LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-6-luna)

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The permutation average is the symmetric-subspace projector

Statement

Let $H$ be finite-dimensional and let $k \geq 1$ . The average $Π_{s y m, k} = \frac{1}{k!} \sum_{π \in S_{k}} U_{π}$ is self-adjoint and idempotent, with range exactly ${Sym}^{k} (H) = {v : U_{τ} v = v \forall τ \in S_{k}}$ . For $k = 0$ on the tensor unit, $Π_{s y m, 0} = I$ and its range is the full tensor unit.

Facts & Assumptions

Given: A finite-dimensional complex Hilbert space $H$ , an integer $k \geq 0$ , and the tensor-factor permutation representation $π \mapsto U_{π}$ .

[L1]

Ordered tensor registers, their inner product, and wire permutations defines each factor permutation by reordering the product basis. In equal factors this is a basis permutation, hence unitary; under the stated output-order convention, $U_{π}^{*} = U_{π^{- 1}}$ and $U_{π} U_{σ} = U_{σ \circ π}$ for ordinary right-to-left composition of permutations.

[F1]

$S_{k}$ is the finite group of permutations of a $k$ -element set, with $k!$ elements. Inversion and left or right multiplication by a fixed element are bijections; for each $τ \in S_{k}$ , exactly $k!$ pairs $(π, σ)$ satisfy $σ \circ π = τ$ , since every $π$ determines the unique $σ = τ \circ π^{- 1}$ .

Proof

1.1givenL1F1algebra

For $k \geq 1$ , $Π^{*} = \frac{1}{k!} \sum_{π} U_{π}^{*} = \frac{1}{k!} \sum_{π} U_{π^{- 1}} = Π$ by [L1] and reindexing by inversion. Also $Π^{2} = \frac{1}{(k!)^{2}} \sum_{π, σ} U_{σ \circ π} = \frac{1}{(k!)^{2}} \sum_{τ \in S_{k}} k! U_{τ} = Π,$ using [F1]. Thus $Π$ is an orthogonal projection.

2.1step 1.1L1F1algebra

For every $τ \in S_{k}$ , $U_{τ} Π = \frac{1}{k!} \sum_{π} U_{π \circ τ} = Π$ by [L1] and reindexing by right multiplication. Hence every vector in $ran Π$ is invariant. Conversely, if $U_{π} v = v$ for all $π$ , then $Π v = \frac{1}{k!} \sum_{π} v = v$ . Therefore $ran Π = {Sym}^{k} (H)$ .

3.1givenalgebra∎

For $k = 0$ , $S_{0}$ has one element and the defining empty tensor power is the one-dimensional tensor unit, so the stated average is $I$ and its range is the whole space.

Depends on

Used by

Optimal Cloner Attains Bound Lemma
Optimal Cloning Worst Case Upper Bound Lemma
Symmetric Power Irreducible Lemma
Symmetric Subspace Dimension And Haar Moment Theorem

Cited to discharge well-definedness by Symmetric Subspace Projector.

Dependency tree · two levels

13 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 Watrous, The Theory of Quantum Information (standard reference, not scraped)