\Omega \end{array}}},{{\renewcommand{\arraystretch}{.7}     \begin{array}[b]{@{}c@{}} {\scriptscriptstyle (s)} \\ \Omega \end{array}}}]_{\mbox{orig}} + [{{\renewcommand{\arraystretch}{.7}     \begin{array}[b]{@{}c@{}} {\scriptscriptstyle (r')} \\ \Omega \end{array}}},
{{\renewcommand{\arraystretch}{.7}     \begin{array}[b]{@{}c@{}} {\scriptscriptstyle (s')} \\ \Omega \end{array}}}]_{{{\cal P}}, \eta}$$ ($r+s = k-1, \; r' + s' = k$), the right hand side is of degree $\leq
-k+2$. Thus $\deg {{\renewcommand{\arraystretch}{.7}     \begin{array}[b]{@{}c@{}} {\scriptscriptstyle (k)} \\ \Omega \end{array}}} \leq -k + 2 - 1 = -k + 1$.

But the element with most negative degree in the algebra is given by the product of all the $\eta$’s, which has degree $-\sum_a(n_a - 1)$. Accordingly, ${{\renewcommand{\arraystretch}{.7}     \begin{array}[b]{@{}c@{}} {\scriptscriptstyle (k)} \\ \Omega \end{array}}}$ is zero whenever $-k+1 >= - \sum_a(n_a - 1)$, which implies $r \leq \sum_a(n_a-1)+1$ as stated in the theorem.

[**Remarks:**]{}

1.  One can improve greatly the bound by observing that the $\eta$’s do not come alone in ${{\renewcommand{\arraystretch}{.7}     \begin{array}[b]{@{}c@{}} {\scriptscriptstyle (k)} \\ \Omega \end{array}}}$. There are also $k$ momenta ${{\cal P}}_a$ which carry positive degree. This remark will, however, not be pursued further here.

2.  One can actually assign degrees smaller than one to the generators $G_a$. For instance, in the case of an Abelian Lie algebra, one may take $deg G_a = 1/2, \; \deg \eta^a = 1/2, \deg {{\cal P}}_a = -
    1/2$. 