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\begin{document}

\title{Inversion of Incidence Mappings}
\date{}
\author{Helmut Kr\"amer\\[5pt]
Mathematisches Seminar der Universit\"at Hamburg\\
Bundesstra{\ss}e 55\\
D--20146 Hamburg}

\maketitle

\bigskip

\begin{abstract}
Denote by $H(t,q),\;t\leq q$, the incidence matrix (with respect to inclusion)
of the $t$--sets versus the $q$--sets of the $n$--set $\{1,2,\ldots,n\}$. This
matrix is considered as a linear map of $\I Q$--vector spaces
$$ C_q(n)\longrightarrow C_t(n), $$
where $C_s(n)$ is the $\I Q$--vector space having the $s$--sets as a basis
$(s\leq n)$.

As a basic tool, we introduce a connection of the vector spaces to a graded
$\I Q$--algebra (which is at the same time an Artinian local ring). We define
mappings $\triangle$ and $\I X$ of this algebra of degree $-1$ and 1, 
respectively. These two mappings correspond up to a scalar factor to the
linear mappings $H(s-1,s)$ and $H(s-1,s)^T$, respectively.

Then, a relation between the algebra maps $\triangle$ and $\I X$
is established. This relation allows to rewrite a term $\triangle^\beta
\I X^\alpha$ with $\alpha,\beta$ non--negative integers (subject to some
restrictions) as a sum $\sum\limits_{k=0}^\beta
\binom{\beta}{k}\cdot\I X^{\alpha-k}\cdot\triangle^{\beta-k}$ (up to some
scalar factors). As a main result of this relation surjectivity of the map
$\triangle^{q-t}$ (related to $H(t,q)$ up to a scalar factor) is proved under
the assumption $\binom{n}{t}\leq\binom{n}{q}$. Moreover, a right inverse for
the matrix $H(t,q)$ is given explicitely.

This result is exploited to give an inverse of the (square) incidence matrix
$H(t,q)$ in the case $t=n-q$.

These results extend some work done by J.B. Graver and W.B. Jurkat.   
\end{abstract}

\bigskip

{\bf 1.} For $n\in\I N$ we put
$$ \underline{\underline{n}}=\{1,2,\ldots,n\}. $$
For $0\leq  t\leq  q\leq  n$ we denote by $H\,(t,q)$ the incidence matrix 
(with respect to inclusion) of the $t$--sets versus the $q$--sets of the 
$n$--set $\underline{\underline{n}}$ and by $C_q(n)$ the $\I Q$--vector space 
having the basis $\big\{[M]\big\}_{M\subseteq\underline{\underline{n}},
|M|=q}$. Then $H\,(t,q)$ defines a linear mapping (``incidence mapping'') 
$$ C_q(n)\longrightarrow C_t(n),\;[M]\longrightarrow
\sum_{N\subseteq M}^{|N|=t}[N], $$
which is denoted by the same symbol $H\,(t,q)$. --

\medskip

The transpose $H\,(t,q)^T$ of $H\,(t,q)$ defines a linear mapping
$$ C_t(n)\longrightarrow C_q(n),\;[N]\longrightarrow\sum_{N\subseteq M}^{|M|=
q}[M], $$
which is denoted by the same symbol $H\,(t,q)^T$. Finally we define the 
``augmentation mapping''
$$ H\,(-1,0):C_0(n)\longrightarrow 0. $$
Assume that $\dim_{\I Q}  C_t(n)=\binom{n}{t}\leq \binom{n}{q}=
\dim_{\I Q} C_q(n)$. Then it is known that the mapping $H\,(t,q)$ is 
surjective ([3], 2.3, 2.4), therefore in case of equality of the two 
dimensions under consideration an isomorphism. In this case moreover we 
exhibit a method to compute explicitely $H\,(t,q)^{-1}$ by defining a 
structure of a graded commutative algebra on the graded $\I Q$--vector space
$C_\ast(n):=\bigoplus\limits_{q=0}^n C_q(n)$ such that at the same time
$C_\ast(n)$ becomes an Artinian local ring.

