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The ordered-basis operator is a Hecke symmetry

Statement

For every $N \geq 1$ , the operator $T$ of The ordered-basis Hecke operator and its normalization is invertible, satisfies $(T - q I) (T + q^{- 1} I) = 0$ , and satisfies the adjacent braid identity $T_{1} T_{2} T_{1} = T_{2} T_{1} T_{2}$ on $V_{N}^{\otimes 3}$ . Its inverse is $T^{- 1} = T - c I$ , where $c = q - q^{- 1}$ . If $S = q T$ and $Q = q^{2}$ , then $(S - Q I) (S + I) = 0$ and $S_{1} S_{2} S_{1} = S_{2} S_{1} S_{2}$ . At $q = 1$ , $T$ specializes to the ordinary flip; at $q = - 1$ its quadratic relation specializes to $T^{2} = I$ , but the operator itself is not the ordinary flip.

Facts & Assumptions

Given: $F = Q (q)$ , the ordered basis $e_{1} < \dots < e_{N}$ , the operator $T$ and $c = q - q^{- 1}$ from The ordered-basis Hecke operator and its normalization.

[F1]

The formula for $T$ on each ordered basis pair is the three-case formula in The ordered-basis Hecke operator and its normalization.

[F2]

The elementary tensors $e_{a} \otimes e_{b}$ and $e_{a} \otimes e_{b} \otimes e_{d}$ form bases of the corresponding finite tensor powers, so equality on these vectors proves equality of linear maps (Scalars, tensor powers, the empty tensor, opposite algebras and finite sums, The elementary tensors of two bases form the product basis of the tensor product).

Proof

Given: The data in [F1]–[F2].

1.1givenF1F2algebra

For $a < b$ , the span of $e_{a} \otimes e_{b}, e_{b} \otimes e_{a}$ is $T$ -invariant and, in that ordered basis, $T$ has matrix $(\begin{array}{cc} 0 & 1 \\ 1 & c \end{array})$ ; hence $T^{2} = c T + I$ there. On $e_{a} \otimes e_{a}$ one has $T = q I$ and $q^{2} = c q + 1$ . These subspaces partition the basis of $V_{N} \otimes V_{N}$ , so $T^{2} - c T - I = 0$ on the whole space. Since $T (T - c I) = (T - c I) T = I$ , it is invertible with inverse $T - c I$ , and $T^{2} - c T - I = (T - q I) (T + q^{- 1} I)$ by expansion.

1.2givenF1F2algebra

Write $a b c$ for $e_{a} \otimes e_{b} \otimes e_{c}$ . For distinct letters assume $a < b < c$ , and for two letters assume $a < b$ . Applying the three-case rule [F1] gives the same value for $T_{1} T_{2} T_{1}$ and $T_{2} T_{1} T_{2}$ in all thirteen weak-order patterns: $a a a \mapsto q^{3} a a a$ ; $a a b \mapsto q b a a$ ; $a b a \mapsto q a b a + q c b a a$ ; $b a a \mapsto q a a b + q c a b a + q^{2} c b a a$ ; $a b b \mapsto q b b a$ ; $b a b \mapsto q b a b + q c b b a$ ; $b b a \mapsto q a b b + q c b a b + q^{2} c b b a$ ; $a b c \mapsto c b a$ ; $a c b \mapsto b c a + c c b a$ ; $b a c \mapsto c a b + c c b a$ ; $b c a \mapsto a c b + c b c a + c c a b + c^{2} c b a$ ; $c a b \mapsto b a c + c b c a + c c a b + c^{2} c b a$ ; and $c b a \mapsto a b c + c a c b + c b a c + c^{2} b c a + c^{2} c a b + (c + c^{3}) c b a$ . The first seven entries cover one or two distinct labels, and the last six cover all orders of three distinct labels. Thus the two braid composites agree on the tensor basis [F2] and hence on the whole tensor cube.

2.1step 1.1step 1.2algebra∎

The normalized quadratic relation in step 1.1 gives $(q T - q^{2} I) (q T + I) = q^{2} (T - q I) (T + q^{- 1} I) = 0$ ; multiplying the braid identity of step 1.2 by $q^{3}$ on both sides gives the braid identity for $S = q T$ . At $q = 1$ , $c = 0$ and each branch of the formula is the flip, including the equal-label branch. At $q = - 1$ , $c = 0$ and the quadratic relation is $T^{2} = I$ , while $T (e_{a} \otimes e_{a}) = - e_{a} \otimes e_{a}$ , so it is not the ordinary flip.

Depends on

Used by

Cited to discharge well-definedness by The ordered-basis Hecke operator and its normalization.

Dependency tree · two levels

15 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

Alexei Davydov and Alexander Molev, A categorical approach to classical and quantum Schur–Weyl duality (standard reference, not scraped)