Alphabeta Math
PropositionStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-09-01
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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.

Smoothness of a tensor field is equivalent to smooth coordinate components

Statement

A type (r,s) tensor field is smooth if and only if, in every smooth chart, its coordinate component functions are smooth.

Facts & Assumptions

Given: A type (r,s) tensor field T on a smooth manifold M.

[F1]

A smooth tensor field is a smooth section of the tensor bundle (A smooth tensor field).

[L1]

The tensor bundle is a smooth vector bundle with the standard tensor-coordinate trivializations (Tensor transition laws define a smooth vector bundle).

[L2]

Smoothness of a section is equivalent to smoothness of its local component functions (Smoothness of a section is equivalent to smooth local components).

Proof

technique · direct
1.1

By [F1] and [L1], a tensor field is a section of a smooth vector bundle whose local bundle coordinates are exactly the tensor coefficients relative to the chart bases /xi and dxi.

F1L1given
2.1

Applying [L2] to those trivializations shows that the section is smooth exactly when each local coefficient function is smooth.

L1L2step 1.1
3.1

Therefore smoothness of a tensor field is equivalent to smoothness of its coordinate components.

step 2.1

Depends on

Used by

Dependency tree · two levels

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Sources