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The differential of the squared-displacement function and the position vector of the reversed geodesic

Statement

Let $(M, g, N)$ be a time-oriented smooth Lorentzian four-manifold with metric-selected connection $\nabla$ (Smooth time-oriented Lorentzian four-manifolds: the geometric objects of the model, Metric compatibility, torsion freedom and the metric-selected connection class), let $p \in M$ and let $U_{p} = \exp_{p} (W_{p})$ be a normal neighborhood of $p$ with $W_{p}$ star-shaped, as in Normal neighborhoods, convex normal sets, the position vector and the squared-displacement function clause (a). Write $v (q) : = \exp_{p}^{- 1} (q)$ for $q \in U_{p}$ , $P (p, q) : = d (\exp_{p})_{v (q)} (v (q))$ , and $D_{p}^{2} (q) : = g_{p} (v (q), v (q))$ .

(a) Differential in the second argument. $D_{p}^{2}$ is smooth on $U_{p}$ and, for every $q \in U_{p}$ and every $w \in T_{q} M$ , $d (D_{p}^{2})_{q} (w) = 2 g_{q} (P (p, q), w) .$

(b) Reversal. Let $q \in U_{p}$ and let $γ : [0, 1] \to M$ , $γ (t) = \exp_{p} (t v (q))$ , be the radial geodesic, so that $γ (1) = q$ . Then the reversed segment $t \mapsto γ (1 - t)$ is the restriction to $[0, 1]$ of the maximal geodesic from $q$ with initial velocity $- P (p, q)$ ; consequently $\exp_{q}^{- 1} (p) = - P (p, q) and P (q, p) = - v (q) = - (\exp_{p})^{- 1} (q)$ whenever the chosen normal inverse at $q$ has a domain containing $- P (p, q)$ . This condition holds in particular when both inverses are those of one strictly convex normal set $C$ containing $p, q$ and the inverse at $p$ used to define $v (q)$ is the one mapping $C$ to its radial domain. Merely having $p, q \in C$ does not identify an independently chosen inverse on $U_{p}$ with the inverse selected by $C$ . Under the inverse-domain condition, $D_{q}^{2} (p) = D_{p}^{2} (q)$ by the Gauss lemma of The Lorentzian Gauss lemma and the radial maximum of proper time.

(c) Differential on pairs in a strictly convex normal set. Use the inverse domains selected by $C$ in this clause. If $C$ is a strictly convex normal set (Existence of strictly convex normal neighborhoods: small coordinate balls are normal neighborhoods of each of their points) and $p, q \in C$ , then $D^{2} : C \times C \to R$ is smooth and for all $u \in T_{p} M$ , $w \in T_{q} M$ , $d D^{2} (p, q) (u, w) = 2 g_{p} (P (q, p), u) + 2 g_{q} (P (p, q), w) .$

Facts & Assumptions

Given: A time-oriented smooth Lorentzian four-manifold $(M, g, N)$ with metric-selected connection $\nabla$ , a point $p \in M$ and a normal neighborhood $U_{p} = \exp_{p} (W_{p})$ with $W_{p}$ star-shaped.

[L1]

The map $\exp_{p}^{- 1} : U_{p} \to W_{p}$ is a diffeomorphism and the Gauss lemma holds: for all $v \in W_{p}$ , $w \in T_{p} M$ , $g_{\exp_{p} v} (d (\exp_{p})_{v} (v), d (\exp_{p})_{v} (w)) = g_{p} (v, w)$ (The Lorentzian Gauss lemma and the radial maximum of proper time clause (a), The geodesic initial-value problem and the exponential map of the metric-selected connection clause (d)).

[L2]

The velocity of the radial geodesic $s \mapsto \exp_{p} (s v)$ is $d (\exp_{p})_{s v} (v)$ ; its radial position vector at parameter $s$ is $d (\exp_{p})_{s v} (s v)$ . The velocity is parallel transported along the geodesic and its direction at $s = 1$ agrees with the direction at $s = 0^{+}$ , i.e. $P (p, q)$ is future timelike whenever $v (q)$ is future timelike, and $P (p, q) = 0$ if and only if $v (q) = 0$ (Covariant differentiation along curves: metric pairing, parallel transport and geodesic speed, The geodesic initial-value problem and the exponential map of the metric-selected connection clause (c), Normal neighborhoods, convex normal sets, the position vector and the squared-displacement function clause (c)).

