$x(\pi)$ are sorted descending from right to left. Order on monomials is then defined as follows: let $$\begin{aligned}
x(\pi) &= x_{\delta_r}(-i_r)x_{\delta_{{r-1}}} (-i_{r-1})\dots x_{\delta_2}(-i_2)x_{\delta_1}(-i_1) \quad \textrm{and} \\
x(\pi') &= x_{\delta_s'}(-i_s)x_{\delta_{s-1}'}(-i_{s-1}') \dots x_{\delta_2'}(-i_2')x_{\delta_1'}(-i_1').\end{aligned}$$ Then $x(\pi) < x(\pi')$ if

-   $s(\pi) < s(\pi')$ (we first compare shapes) or

-   $s(\pi) = s(\pi')$ and there exist $j_0 \in {\mathbb{N}}$ such that $\delta_j = \delta_j'$ for all $j < j_0$ and $\delta_{j_0} < \delta_{j_0}'$ (if monomials are of the same shape, we compare their colors from right to left).

Such order is compatible with multiplication in $S({\tilde{\mathfrak{g}}}_1)$:

\[uredjaj\] Let $x(\delta_1) \leq x (\delta_2)$ and $x(\mu_1) \leq x(\mu_2)$. Then $x(\delta_1) x(\delta_2) \leq x(\mu_1)x(\mu_2)$ and if any of the first two inequalities is strict, then so is the last one.

We divide the proof in three main cases:

-   both inequalities $x(\delta_1) \leq x (\delta_2)$ and $x(\mu_1) \leq x(\mu_2)$ come from shape comparison,

-   one of the inequalities comes from shape and the other from color comparison,

-   both inequalities come from color comparison.

Cases 1) and 2) are straightforward, and the third case is an example of a merge sort, see for instance [@T1].

Vertex operator construction of level one modules
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In the paper we 