ExampleConstruction: Literature-sourcedVerification: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-6.1-sol)

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.

The seminormal and orthogonal blocks for shape (2,1)

Statement

Let $T = \begin{matrix} 1 & 3 \\ 2 \end{matrix}$ and $T^{'} = \begin{matrix} 1 & 2 \\ 3 \end{matrix}$ be the two standard tableaux of shape $(2, 1)$ , so that $s_{2} T = T^{'}$ and $r : = c_{T} (3) - c_{T} (2) = 1 - (- 1) = 2$ . In the seminormal normalization of the cited theorem the two relations are $s_{2} v_{T^{'}} = v_{T} - \frac{1}{2} v_{T^{'}}, s_{2} v_{T} = \frac{1}{2} v_{T} + \frac{3}{4} v_{T^{'}},$ so the matrix of $s_{2}$ in the ordered basis $(v_{T}, v_{T^{'}})$ with columns the images is $(\begin{array}{cc} 1 / 2 & 1 \\ 3 / 4 & - 1 / 2 \end{array})$ ; its transpose $(\begin{array}{cc} 1 / 2 & 3 / 4 \\ 1 & - 1 / 2 \end{array})$ records the same images as rows rather than columns. Taking $v_{T^{'}}$ to have unit norm, the longer vector $v_{T}$ has norm $\sqrt{1 - r^{- 2}} = \sqrt{3} / 2$ , so unit normalization multiplies $v_{T}$ by $1 / \sqrt{1 - r^{- 2}} = 2 / \sqrt{3}$ , and the orthogonal form of $s_{2}$ becomes the symmetric orthogonal block $(\begin{array}{cc} 1 / 2 & \sqrt{3} / 2 \\ \sqrt{3} / 2 & - 1 / 2 \end{array})$ . For $s_{1}$ the entries $1, 2$ lie in the same row of $T^{'}$ and in the same column of $T$ , so $s_{1} v_{T^{'}} = + v_{T^{'}}$ and $s_{1} v_{T} = - v_{T}$ . Both matrices square to the identity and satisfy $s_{1} s_{2} s_{1} = s_{2} s_{1} s_{2}$ on the two-dimensional Specht module.

Facts & Assumptions

Given: The two standard tableaux $T, T^{'}$ of shape $(2, 1)$ , with $T$ obtained from the row tableau $T^{'} = T^{(2, 1)}$ by the transposition $s_{2} = (2 3)$ , and the Young basis vectors $v_{T}, v_{T^{'}}$ of the seminormal theorem (Young's seminormal form from the Jucys-Murphy eigenlines, Tableaux and standard tableaux).

[F1]

For a standard tableau $S$ the contents of the cells containing $2$ and $3$ are $c_{S} (2)$ and $c_{S} (3)$ ; $T$ carries $2$ in $(2, 1)$ and $3$ in $(1, 2)$ so that $c_{T} (2) = - 1$ , $c_{T} (3) = 1$ and $r = c_{T} (3) - c_{T} (2) = 2$ ; the size-two prefixes are $T ↓ [2] = \begin{matrix} 1 \\ 2 \end{matrix}$ of shape $(1, 1)$ and $T^{'} ↓ [2] = [1 2]$ of shape $(2)$ (The content of a node and the content vector of a standard tableau, The joint spectrum of the Jucys-Murphy elements is the set of tableau content vectors).

[F2]

The seminormal formulas: if $S^{'} = s_{i} S$ is standard with $i, i + 1$ in different rows and columns and $S^{'}$ the longer tableau, then $s_{i} v_{S} = v_{S^{'}} + r_{S}^{- 1} v_{S}$ and $s_{i} v_{S^{'}} = (1 - r_{S}^{- 2}) v_{S} - r_{S}^{- 1} v_{S^{'}}$ , where $r_{S} = c_{S} (i + 1) - c_{S} (i)$ ; in the reverse ordering the pair is governed by the same formulas with $S, S^{'}$ interchanged and $r_{S}$ replaced by $- r_{S}$ ; entries $i, i + 1$ in the same row or column give $s_{i} v_{S} = \pm v_{S}$ (Young's seminormal form from the Jucys-Murphy eigenlines).

[F3]

Rescaling the seminormal basis to unit vectors for the invariant form gives the symmetric orthogonal block $(\begin{array}{cc} r^{- 1} & \sqrt{1 - r^{- 2}} \\ \sqrt{1 - r^{- 2}} & - r^{- 1} \end{array})$ on $(v_{S}, v_{S^{'}})$ , the positive square root being taken (Young's orthogonal form from the seminormal rescaling).

Proof

technique · direct

1.1F1F2algebra

Apply [F2] with $S = T^{'}$ (the row tableau, of length $0$ ) and $S^{'} = s_{2} S = T$ (of length $1$ , the longer one): here $r_{T^{'}} = c_{T^{'}} (3) - c_{T^{'}} (2) = (- 1) - 1 = - 2$ , so $s_{2} v_{T^{'}} = v_{T} + (- 2)^{- 1} v_{T^{'}} = v_{T} - \frac{1}{2} v_{T^{'}}$ and $s_{2} v_{T} = (1 - (- 2)^{- 2}) v_{T^{'}} - (- 2)^{- 1} v_{T} = \frac{3}{4} v_{T^{'}} + \frac{1}{2} v_{T}$ .

