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Ricci tensor and volume form on a Lorentzian surface

Statement

Let $(U, q, ϵ_{2})$ be a connected, oriented, time-oriented smooth two-dimensional Lorentzian manifold of signature $(-, +)$ with ${A C}_{ω}$ . In local coordinates $y^{0}, y^{1}$ , define $Γ^{A}_{B C} : = \frac{1}{2} q^{A D} (\partial_{B} q_{D C} + \partial_{C} q_{B D} - \partial_{D} q_{B C}),$ and use the resulting metric-selected derivative $\nabla^{q}$ and the curvature convention $R^{A}_{B C D} : = \partial_{C} Γ^{A}_{D B} - \partial_{D} Γ^{A}_{C B} + Γ^{A}_{C E} Γ^{E}_{D B} - Γ^{A}_{D E} Γ^{E}_{C B} .$ Let $Ric (q)_{B D} : = q^{A C} q_{A E} R^{E}_{B C D}$ and $S_{q} : = q^{B D} Ric (q)_{B D}$ . Then $\nabla^{q} ϵ_{2} = 0, Ric (q) = \frac{1}{2} S_{q} q .$ The only inherited choice assumption is ${A C}_{ω}$ from the four-dimensional Levi-Civita supplier used below; no full axiom of choice is used.

Statement references

Spherical orbit geometry and rotational invariance

Facts & Assumptions

Given: The smooth oriented, time-oriented Lorentzian surface $(U, q, ϵ_{2})$ and ${A C}_{ω}$ .

[F1]

Spherical orbit geometry and rotational invariance supplies a smooth two-dimensional Lorentzian metric $q$ and its chosen metric volume form $ϵ_{2}$ .

[F2]

Existence and uniqueness of the metric-selected connection of a nondegenerate metric, and its extension to tensor fields gives the Christoffel formula, torsion freedom, metric compatibility and ${A C}_{ω}$ for an oriented, time-oriented Lorentzian four-manifold.

[F3]

The Riemann curvature operator and the derived Ricci, scalar and Einstein tensors fixes $R (X, Y) Z = \nabla_{X} \nabla_{Y} Z - \nabla_{Y} \nabla_{X} Z - \nabla_{[X, Y]} Z$ and ${Ric}_{B D} = q^{A C} R_{A B C D}$ .

[F4]

The algebraic symmetries of the curvature tensor and the symmetry of the Ricci and Einstein tensors gives antisymmetry in each lowered curvature pair and exchange of the two pairs for the four-dimensional product metric used in the proof.

[F5]

Tensoriality of the curvature, its coordinate formula and its point-normal metric formula gives the coordinate formula for that product metric's curvature; its coordinate Christoffels determine the base curvature components.

[A1]

The Axiom of Countable Choice ( ${A C}_{ω}$ ): assume only ${A C}_{ω}$ , exactly the choice strength required by [F2].

Proof

1.1F1F2F5A1givenalgebra

Form the four-manifold $\tilde{M} : = U \times R^{2}$ with product metric $\tilde{g} : = q + (d x^{2})^{2} + (d x^{3})^{2}$ , product orientation, and the lift of the chosen future timelike field on $U$ . This is a smooth time-oriented Lorentzian four-manifold. By [F2], its metric-selected connection has the displayed Christoffel formula. In product coordinates its base Christoffels are exactly $Γ [q]^{A}_{B C}$ , every Christoffel with a flat-factor index is zero, and its curvature components with all indices in the base are exactly $R [q]^{A}_{B C D}$ by [F5]. Thus the base Ricci contraction in the product equals the two-dimensional contraction in the Statement. Moreover, in an oriented base chart, $ϵ_{01} = \pm \sqrt{∣ \det q ∣}$ and $Γ^{C}_{A C} = \frac{1}{2} q^{C D} \partial_{A} q_{C D} = \partial_{A} \log \sqrt{∣ \det q ∣}$ . Substituting these identities in $\nabla_{A} ϵ_{B C} = \partial_{A} ϵ_{B C} - Γ^{D}_{A B} ϵ_{D C} - Γ^{D}_{A C} ϵ_{B D}$ gives $\nabla^{q} ϵ_{2} = 0$ ; the only independent component is $B C = 01$ , for which the connection terms sum to $Γ^{C}_{A C} ϵ_{01}$ and cancel the partial derivative.

2.1F3F4step 1.1algebra∎

At any point choose a $q$ -orthonormal basis $(e_{0}, e_{1})$ with $q (e_{0}, e_{0}) = - 1$ and $q (e_{1}, e_{1}) = 1$ . In the product four-manifold the algebraic symmetries [F4] and antisymmetry in each pair force every base component of $R_{A B C D}$ to vanish unless both pairs are $(0, 1)$ up to order. Write $K : = R_{0101}$ . Pair exchange and the two antisymmetries give $R_{1010} = K$ and all components with repeated entries in either pair vanish. Using [F3], ${Ric}_{00} = q^{11} R_{1010} = K$ , ${Ric}_{11} = q^{00} R_{0101} = - K$ , and ${Ric}_{01} = {Ric}_{10} = 0$ . Hence $Ric (q) = - K q$ and $S_{q} = - 2 K$ , so $Ric (q) = \frac{1}{2} S_{q} q$ . The point was arbitrary; together with step 1.1 this proves both identities.

Ricci tensor and volume form on a Lorentzian surface

Statement

Statement references

Facts & Assumptions

Proof

Depends on

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