How statement and proof provenance work
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Symmetric and alternating images are smooth subbundles
Statement
For each , the symmetric and alternating fibrewise parts of form smooth vector subbundles of the covariant tensor bundle.
Facts & Assumptions
Given: A smooth manifold and an integer .
The symmetric and alternating parts are defined fibrewise inside the covariant tensor bundle (Symmetric and alternating covariant tensor subbundles).
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).
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
By [L1], symmetrization and alternation act fibrewise on as smooth bundle endomorphisms over . Their fibres are the usual linear projections onto the symmetric and alternating tensors.
Because a projection has constant rank equal to the dimension of its image, the fibre ranks of these bundle maps are constant on . Therefore [L2] shows that their images are smooth vector subbundles.
Those images are exactly the symmetric and alternating bundles from [F1]. Hence both are smooth vector subbundles of .
Depends on
Used by
- The Euclidean metric as a symmetric two-tensor Example
- Exterior-power transition laws define a smooth vector bundle Theorem
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
- John M. Lee, Introduction to Smooth Manifolds, 2nd ed. (standard reference, not scraped)