ExampleConstruction: AI-generatedVerification: AI-adaptedPipeline-generatedjudge pass (gpt-6-luna)

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A partition orbit-basis element vanishes below the stable range

Example

Take $N = 1$ and $r = 1$ , so $N < 2 r$ . In the partition algebra $P_{1} (1)$ let $d$ be the discrete partition ${{1}, {1^{'}}}$ and let $c = {{1, 1^{'}}}$ be the one-block partition. The orbit-basis element $x_{d} = d - c$ is nonzero in the free diagram module but acts as zero on $W = Q^{1}$ .

Facts & Assumptions

Given: The two partition diagrams and the permutation action on $W^{\otimes 1}$ .

[F1]

Typed Brauer and partition diagram categories: distinct set partitions are a free basis, and composition weights a removed middle block by the parameter.

[F2]

Partition-diagram orbit basis and symmetric-group centralizer: the orbit-basis element for a pattern with more than $N$ blocks maps to zero.

Verification

1.1F1algebra

The discrete partition has two blocks, while the one-block partition coarsens it. Möbius inversion on this two-element interval gives $x_{d} = d - c$ ; by [F1] it is nonzero because $d$ and $c$ are distinct basis vectors.

2.1F2step 1.1algebra∎

The one-dimensional tensor basis has only the index $1$ , so its equality pattern is $c$ and it cannot realize the two-block pattern $d$ . Equivalently, both $d$ and $c$ act by the $1 \times 1$ matrix $(1)$ , so $x_{d}$ acts by zero. Since $2 > N = 1$ , this is the below-stable-range kernel predicted by [F2].

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

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Sources

Halverson and Ram, Partition Algebras, Theorem 3.6(a) (standard reference, not scraped)