Explicit Generators of the Centre of the Quantum Group

Abstract

A finite generating set of the centre of any quantum group is obtained, where the generators are given by an explicit formulae. For the slightly generalised version of the quantum group which we work with, we show that this set of generators is algebraically independent, thus the centre is isomorphic to a polynomial algebra.

Appendix A. Proof of the Harish-Chandra isomorphism

This part is about the algebraic proof of Theorem 2.5, i.e., the quantised Harish-Chandra isomorphism of $\mathrm{U}_q(\mathfrak {g})$. Note that it can be proven in much the same way as the proof in [11, Chapter 6]. However, we can hardly find a proof in detail with the method developed in [11]. Hence, we give some pertinent steps in the following.

Write $\mathrm{U}=\mathrm{U}_q(\mathfrak {g})$. We first show that $\gamma _{-\rho }\circ \pi$ indeed maps $Z(\mathrm{U})$ into the invariant subalgebra $(\mathrm{U}_{ev}^0)^W$.

Observe the following elementary result.

Lemma A.1

Let $\lambda \in P$. Any $u\in Z(\mathrm{U})$ acts on the Verma module $M(\lambda )$ as a scalar multiplication by $\chi _{\lambda }(\pi (u))$.

As an immediate consequence, we have

Lemma A.2

The restriction of $\pi$ to $Z(\mathrm{U})$ is injective, and hence so is $\gamma _{-\rho }\circ \pi$.

Proof

If $\pi (u)=0$, then by Lemma A.1, we have $u.M(\lambda )=0$ and hence $u.V(\lambda )=0$ for all $\lambda \in P^+$. By Proposition 2.4, $u=0$. $\square$

We now show that the image $\gamma _{-\rho }\circ \pi (Z(\mathrm{U}))$ of the centre is invariant under the Weyl group action.

Lemma A.3

The images of $Z(\mathrm{U})$ under the Harish-Chandra isomorphism are all in $(\mathrm{U}^0)^W$, i.e., $\gamma _{-\rho }\circ \pi (Z(\mathrm{U}))\subseteq (\mathrm{U}^0)^W$.

Proof

Fix any central element $u\in Z(\mathrm U)$, we write $h=\gamma _{-\rho }\circ \pi (u)$.

Given any $\lambda \in P$ and $i\in \{1,2,\ldots ,n\}$, we let $\mu =s_{\alpha _i}(\lambda +\rho )-\rho$.

If $(\lambda ,\alpha _i^{\vee })\geqslant 0$, there is a nontrivial homomorphism $M(\mu )\rightarrow M(\lambda )$ [11, Chapter 5.9]. By Lemma A.1,

$$\begin{aligned} \chi _{\lambda +\rho }(h)=\chi _{\mu +\rho }(h)=\chi _{\lambda +\rho }(s_{\alpha _i}h). \end{aligned}$$

(A.1)

If $(\lambda ,\alpha _i^{\vee })<-1$, then $(\mu ,\alpha _i^{\vee })$ is nonnegative, thus we may apply the above arguments to $\mu$ to show that (A.1) still holds.

Then the only other possibility is that $(\lambda ,\alpha _i^{\vee })=-1$. In this case, $\mu =\lambda$, and (A.1) holds trivially.

Since (A.1) holds for all $\lambda$ and i, and $s_{\alpha _i}$ generate W, we have

$$\begin{aligned} \chi _{\lambda }(wh-h)=0,\quad \forall w\in W, \ \lambda \in P. \end{aligned}$$

(A.2)

We can always write $wh-h=\sum _{\eta }a_{\eta }K_{\eta }$. Then (A.2) leads to

$$\begin{aligned} \sum \limits _{\eta }a_{\eta }\chi _{\lambda }(K_\eta )=\sum \limits _{\eta }a_{\eta }q^{(\lambda , \eta )}= \sum \limits _{\eta }a_{\eta }\chi _{\eta }(K_\lambda )=0, \quad \forall \lambda \in P. \end{aligned}$$

Thus, $\sum \limits _{\eta }a_{\eta }\chi _{\eta }=0$. The linear independence of characters then implies $a_{\eta }=0$ for all $\eta$. Hence, $wh-h=0$ for all $w\in W$, i.e., $h\in (\mathrm{U}^0)^W$ as claimed. $\square$

Now the following lemma justifies the range of $\gamma _{-\rho }\circ \pi$ as defined in (2.14).

