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C^2 eta-preserving maps of affine Minkowski space are affine Lorentz maps

Statement

Let $(A, V, η, C^{+})$ be the affine Minkowski space with its inertial smooth structure (The affine atlas makes the Minkowski model a smooth 4-manifold diffeomorphic to R^4), and let $F : A \to A$ be a map whose coordinate expression in some (equivalently, every) inertial chart is of class $C^{2}$ and which satisfies

$η (d F_{p} v, d F_{p} w) = η (v, w) for all p \in A and all v, w \in V,$

where $d F_{p} : T_{p} A \to T_{F (p)} A$ is the total derivative and the tangent spaces are identified with $V$ by the canonical isomorphism of The tangent space of the affine Minkowski model at any event is canonically V. Then $F$ is an affine Lorentz map: there exist $p_{0} \in A$ and a linear Lorentz map $L$ with $F (p) = F (p_{0}) + L (p - p_{0})$ for all $p \in A$ (Linear and affine Lorentz maps, and the Lorentz group). Conversely every affine Lorentz map is a $C^{\infty}$ map of $A$ satisfying the same preservation identity. Consequently the $C^{2}$ isometries of the affine Minkowski model are exactly its affine Lorentz maps, and these form a group under composition.

Facts & Assumptions

Given: An oriented time-oriented affine Minkowski space $(A, V, η, C^{+})$ , a map $F : A \to A$ that is $C^{2}$ in inertial charts and satisfies $η (d F_{p} v, d F_{p} w) = η (v, w)$ for all $p, v, w$ under the canonical tangent identification, and an inertial chart $x$ with frame $(o; e_{0}, e_{1}, e_{2}, e_{3})$ and coordinate isomorphism $E (u) = \sum_{a} u^{a} e_{a}$ .

[L1]

A linear map $L : V \to V$ with $η (L v, L w) = η (v, w)$ for all $v, w$ is a linear Lorentz map; an affine Lorentz map is a map $F : A \to A$ with $F (p) = F (p_{0}) + L (p - p_{0})$ for a linear Lorentz map $L$ , and its linear part $L$ is unique (Linear and affine Lorentz maps, and the Lorentz group).

[L2]

Affine Lorentz maps are bijections and form a group under composition (Lorentz maps are isomorphisms of determinant ±1 that preserve causal character and permute the cones; the Lorentz sets are groups).

[L3]

The model is the quadruple $(A, V, η, C^{+})$ with events and displacements $q - p \in V$ (Oriented time-oriented affine Minkowski space (the spacetime primitives)).

[L4]

For the inertial chart $x$ , $x (p) = E^{- 1} (p - o)$ ; $η (e_{a}, e_{b}) = diag (- 1, 1, 1, 1)_{a b}$ , so $η (E a, E b) = η_{0} (a, b)$ with $η_{0} = diag (- 1, 1, 1, 1)$ ; the chart is a bijection $A \to R^{4}$ (Inertial frames and inertial charts of the affine Minkowski model).

[L5]

The inertial charts form a smooth atlas; the transitions between inertial charts are affine, hence smooth (The affine atlas makes the Minkowski model a smooth 4-manifold diffeomorphic to R^4).

[L6]

The tangent space at every event is canonically identified with $V$ ; for an inertial chart the identification is $Φ_{p} (v) = \sum_{a} v^{a} \partial_{a} ∣_{p}$ for $v = \sum_{a} v^{a} e_{a}$ , and under this identification the total derivative of a chart-represented map is the coordinate total derivative conjugated by $E$ (The tangent space of the affine Minkowski model at any event is canonically V).

[L7]

A map between open subsets of Euclidean space is $C^{k}$ when all iterated partial derivatives of order at most $k$ exist and are continuous ( $C^{k}$ Euclidean maps and diffeomorphisms).

[L8]

$f : U \to R^{n}$ is totally differentiable at $a$ with total derivative $D f (a)$ when $f (a + h) = f (a) + D f (a) h + r (h)$ with $∥ r (h) ∥_{2} / ∥ h ∥_{2} \to 0$ (The total (Fréchet) derivative $D f (a)$ as the linear first-order approximation with $o (∥ h ∥_{2})$ remainder).

