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Pullback of covariant tensors is smooth and functorial
Statement
If is smooth and is a smooth covariant tensor field on , then is a smooth covariant tensor field on . Moreover,
for every composable smooth map .
Facts & Assumptions
Given: Smooth maps and , and a smooth covariant tensor field on the target.
Pullback of a covariant tensor field is defined by inserting the differential into every slot (The pullback of a covariant tensor field).
Tensor pullback commutes with the corresponding fibrewise linear operations, and tensor products and contractions of smooth tensor fields are smooth (Linear pullback respects tensor products and permutations, Tensor products and contractions of smooth tensor fields are smooth).
Differentials satisfy the chain rule (The chain rule for differentials of smooth maps).
Proof
In local coordinates, [F1] expresses each coefficient of as a finite sum of the coefficients of multiplied by partial derivatives of . Those are smooth, so [L1] implies that is smooth.
The identity map has identity differential, so [F1] gives .
For , [F1] and [L2] give which is exactly .
Therefore pullback of covariant tensors is smooth and functorial.
Depends on
Used by
Dependency tree · two levels
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Sources
- John M. Lee, Introduction to Smooth Manifolds, 2nd ed. (standard reference, not scraped)