Lemma A.4

The Harish-Chandra homomorphism $\gamma _{-\rho }\circ \pi$ maps $Z(\mathrm{U})$ to $(\mathrm{U}_{ev}^0)^W$.

Proof

Take an arbitrary $u\in Z(\mathrm{U})$, and write

$$\begin{aligned} \gamma _{-\rho }\circ \pi (u)=\sum \limits _{\mu \in P}a_{\mu }K_{\mu }. \end{aligned}$$

By Lemma A.3, $\gamma _{-\rho }\circ \pi (u)\in (\mathrm{U}^0)^W$. Thus, $a_{w\mu }=a_{\mu }$ for all $w\in W$ and $\mu \in P$. We have to show that $a_{\mu }\ne 0$ only if $\mu \in 2 P.$

Recall from (2.10) that there is an automorphism $\psi _{\sigma }$ of $\mathrm{U}$ associated to each group character $\sigma$ as defined in (2.9). It can be easily verified that $\psi _{\sigma }$ commutes with both $\pi$ and $\gamma _{-\rho }$. Therefore, we have

$$\begin{aligned} \gamma _{-\rho }\circ \pi (\psi _{\sigma }(u))=\psi _{\sigma }(\sum \limits _{\mu }a_{\mu }K_{\mu })=\sum \limits _{\mu }a_{\mu }\sigma (\mu ) K_{\mu }, \end{aligned}$$

which lands in $(\mathrm{U}^0)^W$ since $\psi _{\sigma }(u)$ is central. It follows that

$$\begin{aligned} a_{\mu }\sigma (\mu )=a_{w\mu }\sigma (w\mu )=a_{\mu }\sigma (w\mu )\quad \forall w\in W, \mu \in P. \end{aligned}$$

Since we have assumed that $a_{\mu }\ne 0$, this in particular implies $1=\sigma (\mu -s_{\alpha _i}\mu )$ for $1\leqslant i\leqslant n$. Fixing a group character $\sigma : P\rightarrow \mathbb {C}^{\times }$ such that $\sigma (\alpha _i)=-1$ for all i, we have

$$\begin{aligned} \sigma (\mu -s_{\alpha _i}\mu )=\sigma ((\mu , \alpha _i^{\vee })\alpha _i)=(-1)^{(\mu , \alpha _i^{\vee })}=1. \end{aligned}$$

This implies that $(\mu , \alpha _i^{\vee })$ is even for $1\leqslant i\leqslant n$, i.e., $\mu \in 2P$. $\square$

Now we prove the quantum Harish-Chandra isomorphism following [11, Chapter 6].

1.1 A.1. Proof of the isomorphism

By Lemma A.2, the restriction of $\gamma _{-\rho }\circ \pi$ to $Z(\mathrm{U})$ is injective. Therefore, it suffices to show surjectivity of the map (2.14) in order to prove Theorem 2.5. We do this by showing that each basis element of the invariant subalgebra $(\mathrm{U}_{ev}^0)^W$ has a pre-image in $Z(\mathrm{U})$.

We will follow the strategy of [11] to prove the surjectivity. This relies in an essential way on a non-degenerate bilinear form on $\mathrm{U}$, which can be constructed in exactly the same way as in [11, Chapter 6]. However, the explicit construction is rather involved and technical. We will merely describe the main properties of the form here, and refer to op. cit. for details.

Lemma A.5

[11, Chapter 6] There exists a unique bilinear form

$$\begin{aligned} (\ ,\ ): \mathrm{U}^{\leqslant 0} \times \mathrm{U}^{\geqslant 0} \rightarrow \mathbb {F}\end{aligned}$$

with the following properties:

$$\begin{aligned} \begin{aligned}&(K_{\lambda }, K_{\mu })=q^{-(\lambda , \mu )},&\quad&(K_{\lambda }, E_i)=0,\\&(F_i, E_j)= -\delta _{ij} (q_i-q_i^{-1})^{-1},&\quad&(F_i, K_{\lambda })=0,\\&(x, y_1y_2)=(\Delta (x), y_2\otimes y_1),&\quad&(x_1x_2,y)=(x_1\otimes x_2, \Delta (y)), \end{aligned} \end{aligned}$$

for all $x, x_1, x_2\in \mathrm{U}^{\leqslant 0}$, $y,y_1,y_2\in \mathrm{U}^{\geqslant 0}$, $\lambda , \mu \in P$ and $1\leqslant i,j\leqslant n$.

Proposition A.6

[11, Chapter 6] Let $\lambda ,\eta \in P$, $\mu ,\nu \in Q^+.$

• (1) $(xK_{\lambda },yK_{\eta })=q^{-(\lambda ,\eta )}(x,y)$ for any $x\in \mathrm{U}^-$ and $y\in \mathrm{U}^+$.

• (2) $(\mathrm{U}^{-}_{-\nu } ,\mathrm{U}_{\mu }^+)=0$ for any $\mu \ne \nu$.

• (3) The restriction $(\ ,\ )|_{\mathrm{U}_{-\mu }^-\times \mathrm{U}_{\mu }^+}$ is non-degenerate.

We now define a bilinear form on $\mathrm{U}$ by using Lemma A.5. Recall that $\mathrm{U}^+$ (resp. $\mathrm{U}^-$) is $Q^+$-graded (resp. $Q^-$-graded) vector space with respect to the $\mathrm{U}^0$-action given in (2.11), and the multiplication induces an isomorphism $\mathrm{U}^-\otimes \mathrm{U}^0 \otimes \mathrm{U}^+\cong \mathrm{U}$. Since $K_{\mu }$ is a unit in $\mathrm{U}$, we can rearrange this isomorphism into

$$\begin{aligned} \bigoplus _{\mu ,\nu \in Q^+} \mathrm{U}^-_{-\mu }K_{\mu } \otimes \mathrm{U}^0 \otimes \mathrm{U}^+_{\nu } \cong \mathrm{U}. \end{aligned}$$

Now the bilinear form $\langle \ ,\ \rangle : \mathrm{U}\times \mathrm{U}\rightarrow \mathbb {F}$ is defined on the graded components by

$$\begin{aligned} \langle yK_{\nu }K_{\lambda }x,y'K_{\nu '}K_{\eta }x'\rangle :=(y',x)(y,x')q^{(2\rho ,\nu )}(q^{1/2})^{-(\lambda ,\eta )} \end{aligned}$$

(A.3)

for all $x\in \mathrm{U}_{\mu }^+, x'\in \mathrm{U}_{\mu '}^+$, $y\in \mathrm{U}_{-\nu }^-$, and $y'\in \mathrm{U}_{-\nu '}^-$, with $\lambda ,\eta \in P,\mu ,\mu ',\nu ,\nu '\in Q^+$. It follows immediately from part (2) of Proposition A.6 that

$$\begin{aligned} \langle \mathrm{U}_{-\nu }^-\mathrm{U}^0\mathrm{U}_{-\mu }^+,\mathrm{U}_{-\nu '}^-\mathrm{U}^0\mathrm{U}_{\mu '}^+\rangle =0,\quad \text {unless}\ \mu =\nu ', \nu =\mu '.\end{aligned}$$

The following proposition gives two significant properties for the bilinear form (A.3), which will be used in the proof of surjectivity of the Harish-Chandra homomorphism.

Proposition A.7

[11, Chapter 6]

• (1) If $\langle v,u\rangle =0$ for all $v\in \mathrm{U}$, then $u=0$;

• (2) $\langle \text {ad}(x)u,v\rangle =\langle u,\text {ad}(S(x))v\rangle$ for all $x,u,v\in \mathrm{U}$.

Let M be a finite-dimensional $\mathrm{U}$-module. For any $m\in M$ and $f\in M^*$, let $c_{f,m}\in \mathrm{U}^*$ be the linear form with $c_{f,m}(v)=f(vm)$ for any $v\in \mathrm{U}.$ The following lemma follows from the non-degeneracy of the form $\langle \ , \ \rangle$ [11, Chapter 6.22].

Lemma A.8

Retain notation above. There exists a unique element $u\in \mathrm{U}$, depending on $f\in M^*$, $m\in M$ such that

$$\begin{aligned} c_{f,m}(v)=\langle v,u\rangle , \quad \forall v\in \mathrm{U}. \end{aligned}$$

This leads to the following key lemma.

Lemma A.9

Fix $\lambda \in P^+$, and let $V(\lambda )$ be the finite-dimensional simple $\mathrm{U}$-module with highest weight $\lambda$. Then there exists a unique central element $z_{\lambda }\in Z(\mathrm{U})$ such that

$$\begin{aligned} \langle u,z_{\lambda }\rangle =\mathrm{Tr}(uK_{2\rho }^{-1}), \quad \forall u\in \mathrm{U}, \end{aligned}$$

(A.4)

where $\mathrm{Tr}(x)$ denotes the trace of $x\in \mathrm{U}$ over $V(\lambda )$.

