Spectral sublagebras

## Spectral subalgebras

Then since, if ${l}_{1},{l}_{2}\in {C}_{0}$ and $a\in A$ then $l 2 l 1 a = l 1 ⊗ l 2 Δ op a = l 1 ⊗ l 2 ℛΔ a ℛ -1 = l 1 ⊗ l 2 Δ a ℛ -1 ℛ= l 1 ⊗ l 2 Δ a= l 1 l 2 a ,$ where the third equality uses the definition of ${C}_{0}.$

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

Since $ℛ= ε⊗id⊗id Δ⊗id ℛ = ε⊗id⊗id ℛ 13 ℛ 23 = ε⊗id ℛ .ℛ,$ and $ℛ= id⊗id⊗ε id ⊗ Δ ℛ = id⊗id⊗ε ℛ 13 ℛ 23 = id⊗ε ℛ .ℛ,$ and so $ε⊗id ℛ =1and id⊗ε ℛ =1.$

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

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

Let $\left(A,ℛ\right)$ be a quasitrtiangular Hopf algebra. Then $C= λ∈A*| λ xy =λ y S 2 x is a commutative algebra$ and the map $φ: C → Z A l ↦ id⊗l ℛ 21 ℛ is a well defined algebra homomorphism.$

 Proof. If ${l}_{1},{l}_{2}\in A*$ and $a\in A$ then $l 1 l 2 a = l 1 ⊗ l 2 Δ op a = l 1 ⊗ l 2 ℛΔ a ℛ -1 = l 1 ⊗ l 2 Δ a ℛ -1 S 2 ⊗ S 2 ℛ ,by definiton of C, = l 1 ⊗ l 2 Δ a ℛ -1 ℛ = l 1 ⊗ l 2 Δ a = l 1 l 2 a ,$ and hence $C$ is a commutative algebra. Let $a\in A.$ First note that $a⊗1 = id⊗ε Δ a = id⊗m id⊗ S -1 ⊗id id⊗ Δop Δ a = ∑ a a 1 ⊗ S -1 a 3 a 2 = ∑ a 1⊗ S -1 a 2 a 11 ⊗ a 12 = ∑a 1⊗ S -1 a 2 Δ a ,$ since ${S}^{-1}$ is the antipode of ${A}^{\mathrm{op}}$, and $a⊗1 = id⊗ε Δ a = id⊗m id⊗id⊗S id⊗Δ Δ a = ∑ a a 1 ⊗ a 2 S a 3 = ∑ a a 11 ⊗ a 12 1⊗S a 2 = ∑a Δ a 1 1⊗S a 2 , .$ Then, since $ℛ 21 ℛΔ a = ℛ 21 Δ op a ℛ=Δ a ℛ 21 ℛ,$ and so $\phi \left(l\right)\in Z\left(A\right).$ Since and so $\phi$ is a homomorphism. $\square$