[L3]

Chain rule and smoothness: $v = \exp_{p}^{- 1}$ is smooth, so $D_{p}^{2} = g_{p} (v (\cdot), v (\cdot))$ is smooth as a sum of products of smooth functions with the constant bilinear form $g_{p}$ (The chain rule for total derivatives: $D (g \circ f) (a) = D g (f (a)) \circ D f (a)$ , The differential of a smooth map).

[L4]

Let $C$ be strictly convex normal and $(q, r) \in C \times C$ . The inverse $v (q, r) = \exp_{q}^{- 1} (r)$ is defined because $\exp_{q} : W_{q} \to C$ is a diffeomorphism. The fiber differential $d (\exp_{q})_{v (q, r)}$ is invertible. In local bundle coordinates the joint exponential map $Φ (q, v) = (q, \exp_{q} v)$ has differential at $(q, v (q, r))$ of block form $d Φ_{(q, v)} = (\begin{array}{cc} I & 0 \\ * & d (\exp_{q})_{v} \end{array}),$ so $d Φ_{(q, v)}$ is invertible. By the inverse function theorem, $Φ$ has a smooth local inverse around $(q, r)$ . On its overlap with $C \times C$ this local inverse agrees with $(q^{'}, r^{'}) \mapsto \exp_{q^{'}}^{- 1} (r^{'})$ . Indeed, the geodesic segment for $(q, r)$ has compact image in the open set $C$ ; smooth dependence on initial data implies that the local inverse segments for all sufficiently nearby pairs $(q^{'}, r^{'})$ also lie in $C$ . Strict convexity gives a unique geodesic segment in $C$ between each such pair, so the local inverse segment is the segment selected by $\exp_{q^{'}}^{- 1} (r^{'})$ . These local inverses show that the two-point inverse is smooth on all of $C \times C$ . This is a local argument at every pair; the near-diagonal domain in The joint exponential map: smoothness, invertible differential at the diagonal, and the small-velocity two-point inverse clause (c) alone would not cover all of $C \times C$ (The joint exponential map: smoothness, invertible differential at the diagonal, and the small-velocity two-point inverse clause (a), Existence of strictly convex normal neighborhoods: small coordinate balls are normal neighborhoods of each of their points clause (a), The smooth inverse function theorem on manifolds).

Proof

technique · direct

1.1L1L3givenalgebra

(Smoothness and the differential in the second argument.) By [L3], $D_{p}^{2}$ is smooth on $U_{p}$ . Fix $q \in U_{p}$ , $w \in T_{q} M$ and a smooth curve $c : (- ε, ε) \to U_{p}$ with $c (0) = q$ , $c^{'} (0) = w$ ; put $u (t) : = \exp_{p}^{- 1} (c (t))$ , so $u$ is smooth and $u^{'} (0) = d (\exp_{p}^{- 1})_{q} (w)$ . Differentiating $D_{p}^{2} (c (t)) = g_{p} (u (t), u (t))$ at $t = 0$ gives $d (D_{p}^{2})_{q} (w) = 2 g_{p} (u (0), u^{'} (0)) = 2 g_{p} (v (q), d (\exp_{p}^{- 1})_{q} (w))$ . Since $d (\exp_{p})_{v (q)} \circ d (\exp_{p}^{- 1})_{q} = {i d}_{T_{q} M}$ , the Gauss lemma [L1] gives $g_{p} (v (q), d (\exp_{p}^{- 1})_{q} (w)) = g_{q} (d (\exp_{p})_{v (q)} (v (q)), w) = g_{q} (P (p, q), w)$ . This proves (a).

