ExampleConstruction: AI-adaptedVerification: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-10

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

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Hodge star on euclidean three space

Example

In standard oriented Euclidean $R^{3}$ , $* 1 = d x \land d y \land d z$ , $* d x = d y \land d z$ , $* d y = d z \land d x$ , $* d z = d x \land d y$ , and $*^{2} = i d$ in every degree.

Facts & Assumptions

Given: The orthonormal coframe $(d x, d y, d z)$ with volume $ν = d x \land d y \land d z$ .

[F1]

Hodge star is a smooth bundle isomorphism: The Hodge star exists uniquely and is a smooth bundle isomorphism in every degree $0 \leq k \leq n$ .

[F2]

Hodge star squared sign: On real $k$ -forms, $*^{2} = (- 1)^{k (n - k)} id$ .

Verification

technique · direct

1.1F1given

The complementary-wedge formula gives $* 1 = ν$ . For the one-forms, $d x \land (d y \land d z) = ν$ , $d y \land (d z \land d x) = ν$ , and $d z \land (d x \land d y) = ν$ : the latter two permutations each have two transpositions. These complementary two-forms wedge to zero with either of the other one-form basis vectors because of a repeated factor. Thus they satisfy all pairings in the defining identity and are the displayed stars.

2.1F1step 1.1

Similarly $(d x \land d y) \land d z = ν$ , $(d x \land d z) \land (- d y) = ν$ , and $(d y \land d z) \land d x = ν$ . Distinct two-form basis vectors have zero pairing and wedge to zero with the listed complementary one-form. Consequently $* (d x \land d y) = d z$ , $* (d x \land d z) = - d y$ , $* (d y \land d z) = d x$ , and $* ν = 1$ . For example $* (2 d x + 3 d y) = 2 d y \land d z + 3 d z \land d x$ by linearity.

3.1F2step 1.1step 2.1∎

For $k = 0, 1, 2, 3$ , the exponent $k (3 - k)$ is respectively $0, 2, 2, 0$ , always even. The star-square formula therefore gives $*^{2} = i d$ in all these degrees, in agreement with the table.

Source locator

Lee, Problem 16-18(a–e), pp. 437–438, and Problem 16-19, p. 438, Euclidean Hodge-star computations.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

4 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)