Proof

Let $m_1,m_2,\ldots ,m_{r}$ be a basis of $V(\lambda )$ and $f_1,f_2,\ldots ,f_r$ the dual basis of $V(\lambda )^*$, i.e., $f_i(m_j)=\delta _{ij}$. Then the trace of $uK_{2\rho }^{-1}$ over $V(\lambda )$ is equal to $\sum \nolimits _{i=1}^r c_{f_i,K_{2\rho }^{-1}m_i}(u).$ By Lemma A.8, there is a unique $v_i\in \mathrm{U}$ such that $\langle u,v_i\rangle =c_{f_i,K_{2\rho }^{-1}m_i}(u)$ for all $u\in \mathrm{U}$. Let $z_{\lambda }=v_1+v_2+\cdots +v_r$, then we have $\langle u,z_{\lambda }\rangle =\sum \nolimits _{i=1}^r c_{f_i,K_{2\rho }^{-1}m_i}(u)$, which is the trace of $uK_{2\rho }^{-1}$ over $V(\lambda )$.

It remains to show that $z_{\lambda }$ is central in $\mathrm{U}$, which is equivalent to showing that $\text {ad}(u)z_{\lambda }=\varepsilon (u)z_{\lambda }$ for any $u\in \mathrm{U}$. Then the linear representation $\varsigma _{\lambda }:\mathrm{U}\rightarrow \mathrm{End}(V(\lambda ))$ is a homomorphism of $\mathrm{U}$-modules, where $\mathrm{U}$ acts on itself by the adjoint action, that is, $u.v:=\mathrm{ad}(u)v$ for any $u,v\in \mathrm{U}$. The quantum trace $\text {Tr}_{q}:\mathrm{End}(V(\lambda ))\rightarrow \mathbb {F}$ that takes $\varphi \mapsto \text {Tr}(\varphi \circ K_{2\rho }^{-1})$ is also a $\mathrm{U}$-module homomorphism, where $\mathbb {F}$ is the trivial module such that $u.a=\varepsilon (u)a$ for any $a\in \mathbb {F}$. Let $\theta =\mathrm{Tr}_q\circ \varsigma _{\lambda }$. Then by definition

$$\begin{aligned} \theta (u)= \mathrm{Tr}_q\circ \varsigma _{\lambda }(u)=\mathrm{Tr}(uK_{2\rho }^{-1})=\langle u,z_{\lambda }\rangle , \quad \forall u\in \mathrm{U}. \end{aligned}$$

Since $\theta$ is a $\mathrm{U}$-module homomorphism, we have

$$\begin{aligned} \theta (u.v)=u.\theta (v)= \varepsilon (u)\theta (v)=\varepsilon (u)\langle v,z_{\lambda }\rangle , \end{aligned}$$

On the other hand, using the adjoint structure of $\mathrm{U}$ we have

$$\begin{aligned} \theta (u.v)=\mathrm{Tr}_q\circ \varsigma _{\lambda }(\mathrm{ad}(u)v)=\langle \text {ad}(u)v,z_{\lambda }\rangle =\langle v,\text {ad}(S(u))z_{\lambda }\rangle , \end{aligned}$$

where the last equation follows from part(2) of Proposition A.7. Since the bilinear form is non-degenerate, we have $\text {ad}(S(u))z_{\lambda }=\varepsilon (u)z_{\lambda }$ for all $u\in \mathrm{U}$. Recalling that the antipode S satisfies $\varepsilon \circ S=\varepsilon$, we obtain $\text {ad}(u)z_{\lambda }=\varepsilon (u)z_{\lambda }$ for all $u\in \mathrm{U}$. Therefore, $z_{\lambda }\in Z(\mathrm{U})$. $\square$

Lemma A.10

Let $\lambda \in P^+$, and $V(\lambda )$ the finite-dimensional simple module of $\mathrm{U}$. Let $z_{\lambda }\in Z(\mathrm{U})$ be the central element defined in (A.4). Then

$$\begin{aligned} \gamma _{-\rho }\circ \pi (z_{\lambda })=\sum \limits _{\eta \in \Pi (\lambda )}m_{\lambda }(\eta )K_{-2\eta }, \end{aligned}$$

where $\Pi (\lambda )$ is the set of weights of $V(\lambda )$ and $m_{\lambda }(\eta )$ denotes the dimension of the weight space $V(\lambda )_{\eta }$.

