Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-09-01
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Symmetric and alternating images are smooth subbundles

Statement

For each k0, the symmetric and alternating fibrewise parts of Tk0M form smooth vector subbundles of the covariant tensor bundle.

Facts & Assumptions

Given: A smooth manifold M and an integer k0.

[F1]

The symmetric and alternating parts are defined fibrewise inside the covariant tensor bundle (Symmetric and alternating covariant tensor subbundles).

[L1]

The covariant tensor bundle is a smooth vector bundle, and the fibrewise symmetrization and alternation operators are projections (Tensor transition laws define a smooth vector bundle, Symmetrization and alternation are projections).

[L2]

The image of a constant-rank bundle map over one base is a smooth vector subbundle (Constant-rank kernels and images of bundle maps over one base are subbundles).

Proof

technique · direct
1.1

By [L1], symmetrization and alternation act fibrewise on Tk0M as smooth bundle endomorphisms over idM. Their fibres are the usual linear projections onto the symmetric and alternating tensors.

F1L1given
2.1

Because a projection has constant rank equal to the dimension of its image, the fibre ranks of these bundle maps are constant on M. Therefore [L2] shows that their images are smooth vector subbundles.

L1L2step 1.1algebra
3.1

Those images are exactly the symmetric and alternating bundles from [F1]. Hence both are smooth vector subbundles of Tk0M.

F1step 2.1

Depends on

Used by

Cited to discharge well-definedness by Symmetric and alternating covariant tensor subbundles.

Dependency tree · two levels

14 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources