Alphabeta Math
LemmaStatement: AI-adaptedProof: AI-generatedPipeline-generatedjudge pass (gpt-6-luna)
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Operator support weight is invariant under wire permutation

Statement

Let the source register have ordered factors H0,…,Hn−1, and let the target register have factors in the QC02 order Hσ(0),…,Hσ(n−1). Let Πσ be the corresponding wire bijection and set E′:=ΠσEΠσ−1. An original factor labeled i is in target position σ−1(i). Thus E is supported in S exactly when E′ is supported in σ−1(S); consequently wt⁡(E′)=wt⁡(E).

Facts & Assumptions

Given: The source and target ordered tensor registers in the Statement, a permutation σ, its wire bijection Πσ, and E∈End⁡(HP).

[L1]

Support and weight of an operator on a finite register: Support means conjugating by ΠS and its inverse gives a local operator tensored with identity on the complementary factors; weight is the minimum support size.

[F1]

Ordered tensor registers, their inner product, and wire permutations: A wire permutation sends a product-basis tensor to the tensor with the same factor labels in the declared output order and is extended linearly.

[F2]

Finite wire permutations are unitary basis permutations: Every finite wire permutation is a bijective linear isometry, with inverse the wire permutation for the inverse ordering.

[F3]

Tensor products of finitely many linear maps: A tensor map is the unique linear extension of its factorwise action on elementary tensors.

[F4]

The product basis is orthonormal and tensor dimensions multiply: Elementary product tensors form a basis and span each finite ordered tensor register.

Proof

1.1F1F2F3F4given

Put Sσ:=σ−1(S), the target positions occupied by source factors from S. Let αS reorder those source factors into their target order, and let αSc reorder the complementary factors likewise; these are wire bijections by [F2]. On each elementary tensor, both ΠSσtargetΠσ and (αS⊗αSc)ΠSsource place each factor vector in the same target order. Elementary tensors span by [F4], and the maps agree on them by [F1], so ΠSσtargetΠσ=(αS⊗αSc)ΠSsource.

2.1L1F2F3step 1.1

If E is supported in S, [L1] gives AS with ΠSsourceE(ΠSsource)−1=AS⊗I. Conjugating by step 1.1 gives ΠSσtargetE′(ΠSσtarget)−1=(αS⊗αSc)(AS⊗I)(αS⊗αSc)−1. On y⊗z the right side is αSASαS−1y⊗z, hence equals (αSASαS−1)⊗I by [F3]; thus E′ is supported in Sσ.

3.1F2L1step 1.1step 2.1algebra∎

Apply the same implication to E′ and the inverse wire permutation. It gives the converse: if E′ is supported in Sσ, then E is supported in S. Thus S↦σ−1(S) bijects the support sets of E and E′, preserving cardinality; their minimum support sizes agree.

Remarks

The QC02 bodies were inspected. This proof uses their declared factor-order action and bijective inverse; it does not identify differently ordered registers without a map.

Depends on

Used by

Nothing in the library uses this result yet.

Cited to discharge well-definedness by Support and weight of an operator on a finite register.

Dependency tree · two levels

18 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