The quantum group as a shuffle algebra

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

Last updates: 4 March 2012

The shuffle algebra and quantum group $U_q{𝔫}^{-}$

Let $ℱ$ be the free algebra generated by $f_1,\dots ,f_n$. Define a new product on $ℱ$ by $$u\circ v=\sum _{\sigma \in S_{k+l}/\left(S_k\times S_l\right)}{q}^{\mathrm{wt}(\sigma ,u,v)}\sigma \left(uv\right)\text{where}k=l\left(u\right),\ell =l\left(v\right),$$ the sum is over all minimal length coset representatives of the cosets in $S_{k+l}/\left(S_k\times S_l\right)$ and $$\mathrm{wt}(\sigma ,u,v)=\sum _{\genfrac{}{}{0ex}{}{1\le i<j\le k+\ell }{\sigma \left(i\right)>\sigma \left(j\right)}}-⟨u_i,v_{j-k}⟩,$$ where $u_i$ is the $i^{\mathrm{th}}$letter in the word $u$ and $v_{j-k}$is the ${(j-k)}^{\mathrm{th}}$ letter in $v$. By [Le, prop. 1], $\circ$ is associative. Note that the powers of $q$ in $u\circ v$ depend on the choice of the matrix of the form.

Remark. Leclerc [Le, §2.5 (8)] recursively defines a product by $$wa*xb=(w*xb)a+q^{-⟨\mathrm{deg}\left(wa\right),\mathrm{deg}\left(b\right)⟩}(wa*x)b,$$ for words $w$ and $x$ and letters $a$ and $b$. By a straightforward check $wa*xb=xb\circ wa$ so that the product $\circ$ is the opposite of $*$.

The quantum group $$U_q{𝔫}^{-}\text{is the subalgebra of}(ℱ,\circ )\text{generated by}{f}_1,\dots ,f_r.$$

Let $a\in ℱ$ let ${a|}_u$ denote the coefficient of the word $u$ in the expansion of $a$. Then $a\in U_q{𝔫}^{-}$ if and only if $$\sum _{k=0}^{-⟨{\alpha }_i,{\alpha }_j^{\vee }⟩+1}{(-1)}^k{\left[\begin{array}{c}1-⟨{\alpha }_i,{\alpha }_j^{\vee }⟩\\ k\end{array}\right]}_{q^{2/⟨{\alpha }_i,{\alpha }_i⟩}}{\underset{\phantom{T}}{\overset{\phantom{T}}{a}}|}_{z{f}_i^k{f}_j{f}_i^{1-⟨{\alpha }_i,{\alpha }_j^{\vee }⟩-k}t}=0,$$ for all $i\ne j$ with $⟨{\alpha }_i,{\alpha }_j^{\vee }⟩\ne 0$ and all words $z,t$.

Example. (Type $A_2$) If $$C=\left(\begin{array}{cc}2& -1\\ -1& 2\end{array}\right),\text{with}{\alpha }_1^{\vee }={\alpha }_1\text{and}{\alpha }_2^{\vee }={\alpha }_2,$$ then $U_q{𝔫}^{-}$ contains $1,f_1,f_2$, $$f_1\circ f_2=f_1{f}_2+q{f}_2{f}_1\text{,}{f}_2\circ f_1=f_2{f}_1+q{f}_1{f}_2$$ and $$\begin{array}{rcl}{f}_1\circ (f_1\circ f_2)& =& (f_1{f}_1{f}_2+q^{-2}{f}_1{f}_1{f}_2+q^{-1}{f}_1{f}_2{f}_1)+q(f_1{f}_2{f}_1+q{f}_2{f}_1{f}_1+q^{-1}{f}_2{f}_1{f}_1)\\ & =& (1+q^{-2})f_1{f}_1{f}_2+(q+q^{-1})f_1{f}_2{f}_1+(1+q^2)f_2{f}_1{f}_1,\end{array}$$ and the Serre relation for this element (Proposition ??? above) is $$\left[\begin{array}{c}1+1\\ 0\end{array}\right](1+q^{-2})-\left[\begin{array}{c}1+1\\ 1\end{array}\right](q+q^{-1})+\left[\begin{array}{c}1+1\\ 2\end{array}\right](1+q^2)=(1+q^{-2})-(q+q^{-1})(q+q^{-1})+(1+q^2)=0.$$

Notes and References

These notes are a presentation of the point of view of J.A. Green [Gr], which views the quantum group as a subalgebra of the free associative algebra instead of as a quotient. This version of the quantum group $U_q{𝔫}^{-}$ is the one which arises naturally as the character ring of the category of finite dimensional graded quiver Hecke algebra modules.

References

[Le] B. Leclerc, Dual canonical bases, quantum shuffles and $q$-characters, Math. Zeitschrift 246 (2004) 691-732 MR2045836 arXiv:math/0209133v3

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