ExampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passaudited 2026-08-31

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.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

The special linear group is a codimension-one embedded submanifold

Example

For $n \geq 1$ , the special linear group

$S L (n, R) = {A \in M_{n} (R) : \det A = 1}$

is an embedded codimension-one submanifold of the Euclidean space $M_{n} (R) ≅ R^{n^{2}}$ .

Facts & Assumptions

Given: The determinant map $\det : M_{n} (R) \to R$ .

[L1]

A nonempty regular level set of a map into an $n$ -manifold is an embedded submanifold of codimension $n$ (A regular level set is an embedded submanifold).

[F1]

The determinant is multiplicative, and an invertible matrix has inverse given by the adjugate formula (For $n \geq 1$ , the determinant over a commutative ring by the Leibniz formula, and $∣ \det A ∣$ for a real matrix, For same-sized finite square matrices over a commutative ring, $\det (A B) = \det (A) \det (B)$ , If $\det (A)$ is a unit, then $A^{- 1} = \det (A)^{- 1} adj (A)$ ).

[L2]

Euclidean differentials are computed by one-variable directional derivatives and the usual derivative algebra (A total derivative computes every directional derivative, and its matrix is the Jacobian, Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$ ).

Verification

technique · direct

1.1

Let $A \in S L (n, R)$ and $H \in M_{n} (R)$ . Using [F1], for small $t$ one has $\det (A + t H) = \det (A) \det (I + t A^{- 1} H) = \det (I + t A^{- 1} H)$ because $\det (A) = 1$ . Differentiating at $t = 0$ with [L2] gives $D (\det)_{A} (H) = tr (A^{- 1} H)$ .

F1L2given

2.1

This linear functional is surjective, because $D (\det)_{A} (A / n) = tr (I / n) = 1$ . Hence $1$ is a regular value of the determinant.

step 1.1

3.1

The level set is nonempty because $\det I_{n} = 1$ . By [L1], $\det^{- 1} (1) = S L (n, R)$ is therefore an embedded codimension-one submanifold.

L1step 2.1algebra∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

39 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, Level Sets (standard reference, not scraped)