[L9]

The $j$ -th partial derivative of $f$ at $a$ is $\partial_{j} f (a) = D_{e_{j}} f (a)$ , the directional derivative along $e_{j}$ (Directional derivatives and partial derivatives of a map $U \subseteq R^{m} \to R^{n}$ ).

[L10]

If $f$ is totally differentiable at $a$ then $D_{v} f (a) = D f (a) v$ for every $v$ , in particular $\partial_{j} f (a) = D f (a) e_{j}$ , and the matrix of $D f (a)$ in the standard bases is the Jacobian matrix $(\partial_{j} f_{i} (a))_{i, j}$ (A total derivative computes every directional derivative, and its matrix is the Jacobian).

[L11]

If $f$ is $C^{2}$ on an open subset of $R^{m}$ , then $\partial_{i} \partial_{j} f = \partial_{j} \partial_{i} f$ for all coordinate indices (Clairaut--Schwarz theorem for continuous second partial derivatives).

[L12]

$D (g \circ f) (a) = D g (f (a)) \circ D f (a)$ whenever the two total derivatives exist (The chain rule for total derivatives: $D (g \circ f) (a) = D g (f (a)) \circ D f (a)$ ).

[L13]

Sums, scalar multiples and differences of totally differentiable maps are totally differentiable with the expected total derivatives (Sums and scalar multiples of totally differentiable maps are totally differentiable with the expected derivatives).

[L14]

If $U \subseteq R^{m}$ is convex and open and $f : U \to R^{n}$ is totally differentiable at every point with $D f (z) = 0$ for every $z \in U$ , then $f$ is constant on $U$ (A totally differentiable map with zero derivative on a convex open set is constant).

[L15]

A subset $U \subseteq R^{m}$ is convex when $(1 - t) x + t y \in U$ for all $x, y \in U$ and $t \in [0, 1]$ ; $R^{m}$ is convex and open (A convex subset of $R^{m}$ contains every line segment between two of its points).

[L16]

$η$ is a symmetric bilinear form whose radicals are ${0}$ , so $η (u, z) = 0$ for all $z \in V$ implies $u = 0$ (Lorentz forms of inertia (3,1,0) on a 4-dimensional real vector space); in coordinates $η_{0} (a, b) = - a^{0} b^{0} + a^{1} b^{1} + a^{2} b^{2} + a^{3} b^{3}$ by [L4].

[L17]

A linear map $T : R^{4} \to R^{4}$ with $\ker T = {0}$ satisfies $\dim (im T) = \dim R^{4}$ by rank-nullity (Rank-nullity: $\dim_{F} V = nullity T + rank T$ ), so its image is all of $R^{4}$ (If $\dim_{F} V = n$ and $U$ is a linear subspace of $V$ , then $U$ is finite-dimensional, $\dim_{F} U \leq n$ , and $\dim_{F} U = n$ if and only if $U = V$ ).

Proof

technique · direct

1.1L4L5L6L12L16L17algebra

Put $Φ : = x \circ F \circ x^{- 1} : R^{4} \to R^{4}$ , the coordinate expression of $F$ in the chart $x$ . By hypothesis and [L5], $Φ$ is $C^{2}$ on the open set $R^{4}$ ; by the chain rule [L12] and the tangent identification [L6], at $u = x (p)$ the total derivative of $Φ$ is the conjugate $D Φ_{u} = E^{- 1} \circ d F_{p} \circ E$ of the total derivative of $F$ . Hence for all $u \in R^{4}$ and all $a, b \in R^{4}$ , using $η (E c, E d) = η_{0} (c, d)$ from [L4] and the preservation identity of the Given, $η_{0} (D Φ_{u} a, D Φ_{u} b) = η (d F_{p} (E a), d F_{p} (E b)) = η (E a, E b) = η_{0} (a, b)$ . In particular, if $D Φ_{u} a = 0$ for some $a$ , then $η_{0} (a, b) = η_{0} (D Φ_{u} a, D Φ_{u} b) = 0$ for every $b$ ; testing $b = e_{0}, e_{1}, e_{2}, e_{3}$ gives $a^{0} = 0$ and $a^{1} = a^{2} = a^{3} = 0$ , so $a = 0$ by [L16], and $D Φ_{u}$ has trivial kernel; by [L17] it is invertible, so the vectors $D Φ_{u} e_{0}, \dots, D Φ_{u} e_{3}$ form a basis of $R^{4}$ .

