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The rank-two ordered-basis Hecke operator and all braid cases

Example

Let $F = Q (q)$ , put $η = q - q^{- 1}$ , and use the operator $T$ of The ordered-basis Hecke operator and its normalization. Write $e_{a} e_{b}$ for $e_{a} \otimes e_{b}$ . For $N = 2$ , in the ordered basis $(e_{1} e_{1}, e_{1} e_{2}, e_{2} e_{1}, e_{2} e_{2})$ its matrix is

$(\begin{array}{cccc} q & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & η & 0 \\ 0 & 0 & 0 & q \end{array}) .$

The complete weak-order case table for three tensor positions is below. In rows with two distinct labels assume $a < b$ ; in rows with three distinct labels assume $a < b < c$ .

Input	$T_{1} T_{2} T_{1}$	$T_{2} T_{1} T_{2}$
$a a a$	$q^{3} a a a$	$q^{3} a a a$
$a a b$	$q b a a$	$q b a a$
$a b a$	$q a b a + q η b a a$	$q a b a + q η b a a$
$b a a$	$q a a b + q η a b a + q^{2} η b a a$	$q a a b + q η a b a + q^{2} η b a a$
$a b b$	$q b b a$	$q b b a$
$b a b$	$q b a b + q η b b a$	$q b a b + q η b b a$
$b b a$	$q a b b + q η b a b + q^{2} η b b a$	$q a b b + q η b a b + q^{2} η b b a$
$a b c$	$c b a$	$c b a$
$a c b$	$b c a + η c b a$	$b c a + η c b a$
$b a c$	$c a b + η c b a$	$c a b + η c b a$
$b c a$	$a c b + η b c a + η c a b + η^{2} c b a$	$a c b + η b c a + η c a b + η^{2} c b a$
$c a b$	$b a c + η b c a + η c a b + η^{2} c b a$	$b a c + η b c a + η c a b + η^{2} c b a$
$c b a$	$a b c + η a c b + η b a c + η^{2} b c a + η^{2} c a b + (η + η^{3}) c b a$	$a b c + η a c b + η b a c + η^{2} b c a + η^{2} c a b + (η + η^{3}) c b a$

At $q = 1$ , $T$ is the ordinary flip. At $q = - 1$ , $T^{2} = I$ , but $T (e_{a} \otimes e_{a}) = - e_{a} \otimes e_{a}$ , so $T$ is not the ordinary flip.

Facts & Assumptions

Given: $F = Q (q)$ , an ordered basis $e_{1} < \dots < e_{N}$ , the operator $T$ , and the standard elementary-tensor bases of the finite tensor powers.

[F1]

The operator fixes an equal pair with coefficient $q$ , flips an increasing pair, and sends a decreasing pair $e_{a} \otimes e_{b}$ to $e_{b} \otimes e_{a} + (q - q^{- 1}) e_{a} \otimes e_{b}$ (The ordered-basis Hecke operator and its normalization).

[F2]

If two vector spaces are free with specified bases, the elementary tensors of those bases form a basis of their tensor product; iterating this gives the displayed pair and triple bases (The elementary tensors of two bases form the product basis of the tensor product).

Verification

Given: The data in [F1]–[F2].

1.1givenF1F2algebra

Applying [F1] to the four ordered pairs for $N = 2$ gives the displayed matrix: the unequal-pair block on $(e_{1} e_{2}, e_{2} e_{1})$ is $(\begin{array}{cc} 0 & 1 \\ 1 & η \end{array})$ and the equal pairs have eigenvalue $q$ . The elementary tensor basis from [F2] makes these blocks a full decomposition of the operator. The unequal-pair block satisfies $T^{2} = η T + I$ , while the equal-pair identity is $q^{2} = η q + 1$ ; since $T^{2} - η T - I = (T - q I) (T + q^{- 1} I)$ , the quadratic relation holds. The same block has inverse $T - η I$ , and the equal-pair inverse is $q^{- 1} = q - η$ .

1.2givenF1F2algebra

Every entry in the table is obtained by applying the three cases of [F1] successively to the indicated adjacent pair; the thirteen rows exhaust the weak-order patterns (one all-equal, six with exactly two distinct labels, and six with three distinct labels). The two computed columns agree in every row, and the triple basis in [F2] spans $V_{N}^{\otimes 3}$ , so $T_{1} T_{2} T_{1} = T_{2} T_{1} T_{2}$ on the whole tensor cube.

2.1givenF1step 1.1algebra∎

At $q = 1$ , $η = 0$ and the three defining branches reduce to the flip, including the equal-label branch. At $q = - 1$ , $η = 0$ , so the quadratic relation becomes $T^{2} = I$ , while the equal-label formula gives $T (e_{a} \otimes e_{a}) = - e_{a} \otimes e_{a}$ ; this distinguishes that specialization from the ordinary flip.

Depends on

Used by

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Sources

Li Luo and Weiqiang Wang, Lectures on dualities ABC in representation theory (standard reference, not scraped)