is analogous to Corollary 4.8 of [@LI], and the proof is almost the same. First, we claim that for all $j,k,l,m\geq 0$, $\Omega_{j,k}\circ_0\Omega_{l,m}$ is a total derivative. In view of the decomposition (\[deca\]) which holds in $\cV_n$ as well as $\cH(n)^{O(n)}$, it suffices to show that $J^{2k}\circ_0 J^{2l}$ is a total derivative for all $k,l$. This is clear because $J^{2k}\circ_0 J^{2l}$ lies in $A_m$ for $m=2k+2l+1$, and $A_m = \partial(A_{m-1})$.

Next, let $\mu = ~:a_1 \cdots a_m:$ be a normally ordered monomial in $(\cV_{n})_{(2m)}$, where each $a_i$ is one of the generators $\Omega_{a,b}$. Let $\tilde{\mu} = :a_{i_1} \cdots a_{i_m}:$, where $(i_1,\dots, i_m)$ is some permutation of $(1,\dots,m)$. We claim that for any decomposition $\mu - \tilde{\mu} = \sum_{k=1}^{m-1} E^{2k}$ of the difference $\mu - \tilde{\mu}\in (\cV_{n})_{(2m-2)}$, the term $E^2$ is a second derivative. To prove this statement, we proceed by induction on $m$. For $m=1$, there is nothing to prove since $\mu - \tilde{\mu} =0$. For $m=2$, and $\mu = ~:\Omega_{a,b}\Omega_{c,d}:$, we have $$\label{reari} \mu - \tilde{\mu} = ~:\Omega_{a,b}\Omega_{c,d}: ~- ~: \Omega_{c,d}\Omega_{a,b}:~ = \sum_{i\geq 0} \frac{(-1)^i}{(i+1)!}\partial^{i+1}(\Omega_{a,b}\circ_{i} \Omega_{c,d}),$$ by (\[vaidii\]). Since $\Omega_{a,b}\circ_{0} \Omega_{c,d}$ is already a total derivative, it follows that $\mu -\tilde{\mu}$ is a second derivative, as claimed. Next, we 