* 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](http://bookstore.ams.org/cbms-85) by Jimbo and Miwa (chapter 6).
* There is a [paper](https://arxiv.org/abs/q-alg/9504013v1) 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](https://arxiv.org/abs/q-alg/9602025v1), 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)?



Does the Leclerc-Thibon involution exchange vertex operators of the first and second type?
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 
$$x\_0^+ \rightarrow x\_0^+ \quad x\_0^- \rightarrow x\_0^-$$
$$x\_{-1}^+ \rightarrow K^{2} x\_{-1}^+ \quad x\_{1}^- \rightarrow x\_{1}^- K^{-2}$$


