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].
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 \(\text {av}(-\lambda ).\) \(\square \)