Details to the Proof of 13.1

We prove a more general statement concerning elements $D_{a,m\backslash l}$ defined similar to the elements $D_{a,l}$ of section 14. For the set of subsets $K \subseteq \underline{m}\backslash \underline{l-1}$ which have $a$ elements we will write $P({lm},{a})$. We set

$${D_{a,lm}}$$$$\mbox{\index{${D_{a,lm}}$}}$$$$:=\sum_{K\in P({lm},{a})} d_K.$$

The more general statement of 13.1 reads:

Let $a, l\in \underline{m}$. If $\underline{lm}:=\underline{m}\backslash \underline{l-1}=L\cup M$ is a partition of $\underline{lm}$ into disjoint subsets $L$ and $M$ then to each $K \in P({lm},{a})$ there is an integer $s(K,L,l)$ such that

$$D_{a,lm}=\sum_{K \in P({lm},{a})} y^{s(K,L,l)}c_{K\cap L}d_{K\cap M}.$$

We will prove this by induction on $m-l$. If $m=l$ and $a>1$ then $D_{a,mm}=0$ and there is no $K \in P({mm},{a})$. If $a=1$ we have $D_{1, mm}=d_m$ and $K=\{m\}$. Thus $s(K, \emptyset, m)=1, s(K,\{m\}, m)=0$ leads to a solution.

For the induction step we first consider the case $l\in M$ and calculate

$$D_{a,lm}= d_l D_{a-1,(l+1)m} + D_{a, (l+1)m}$$

$$= d_l\sum_{K \in P({(l+1)m},{a-1})}y^{s(K,L,l+1)} c_{K\cap L}d_{K\cap M}+\sum_{K \in P({(l+1)m},{a})}y^{s(K,L,l+1)} c_{K\cap L}d_{K\cap M}$$

$$= \sum_{K \in P({lm},{a}), l \in K}y^{s(K\backslash \{l\},L,l+1)} ... ...\sum_{K \in P({lm},{a}), l \not \in K} y^{s(K,L,l+1)} c_{K\cap L}d_{K\cap M},$$

Setting $s(K,L,l):=s(K\backslash \{l\},L,l+1)$ leads to a solution. If $l \in L$ we apply relation (14) of the exterior algebra to obtain

$$D_{a,lm}= d_l D_{a-1,(l+1)m} + D_{a, (l+1)m}$$

$$= y^{1-l}c_l D_{a-1,(l+1)m} + (y-1)D_{1,(l+1)m}D_{a-1,(l+1)m} + D_{a, (l+1)m}$$

Now, in a similar way as in the proof of Lemma 14.1 we see that $D_{1,(l+1)m}D_{a-1,(l+1)m}= \{a\}_{y}D_{a,(l+1)m}$ where ${\{k\}_{y}}$$:=1+y+y^{2}+\ldots +y^{k-1}\in R$. Since $(y-1)\{a\}_{y} +1=y^a$ we obtain $D_{a,lm}= y^{1-l}c_l D_{a-1,(l+1)m} + y^{a}D_{a, (l+1)m}$. Thus, setting $s(K,L,l):=1-l+s(K\backslash \{l\},L\backslash \{l\},l+1)$ if $l\in K$ and $s(K,L,l):=a+s(K,L\backslash \{l\},l+1)$ elsewise leads to a solution.

It remains to check that $s(K,L,1)=v(K,L)$ if $K\subseteq M$. More generally we prove that

$$s(K,L,l)=(1-l)a + v(K,\underline{m} \backslash M)$$

by induction on $m-l$ again. If $l=m$ we must have $K=\{m\}=M$ or $K=\emptyset$. In both cases both sides of the equation equal zero. For the induction step let us first consider the case $l\in K$. By the above calculation this gives

$$s(K,L,l) = s(K\backslash \{l\},L,l+1) = (1-(l+1))(a-1) + v(K\backslash \{l\}, \underline{m}\backslash M \cup \{l\} )$$

Since $v(K\backslash \{l\}, \underline{m}\backslash M \cup \{l\} ) = v(K, \underline{m}\backslash M ) + a -l$ the assertion follows in the first case. Next we consider $l\in M\backslash K$. Here we have

$$s(K,L,l) = s(K,L,l+1) = (1-(l+1))a + v(K, \underline{m}\backslash M \cup \{l\})$$

and the assertion follows since $v(K, \underline{m}\backslash M \cup \{l\} ) = v(K, \underline{m}\backslash M ) + a$. Finally we have to consider $l \in L$. From the calculation above we get

$$s(K,L,l) =a+s(K,L\backslash \{l\},l+1)= a + (1-(l+1))a + v(K, \underline{m}\backslash M )$$

which directly gives the result.