PropositionStatement: AI-adaptedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-10

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Riemannian metrics induce metrics on dual tensor and exterior bundles

Statement

A Riemannian metric induces smooth metrics on dual, tensor and exterior bundles. On decomposable covectors, $⟨ α_{1} \land \dots \land α_{k}, β_{1} \land \dots \land β_{k} ⟩ = \det (⟨ α_{i}, β_{j} ⟩)$ ; increasing orthonormal wedge monomials have norm one.

Facts & Assumptions

Given: A Riemannian metric $g$ on a smooth manifold.

[F1]

The musical maps are smooth inverse bundle isomorphisms: $♭ : T M \to T^{*} M$ and $♯ : T^{*} M \to T M$ are smooth inverse bundle isomorphisms.

[F2]

Universal property of the finite-dimensional exterior power: Let $A : V^{k} \to W$ be an alternating $k$ -linear map into a real vector space $W$ . Then there is a unique linear map $\tilde{A} : ⋀^{k} V \to W$ such that $\tilde{A} (v_{1} \land \dots \land v_{k}) = A (v_{1}, \dots, v_{k})$ for all $v_{1}, \dots, v_{k} \in V$ .

[F3]

Wedge monomials in a dual basis form a basis: Let $e_{1}, \dots, e_{n}$ be a basis of $V$ , with dual basis $e^{1}, \dots, e^{n}$ . Then the wedges $e^{i_{1}} \land \dots \land e^{i_{k}} (1 \leq i_{1} < \dots < i_{k} \leq n)$ form a basis of ${Alt}^{k} (V)$ .

[F4]

Universal property of the tensor product for balanced maps into abelian groups: Let $R$ be a unital ring, $M$ a right $R$ -module, $N$ a left $R$ -module, and $τ : M \times N ⟶ M \otimes_{R} N, τ (m, n) = m \otimes n .$ The map $τ$ is balanced (def-balanced-and-bilinear-maps). For every abelian group $A$ and every balanced map $b : M \times N \to A$ , there is a unique group homomorphism $\overline{b} : M \otimes_{R} N ⟶ A$ such that $\overline{b} (m \otimes n) = b (m, n)$ for all $m, n$ . Consequently composition with $τ$ is a bijection ${Hom}_{A b} (M \otimes_{R} N, A) ≅ {Bal}_{R} (M, N; A) .$

[F5]

The elementary tensors of two bases form the product basis of the tensor product: Let $R$ be a commutative ring. If $M$ is free with basis $(e_{i})_{i \in I}$ and $N$ is free with basis $(f_{j})_{j \in J}$ , then $M \otimes_{R} N$ is free with basis $(e_{i} \otimes f_{j})_{(i, j) \in I \times J} .$ Equivalently, the canonical map $R^{(I \times J)} \to M \otimes_{R} N$ sending the standard basis vector at $(i, j)$ to $e_{i} \otimes f_{j}$ is an isomorphism. This includes an empty basis in either factor.

Proof

technique · direct

1.1

Define the dual pairing by $⟨ α, β ⟩ = g (α^{♯}, β^{♯})$ . It is positive definite and smooth because $♯$ is a smooth isomorphism. Define the tensor pairing on pure tensors by the product of the pairings of the factors and extend multilinearly; the tensor universal property applied successively in each list makes this well defined. Here the base ring is $R$ , and multilinearity implies balance in each adjacent pair of factors. The descended maps are real linear because scaling an elementary tensor scales the product, and elementary tensors generate. The tensor-product-basis theorem, iterated over the finitely many factors, gives a basis of products of orthonormal basis vectors. Its Gram matrix is the identity, so the pairing is positive definite.

F1F4F5given

2.1

The determinant is multilinear and alternating in each of its two lists, so the exterior universal property, applied twice, gives a bilinear pairing on the two exterior powers. In an orthonormal covector basis, its matrix on increasing wedge monomials is the identity: equal index lists give determinant one, and different lists give a zero row. The wedge-basis theorem therefore proves positive definiteness and the stated normalization, with no factor $k!$ .

F2F3step 1.1

3.1

Local smooth orthonormal frames are obtained from a coordinate frame by $w_{j} = e_{j} - \sum_{i < j} g (e_{j}, u_{i}) u_{i}$ , $u_{j} = w_{j} / \sqrt{g (w_{j}, w_{j})}$ . Linear independence makes each denominator positive; induction makes all coefficients smooth. In these frames the constructed metrics have constant matrices, hence are smooth. Their intrinsic pairing formulas prove agreement on overlaps. For $k = 0$ the empty determinant is one; for zero exterior spaces the metric is vacuous.

step 1.1step 2.1∎

Source locator

Lee, pp.330 and 341–342, local orthonormal frames and dual metrics; Problem 16-18(a), pp.437–438, determinant pairing on exterior powers. Tensor existence and product bases use the two declared algebra theorems over the field of real numbers.

Depends on

Used by

Riemannian hodge star Definition
The riemannian volume form is the unique positive unit top form Proposition
Riemannian divergence theorem Theorem

Dependency tree · two levels

19 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, second edition (standard reference, not scraped)
Ved Datar, Lectures on Riemannian Geometry, September 2025 (standard reference, not scraped)