LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-09-01

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Contraction is independent of the basis formula

Statement

The contraction formula

$\sum_{i} T (e^{i}, α_{2}, \dots, α_{r}, e_{i}, v_{2}, \dots, v_{s})$

has the same value for every basis $(e_{i})$ and its dual basis $(e^{i})$ .

Facts & Assumptions

Given: A type $(r, s)$ tensor $T$ with $r, s \geq 1$ , fixed arguments $α_{2}, \dots, α_{r}, v_{2}, \dots, v_{s}$ , and two bases $(e_{i})$ and $(b_{j})$ with dual families $(e^{i})$ and $(b^{j})$ .

[F1]

Contraction is given by the displayed dual-basis sum (The contraction of a mixed tensor).

[L1]

Every vector and every covector expand in a basis and its dual family (The dual family $(b^{*})_{b \in B}$ associated to a Hamel basis $B$ , defined by $b^{*} (c) = δ_{b c}$ ).

Proof

technique · direct

1.1F1givenconstruct

Define the bilinear map $B : V^{*} \times V \to R$ by $B (β, w) = T (β, α_{2}, \dots, α_{r}, w, v_{2}, \dots, v_{s})$ . Then the two contraction sums are $\sum_{i} B (e^{i}, e_{i})$ and $\sum_{j} B (b^{j}, b_{j})$ .

2.1L1step 1.1algebra

By [L1], any vector $w$ satisfies $w = \sum_{i} e^{i} (w) e_{i} = \sum_{j} b^{j} (w) b_{j}$ , and any covector $β$ satisfies $β = \sum_{i} β (e_{i}) e^{i} = \sum_{j} β (b_{j}) b^{j}$ . Bilinearity of $B$ therefore gives $\sum_{i} B (e^{i}, e_{i}) = \sum_{i, j} b^{j} (e_{i}) B (e^{i}, b_{j}) = \sum_{j} B (\sum_{i} b^{j} (e_{i}) e^{i}, b_{j}) = \sum_{j} B (b^{j}, b_{j}) .$

3.1step 2.1∎

The displayed sum is therefore basis-independent, so contraction is intrinsically defined.

Depends on

Used by

Mean curvature vector Definition
Tensor product and contraction in a basis Example
Ricci curvature is symmetric and basis independent Lemma
Ricci decomposition of the Riemann tensor in dimension at least three Proposition
Tensor products and contractions of smooth tensor fields are smooth Proposition
Contracted second Bianchi identity Theorem

Dependency tree · two levels

16 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

Will J. Merry, Differential Geometry (standard reference, not scraped)