})^{2} = 1 / D_{q}$ and $D_{q} < 0$ (Invertible matrices and the general linear group ${GL}_{n} (F)$ , algebra).

[L3]

The exterior derivative of a smooth function is the covector of its partial derivatives; the chain, product and quotient rules hold, and a smooth function on a connected nonempty open Euclidean set whose differential is zero is constant (A differentiable map on a connected open Euclidean set has zero derivative exactly when it is constant, with $m = 2$ , $q = 1$ ; The derivative $f^{'} (c) = \lim_{x \to c} \frac{f (x) - f (c)}{x - c}$ of $f : A \to R$ at a point $c \in A$ that is a limit point of $A$ , and differentiability on a set, The chain rule, in one line from Carathéodory: if $g$ is differentiable at $c$ and $f$ is differentiable at $g (c)$ , then $f \circ g$ is differentiable at $c$ with $(f \circ g)^{'} (c) = f^{'} (g (c)) g^{'} (c)$ , Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$ , $C^{r}$ and smooth maps between smooth manifolds, Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length).

[L4]

The base Killing definition is $L_{ξ} q = 0$ ; its coordinate Lie-derivative formula is the one in the Statement, valid in every dimension (The Lie derivative of a tensor field, The coordinate formula for the Lie derivative of a covariant tensor). Step 4.2 proves its equivalence with the symmetrized base derivative using [L1].

[L5]

The square root $\sqrt{- D_{q}} > 0$ is smooth because $D_{q} < 0$ and $D_{q}$ is smooth, and $\partial_{C} \log (- D_{q}) = \sum_{A, B} q^{A B} \partial_{C} q_{A B} = 2 \sum_{D} Γ^{D}_{C D}$ (Rational powers $a^{r}$ of a positive base, The chain rule, in one line from Carathéodory: if $g$ is differentiable at $c$ and $f$ is differentiable at $g (c)$ , then $f \circ g$ is differentiable at $c$ with $(f \circ g)^{'} (c) = f^{'} (g (c)) g^{'} (c)$ , The derivative $f^{'} (c) = \lim_{x \to c} \frac{f (x) - f (c)}{x - c}$ of $f : A \to R$ at a point $c \in A$ that is a limit point of $A$ , and differentiability on a set, Continuity and derivatives of positive-base real powers, The natural logarithm has derivative 1/x and equals the integral from 1 to x of 1/t, The Axiom of Countable Choice ( ${A C}_{ω}$ )).

Proof

technique · compute the lowered curvature formula and its pair antisymmetries, derive the two-dimensional pure-trace identity from them, reduce the vacuum equations to the Hessian identity, differentiate $z$, and verify the Killing and component claims for the volume-form field

1.1L1algebra

(The two curvature antisymmetries.) Let $X, Y, Z, W$ be coordinate vector fields on $V$ , so $[X, Y] = 0$ by Coordinate vector fields commute. The compatibility identity in [L1] says $X q (Z, W) = q (\nabla_{X} Z, W) + q (Z, \nabla_{X} W)$ for the base connection defined by the displayed Christoffel symbols. Apply $X$ to the compatibility identity for $Y q (Z, W)$ , apply $Y$ to the one for $X q (Z, W)$ , and subtract. The mixed terms cancel in pairs, and $X Y q (Z, W) - Y X q (Z, W) = 0$ ; using the curvature definition in [L1] gives [L1, algebra] $q (R (X, Y) Z, W) + q (Z, R (X, Y) W) = 0.$ With $X = \partial_{C}$ , $Y = \partial_{D}$ , $Z = \partial_{B}$ , $W = \partial_{A}$ , this is $R_{A B C D} = - R_{B A C D}$ . The defining curvature formula also changes sign when $C, D$ are exchanged, so $R_{A B C D} = - R_{A B D C}$ . In dimension two, a component can therefore be nonzero only if each of $(A, B)$ and $(C, D)$ is $(0, 1)$ or $(1, 0)$ ; the only possible components are $R_{0101}$ and its three skew variants. No pair-exchange identity is used.

