Alphabeta Math
ExampleConstruction: AI-adaptedVerification: AI-adaptedPipeline-generatedaudited 2026-09-14
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.

Unitary and special unitary Lie groups

Example

Assume ACω and let n≥1. Regarded as real Lie groups,

U(n)={A∈GL⁡n(C):A∗A=I},SU(n)={A∈U(n):det⁡A=1}

have Lie algebras

u(n)={X:X∗+X=0},su(n)={X∈u(n):tr⁡X=0}.

Facts & Assumptions

Given: An integer n≥1 and complex matrices viewed as a finite-dimensional real vector space.

[F1]

A Lie group has smooth multiplication and inversion, and its tangent bracket is the bracket of left-invariant fields. Lie group. Lie bracket on the tangent space of a Lie group.

[F2]

Over the field C, a positive-sized matrix is invertible exactly when its determinant is nonzero, and its inverse is its adjugate divided by that determinant. A positive-sized square matrix over a commutative ring is invertible if and only if its determinant is a unit. If det⁡(A) is a unit, then A−1=det⁡(A)−1adj⁡(A).

[F3]

Complex conjugation supplies the conjugate transpose A∗. Real and imaginary parts, complex conjugation, and modulus.

[F4]

Constant-rank level sets are embedded with tangent kernel. The constant-rank theorem for manifolds.

[F6]

Countable choice is inherited through the tangent-bracket supplier in [F1]; the finite matrix and level-set calculations need no further choice. The Axiom of Countable Choice (ACω).

Verification

technique · direct
1.1F1F2F3F4F5algebra

Regard Mn(C) as R2n2. The complex determinant is a finite polynomial in matrix entries by [F5], hence its real and imaginary parts are real polynomials. By [F2], GL⁡n(C)={A:det⁡A≠0} is open in this real vector space; multiplication is polynomial and inversion is real smooth there by the adjugate formula. Thus it is a real Lie group by [F1]. Its left-invariant field with identity value X is A↦AX; differentiating these linear fields gives the tangent bracket [X,Y]=XY−YX in the convention of [F1]. Now F(A)=A∗A maps this open group smoothly into the real vector space of Hermitian matrices and has dFA(X)=X∗A+A∗X. For Hermitian S, X=12A−∗S maps to S, so [F4] makes U(n)=F−1(I) embedded; the adjoint-product identities make it a subgroup. At I its tangent kernel is X∗+X=0.

2.1F4F5step 1.1algebra

For A∈U(n), ∣det⁡A∣2=det⁡(A∗A)=1, so determinant maps U(n) into the unit circle. Near 1 that circle has the real coordinate z↦Im⁡z on the arc Re⁡z>0. Differentiating the finite determinant formula at I gives d(det⁡)I(X)=tr⁡X; on skew-Hermitian X this is imaginary and every imaginary scalar occurs from a diagonal X. Left multiplication by any A∈SU(n) transports this surjectivity to A. Hence 1 is a regular value of the circle-valued determinant map and [F4] makes SU(n) embedded in U(n), with tangent kernel tr⁡X=0 at I. Its subgroup operations are smooth by restriction.

3.1F1F3step 1.1step 2.1algebra

Both tangent spaces are closed under commutator: adjoint reverses products, and trace of a commutator vanishes by finite reindexing. Hence they are the asserted Lie algebras.

4.1F1F2F3F4F5F6step 1.1step 2.1step 3.1∎

At n=1, u(1)=iR and su(1)=0; n=0 is excluded by the Statement. No interval, endpoint, arbitrary metric choice, or biconditional occurs. The ambient determinant is nonzero exactly on GL⁡n(C) by [F2]. ACω is inherited through [F1]; finite coordinates add no choice.

Depends on

Dependency tree · 2 levels

Sources