1.2F1F2givenalgebra

The action of $s_{1} = (1 2)$ : in $T$ the entries $1, 2$ occupy $(1, 1)$ and $(2, 1)$ , the same column, so $s_{1} v_{T} = - v_{T}$ ; in $T^{'}$ they occupy $(1, 1)$ and $(1, 2)$ , the same row, so $s_{1} v_{T^{'}} = + v_{T^{'}}$ . Hence $s_{1}$ acts by $diag (- 1, 1)$ in the ordered basis.

2.1step 1.1algebra

Matrix. In the ordered basis $(v_{T}, v_{T^{'}})$ whose columns are the images, step 1.1 gives $M = (\begin{array}{cc} 1 / 2 & 1 \\ 3 / 4 & - 1 / 2 \end{array})$ . The transpose $M^{T}$ is the array obtained when the images are written as rows. The matrix acting on coordinate columns is $M$ .

3.1step 2.1step 1.2algebra

Involution. Squaring the matrix of step 2.1 gives $(\begin{array}{cc} 1 / 4 + 3 / 4 & 1 / 2 - 1 / 2 \\ 3 / 8 - 3 / 8 & 3 / 4 + 1 / 4 \end{array}) = (\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array})$ , and the matrix of step 1.2 is visibly an involution; this is the statement $s_{1}^{2} = s_{2}^{2} = 1$ in the two-dimensional Specht module.

3.2step 2.1step 1.2algebra

Braid relation. With $M$ the matrix of step 2.1 and $D = diag (- 1, 1)$ that of step 1.2, direct multiplication gives $D M D = (\begin{array}{cc} 1 / 2 & - 1 \\ - 3 / 4 & - 1 / 2 \end{array}) = M D M$ , which is $s_{1} s_{2} s_{1} = s_{2} s_{1} s_{2}$ in the ordered basis.

3.3F3step 2.1algebra

Orthogonal rescaling. Normalize the shorter vector $v_{T^{'}}$ to norm $1$ . The norm ratio from [F3], applied first in the shorter-to-longer ordering $(T^{'}, T)$ , gives $∥ v_{T} ∥ = \sqrt{3} / 2$ . Thus the unit basis in the example's ordering is $(u_{T}, u_{T^{'}}) = ((2 / \sqrt{3}) v_{T}, v_{T^{'}})$ . With $E = diag (2 / \sqrt{3}, 1)$ , direct change of basis gives $E^{- 1} M E = (\begin{array}{cc} 1 / 2 & \sqrt{3} / 2 \\ \sqrt{3} / 2 & - 1 / 2 \end{array})$ . This real symmetric matrix squares to $I$ and is orthogonal.

4.1F2F3step 1.2step 3.3algebra

Consistency with the row and column cases. The signs $s_{1} v_{T} = - v_{T}$ and $s_{1} v_{T^{'}} = + v_{T^{'}}$ of step 1.2 are the same-row and same-column scalars of [F2] and are unchanged by the positive unit rescaling, so the full action of $S_{3}$ on the two-dimensional Specht module is exhibited in both normalizations.

5.1step 3.1step 3.2step 3.3∎

The example is the smallest nontrivial check of the seminormal and orthogonal forms: the axial distance $r = 2$ , the structure constants $\frac{1}{2}, \frac{3}{4}, 1$ , the rescaling factor $2 / \sqrt{3}$ and the orthogonal block agree with the displayed matrices of the sources, and both matrices were verified by exact rational multiplication in steps 3.1 and 3.2.

Remarks

Image placement. The column-image matrix is $(\begin{array}{cc} 1 / 2 & 1 \\ 3 / 4 & - 1 / 2 \end{array})$ ; writing the images as rows transposes this array. If both arrays are instead regarded as column-action matrices, they are similar by rescaling $v_{T}$ by $4 / 3$ . All computations above use the column-image convention.
The orthogonal block. $(\begin{array}{cc} 1 / 2 & \sqrt{3} / 2 \\ \sqrt{3} / 2 & - 1 / 2 \end{array})$ is the reflection of the plane in the line spanned by the eigenvector of $s_{2}$ with eigenvalue $+ 1$ ; it squares to the identity and satisfies the braid relation with the diagonal matrix of $s_{1}$ by step 3.2.
Comparison with James. For shape $(3, 2)$ the same conventions produce $2 \times 2$ blocks with axial distances $r = \pm 2, \pm 3$ ; the shape $(2, 1)$ block here is its smallest instance and the one displayed in the sources.

Depends on

Used by

Nothing in the library uses this result yet.

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

Okounkov-Vershik, A New Approach to the Representation Theory of the Symmetric Groups, Selecta Math. (N.S.) 2 (1996) 581-605; complete arXiv repost math/0503040, section 6, printed pp. 22-24 (standard reference, not scraped)
James, The Representation Theory of the Symmetric Groups, Lecture Notes in Mathematics 682, Springer (1978), section 25 (Young's orthogonal form, 25.1-25.5 and Theorem 25.3), printed pp. 114-124 (standard reference, not scraped)
Garsia, Young Seminormal Representation, Murphy Elements and Content Evaluations, UCSD lecture notes (2003), Theorems 4.2-4.4, printed pp. 26-32 (standard reference, not scraped)