\[reductionpr\]

1.  Let $\xi_0\in {\mathbb N}$ and $\xi=(\xi_1\geq \xi_2\geq \cdots \geq \xi_{\ell})$, $\ell\geq 2$, a partition such that $\xi_0\geq \xi_1$. Then $$\mathcal{V}_{|\xi|-2k}^{\xi\rightarrow \xi_0}(q)=\sum_{ 0=p_0\leq p_1\leq \cdots \leq p_{\ell-2}\leq p_{\ell-1}=k} q^{p_1+\cdots+p_{\ell-2}}\ \prod_{i=1}^{\ell-1} \mathcal{V}_{|\xi|_i-2p_{\ell-i-1},|\xi|_i-2p_{\ell-i}}^{\xi_{i}\rightarrow \xi_{i-1}}(q).$$

2.  If $m,m'\in {\mathbb N}$ and $s,n\in{\mathbb Z}_+$ are such that $m \geq m'$, then $$\mathcal{V}^{m'\rightarrow m}_{s,n}(q)=\sum_{j=0}^{s}\left( \sum_{\ell=0}^j(-1)^{\ell+1}\hspace{-0,5cm} \sum_{0=p_0<p_1<\cdots<p_{\ell}<p_{\ell+1}=j}\ \prod_{i=0}^{\ell}\ \mathcal{V}^{1\rightarrow m'}_{s-2p_i,s-2p_{i+1}}(q)\right)\  \mathcal{V}^{1\rightarrow m}_{s-2j,n}(q).$$

We prove the first part by an induction on $\ell$. If $\ell=2$, the discussion in Section \[section23\] implies that $V(\xi)\cong D(\xi_1,|\xi|)$. Hence $\mathcal{V}_{|\xi|-2k}^{\xi\rightarrow \xi_0}(q)=\mathcal{V}_{|\xi|,|\xi|-2k}^{\xi_{1}\rightarrow \xi_0}(q),$ and the induction begins. If $\ell>2$, we have $$\begin{aligned}
&& \mathcal{V}_{|\xi|-2k}^{\xi\rightarrow \xi_0}(q)&=\sum_{p=0}^k \mathcal{V}_{|\xi|-2p}^{\xi\rightarrow \xi_1}(q)\ \mathcal{V}_{|\xi|-2p,|\xi|-2k}^{\xi_1\rightarrow \xi_0}(q)&& \text{by \eqref{mat}}\\
&& &=\sum_{p=0}^k q^p\hspace{0.05cm}\mathcal{V}_{|\xi|_2-2p}^{\xi'\rightarrow \xi_1}(q)\ \mathcal{V}_{|\xi|-2p,|\xi|-2k}^{\xi_1\rightarrow \xi_0}(q)&& \text{by Lemma~\ref{ausc}}.\end{aligned}$$ The claim now follows by a trivial application of the induction hypothesis.

Now we prove the second part of the proposition. If $s-n\notin 2{\mathbb Z}_+$ both sides of the equation are zero and the statment is trivial. Otherwise assume that $s=n+2k$ for some $k\in{\mathbb Z}_+$. We can rewrite as a system of linear equations $Uv=b$, where $U=(\mathcal{V}_{i,j})_{0\leq i,j\leq k}$ is an upper triangular unipotent matrix of size $(k+1)\times (k+1)$ and $$\mathcal{V}_{i,j}=\mathcal{V}^{1\rightarrow m'}_{s-2i,s-2j}(q),\ \ b_i=\mathcal{V}^{1\rightarrow m}_{s-2i,n}(q),\ \ v_i=\mathcal{V}^{m'\rightarrow m}_{s-2i,n}(q).$$ Now it is straightforward to check that the inverse of $U$ is again upper 