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Smoothness and transformation of the inverse-metric components

Statement

Let $(M, g, N)$ be a time-oriented smooth Lorentzian four-manifold and let $g_{p}^{- 1}$ and $g^{a b}$ be the inverse metric and its chart components as in The inverse metric, musical isomorphisms and index raising. Then:

For every smooth chart $(U, x)$ the component functions $g^{a b} : U \to R$ are $C^{\infty}$ on $U$ .
For overlapping smooth charts $(U, x)$ , $(V, y)$ the components satisfy the contravariant transformation law

$g^{' a b} (u) = \sum_{c, d = 0}^{3} \frac{\partial y^{a}}{\partial x^{c}} (x (u)) \frac{\partial y^{b}}{\partial x^{d}} (x (u)) g^{c d} (x (u)), u \in y (U \cap V),$

where every component evaluated at a coordinate tuple is its coordinate representative, as in Smooth covariant 2-tensor fields on a smooth manifold, $x (u)$ denotes $(x \circ y^{- 1}) (u)$ , and $\partial y^{a} / \partial x^{c}$ are the entries of $D (y \circ x^{- 1})$ .

Consequently the pointwise bilinear forms $g_{p}^{- 1}$ on $T_{p}^{*} M$ are presented by component functions that are smooth in every chart and glue by the contravariant law; the inverse metric is a well-defined smooth component family determined by $g$ alone.

Facts & Assumptions

Given: A time-oriented smooth Lorentzian four-manifold $(M, g, N)$ and smooth charts $(U, x)$ , $(V, y)$ with $U \cap V \neq \emptyset$ .

[L1]

In a chart, $g^{a b} (q)$ are the entries of the inverse matrix of $(g_{a b} (q))$ , so that $g^{a c} g_{c b} = δ^{a}_{b}$ , and $g_{a b}$ are $C^{\infty}$ on $U$ with $\det (g_{a b} (q)) \neq 0$ for every $q$ (The inverse metric, musical isomorphisms and index raising, Smooth covariant 2-tensor fields on a smooth manifold).

[L2]

If the entries of a square matrix-valued map $A : U \to M_{4} (R)$ on an open set are $C^{r}$ and $\det A$ never vanishes, then the entries of $A^{- 1}$ are $C^{r}$ ; for an invertible matrix $A^{- 1} = \det (A)^{- 1} adj (A)$ (Matrix inversion preserves $C^{k}$ regularity where the determinant is nonzero, If $\det (A)$ is a unit, then $A^{- 1} = \det (A)^{- 1} adj (A)$ ).

[L3]

For overlapping smooth charts, the covariant components of a smooth covariant $2$ -tensor field satisfy $g_{a b}^{'} (y) = \sum_{c, d} \frac{\partial x^{c}}{\partial y^{a}} (y) \frac{\partial x^{d}}{\partial y^{b}} (y) g_{c d} (x (y))$ , where the entries are those of $D (x \circ y^{- 1}) (y)$ (Chart independence and the transformation law of a smooth covariant 2-tensor field).

[L4]

For composable totally differentiable maps, $D (g \circ f) = D g \circ D f$ (The chain rule for total derivatives: $D (g \circ f) (a) = D g (f (a)) \circ D f (a)$ ); the identity map is totally differentiable with $D (id) = I$ , since the identity is a linear map satisfying the defining limit (The total (Fréchet) derivative $D f (a)$ as the linear first-order approximation with $o (∥ h ∥_{2})$ remainder).

[L5]

Overlapping smooth charts have smooth transition maps $x \circ y^{- 1}$ and $y \circ x^{- 1}$ , whose total derivatives are computed from the coordinate functions (Smoothly compatible charts and the smoothness of Euclidean transition maps).

Proof

technique · direct

1.1L1L2given

Let $(U, x)$ be a smooth chart. By [L1] the component functions $g_{a b}$ are $C^{\infty}$ on $U$ and $\det (g_{a b} (q)) \neq 0$ for every $q \in U$ , because $g_{q}$ is nondegenerate; hence the coordinate representative $z \mapsto (g_{a b} (x^{- 1} (z)))$ on the open Euclidean set $x (U)$ is $C^{\infty}$ with nowhere-vanishing determinant, and by [L2] the entries of its inverse at each $q$ , namely $g^{a b} (q)$ by [L1], are $C^{\infty}$ on $U$ .

1.2L3algebra

Let $p \in U \cap V$ and put $y_{0} : = y (p)$ . Write $J : = D (x \circ y^{- 1}) (y_{0})$ for the Jacobian of the transition with entries $J^{c}_{a} = \frac{\partial x^{c}}{\partial y^{a}} (y_{0})$ , so that by [L3] the covariant components are related by the matrix equation $G^{'} (y_{0}) = J^{T} G (x (p)) J$ with $G = (g_{c d})$ , $G^{'} = (g_{a b}^{'})$ .

1.3L4L5algebra

The transition $y \circ x^{- 1}$ is smooth by [L5] and $(y \circ x^{- 1}) \circ (x \circ y^{- 1}) = id$ on $y (U \cap V)$ ; applying [L4] at $y_{0}$ gives $D (y \circ x^{- 1}) (x (p)) \circ D (x \circ y^{- 1}) (y_{0}) = D (id) (y_{0}) = I$ , so $J$ is invertible with inverse the matrix of entries $\frac{\partial y^{a}}{\partial x^{c}} (x (p))$ .

2.1step 1.2step 1.3algebra

Inverting the matrix equation of step 1.2 and using step 1.3 gives $G^{'} (y_{0})^{- 1} = J^{- 1} G (x (p))^{- 1} J^{- T}$ , that is $g^{' a b} (y_{0}) = \sum_{c, d} \frac{\partial y^{a}}{\partial x^{c}} (x (p)) \frac{\partial y^{b}}{\partial x^{d}} (x (p)) g^{c d} (x (p))$ , which is the contravariant transformation law of claim 2.

3.1step 1.1step 2.1given∎

The chart $(U, x)$ in step 1.1 was arbitrary, so claim 1 holds in every chart, and step 2.1 holds for every pair of overlapping charts, so claim 2 holds as well. Since the functions $g^{a b}$ were defined from the pointwise bilinear form $g_{q}^{- 1}$ by evaluation on the dual coordinate covectors, and the two claims show that the resulting component functions are smooth and related by the contravariant law in all charts, the inverse metric is a well-defined smooth component family determined by $g$ alone, which is claim 3.

Smoothness and transformation of the inverse-metric components

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

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