1.2L4L6L7L8L12algebra

Conversely, let $G$ be an affine Lorentz map with linear part $\tilde{L}$ and $G (p) = G (p_{0}) + \tilde{L} (p - p_{0})$ . Its coordinate expression in $x$ is $Φ_{G} (u) = x (G (x^{- 1} (u))) = x (G (p_{0})) + {\tilde{L}}_{x} (u - x (p_{0}))$ with ${\tilde{L}}_{x} : = E^{- 1} \circ \tilde{L} \circ E$ a linear map of $R^{4}$ , hence a polynomial map of degree at most one and consequently $C^{\infty}$ ; its total derivative at every point is the constant ${\tilde{L}}_{x}$ , and for $a, b \in R^{4}$ the Lorentz property of $\tilde{L}$ together with [L4] gives $η_{0} ({\tilde{L}}_{x} a, {\tilde{L}}_{x} b) = η (\tilde{L} (E a), \tilde{L} (E b)) = η (E a, E b) = η_{0} (a, b)$ , which under the identification [L6] is exactly the preservation identity $η (d G_{p} v, d G_{p} w) = η (v, w)$ for all $p, v, w$ .

2.1L8L9L10L11L16step 1.1algebra

For $j \in {0, 1, 2, 3}$ define the vector-valued map $A_{j} : R^{4} \to R^{4}$ by $A_{j} (u) : = \partial_{j} Φ (u) = D Φ_{u} e_{j}$ , the equality being [L10]; each $A_{j}$ is $C^{1}$ because $Φ$ is $C^{2}$ . For $i, j, k$ define the real-valued function $C_{i j k} (u) : = η_{0} (\partial_{i} A_{j} (u), A_{k} (u)) = η_{0} (\partial_{i} \partial_{j} Φ (u), \partial_{k} Φ (u))$ . The identity of step 1.1, $η_{0} (A_{j} (u), A_{k} (u)) = η_{j k}$ with the constant $η_{j k} : = η_{0} (e_{j}, e_{k})$ , says that $u \mapsto η_{0} (A_{j} (u), A_{k} (u))$ is constant; differentiating it with the product rule for the bilinear form $η_{0}$ (which in coordinates is the displayed polynomial of [L16]) gives $C_{i j k} (u) + C_{i k j} (u) = 0$ for all $u$ . Clairaut's theorem [L11] applied to the components of the $C^{2}$ map $Φ$ gives $\partial_{i} \partial_{j} Φ = \partial_{j} \partial_{i} Φ$ pointwise, hence $C_{i j k} = η_{0} (\partial_{j} \partial_{i} Φ, \partial_{k} Φ) = C_{j i k}$ .

3.1step 2.1algebra

We show $C_{i j k} = 0$ for all $i, j, k$ . If $i, j, k$ are pairwise distinct, the two relations of step 2.1 give $C_{i j k} = C_{j i k} = - C_{j k i} = - C_{k j i} = C_{k i j} = C_{i k j} = - C_{i j k}$ , where each equality is an instance of $C_{a b c} = C_{b a c}$ (first two indices) or $C_{a b c} = - C_{a c b}$ (last two indices); hence $2 C_{i j k} = 0$ and $C_{i j k} = 0$ in $R$ . If exactly two of $i, j, k$ coincide, consider the three possible coincidences: $C_{i i k} = - C_{i k i}$ and $C_{i k i} = C_{k i i}$ and $C_{k i i} = - C_{k i i}$ , so $C_{k i i} = 0$ , then $C_{i k i} = 0$ , then $C_{i i k} = 0$ ; the remaining coincidences are the same three after renaming. If $i = j = k$ , then $C_{i i i} = - C_{i i i}$ by the last-two relation, so $C_{i i i} = 0$ . Thus $C_{i j k} = 0$ for every triple.

