Alphabeta Math
TheoremStatement: AI-adaptedProof: AI-generatedPipeline-generatedjudge pass (gpt-6-luna)
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.

Tensor maps preserve composition, identities, and adjoints

Statement

Let (Tk:Hk→Kk)k<n and (Sk:Kk→Lk)k<n be finite ordered families of linear maps between finite-dimensional complex inner-product spaces. Then

(⨂k<nSk)(⨂k<nTk)=⨂k<n(SkTk),⨂k<nIHk=I⊗k<nHk,(⨂k<nTk)∗=⨂k<nTk∗.

In the first formula, SkTk means Sk∘Tk. Here IHk denotes the identity map x↦x. All adjoints use the repository pairing, linear in the first argument. The empty tensor-map identity is the identity map z↦z on C.

Facts & Assumptions

Given: The finite ordered families of spaces and linear maps in the Statement.

[F1]

The tensor map is the unique linear extension of its action on elementary tensors (Tensor products of finitely many linear maps).

[F2]

Elementary tensors in the specified product basis form an orthonormal basis and span (The product basis is orthonormal and tensor dimensions multiply).

[F3]

Every linear map between finite-dimensional inner-product spaces has a unique adjoint (Every linear map between finite-dimensional inner product spaces has a unique adjoint).

[F4]

An adjoint T∗ satisfies ⟨Tx,y⟩=⟨x,T∗y⟩ (The adjoint T∗:W→V is characterised by ⟨Tv,w⟩W=⟨v,T∗w⟩V).

[F5]

Each root inner product is linear in the first argument and conjugate-linear in the second (Real and complex inner-product spaces and their induced length).

[F6]

Every finite-dimensional real or complex inner-product space has an orthonormal basis, with the empty basis in dimension zero (Every finite-dimensional real or complex inner product space has an orthonormal basis).

[F7]

Identity maps and compatible composites of linear maps are linear (Identity maps and composites of linear maps are linear).

Proof

1.1givenF1F2F7

By [F7], each SkTk is linear and the composite of the two tensor maps is linear, so [F1] applies to both sides. On an elementary tensor ⨂k<nxk, the left side of the composition identity equals ⨂k<nSk(Tkxk) by [F1], which is the value of the right side. Since elementary tensors span by [F2], the two linear maps are equal.

1.2F2F3F4F5F6choose

By [F6], choose orthonormal bases for the finite factor spaces. For elementary tensors x=⨂k<nxk and y=⨂k<nyk, the product pairing in [F2] and the factor adjoint identities in [F4] give ⟨(⨂kTk)x,y⟩=∏k⟨Tkxk,yk⟩=∏k⟨xk,Tk∗yk⟩=⟨x,⨂kTk∗yk⟩. The pairing is extended using the first-linear convention [F5].

2.1step 1.1F1F2F7algebra

Each factor identity and the identity on the tensor space are linear by [F7]. The tensor map ⨂k<nIHk fixes every elementary tensor by [F1]. The identity map does too; spanning in [F2] gives equality. If n=0, both maps are the identity on C.

3.1step 1.2F2F3F4F5F6algebra∎

Sesquilinearity and spanning extend the identity in step 1.2 to all x,y. By existence and uniqueness in [F3] and the defining condition [F4], (⨂kTk)∗=⨂kTk∗. For n=0 both sides are the identity on C; zero-dimensional factors are covered because the product bases may be empty.

Depends on

Used by

Dependency tree · two levels

21 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