Spectral sublagebras

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

Department of Mathematics
University of Wisconsin, Madison
Madison, WI 53706 USA
ram@math.wisc.edu

Last updates: 10 April 2010

Spectral subalgebras

Then $C_{0} = \{μ \in A * | μ (x y) = μ (y x)\} is a commutative algebra,$ since, if $l_{1}, l_{2} \in C_{0}$ and $a \in A$ then $\begin{array}{rcl} (l_{2} l_{1}) (a) & = & (l_{1} \otimes l_{2}) Δ^{op} (a) = (l_{1} \otimes l_{2}) ℛ Δ (a) ℛ^{- 1} & = & (l_{1} \otimes l_{2}) Δ (a) ℛ^{- 1} ℛ = (l_{1} \otimes l_{2}) Δ (a) = (l_{1} l_{2}) (a), \end{array}$ where the third equality uses the definition of $C_{0} .$

If $(A, ℛ)$ is a quasitriangular Hopf algebra the $ℛ$ satisfies the quantum Yang-Baxter equation, (QYBE), $ℛ^{12} ℛ^{13} ℛ^{23} = ℛ^{12} (Δ \otimes id) (ℛ) = (Δ^{op} \otimes id) (ℛ) ℛ^{12} = ℛ^{23} ℛ^{13} ℛ^{12} .$

Since $ℛ = (ε \otimes id \otimes id) (Δ \otimes id) (ℛ) = (ε \otimes id \otimes id) ℛ^{13} ℛ^{23} = (ε \otimes id) (ℛ) . ℛ,$ and $ℛ = (id \otimes id \otimes ε) (id \otimes Δ) (ℛ) = (id \otimes id \otimes ε) ℛ^{13} ℛ^{23} = (id \otimes ε) (ℛ) . ℛ,$ and so $(ε \otimes id) (ℛ) = 1 and (id \otimes ε) (ℛ) = 1 .$

Then, since $ℛ (S \otimes id) (ℛ) = (m \otimes id) (id \otimes S \otimes id) (ℛ^{13} ℛ^{23}) = (m \otimes id) (id \otimes S \otimes id) (Δ \otimes id) (ℛ) = (ε \otimes id) (ℛ) = 1,$ it follows that $(S \otimes id) (ℛ) = ℛ^{- 1} .$ Applying this to the pair $(A^{op}, ℛ^{21})$ gives $(S^{- 1} \otimes id) (ℛ^{21}) = {(ℛ^{21})}^{op},$ and so $(id \otimes S^{- 1}) (ℛ) = ℛ^{- 1} .$ Then $(S \otimes S) (ℛ) = (id \otimes S) (S \otimes id) (ℛ) = (id \otimes S) (ℛ^{- 1}) = (id \otimes S) (id \otimes S^{- 1}) (ℛ) = ℛ .$

The map $φ : C \to Z (A)$ in the following proposition is ananalogue of the Harish-Chandra homomorphism.

Let $(A, ℛ)$ be a quasitrtiangular Hopf algebra. Then $C = \{λ \in A * | λ (x y) = λ (y S^{2} (x))\} is a commutative algebra$ and the map $\begin{array}{rcl} φ : & C & \to & Z (A) \\ l & \mapsto & (id \otimes l) (ℛ_{21}, ℛ) \end{array} is a well defined algebra homomorphism.$

Proof.

l_{1}, l_{2} \in A *

and

a \in A

then

\begin{array}{rcl} (l_{1} l_{2}) (a) & = & (l_{1} \otimes l_{2}) Δ^{op} (a) = (l_{1} \otimes l_{2}) (ℛ Δ (a) ℛ^{- 1}) \\ = & (l_{1} \otimes l_{2}) (Δ (a) ℛ^{- 1} (S^{2} \otimes S^{2}) (ℛ)), by definiton of C, \\ = & (l_{1} \otimes l_{2}) (Δ (a) ℛ^{- 1}, ℛ) \\ = & (l_{1} \otimes l_{2}) (Δ (a)) \\ = & (l_{1} l_{2}) (a), \end{array}

and hence

C

is a commutative algebra.

Let $a \in A .$ First note that $\begin{array}{rcl} a \otimes 1 & = & (id \otimes ε) Δ (a) = (id \otimes m) (id \otimes S^{- 1} \otimes id) (id \otimes Δ^{op}) Δ (a) \\ = & \sum_{a} a_{(1)} \otimes S^{- 1} (a_{(3)}) a_{(2)} = \sum_{a} (1 \otimes S^{- 1} (a_{(2)}) (a_{(11)} \otimes a_{(12)})) \\ = & \sum_{a} (1 \otimes S^{- 1} (a_{(2)})) Δ (a), \end{array}$ since $S^{- 1}$ is the antipode of $A^{op}$ , and $\begin{array}{rcl} a \otimes 1 & = & (id \otimes ε) Δ (a) = (id \otimes m) (id \otimes id \otimes S) (id \otimes Δ) Δ (a) \\ = & \sum_{a} a_{(1)} \otimes a_{(2)} S^{(a_{(3)})} = \sum_{a} (a_{(11)} \otimes a_{(12)}) (1 \otimes S (a_{(2)})) \\ = & \sum_{a} Δ (a_{(1)}) (1 \otimes S (a_{(2)})), \end{array} .$

Then, since $ℛ^{21} ℛ Δ (a) = ℛ^{21} Δ^{op} (a) ℛ = Δ (a) ℛ^{21} ℛ,$ $\begin{array}{rcl} a φ (l) & = & a (id \otimes l) (ℛ^{21} ℛ) = (id \otimes l) ((a \otimes 1) ℛ^{21} ℛ^{12}) \\ = & (id \otimes l) (\sum_{a} (1 \otimes S^{- 1} (a_{(2)})) Δ (a_{(1)}) ℛ^{21} ℛ) \\ = & (id \otimes l) (\sum_{a} Δ (a_{(1)}) ℛ^{21} ℛ (1 \otimes S (a_{(2)}))), by definition of C, \\ = & (id \otimes l) (ℛ^{21} ℛ \sum_{a} Δ (a_{(1)}) (1 \otimes S (a_{(2)}))) \\ = & (id \otimes l) (ℛ^{21} ℛ (a \otimes 1)) = (id \otimes l) (ℛ^{21} ℛ) a = φ (l) a, \end{array}$ and so $φ (l) \in Z (A) .$ Since $\begin{array}{rcl} φ (l_{1} l_{2}) & = & (id \otimes l_{1} l_{2}) (ℛ^{21} ℛ) = (id \otimes l_{1} \otimes l_{2}) ((id \otimes Δ) (ℛ^{21} ℛ)) \\ = & (id \otimes l_{1} \otimes l_{2}) (ℛ^{21} ℛ^{31} ℛ^{13} ℛ^{12}) = (id \otimes l_{1}) (ℛ^{21} (φ (l_{2}) \otimes 1) ℛ^{12}) \\ = & (id \otimes l_{1}) (ℛ^{21} ℛ (φ (l_{2}) \otimes 1)), since φ (l_{2}) \in Z (A), \\ = & φ (l_{1}) φ (l_{2}), \end{array}$ and so $φ$ is a homomorphism. $□$

References [PLACEHOLDER]

[BG] A. Braverman and D. Gaitsgory, Crystals via the affine Grassmanian, Duke Math. J. 107 no. 3, (2001), 561-575; arXiv:math/9909077v2, MR1828302 (2002e:20083)

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