4.1L10L16step 1.1step 3.1algebra

Fix $i, j$ and $u$ , and put $w : = \partial_{i} A_{j} (u) \in R^{4}$ . By step 3.1, $η_{0} (w, A_{k} (u)) = C_{i j k} (u) = 0$ for $k = 0, 1, 2, 3$ ; since $(A_{0} (u), \dots, A_{3} (u))$ is a basis of $R^{4}$ by step 1.1, expanding an arbitrary $z \in R^{4}$ in that basis and using bilinearity of $η_{0}$ gives $η_{0} (w, z) = 0$ for every $z$ , whence $w = 0$ by [L16]. Therefore $\partial_{i} A_{j} (u) = 0$ for all $i, j, u$ ; equivalently the total derivative $D A_{j} (u)$ , whose matrix is the Jacobian $(\partial_{i} (A_{j})_{m} (u))$ by [L10], is the zero linear map for every $u$ .

5.1L14L15step 4.1

By step 4.1 and [L14], applied on the convex open set $R^{4}$ of [L15], each map $A_{j} : R^{4} \to R^{4}$ is constant on $R^{4}$ : there are vectors $L_{j} \in R^{4}$ with $A_{j} (u) = L_{j}$ for all $u$ and all $j$ .

6.1L8L13L14L15step 5.1algebra

Let $L : R^{4} \to R^{4}$ be the linear map defined by $L e_{j} : = L_{j}$ for $j = 0, 1, 2, 3$ (Linear map between vector spaces over the same field); it is well defined because $(e_{0}, \dots, e_{3})$ is a basis of $R^{4}$ . By step 5.1 and step 2.1, $D Φ_{u} e_{j} = A_{j} (u) = L_{j} = L e_{j}$ for all $u$ and all $j$ , so the two linear maps $D Φ_{u}$ and $L$ agree on a basis and $D Φ_{u} = L$ for every $u$ . Define $Ψ (u) : = Φ (u) - L u$ ; by [L13] $Ψ$ is totally differentiable at every point, and since the linear map $L$ is totally differentiable with $D L (u) = L$ (substituting $L (u + h) - L (u) - L h = 0$ into [L8] verifies the defining limit), [L13] gives $D Ψ (u) = D Φ_{u} - L = L - L = 0$ for every $u$ . Applying [L14] on the convex open set $R^{4}$ again gives a vector $b \in R^{4}$ with $Ψ (u) = b$ for all $u$ ; hence $Φ (u) = L u + b$ for all $u \in R^{4}$ .

7.1L1L4step 1.1step 6.1algebraL3

Evaluating the preservation identity of step 1.1 at $u = 0$ and using $D Φ_{0} = L$ from step 6.1 gives $η_{0} (L a, L b) = η_{0} (a, b)$ for all $a, b \in R^{4}$ . Define the linear map $\tilde{L} : V \to V$ by $\tilde{L} : = E \circ L \circ E^{- 1}$ ; for $v = E a$ , $w = E b$ with $a, b \in R^{4}$ , [L4] gives $η (\tilde{L} v, \tilde{L} w) = η (E (L a), E (L b)) = η_{0} (L a, L b) = η_{0} (a, b) = η (v, w)$ , so $\tilde{L}$ is a linear Lorentz map by [L1]. Moreover, for every event $p$ with $u : = x (p)$ , step 6.1 and the chart identity $x^{- 1} (u) = o + E (u)$ of [L4] give $F (p) = x^{- 1} (Φ (u)) = x^{- 1} (L u + b) = o + E (L u + b) = o + E b + E (L u) = F (o) + \tilde{L} (E (u))$ , and since $x (p) = E^{- 1} (p - o)$ this reads $F (p) = F (o) + \tilde{L} (p - o)$ for every $p \in A$ : $F$ is an affine Lorentz map with linear part $\tilde{L}$ and $p_{0} : = o$ .

8.1step 1.1step 3.1step 4.1step 5.1step 6.1step 7.1step 1.2L3L2∎

Step 1.1 sets up the coordinate expression and proves $D Φ_{u}$ injective; steps 3.1 through 7.1 construct the constant linear part $\tilde{L}$ and prove that it is a linear Lorentz map; step 1.2 is the converse. Hence the $C^{2}$ isometries of the model are exactly its affine Lorentz maps, and these form a group under composition by [L2] (with every affine Lorentz map being a bijection).

C^2 eta-preserving maps of affine Minkowski space are affine Lorentz maps

Statement

Facts & Assumptions

Proof

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