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Jordan Hall–Littlewood basis and pairing in degrees zero, one and two

Example

For $q > 1$ a finite-field cardinality, set $s = q^{- 1}$ and use the exact Jordan map $Φ_{q}$ of Jordan Hall classes are Hall–Littlewood functions with the transported Green pairing. Then $P_{\emptyset} = 1, P_{(1)} = m_{(1)}, P_{(1, 1)} = m_{(1, 1)}, P_{(2)} = m_{(2)} + (1 - s) m_{(1, 1)} .$ The images of $u_{(1, 1)}$ and $u_{(2)}$ are respectively $q^{- 1} P_{(1, 1)}$ and $P_{(2)}$ . Their Hall–Littlewood norms are $1 / ((1 - s) (1 - s^{2}))$ and $1 / (1 - s)$ . The transported Jordan pairing has $B_{q} (p_{2}, p_{2}) = 2 / (q^{2} - 1)$ ; this differs from both the ordinary Hall norm $2$ and the standard Hall–Littlewood norm $2 / (1 - q^{- 2})$ .

Facts & Assumptions

Given: The finite-field parameter $q > 1$ , its reciprocal $s$ , and partitions of sizes at most two.

[F1]

The Jordan square is $u_{(1)}^{2} = (q + 1) u_{(1, 1)} + u_{(2)}$ (The degree-two Jordan-quiver Hall product).

[F2]

The map, all transported pairing formulas and the Hall–Littlewood norms have the conventions and proofs of Jordan Hall classes are Hall–Littlewood functions with the transported Green pairing and Hall–Littlewood scalar product and monic functions.

[F3]

The ordinary symmetric-function Hall pairing is a separate form, with $⟨ p_{2}, p_{2} ⟩_{H} = 2$ (The Hall inner product on symmetric functions, Power sums are orthogonal for the Hall form).

Verification

1.1F1F2algebra

The normalized generator map gives $Φ_{q} (u_{(1)}) = e_{1} = m_{(1)}$ and $Φ_{q} (u_{(1, 1)}) = q^{- 1} e_{2} = q^{- 1} m_{(1, 1)}$ . Since $e_{1}^{2} = m_{(2)} + 2 m_{(1, 1)}$ , applying it to [F1] yields $Φ_{q} (u_{(2)}) = e_{1}^{2} - (1 + q^{- 1}) e_{2} = m_{(2)} + (1 - q^{- 1}) m_{(1, 1)}$ . The exponents are $n ((1)) = n ((2)) = 0$ , $n ((1, 1)) = 1$ , so [F2] identifies exactly the four asserted $P$ functions, including the unit.

2.1F2step 1.1algebra

In the power-sum basis, $m_{(2)} = p_{2}$ and $m_{(1, 1)} = (p_{1}^{2} - p_{2}) / 2$ , hence $P_{(2)} = ((1 - s) p_{1}^{2} + (1 + s) p_{2}) / 2$ . The diagonal Hall–Littlewood norms are $⟨ p_{1}^{2}, p_{1}^{2} ⟩_{s} = 2 / (1 - s)^{2}$ and $⟨ p_{2}, p_{2} ⟩_{s} = 2 / (1 - s^{2})$ . Substitution gives $⟨ P_{(2)}, P_{(1, 1)} ⟩_{s} = \frac{1}{4} ((1 - s) 2 / (1 - s)^{2} - (1 + s) 2 / (1 - s^{2})) = 0$ , $⟨ P_{(2)}, P_{(2)} ⟩_{s} = 1 / (1 - s)$ and $⟨ P_{(1, 1)}, P_{(1, 1)} ⟩_{s} = 1 / ((1 - s) (1 - s^{2}))$ . These independently check the theorem's degree-two scalar convention.

3.1F2F3step 1.1step 2.1algebra∎

The automorphism factors from [F2] are $a_{(2)} = q^{2} (1 - q^{- 1})$ and $a_{(1, 1)} = q^{4} (1 - q^{- 1}) (1 - q^{- 2})$ . Since $Φ_{q} (u_{(1, 1)}) = q^{- 1} P_{(1, 1)}$ , the rule $B_{q} = q^{- 2} ⟨, ⟩_{q^{- 1}}$ gives exactly $a_{(1, 1)}^{- 1}$ on that image, and gives $a_{(2)}^{- 1}$ on the other image. In particular $B_{q} (p_{2}, p_{2}) = q^{- 2} 2 / (1 - q^{- 2}) = 2 / (q^{2} - 1)$ , whereas [F3] gives the ordinary Hall value $2$ . The factors $q^{- 1}$ on the column class and $q^{- 2}$ on the degree-two form are both necessary.

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Sources

Olivier Schiffmann, Lectures on Hall Algebras, Example 2.2 and §2.4 (standard reference, not scraped)