1.2L1L5algebra

(The base volume form is parallel.) Set $ε_{01} = λ$ , $ε_{10} = - λ$ , $ε_{00} = ε_{11} = 0$ , where $λ : = \sqrt{- D_{q}} > 0$ . By [L5], $\partial_{C} λ = λ (Γ^{0}_{C 0} + Γ^{1}_{C 1})$ . The coordinate formula for the covariant derivative of a covariant two-tensor gives $\nabla_{C} ε_{A B} = \partial_{C} ε_{A B} - Γ^{D}_{C A} ε_{D B} - Γ^{D}_{C B} ε_{A D}$ . For $(A, B) = (0, 0)$ and $(1, 1)$ the two connection terms cancel because the only nonzero entries of $ε$ are opposite; for $(A, B) = (0, 1)$ the value is $\partial_{C} λ - Γ^{0}_{C 0} λ - Γ^{1}_{C 1} λ = 0$ , and for $(A, B) = (1, 0)$ it is the negative of this. Thus $\nabla_{C} ε_{A B} = 0$ for every choice of indices.

2.1step 1.1L2algebra

(Pure trace.) By step 1.1 the only possibly nonzero lowered components are $R_{0101}$ , $R_{0110} = - R_{0101}$ , $R_{1001} = - R_{0101}$ and $R_{1010} = R_{0101}$ ; write $ρ : = R_{0101}$ . In the sums $Ric (q)_{B D} = \sum_{A, C} q^{A C} R_{A B C D}$ every term has an index repeated inside a pair except those with ${A, B} = {C, D} = {0, 1}$ , so $Ric (q)_{00} = q^{11} R_{1010} = q^{11} ρ$ , $Ric (q)_{11} = q^{00} R_{0101} = q^{00} ρ$ and $Ric (q)_{01} = Ric (q)_{10} = q^{10} R_{1001} = - q^{01} ρ$ . Cramer's rule [L2] gives $q^{00} q^{11} - (q^{01})^{2} = 1 / D_{q}$ , hence $S_{q} = q^{00} Ric (q)_{00} + 2 q^{01} Ric (q)_{01} + q^{11} Ric (q)_{11} = ρ (2 q^{00} q^{11} - 2 (q^{01})^{2}) = 2 ρ / D_{q}$ . Therefore $\frac{1}{2} S_{q} q_{00} = ρ q_{00} / D_{q} = ρ q^{11} = Ric (q)_{00}$ by [L2], $\frac{1}{2} S_{q} q_{11} = ρ q_{11} / D_{q} = ρ q^{00} = Ric (q)_{11}$ , and $\frac{1}{2} S_{q} q_{01} = ρ q_{01} / D_{q} = - ρ q^{01} = Ric (q)_{01}$ ; this proves (a).

3.1step 2.1givenalgebra

(Hessian reduction.) Substituting $Ric (q) = \frac{1}{2} S_{q} q$ of step 2.1 into the first warped vacuum equation gives ${Hess}_{A B} r = \frac{r S_{q}}{4} q_{A B}$ ; tracing with $q^{A B}$ and using symmetry gives $Δ_{q} r = \frac{r S_{q}}{4} \cdot 2 = \frac{r S_{q}}{2}$ , hence $S_{q} = 2 Δ_{q} r / r$ and ${Hess}_{A B} r = \frac{1}{2} Δ_{q} r q_{A B}$ . The second warped vacuum equation gives $r Δ_{q} r = 1 - z$ and hence $Δ_{q} r = (1 - z) / r$ and $S_{q} = 2 (1 - z) / r^{2}$ . This proves (b).

4.1step 3.1L1L2L3algebra

(Derivative of $z$ .) Differentiating $q^{A E} q_{E B} = δ^{A}_{B}$ and using the compatibility identity of [L1] gives [step 3.1, L1, L2, L3, algebra] $\partial_{C} q^{A B} = - Γ^{B}_{C D} q^{A D} - Γ^{A}_{C D} q^{D B} .$ Differentiate $z = \sum_{A, B} q^{A B} \partial_{A} r \partial_{B} r$ and substitute this identity. The two connection sums are equal after exchanging the dummy indices $A, B$ , so $\partial_{C} z = 2 \sum_{A, B} q^{A B} \partial_{A} r (\partial_{C} \partial_{B} r - \sum_{D} Γ^{D}_{C B} \partial_{D} r) = 2 \sum_{A, B} q^{A B} \partial_{A} r {Hess}_{C B} r .$ By step 3.1, ${Hess}_{C B} r = \frac{1}{2} Δ_{q} r q_{C B}$ , so $\partial_{C} z = Δ_{q} r \partial_{C} r = \frac{1 - z}{r} \partial_{C} r$ . Hence $\partial_{C} (r (1 - z)) = (1 - z) \partial_{C} r - r \partial_{C} z = 0$ for $C = 0, 1$ ; since $V$ is connected and open and $r (1 - z)$ is smooth, [L3] makes it constant, and writing this constant as $s$ gives $z = 1 - s / r$ because $r > 0$ . This proves (c).