2.1step 1.1L1L2L4givenalgebra

(Reversal.) Let $v : = v (q)$ and let $η (t) : = γ (1 - t) = \exp_{p} ((1 - t) v)$ for $t \in [0, 1]$ . Since $W_{p}$ is star-shaped, $(1 - t) v \in W_{p}$ and $η (t) = \exp_{p} ((1 - t) v)$ for $t \in [0, 1]$ ; as $s \mapsto \exp_{p} (s v)$ is a geodesic and $s = 1 - t$ is an affine reparametrization with nonzero derivative, $η$ is a geodesic with $η (0) = q$ and $η^{'} (0) = - γ^{'} (1) = - P (p, q)$ , where $γ^{'} (1) = d (\exp_{p})_{v} (v)$ . Uniqueness of the maximal initial-value solution implies that $η$ is its restriction to $[0, 1]$ , and in particular $\exp_{q} (- P (p, q)) = η (1) = p$ . Whenever the chosen normal inverse at $q$ has a domain containing $- P (p, q)$ , injectivity of that domain's exponential map gives $\exp_{q}^{- 1} (p) = - P (p, q)$ . For that same chosen inverse domain, the reversed segment has terminal tangent $η^{'} (1) = - γ^{'} (0) = - v (q)$ , so $P (q, p) = - v (q)$ . If both inverse domains map onto the same strictly convex normal set $C$ , the reversed segment stays in $C$ ; uniqueness in $C$ identifies it with the radial segment selected at $q$ , so its initial tangent $- P (p, q)$ belongs to that inverse domain. Finally, by [L1] at base point $p$ with $w = v$ , $D_{q}^{2} (p) = g_{q} (\exp_{q}^{- 1} (p), \exp_{q}^{- 1} (p)) = g_{q} (P (p, q), P (p, q)) = g_{p} (v, v) = D_{p}^{2} (q)$ under that same inverse-domain condition. This proves (b).

3.1step 1.1step 2.1L4algebra

(Differential on pairs.) Let $C$ be strictly convex normal and $p, q \in C$ . By [L4] the map $(p^{'}, q^{'}) \mapsto D^{2} (p^{'}, q^{'})$ is smooth on $C \times C$ , and by step 2.1 its values are symmetric, $D^{2} (p^{'}, q^{'}) = D^{2} (q^{'}, p^{'})$ . Fix $u \in T_{p} M$ and $w \in T_{q} M$ , choose smooth curves $p (s), q (t)$ in $C$ with $p (0) = p$ , $p^{'} (0) = u$ , $q (0) = q$ , $q^{'} (0) = w$ , and put $F (s, t) : = D^{2} (p (s), q (t)) = D^{2} (q (t), p (s))$ . By part (a) applied with base point $q (t)$ , $\partial_{s} F (0, t) = d (D^{2} (q (t), \cdot))_{p} (u) = 2 g_{p} (P (q (t), p), u)$ ; and by part (a) with base point $p (s)$ , $\partial_{t} F (s, 0) = 2 g_{q} (P (p (s), q), w)$ . The mixed partial derivatives of the smooth function $F$ at $(0, 0)$ agree (Clairaut--Schwarz theorem for continuous second partial derivatives), and $P (q (t), p) \to P (q, p)$ as $t \to 0$ by the smoothness of [L4]; hence $d D^{2} (p, q) (u, w) = \partial_{s} F (0, 0) + \partial_{t} F (0, 0) = 2 g_{p} (P (q, p), u) + 2 g_{q} (P (p, q), w)$ . This proves (c).

4.1step 1.1step 2.1step 3.1∎

Claims (a), (b) and (c) are steps 1.1, 2.1 and 3.1.

Remarks

Consistency check. For $q = p$ both sides of (c) vanish because $P (p, p) = 0$ ; the formula is therefore compatible with $D^{2} (p, p) = 0$ and with the symmetry $D^{2} (p, q) = D^{2} (q, p)$ .
Scope. Part (a) is stated on a normal neighborhood with a chosen inverse. Part (b) requires compatibility of the chosen inverse domains with the reversed radial segment; part (c) uses the inverse domains of one strictly convex normal set, which satisfy that compatibility by Existence of strictly convex normal neighborhoods: small coordinate balls are normal neighborhoods of each of their points. No completeness and no global injectivity of $\exp$ is used.

The differential of the squared-displacement function and the position vector of the reversed geodesic

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources