Affine Braids, Markov Traces and the Category $𝒪$

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

Last update: 22 December 2013

Markov Traces

A Markov trace on the affine braid group is a trace functional which respects the inclusions ${\stackrel{\sim }{ℬ}}_1\subseteq {\stackrel{\sim }{ℬ}}_2\subseteq \cdots$ where $$\begin{array}{cc}\begin{array}{ccc}{\stackrel{\sim }{ℬ}}_k& \hookrightarrow & {\stackrel{\sim }{ℬ}}_{k+1}\\ \begin{array}{c} b 1 ⋯ k \end{array}& ⟼& \begin{array}{c} b 1 ⋯ k k+1 \end{array}\end{array}& \text{(5.1)}\end{array}$$ More precisely, a Markov trace on the affine braid group with parameters $z,Q_1,Q_2,\dots \in ℂ$ is a sequence of functions ${\text{mt}}_k:{\stackrel{\sim }{ℬ}}_k⟶ℂ$ such that

(1)	${\text{mt}}_1\left(1\right)=1,$
(2)	${\text{mt}}_{k+1}\left(b\right)={\text{mt}}_k\left(b\right),$ for $b\in {\stackrel{\sim }{ℬ}}_k,$
(3)	${\text{mt}}_k\left(b_1{b}_2\right)={\text{mt}}_k\left(b_2{b}_1\right),$ for $b_1,b_2\in {\stackrel{\sim }{ℬ}}_k,$
(4)	${\text{mt}}_{k+1}\left(b{T}_k\right)=z{\text{mt}}_k\left(b\right),$ for $b\in {\stackrel{\sim }{ℬ}}_k,$
(5)	${\text{mt}}_{k+1}\left(b{\left({\stackrel{\sim }{X}}^{{\epsilon }_{k+1}}\right)}^r\right)=Q_r{\text{mt}}_k\left(b\right),$ for $b\in {\stackrel{\sim }{ℬ}}_k,$

where $${\stackrel{\sim }{X}}^{{\epsilon }_{k+1}}=T_k{T}_{k-1}\cdots T_2{X}^{{\epsilon }_1}{T}_2^{-1}\cdots T_{k-1}^{-1}{T}_k^{-1}=\begin{array}{c} 1 2 ⋯ k+1 \end{array}$$

If $M$ is a finite dimensional $U=U_h𝔤$ module and $a\in {\text{End}}_U\left(M\right)$ the quantum trace of $a$ on $M$ (see [LRa1977, §3] and [CPr1994, Def. 4.2.9]) is the trace of the action of $e^{h\rho }a$ on $M,$ $$\begin{array}{cc}{\text{tr}}_q\left(a\right)=\text{Tr}(e^{h\rho }a,M),\text{and}{\text{dim}}_q\left(M\right)={\text{tr}}_q\left({\text{id}}_M\right)=\text{Tr}(e^{h\rho },M)& \text{(5.2)}\end{array}$$ is the quantum dimension of $M\text{.}$ The first step of the standard argument for proving Weyl’s dimension formula [BtD1985, VI Lemma 1.19], shows that the quantum dimension of the finite dimensional irreducible $U_h𝔤\text{-module}$ $L\left(\mu \right)$ is $$\begin{array}{cc}\begin{array}{ccc}{\text{dim}}_q\left(L\left(\mu \right)\right)=\text{Tr}(e^{h\rho },L\left(\mu \right))& =& \prod _{\alpha >0}\frac{e^{\frac{h}{2}⟨\mu +\rho ,{\alpha }^{\vee }⟩}-e^{-\frac{h}{2}⟨\mu +\rho ,{\alpha }^{\vee }⟩}}{e^{\frac{h}{2}⟨\rho ,{\alpha }^{\vee }⟩}-e^{-\frac{h}{2}⟨\rho ,{\alpha }^{\vee }⟩}}\\ & =& \prod _{\alpha >0}\frac{\left[⟨\mu +\rho ,{\alpha }^{\vee }⟩\right]}{\left[⟨\rho ,{\alpha }^{\vee }⟩\right]},\end{array}& \text{(5.3)}\end{array}$$ where $q=e^{h/2}$ and $\left[d\right]=(q^d-q^{-d})/(q-q^{-1})$ for a positive integer $d\text{.}$

Let $\mu ,\nu \in P^{+}$ be dominant integral weights. Let $M=L\left(\mu \right)$ and $V=L\left(\nu \right)$ and let ${\Phi }_k$ be the representation of ${\stackrel{\sim }{ℬ}}_k$ defined in Proposition 3.1. Then the functions $$\begin{array}{cccc}{\text{mt}}_k:& {\stackrel{\sim }{ℬ}}_k& ⟶& ℂ\\ & b& ⟼& \frac{{\text{tr}}_q\left({\Phi }_k\left(b\right)\right)}{{\text{dim}}_q\left(M\right){\text{dim}}_q{\left(V\right)}^k}\end{array}$$ form a Markov trace on the affine braid group with parameters $$z=\frac{q^{⟨\nu ,\nu +2\rho ⟩}}{{\text{dim}}_q\left(V\right)}$$ and $$Q_r=\sum _{\lambda }{q}^{r(⟨\lambda ,\lambda +2\rho ⟩-⟨\mu ,\mu +2\rho ⟩-⟨\nu ,\nu +2\rho ⟩)}\frac{{\text{dim}}_q\left(L\left(\lambda \right)\right)c_{\mu \nu }^{\lambda }}{{\text{dim}}_q\left(L\left(\mu \right)\right){\text{dim}}_q\left(L\left(\nu \right)\right)},$$ where the positive integers $c_{\mu \nu }^{\lambda }$ and the sum in the expression for $Q_r$ are as in the tensor product decomposition $$L\left(\mu \right)\otimes L\left(\nu \right)=\underset{\lambda }{⨁}{L\left(\lambda \right)}^{\oplus c_{\mu \nu }^{\lambda }}\text{.}$$

		Proof.
	The fact that ${\text{mt}}_k$ as defined in the statement of the Theorem satisfies (1)–(4) in the definition of a Markov trace follows exactly as in [LRa1977, Theorem 3.10c]. The formula for the parameter $z$ is derived in [LRa1977, (3.9) and Thm. 3.10(2)]. It remains to check (5). The proof is a combination of the argument used in [Ore1999, Theorem 5.3] and the argument in the proof of [LRa1977, Theorem 3.10c]. Let ${\epsilon }_k:{\text{End}}_U(M\otimes V^{\otimes k})\to {\text{End}}_U(M\otimes V^{\otimes (k-1)})$ be given by $$\begin{array}{cc}{\epsilon }_k\left(z\right)=({\text{id}}_{M\otimes V^{\otimes (k-1)}}\otimes ě)\circ (z\otimes \text{id})& \text{(5.4)}\end{array}$$ where $$\begin{array}{cccc}ě:& V\otimes V^{*}& ⟶& ℂ\\ & x\otimes \varphi & ⟼& {\text{dim}}_q{\left(V\right)}^{-1}\varphi \left(e^{h\rho }x\right)\text{.}\end{array}$$ If $V$ is simple then $ě$ is the unique $U_h𝔤\text{-invariant}$ projection onto the invariants in $V\otimes V^{*}\text{.}$ Pictorially, $${\epsilon }_k\left(\begin{array}{c} z 1 ⋯ k \end{array}\right)=\begin{array}{c} z \end{array}=\begin{array}{c} εk(z) 1 ⋯ k-1 \end{array}\text{.}$$ The argument of [LRa1977, Theorem 3.10b] shows that $$\begin{array}{cc}{\text{mt}}_k\left(b\right)={\text{mt}}_{k-1}\left({\epsilon }_{k-1}\left(b\right)\right),\text{if}\hspace{0.17em}b\in {\stackrel{\sim }{ℬ}}_k\text{.}& \text{(5.5)}\end{array}$$ Since ${\epsilon }_1\left({\left(X^{{\epsilon }_1}\right)}^r\right)$ is a $U_h𝔤\text{-module}$ homomorphism from $M$ to $M$ and, since $M$ is simple, Schur’s lemma implies that $$r\hspace{0.17em}\text{loops}\hspace{0.17em}\{\begin{array}{c}\\ \end{array}\begin{array}{c} \end{array}={\epsilon }_1\left({\left(X^{{\epsilon }_1}\right)}^r\right)=\xi ·{\text{id}}_M,\text{for some}\hspace{0.17em}\xi \in ℂ\text{.}$$ Let ${\stackrel{\sim }{R}}_i={\text{id}}_V^{\otimes (i-1)}\otimes {Ř}_{VM}\otimes {\text{id}}_V^{(k+1)-i}\text{.}$ Then $${\left({\stackrel{\sim }{X}}^{{\epsilon }_{k+1}}\right)}^r={\left({\stackrel{\sim }{R}}_k\cdots {\stackrel{\sim }{R}}_1\right)}^{-1}{\left(X^{{\epsilon }_1}\right)}^r\left({\stackrel{\sim }{R}}_k\cdots {\stackrel{\sim }{R}}_1\right)$$ and $$\begin{array}{ccc}{\text{mt}}_{k+1}\left(b{\left({\stackrel{\sim }{X}}^{{\epsilon }_{k+1}}\right)}^r\right)& =& {\text{mt}}_k\left({\epsilon }_k\left(b{\left({\stackrel{\sim }{X}}^{{\epsilon }_{k+1}}\right)}^r\right)\right)\\ & =& {\text{mt}}_k\left({\epsilon }_k\left(b{\left({\stackrel{\sim }{R}}_k\cdots {\stackrel{\sim }{R}}_1\right)}^{-1}{\left(X^{{\epsilon }_1}\right)}^r\left({\stackrel{\sim }{R}}_k\cdots {\stackrel{\sim }{R}}_1\right)\right)\right)\\ & =& {\text{mt}}_k\left(b{\left({\stackrel{\sim }{R}}_k\cdots {\stackrel{\sim }{R}}_1\right)}^{-1}{\epsilon }_1\left({\left(X^{{\epsilon }_1}\right)}^r\right)\left({\stackrel{\sim }{R}}_k\cdots {\stackrel{\sim }{R}}_1\right)\right)\\ & =& {\text{mt}}_k(b{\left({\stackrel{\sim }{R}}_k\cdots {\stackrel{\sim }{R}}_1\right)}^{-1}\xi ·{\text{id}}_M{\stackrel{\sim }{R}}_k\cdots {\stackrel{\sim }{R}}_1)=\xi ·{\text{mt}}_k\left(b\right)\text{.}\end{array}$$ This last calculation is more palatable in a pictorial format, $$\begin{array}{ccc}{\text{mt}}_{k+1}\left(\begin{array}{c} b (X∼εk+1)r 1 ⋯ k \end{array}\right)& =& {\text{mt}}_{k+1}\left(\begin{array}{c} b (X∼εk+1)r 1 ⋯ k \end{array}\right)\\ & =& {\text{mt}}_k\left(\begin{array}{c} b (X∼εk+1)r 1 ⋯ k \end{array}\right)\\ & =& \xi ·{\text{mt}}_k\left(\begin{array}{c} b 1 ⋯ k \end{array}\right)\\ & =& \xi ·{\text{mt}}_k\left(\begin{array}{c} b 1 ⋯ k \end{array}\right)\text{.}\end{array}$$ It remains to calculate the constant $\xi \text{.}$ By (2.14), $${\left(X^{{\epsilon }_1}\right)}^r={\left({Ř}_0^2\right)}^r={\left(\sum _{\lambda }{q}^{c\left(\lambda \right)P_{\mu \nu }^{\lambda }}\right)}^r=\sum _{\lambda }{q}^{rc\left(\lambda \right)}{P}_{\mu \nu }^{\lambda },$$ where $c\left(\lambda \right)=⟨\lambda ,\lambda +2\rho ⟩-⟨\mu ,\mu +2\rho ⟩-⟨\nu ,\nu +2\rho ⟩$ and $P_{\mu \nu }^{\lambda }$ is the projection onto the ${L\left(\lambda \right)}^{\oplus c_{\mu \nu }^{\lambda }}$ component in the decomposition of $M\otimes V=L\left(\mu \right)\otimes L\left(\nu \right)\text{.}$ Thus $$\begin{array}{ccc}\xi & =& {\text{mt}}_0(\xi ·{\text{id}}_M)={\text{mt}}_1\left({\left(X^{{\epsilon }_1}\right)}^r\right)=\frac{1}{{\text{dim}}_q\left(M\right){\text{dim}}_q\left(V\right)}{\text{tr}}_q\left(\sum _{\lambda }{q}^{rc\left(\lambda \right)}{P}_{\mu \nu }^{\lambda }\right)\\ & =& \frac{1}{{\text{dim}}_q\left(M\right){\text{dim}}_q\left(V\right)}{\text{tr}}_q\left(\sum _{\lambda }{q}^{rc\left(\lambda \right)}{c}_{\mu \nu }^{\lambda }{\text{id}}_{L\left(\lambda \right)}\right)\\ & =& \sum _{\lambda }{q}^{rc\left(\lambda \right)}{c}_{\mu \nu }^{\lambda }\frac{{\text{dim}}_q\left(L\left(\lambda \right)\right)}{{\text{dim}}_q\left(L\left(\mu \right)\right){\text{dim}}_q\left(L\left(\nu \right)\right)}\text{.}\end{array}$$ $\square$

There is another formula [TWe1217386, Lemma (3.51)], for the constant $Q_1$ in Theorem 5.1, namely, $$\begin{array}{cc}{Q}_1=\frac{\sum _{w\in W}{\left(-1\right)}^{\ell \left(w\right)}{q}^{⟨\mu +\rho ,w(\nu +\rho )⟩}}{\sum _{w\in W}{(-1)}^{\ell \left(w\right)}{q}^{⟨\mu +\rho ,w\rho ⟩}},& \text{(5.6)}\end{array}$$ where $W$ is the Weyl group of $𝔤\text{.}$

Let ${\text{mt}}_k$ be as in Theorem 5.1 and let ${\stackrel{\sim }{𝒵}}_k={\text{End}}_U(M\otimes V^{\otimes k})\text{.}$ Then ${\text{mt}}_k$ is the restriction of the linear functional $$\begin{array}{cc}\begin{array}{cccc}{\text{mt}}_k:& {\stackrel{\sim }{𝒵}}_k& ⟶& ℂ\\ & a& ⟼& \frac{{\text{tr}}_q\left(a\right)}{{\text{dim}}_q\left(M\right){\text{dim}}_q\left(V^{\otimes k}\right)}\end{array}& \text{(5.7)}\end{array}$$ to ${\Phi }_k\left({\stackrel{\sim }{ℬ}}_k\right)\text{.}$ Since $M\otimes V^{\otimes k}$ is a finite dimensional semisimple module ${\stackrel{\sim }{𝒵}}_k$ is a finite dimensional semisimple algebra. The weights of the Markov trace mt are the constants $t_{\lambda /\mu }$ defined by $$\begin{array}{cc}{\text{mt}}_k=\sum _{\lambda }{t}_{\lambda /\mu }{\chi }_{{\stackrel{\sim }{𝒵}}_k}^{\lambda /\mu },& \text{(5.8)}\end{array}$$ where ${\chi }_{{\stackrel{\sim }{𝒵}}_k}^{\lambda /\mu }$ are the irreducible characters of ${\stackrel{\sim }{𝒵}}_k\text{.}$

Let $M=L\left(\mu \right)$ and $V=L\left(\nu \right)$ be finite dimensional irreducible $U_h𝔤\text{-modules.}$ The weights of the Markov trace on the affine braid group defined in Theorem 5.1 are $$t_{\lambda /\mu }=\frac{{\text{dim}}_q\left(L\left(\lambda \right)\right)}{{\text{dim}}_q\left(L\left(\mu \right)\right){\text{dim}}_q{\left(V\right)}^k}\text{.}$$

		Proof.
	Since $M\otimes V^{\otimes k}$ is finite dimensional and semisimple the algebra ${\stackrel{\sim }{𝒵}}_k={\text{End}}_U(M\otimes V^{\otimes k})$ is a finite dimensional semisimple algebra. Schur’s lemma can be used to show that, as a $(U_h𝔤,{\stackrel{\sim }{𝒵}}_k)$ bimodule, $$\begin{array}{cc}M\otimes V^{\otimes k}\cong \underset{\lambda }{⨁}L\left(\lambda \right)\otimes {ℒ}^{\lambda /\mu },& \text{(5.9)}\end{array}$$ where the ${ℒ}^{\lambda /\mu }$ are the irreducible ${\stackrel{\sim }{𝒵}}_k$ modules. In the notation of (4.1), ${ℒ}^{\lambda /\mu }=F_{\lambda }\left(L\left(\mu \right)\right)$ and ${\chi }_{{\stackrel{\sim }{𝒵}}_k}^{\lambda /\mu }$ is the character of ${ℒ}^{\lambda /\mu }\text{.}$ Taking the quantum trace on both sides of (5.9) gives $${\text{tr}}_1\left(a\right)=\text{Tr}\left(e^{-h\rho }a\right)=\sum _{\lambda }\text{Tr}(e^{-h\rho },L\left(\lambda \right)){\chi }_{{\stackrel{\sim }{𝒵}}_k}^{\lambda /\mu }\left(a\right)=\sum _{\lambda }{\text{tr}}_{\prime }\left(L\left(\lambda \right)\right){\chi }_{{\stackrel{\sim }{𝒵}}_k}^{\lambda /\mu }\left(a\right)\text{.}$$ The result follows by dividing both sides by ${\text{dim}}_q\left(L\left(\mu \right)\right){\text{dim}}_q{\left(V\right)}^k\text{.}$ $\square$

Notes and references

This is a typed version of the paper Affine Braids, Markov Traces and the Category $𝒪$ by Rosa Orellana and Arun Ram*.

*Research supported in part by National Science Foundation grant DMS-9971099, the National Security Agency and EPSRC grant GR K99015.

This paper is a slightly revised version of a preprint of 2001. We thank F. Goodman, A. Henderson and an anonymous referee for their very helpful comments on the original preprint.

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