4.2step 1.2step 3.1L2L4algebra

(The Killing field.) By step 1.2, $\nabla_{C} ε_{A B} = 0$ , so $\nabla_{C} K_{A} = \sum_{B} ε_{A B} q^{B D} {Hess}_{C D} r$ ; with $Hess = \frac{1}{2} Δ_{q} r q$ of step 3.1 this is $\frac{1}{2} Δ_{q} r ε_{A C}$ , and adding $\nabla_{A} K_{C} = \frac{1}{2} Δ_{q} r ε_{C A}$ gives zero by antisymmetry of $ε$ . For any smooth $ξ$ , expanding the locally defined covariant derivatives gives $\nabla_{A} ξ_{B} + \nabla_{B} ξ_{A} = q_{B C} \partial_{A} ξ^{C} + q_{A C} \partial_{B} ξ^{C} + ξ^{C} (\partial_{A} q_{B C} + \partial_{B} q_{A C} - 2 Γ^{D}_{A B} q_{D C})$ . The compatibility identity in [L1] and lower-index symmetry of $Γ$ reduce the parenthesis to $\partial_{C} q_{A B}$ . This is exactly $(L_{ξ} q)_{A B}$ in [L4]. Applying it to $K$ proves $L_{K} q = 0$ , so $K$ is a Killing field under the stated two-dimensional definition. Put $w^{A} : = \partial^{A} r$ . From $K^{A} = q^{A C} ε_{C B} w^{B}$ we have $K (r) = K^{A} \partial_{A} r = w^{C} ε_{C B} w^{B} = 0$ , since $w^{C} w^{B}$ is symmetric and $ε_{C B}$ antisymmetric. Also $K \neq 0$ because $q (K, K) = - z \neq 0$ . The components $K^{0} = - \partial_{1} r / λ$ , $K^{1} = \partial_{0} r / λ$ follow from the definition of $K_{A}$ , Cramer's rule [L2] and $λ^{2} = - D_{q}$ ; direct substitution gives $q (K, K) = λ^{- 2} (q_{00} (\partial_{1} r)^{2} - 2 q_{01} \partial_{0} r \partial_{1} r + q_{11} (\partial_{0} r)^{2}) = λ^{- 2} D_{q} z = - z$ . Since $z \neq 0$ , $\nabla r \neq 0$ ; a linear relation $a K + b \nabla r = 0$ would give $0 = q (K, a K + b \nabla r) = a (- z)$ and $0 = q (\nabla r, a K + b \nabla r) = b z$ , hence $a = b = 0$ . This proves (d) and (e).

5.1step 1.1step 2.1step 3.1step 4.1step 4.2given∎

(Conclusion.) Step 1.1 supplies the pair antisymmetries, step 2.1 the pure trace (a), step 3.1 the reduction (b), step 4.1 the constancy (c), and step 4.2 the Killing field with its properties (d)-(e); clause (f) records the exact hypotheses, in particular that only the two displayed components of the four-dimensional vacuum equations, the positivity of $r$ and $z \neq 0$ were used, and that the constant $s$ is produced on the connected set $V$ .

Remarks

The role of $z \neq 0$ . The hypothesis makes $\nabla r$ non-null and nowhere zero, so that $K$ and $\nabla r$ are independent by step 4.2 and the coordinate construction of the Birkhoff theorem of this page can be carried out. The vacuum equations force $z = 1 - s / r$ , so $z = 0$ precisely at points with $r = s$ . In a horizon-crossing chart this can occur along a level set while $r$ varies; the hypothesis excludes such points from the open set under consideration, not only a degenerate chart with constant $r$ .
Sign of $s$ . Nothing in this item fixes the sign of the constant $s$ : $s = 0$ gives $z = 1$ and the flat warped form, and the proof uses $1 - z = s / r$ only through the smoothness of $r (1 - z)$ . The positivity of a mass parameter requires separate physical hypotheses or regular-center conditions.

Two-dimensional base identities for the warped spherical metric: pure-trace Ricci, the vacuum reduction and the extra Killing field

Statement

Facts & Assumptions

Proof

Remarks

Depends on

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