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The rank-one tensor action of the two-strand Hecke algebra is not faithful

Statement refuted

For every $N \geq 1$ and $n \geq 0$ , the normalized type-A Hecke algebra $H_{n} (Q (q))$ acts faithfully on $V_{N}^{\otimes n}$ through the ordered-basis Hecke operator.

Facts & Assumptions

Given: $F = Q (q)$ , the type-A Hecke algebra with quadratic relation $(T_{1} - q) (T_{1} + q^{- 1}) = 0$ , and the tensor-power action defined from the ordered-basis operator.

[F1]

The generic Hecke algebra has standard basis ${T_{w} : w \in W}$ , and this basis remains a basis after base change (The standard basis of the generic Hecke algebra and base change).

[F2]

The type-A Hecke algebra is the specialization of the universal Coxeter Hecke presentation at common parameter $q$ (Universal parameters, the generic Coxeter Hecke algebra and generator conjugacy).

[F3]

Sending $T_{i}$ to the ordered-basis operator on positions $i, i + 1$ defines an algebra action on every tensor power (The ordered-basis symmetry gives the normalized Hecke tensor action).

[F4]

On equal labels the ordered-basis operator is $T (e_{a} \otimes e_{a}) = q (e_{a} \otimes e_{a})$ (The ordered-basis Hecke operator and its normalization).

Counterexample

Given: Take $N = 1$ , $n = 2$ , $V_{1} = F e_{1}$ , and the Hecke action $ρ_{2}$ on $V_{1}^{\otimes 2}$ .

1.1givenF1F2F3F4algebra

In type $A_{1}$ the Coxeter presentation has one generator $s$ with $s^{2} = 1$ , so every word reduces to $1$ or $s$ ; the map $s \mapsto \overset{ˉ}{1}$ to the two-element group is well-defined and nontrivial, so these are distinct and the Coxeter group has exactly two elements. Thus [F1]–[F2] give the $F$ -basis $1, T_{1}$ of $H_{2} (F)$ . By [F3]–[F4], $T_{1}$ acts on the one-dimensional tensor square by $ρ_{2} (T_{1}) (e_{1} \otimes e_{1}) = q (e_{1} \otimes e_{1})$ ; hence $T_{1} - q 1$ lies in the kernel of $ρ_{2}$ .

2.1givenF1step 1.1algebra∎

In the basis $1, T_{1}$ , the element $T_{1} - q 1$ has coefficients $(- q, 1)$ , which are not both zero in $F$ . Thus it is a nonzero element of $H_{2} (F)$ mapped to zero, so $ρ_{2}$ is not injective and the universal faithfulness statement is false.

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Sources

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