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DefinitionDefinition: AI-adaptedProof: AI-generatedPipeline-generatedjudge pass (gpt-6-luna)
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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.

Partial trace over a finite tensor factor

Definition

For a complex vector space H, write End⁡(H) for the set of linear maps H→H (Linear map between vector spaces over the same field).

Let HA,HB be finite-dimensional complex inner-product spaces with fixed ordered orthonormal bases (ei)i<a and (fj)j<b, respectively. The ordered product basis of HA⊗HB has the A index first. For R∈End⁡(HA⊗HB) define Tr⁡B(R)∈End⁡(HA) by

(Tr⁡B(R))ii′:=∑j<bR(i,j),(i′,j).

Rows are output coordinates and columns are input coordinates, as in Coordinate columns [v]B and matrices [T]BC of linear maps relative to ordered bases. If b=0, the sum is empty and equals 0. The notation Tr⁡B means that the second factor is contracted; the first factor and its order are retained.

Facts & Assumptions

Given: Finite-dimensional complex inner-product spaces HA,HB, the specified ordered orthonormal bases, and R∈End⁡(HA⊗HB).

[F1]

The ordered product basis identifies the coordinates of HA⊗HB with pairs (i,j) in the displayed factor order (Ordered tensor registers, their inner product, and wire permutations).

Proof

1.1givenF1algebra

For each basis vector ei′ of HA, prescribe Sei′=∑i<a(∑j<bR(i,j),(i′,j))ei. Each inner and outer sum is finite; when b=0 its inner sum is 0.

2.1step 1.1F1F2algebra∎

By the ordered-basis coordinate property [F2], these prescribed images extend uniquely to a linear map S:HA→HA, and its matrix has exactly the displayed entries. Each entry is linear in the entries of R, so the resulting assignment R↦S is linear. This defines Tr⁡B even when either basis is empty.

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