Proof

Since $z_{\lambda }$ is central and $Z(\mathrm{U}) \subseteq \mathrm{U}_0=\mathrm{U}^0\oplus \bigoplus \limits _{\nu >0}\mathrm{U}_{-\nu }^-\mathrm{U}^0\mathrm{U}_{\nu }^+$, we may write

$$\begin{aligned} z_{\lambda }=z_{\lambda ,0} + \sum _{\nu > 0}z_{\lambda ,\nu },\quad \text {with } z_{\lambda ,0}\in \mathrm{U}^0, \, z_{\lambda ,\nu }\in \mathrm{U}_{-\nu }^-\mathrm{U}^0\mathrm{U}_{\nu }^+. \end{aligned}$$

It follows that $\pi (z_{\lambda })= z_{\lambda ,0}$. By (A.3), we have

$$\begin{aligned} \langle K_{\mu },z_{\lambda }\rangle =\langle K_{\mu },z_{\lambda ,0}\rangle = \langle K_{\mu }, \pi (z_{\lambda })\rangle , \quad \forall \mu \in P. \end{aligned}$$

(A.5)

On the other hand, using Lemma A.9 we obtain

$$\begin{aligned} \begin{aligned} \langle K_{\mu },z_{\lambda }\rangle&=\mathrm{Tr}(K_{\mu -2\rho })= \sum \limits _{\eta \in \Pi (\lambda )}m_{\lambda }(\eta )q^{(\eta ,\mu -2\rho )}\\&=\sum \limits _{\eta \in \Pi (\lambda )}m_{\lambda }(\eta )q^{-(2\eta ,\rho )}q^{(\mu ,\eta )}\\&= \sum \limits _{\eta \in \Pi (\lambda )}m_{\lambda }(\eta )q^{-(2\eta ,\rho )}\langle K_{\mu }, K_{-2\eta }\rangle . \end{aligned} \end{aligned}$$

(A.6)

Comparing (A.5) and (A.6) and using the non-degeneracy of the bilinear form, we have

$$\begin{aligned} \gamma _{-\rho }\circ \pi (z_{\lambda })= \sum \limits _{\eta \in \Pi (\lambda )}m_{\lambda }(\eta )K_{-2\eta }. \end{aligned}$$

This completes the proof. $\square$

Now we are ready to prove Theorem 2.5.

Proof of Theorem 2.5

We know that $\gamma _{-\rho }\circ \pi$ is injective from Lemma A.2. It remains to show that $\gamma _{-\rho }\circ \pi$ is surjective. By Lemma 4.6, the elements $\mathrm{av}(-\mu )= \sum _{\eta \in W\mu }K_{-2\eta }$ with $\mu \in P^+$ form a basis for $(\mathrm{U}_{ev}^0)^W$, since each group orbit $W\mu$ in P contains exactly one $-\mu$ such that $\mu$ is dominant.

We use induction on $\mu$ to show that the basis elements $\mathrm{av}(-\mu )$ are in the image of $\gamma _{-\rho }\circ \pi$. Endow P with the standard partial order such that $\mu \leqslant \lambda$ if and only if $\lambda -\mu$ is a nonnegative integral linear combination of positive roots. For the base case $\nu =0$, we have $\text {av}(0)=1=\gamma _{-\rho }\circ \pi (1)$. For any $\lambda \in P^+$, we may apply Lemma A.9 and then obtain the element $z_{\lambda }\in Z(\mathrm{U})$, which by Lemma A.10 has the image

$$\begin{aligned} \gamma _{-\rho }\circ \pi (z_{\lambda })=\sum \limits _{\eta \in \Pi (\lambda )}m_{\lambda }(\eta )K_{-2\eta }=\text {av}(-\lambda )+\sum \limits _{\mu <\lambda ,\mu \in P^+}m_{\lambda }(\mu )\text {av}(-\mu ), \end{aligned}$$

where the second equality follows from the fact that $m_{\lambda }(\lambda )=m_{\lambda }(w\lambda )=1$ for any $w\in W$. The left-hand side of the above equation belongs to $\gamma _{-\rho }\circ \pi (Z(\mathrm{U}))$. By induction hypothesis, all $\text {av}(-\mu )$ with $\mu <\lambda$ are in the image of $\gamma _{-\rho }\circ \pi$, hence so is $\text {av}(-\lambda ).$ $\square$