\dots \ \in z_0^{\langle \lambda, \nu \rangle} V_P [[z_0]]\end{aligned}$$ show that $$\begin{aligned}
\label{komutiranje}
[Y(e^{\gamma},z_1) , \mathcal{Y}(e^{\mu},z_2)] = 0\end{aligned}$$ if and only if $$\begin{aligned}
\label{uvjet_kom}
 \langle \gamma, \mu \rangle \geq 0 \qquad \textrm{for all} \ \gamma \in \Gamma.\end{aligned}$$

Simple current operator {#op_pr_struje}
=======================

For $\lambda \in P$ we denoted by $e^{\lambda}$ the multiplication operator $1 \otimes e^{\lambda}$ on $V_P= M(1) \otimes {\mathbb{C}}[P]$. Let

$$\begin{aligned}
e(\lambda): V_P \longrightarrow V_P \\
e(\lambda) = e^{\lambda} \epsilon(\cdot, \lambda).\end{aligned}$$

Then $e(\lambda)$ is obviously a linear bijection and using the vertex operator formula we have the commutation relation: $$Y(e^{\alpha},z)e(\lambda)=e(\lambda)z^{\langle \lambda, \alpha \rangle}Y(e^{\alpha},z), \qquad \textrm{for} \ \alpha \in R$$ which, in terms of components, gives: $$\begin{aligned}
\label{omega}
x_{\alpha}(n)e(\lambda)=e(\lambda)x_{\alpha}(n + \langle \lambda, \alpha\rangle), \qquad n \in {\mathbb{Z}}.\end{aligned}$$ If we put $\lambda = \omega$ and $\gamma \in \Gamma$ in , we get $$x_{\gamma}(n)e(\omega)=e(\omega)x_{\gamma}(n+1).$$ i.e. commuting of the $e(\omega)$ raises degree of factor $x_{\gamma}(n)$ by one. We will call $e(\omega)$ a simple current operator ([@DLM]). The restricitions of this operator give bijections between the level 1 modules: $$\begin{aligned}
e(\omega) &: L(\Lambda_0) \longrightarrow L(\Lambda_1), \\
e(\omega) &: L(\Lambda_1) \longrightarrow L(\Lambda_0), \\
e(\omega) &: L(\Lambda_{\ell-1}) \longrightarrow L(\Lambda_{\ell}) \quad \textrm{and} \\
e(\omega) &: L(\Lambda_{\ell}) \longrightarrow L(\Lambda_{\ell-1}).\end{aligned}$$ Namely, we have the following relations: $$\begin{aligned}
e(\omega) 