LemmaStatement: AI-adaptedProof: AI-adaptedprecheck passjudge pass (gpt-6.1-sol)

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Existence of future-oriented inertial frames at every event

Statement

In every oriented time-oriented affine Minkowski space $(A, V, η, C^{+})$ and for every event $o \in A$ there exists an inertial frame with origin $o$ ; consequently inertial charts are defined at every event. Explicitly: start from any $η$ -orthonormal ordered basis $(f_{0}, f_{1}, f_{2}, f_{3})$ of $V$ ; if $f_{0}$ is not in $C^{+}$ then replacing $f_{0}$ by $- f_{0}$ makes the first vector future, because the timelike set is the disjoint union of $C^{+}$ and $C^{-}$ and $η (- f_{0}, - f_{0}) = η (f_{0}, f_{0})$ ; the resulting ordered basis has some orientation class, and if that class is not the positive class selected by the fixed orientation of $V$ , then replacing $e_{1}$ by $- e_{1}$ (or interchanging $e_{1}$ and $e_{2}$ ) produces a positively oriented ordered basis, without destroying $η$ -orthonormality and without changing the first, future vector.

Facts & Assumptions

Given: An oriented time-oriented affine Minkowski space $(A, V, η, C^{+})$ and an event $o \in A$ .

[L1]

There is an ordered basis $(f_{0}, f_{1}, f_{2}, f_{3})$ of $V$ with $η (f_{a}, f_{b}) = diag (- 1, 1, 1, 1)_{a b}$ (Every Lorentz form admits an ordered basis with matrix diag(-1,1,1,1)).

[L2]

The timelike set $T$ is the disjoint union $T = C^{+} ⊔ C^{-}$ , and a vector is future timelike exactly when it lies in $C^{+}$ (Time orientation, future and past timelike/null/causal vectors, and the closed cones).

[L3]

For a finite-dimensional real vector space of dimension at least $1$ there are exactly two orientations, described as equivalence classes of ordered bases under the relation $\det P > 0$ ; a basis in the chosen class is positively oriented (Orientation of a finite-dimensional real vector space).

[L4]

An inertial frame is a tuple $(o; e_{0}, e_{1}, e_{2}, e_{3})$ with $o \in A$ and an ordered basis that is $η$ -orthonormal, future and positively oriented (Inertial frames and inertial charts of the affine Minkowski model).

[L5]

If $T : V \to V$ is a linear isomorphism of an oriented real vector space with $\dim V = n \geq 1$ , then $\det T > 0$ means that $T$ carries every positively oriented basis to a positively oriented basis, and $\det T < 0$ means that it carries every positively oriented basis to a negatively oriented one; the sign of $\det T$ is exactly whether $T$ preserves or reverses the orientation (A real linear isomorphism preserves or reverses orientation according to the sign of its determinant).

[L6]

$η$ is a symmetric bilinear form, so $η (- u, - w) = η (u, w)$ and $η (- u, w) = - η (u, w)$ for all $u, w \in V$ (Lorentz forms of inertia (3,1,0) on a 4-dimensional real vector space).

[L7]

The two components $C^{+}$ and $C^{-}$ of the timelike set are convex and satisfy $T = C^{+} ⊔ C^{-}$ and $0 \notin T$ (The timelike set has exactly two convex components, described in orthonormal coordinates).

Proof

1.1L1choose

By [L1] choose an $η$ -orthonormal ordered basis $(f_{0}, f_{1}, f_{2}, f_{3})$ of $V$ .

2.1step 1.1L1L2L6L7algebra

By [L6], $η (- f_{0}, - f_{0}) = η (f_{0}, f_{0}) = - 1 < 0$ , so both $f_{0}$ and $- f_{0}$ lie in the timelike set $T$ ; by [L2] the set $T$ is the disjoint union of $C^{+}$ and $C^{-}$ , so exactly one of $f_{0}$ , $- f_{0}$ lies in $C^{+}$ : if both lay in the same component, its convexity [L7] would put $0 = \frac{1}{2} f_{0} + \frac{1}{2} (- f_{0})$ in that component, contradicting $0 \notin T$ , and if one lay in each component then $- f_{0}$ is the required future vector. Define $g_{0} : = f_{0}$ if $f_{0} \in C^{+}$ and $g_{0} : = - f_{0}$ otherwise, and $g_{i} : = f_{i}$ for $i = 1, 2, 3$ . Then $g_{0} \in C^{+}$ , and $(g_{0}, g_{1}, g_{2}, g_{3})$ is $η$ -orthonormal: $g_{0}$ contributes $η (g_{0}, g_{0}) = η (f_{0}, f_{0}) = - 1$ by [L6], the remaining diagonal entries are $η (f_{i}, f_{i}) = 1$ , and each off-diagonal entry is either $η (f_{a}, f_{b}) = 0$ or $- η (f_{a}, f_{b}) = 0$ .

3.1step 2.1L3

By [L3] the basis $(g_{0}, g_{1}, g_{2}, g_{3})$ lies in exactly one of the two orientation classes of $V$ : either it is positively oriented or it is not.

4.1step 2.1step 3.1L4

Suppose first that $(g_{0}, g_{1}, g_{2}, g_{3})$ is positively oriented. Then $(o; g_{0}, g_{1}, g_{2}, g_{3})$ is $η$ -orthonormal, its first vector is future, and its basis is positively oriented, so it is an inertial frame with origin $o$ by [L4].

4.2step 2.1step 3.1L3L5L6algebra

Suppose instead that $(g_{0}, g_{1}, g_{2}, g_{3})$ is not positively oriented. Put $e_{0} : = g_{0}$ , $e_{1} : = - g_{1}$ , $e_{2} : = g_{2}$ , $e_{3} : = g_{3}$ . Orthonormality is preserved exactly as in step 2.1, and $e_{0} = g_{0} \in C^{+}$ is still future. The change-of-basis linear map carrying $(g_{0}, g_{1}, g_{2}, g_{3})$ to $(e_{0}, e_{1}, e_{2}, e_{3})$ has matrix $diag (1, - 1, 1, 1)$ , of determinant $- 1$ ; by [L5] it carries the positive class onto the negative class, and since it is a bijection of the set of ordered bases and there are exactly two classes by [L3], it carries the negative class, in which $(g_{a})$ lies, onto the positive class; hence $(e_{0}, e_{1}, e_{2}, e_{3})$ is positively oriented. Therefore $(o; e_{0}, e_{1}, e_{2}, e_{3})$ is an inertial frame with origin $o$ by [L4]. (Interchanging the two basis vectors $g_{1}, g_{2}$ would also reverse the orientation and preserve orthonormality and the first vector.)

5.1step 4.1step 4.2L4∎

In either case an inertial frame with origin $o$ exists. The origin is arbitrary in $A$ , so taking $o$ any event gives an inertial frame at that event, and the construction uses at most the replacement of $f_{0}$ by $- f_{0}$ and one further reversal of orientation; both operations preserve $η$ -orthonormality and the future character of the first vector.

Existence of future-oriented inertial frames at every event

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources