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The double-edge $(e = 2)$ KLR braid correction

Example

Take the cyclic affine datum of type $A_{1}^{(1)}$ : $I = {i, j}$ , $a_{i j} = a_{j i} = - 2$ , so $b_{i j} = - 2$ and $d_{i} = d_{j} = 1$ , with the KLR datum $Q_{i j} (u, v) = - (u - v)^{2} = Q_{j i} (v, u)$ (the quiver-with-automorphism specialization of The quiver Hecke (KLR) algebra of a symmetrizable Cartan datum). Then:

(a) On equal-colour pairs $ν_{r} = ν_{r + 1}$ the quadratic relation is $τ_{r}^{2} e (ν) = 0$ , while on double-edge pairs $ν_{r} = i, ν_{r + 1} = j$ it is $τ_{r}^{2} e (ν) = - (x_{r} - x_{r + 1})^{2} e (ν)$ (and symmetrically for $j i$ ).

(b) On $ν = (i, j, i)$ the braid relation has the nonzero correction $τ_{2} τ_{1} τ_{2} e (ν) - τ_{1} τ_{2} τ_{1} e (ν) = (x_{1} + x_{3} - 2 x_{2}) e (ν),$ so the naive simply-laced braid relation fails; the correction is the divided difference $\frac{Q_{i j} (x_{1}, x_{2}) - Q_{i j} (x_{3}, x_{2})}{x_{3} - x_{1}}$ of relation (7), the first divided difference of the quadratic parameter in its first variable, with the displayed sign.

(c) Degree check: the three factors of $τ_{2} τ_{1} τ_{2} e (ν)$ act successively on the pairs $(j, i)$ , then $(i, i)$ , then $(i, j)$ of the two-step permuted sequence, so their degrees are $- b_{j i} = 2$ , $- b_{i i} = - 2$ , $- b_{i j} = 2$ , totalling $2$ ; the right-hand side $x_{1} + x_{3} - 2 x_{2}$ also has degree $2$ . Thus the correction is homogeneous, of degree $- 2 b_{i j} - 2 d_{i} = 2$ , while for a simple edge the corresponding correction is a degree- $0$ constant; the doubled Cartan entry raises the degree of the quadratic parameter $Q_{i j}$ , hence that of the correction, by $2$ .

Facts & Assumptions

Given: The datum $I = {i, j}$ , $a_{i j} = a_{j i} = - 2$ , $d_{i} = d_{j} = 1$ , $b_{i j} = - 2$ , and the KLR datum $Q_{i j} (u, v) = - (u - v)^{2}$ , $Q_{j i} (u, v) = - (u - v)^{2}$ , with $Q_{i i} = Q_{j j} = 0$ , on the algebra $R_{3} (Q)$ of The quiver Hecke (KLR) algebra of a symmetrizable Cartan datum.

[F1]

Relations (4) and (7) of The quiver Hecke (KLR) algebra of a symmetrizable Cartan datum: $τ_{a}^{2} e (ν) = Q_{ν_{a} ν_{a + 1}} (x_{a}, x_{a + 1}) e (ν)$ , and the braid relation with correction $(Q_{ν_{a} ν_{a + 1}} (x_{a}, x_{a + 1}) - Q_{ν_{a} ν_{a + 1}} (x_{a + 2}, x_{a + 1})) / (x_{a + 2} - x_{a})$ when $ν_{a} = ν_{a + 2}$ .

[F2]

Grading (The KLR grading and the diagrammatic multiplication): $\deg x_{a} = 2 d_{ν_{a}}$ , $\deg τ_{a} = - b_{ν_{a} ν_{a + 1}}$ , and every defining relation is homogeneous.

[F3]

PBW basis (The PBW (monomial) basis theorem for KLR algebras): in each block $e (ν) R_{3} (Q) e (ν^{'})$ the monomials $τ_{w} x_{1}^{a_{1}} x_{2}^{a_{2}} x_{3}^{a_{3}} e (ν^{'})$ with $w ν^{'} = ν$ and $a_{1}, a_{2}, a_{3} \geq 0$ form a $k$ -basis, so a computation of an element on the spanning monomials determines it.

Verification

technique · direct computation from the defining relations, then degree bookkeeping

1.1givenF1algebra

(a) Specialize relation (4) of [F1]. On an equal-colour pair $ν_{r} = ν_{r + 1}$ one has $Q_{ν_{r} ν_{r + 1}} = Q_{i i} = 0$ or $Q_{j j} = 0$ , so $τ_{r}^{2} e (ν) = 0$ . On $(i, j)$ one has $Q_{i j} (x_{r}, x_{r + 1}) = - (x_{r} - x_{r + 1})^{2}$ , and on $(j, i)$ the symmetric formula $Q_{j i} (u, v) = - (u - v)^{2}$ gives the same expression, so $τ_{r}^{2} e (ν) = - (x_{r} - x_{r + 1})^{2} e (ν)$ in both orientations.

1.2givenF1F3algebra

(b) On $ν = (i, j, i)$ the outer colours $i, i$ agree and the middle colour differs, so relation (7) of [F1] applies with $a = 1$ : the left side is $τ_{2} τ_{1} τ_{2} e (ν) - τ_{1} τ_{2} τ_{1} e (ν)$ in the sign normalization fixed there (relation (6) of the definition, in which $τ_{1} x_{1} - x_{2} τ_{1} = e (ν)$ on the equal-colour component), and the right side is $(Q_{i j} (x_{1}, x_{2}) - Q_{i j} (x_{3}, x_{2})) / (x_{3} - x_{1})$ . Substituting $Q_{i j} (u, v) = - (u - v)^{2}$ gives $((x_{3} - x_{2})^{2} - (x_{1} - x_{2})^{2}) / (x_{3} - x_{1}) = (x_{3} - x_{1}) (x_{3} + x_{1} - 2 x_{2}) / (x_{3} - x_{1}) = x_{1} + x_{3} - 2 x_{2}$ , a polynomial. The element $(x_{1} + x_{3} - 2 x_{2}) e (ν)$ is nonzero by [F3], since its distinct dot monomials have coefficients $1, 1, - 2$ , and the first two remain nonzero in every characteristic.

1.3givenF2algebra

(c) In the word $τ_{2} τ_{1} τ_{2} e (ν)$ the factors are applied right to left: the rightmost $τ_{2}$ acts on the pair of positions $(2, 3)$ of $ν = (i, j, i)$ , which carries colours $(j, i)$ , so its degree is $- b_{j i} = 2$ ; the middle $τ_{1}$ then acts on the pair of positions $(1, 2)$ of the permuted sequence $(i, i, j)$ , which carries equal colours $(i, i)$ , so its degree is $- b_{i i} = - 2$ ; the leftmost $τ_{2}$ acts on the pair of positions $(2, 3)$ of $(i, i, j)$ , which carries colours $(i, j)$ , so its degree is $- b_{i j} = 2$ . The composite has degree $2 + (- 2) + 2 = 2$ .

2.1step 1.2step 1.3F1F2algebra∎

The right-hand side $x_{1} + x_{3} - 2 x_{2}$ is a sum of degree- $2$ monomials, so the correction in (b) is homogeneous of degree $2 = - 2 b_{i j} - 2 d_{i}$ (with $b_{i j} = - 2$ , $d_{i} = 1$ ); on a simple edge $a_{i j} = - 1$ with $d_{i} = d_{j} = 1$ the analogous correction is the degree- $0$ constant $Q_{i j} (x_{1}, x_{2}) - Q_{i j} (x_{3}, x_{2})$ divided by $x_{3} - x_{1}$ , so the doubled Cartan entry is exactly what raises the degree of the correction by $2$ .

Depends on

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Sources

Raphael Rouquier, 2-Kac-Moody Algebras, arXiv:0812.5023v3 (66 pages) (standard reference, not scraped)
Alexander Kleshchev, Representation theory of symmetric groups and related Hecke algebras, Bulletin of the AMS 47 (2010) 419-481 (survey), arXiv:0909.4844 (standard reference, not scraped)
Mikhail Khovanov and Aaron D. Lauda, A diagrammatic approach to categorification of quantum groups II, Transactions of the AMS 363 (2011) 2685-2700, arXiv:0804.2080v1 (standard reference, not scraped)

The double-edge (e=2)(e=2) KLR braid correction