Does the Leclerc-Thibon involution exchange vertex operators of the first and second type?

• This question is about $U_q ( \hat{\mathfrak{sl}}_2 )$ representation theory. There is a notion of vertex operators $\Phi_{\pm }(z)$ of first and $\Psi_{\pm}(z)$ of the second type. They are defined to be intertwiners $$\Phi(z): V(\Lambda_i) \rightarrow V(\Lambda_{1-i}) \otimes V_z$$ $$\Psi(z): V(\Lambda_i) \rightarrow V_z \otimes V(\Lambda_{1-i})$$ here $V_z$ -- evaluation module corresponding to two-dimensional standard representation. $V(\Lambda_0)$ and $V(\Lambda_1)$ are the only two integrable representations of $U_q ( \hat{\mathfrak{sl}}_2 )$. Also let us define $\Phi_{\pm} (z): V(\Lambda_i) \rightarrow V(\Lambda_{1-i})$ by formula $\Phi(z) = \Phi_{+}(z) \otimes v_{+} + \Phi_{-}(z) \otimes v_-$ (here $v_{\pm}$ -- standard basis of two dimensional representation). The details can be found in a classical textbook by Jimbo and Miwa (chapter 6).

• There is a paper by Stern. It studies $V(\Lambda_i)$ and one kind of vertex operators in terms of semi-infinite power of evaluation representation. I believe it is the second kind. In Stern's notation semi-infinite product is infinite to the right and truncated to the left. So it is easy to tensor with another evaluation representation on the left.

• There is an involution defined be Leclerc and Thibon (one can find definition here, section 3). The involution here is defined again in terms of semi-infinite tensor power of evaluation representation. $$u_{i_1} \wedge_q u_{i_2} \wedge_q \dots \wedge_q u_{k_i} \wedge_q u_{k_{i+1}} \wedge_q \dots \rightarrow (-1)^{{k}\choose{2} } q^{\alpha_{n,k}(I)} u_{i_k} \wedge_q u_{i_{k-1}} \wedge_q \dots \wedge_q u_{i_1} \wedge_q u_{i_{i+1}} \wedge_q \dots$$ This involution is antilinear. It means that it sends $q \rightarrow q^{-1}$.

Consider $\Psi_{\pm} (z)$ conjugated by involution. It is natural to conjecture, that this operator is $\Phi_{\pm}(z)$ because the conjugation "exchange left and right multiplication". But maybe it is true up to some renormalization... I tried to verify some defining relation (from Jimbo-Miwa) to check this and did not succeed. I do not have an idea, what is precisely the conjugated $\Psi$.

By the way, I suppose, that if these conjecture true, then it must be published (all my references are from the nineties). But I did not find anything.

Questions

• What is conjugated by Leclerc-Thibon involution to $\Psi_{\pm}(z)$?
• Do you know any reference for this?
• Is there any good way to find this (using $R$-matrix or semi-infinite wedge)?

I found the answer to the first question. Second and third questions remain.

\begin{align} \Phi_+ (z) \rightarrow K^{-1/2} \Psi_+ (q^{-1} z) \\ \Phi_- (z) \rightarrow K^{1/2} \Psi_- (q^{-1} z) \end{align}

Equivalently

\begin{align} \Psi_+ (z) \rightarrow K^{-1/2} \Phi_+ (q^{-1} z) \\ \Psi_- (z) \rightarrow K^{1/2} \Phi_- (q^{-1} z) \end{align}

I have checked directly that defining relations for $\Phi_{\pm} (z)$ gives defining relations for $\Psi_{\pm}(z)$.

Just for your entertainment. The dining relations for $\Phi$

\begin{align} \Phi_+ (z) x_0^- - q^{-1} x_0^- \Phi_+ (z) = 0 \\ \Phi_+(z) = \Phi_-(z) x_0^- - q x_0^- \Phi_-(z)\\ K \Phi_- (z) = \Phi_+ (z) x_0^+ - x_0^+ \Phi_+ (z) \\ \Phi_- (z) x_0^+ - x_0^+ \Phi_- (z) = 0\\ (qzK)^{-1} \Phi_-(z) = \Phi_+ (z) x_{-1}^+ - x_{-1}^+ \Phi_+(z)\\ \Phi_-(z) x_{-1}^+ - x_{-1}^+ \Phi_-(z)=0\\ \Phi_+(z) x_1^{-} - q x_1^{-} \Phi_+ (z) = 0\\ q^2 z \Phi_+ (z) = \Phi_-(z) x_1^- - q^{-1} x_1^{-} \Phi_-(z). \end{align}

The defining relations for $\Psi$

\begin{align} \Psi_+ (z) x_0^- - x_0^- \Psi_+ (z) = 0 \\ K^{-1} \Psi_+ (z) = \Psi_- (z) x_0^- - x_0^- \Psi_- (z)\\ \Psi_- (z) = \Psi_+ (z) x_0^+ - q x_0^+ \Psi_+ (z) \\ \Psi_- (z) x_0^+ - q^{-1} x_0^+ \Psi_- (z) = 0\\ (qz)^{-1} \Psi_- (z) = q \Psi_+ (z) x_{-1}^+ - x_{-1}^+ \Psi_+ (z)\\ q^{-1} \Psi_- (z) x_{-1}^+ - x_{-1}^+ \Psi_- (z)=0\\ \Psi_+ (z) x_1^{-} - x_1^{-} \Psi_+ (z) = 0\\ q z K \Psi_+ (z) = \Psi_- (z) x_1^- - x_1^{-} \Psi_- (z) \end{align}

Leclerc-Thibon involution for algebra generators $$x_0^+ \rightarrow x_0^+ \quad x_0^- \rightarrow x_0^-$$ $$x_{-1}^+ \rightarrow K^{2} x_{-1}^+ \quad x_{1}^- \rightarrow x_{1}^- K^{-2}$$