\medskip

We provide an example for the computation of $H\,(n-q,q)^{-1}$:

\begin{itemize}
\item[$\bullet$] {\it Assume $n$ odd and $q=\lfloor\frac{n}{2}\rfloor+1$. 
Then one has}
$$ H\,(q-1,q)^{-1} =\sum_{j=1}^q\frac{(-1)^{j+1}}{j}\cdot H\,(q-j,q)^T\circ
H\,(q-j,q-1). $$
\end{itemize}

There is the following generalization:

\begin{itemize}
\item[$\bullet\bullet$] {\it Assume $q\leq \frac{n}{2}$ if $n$ is even or 
$q\leq\lfloor\frac{n}{2}\rfloor+1$ if $n$ is odd. Then the mapping 
$$ K\,(q,q-1):C_{q-1}(n)\longrightarrow C_q(n) $$
defined by
$$ K\,(q,q-1)=\sum_{j=1}^q\frac{(-1)^{j+1}}{j}\binom{n-2q+j+1}{j}^{-1}\cdot H
\,(q-j,q)^T\circ H\,(q-j,q-1) $$
is a right inverse of} $H\,(q-1,q)$.
\end{itemize}

\bigskip

{\bf 2.} One defines the structure of a commutative $\I Q$--algebra on $C_\ast
(n)=\bigoplus\limits_{q=0}^n C_q(n)$ by setting for subsets $M,N$ of 
$\underline{\underline{n}}$
$$ [M]\cdot[N]=\left\{ \begin{array}{cl}
[M\cup N], & \mbox{if } M\cap N=\emptyset,\\
0, & \mbox{otherwise}.
\end{array} \right. $$
This $\I Q$--algebra which we denote by ${\frak C}_\ast(n)$ is isomorphic to
$$ {\cal A}\,(n):=\I Q\,[T_1,\ldots,T_n]/(T_1^2,T_2^2,\ldots,T_n^2)
=\I Q\,[X_1,\ldots,X_n], $$
where $T_1,\ldots,T_n$ are algebraically independent elements and $X_j=T_j$ 
mod $(T_1^2,\ldots,T^2_n)$. The isomorphism ${\cal C}_\ast(n)\longrightarrow
{\cal A}\,(n)$ is induced by
$$ \begin{array}{rcl}
[M]=\big[\{j_1,\ldots,j_q\}\big] & \longrightarrow & X_{j_1}X_{j_2}\cdot\ldots
\cdot X_{j_q},\text{ if }|M|=q\geq 1,\\[2mm]
[\emptyset] & \longrightarrow & 1\in\I Q.
\end{array} $$

In the sequel we identify ${\frak C}_\ast(n)$ and ${\cal A}\,(n)$.

\bigskip

Let ${\frak C}_\ast(n)_p$ denote the $\I Q$--module of the elements of degree
$p$ of the graded algebra ${\frak C}_\ast(n)$. Then one has
$$ {\frak C}_\ast(n)_p=\left\{ \begin{array}{cl}
C_p(n), & 0\leq  p\leq n,\\
0, & p>n.
\end{array} \right. $$

To the mappings $H\,(q-1,q),\,0\leq  q\leq  n$, corresponds the map $\triangle: 
{\frak C}_\ast(n)\longrightarrow{\frak C}_\ast(n)$ of degree $-1$ defined by 
$$ \begin{array}{rcl}
\triangle\Big|_{\I Q}=0,\;\triangle X_j & = & 1,\,j\in
\underline{\underline{n}},\\[4mm]
\triangle\,(X_{j_1}\cdot X_{j_2}\cdot\ldots\cdot X_{j_q}) & = 
& \sum\limits_{k=1}^q X_{j_1}\cdot\ldots\cdot\widehat X_{j_k} \cdot\ldots\cdot 
X_{j_q},\;2\leq q\leq n,
\end{array} $$
$(1\leq  j_1<j_2<\ldots<j_q\leq n,\;\widehat{}\,$ denotes the deleting 
operator). 

\smallskip

Finally we define $\I X=\sum\limits_{j=1}^q X_j$.

\bigskip

\begin{prop} 
Assume $0\leq  t\leq  q\leq  n$. Then with the agreement $\triangle^\circ=
\text{id},\;\I X^\circ=1$ the following identities hold
$$ \begin{array}{rcl}
(q-t)!\cdot H\,(t,q) & = & \triangle^{q-t}\Big|_{C_q(n)},\\[4mm]
(q-t)!\,H\,(t,q)^T(w) & = &\I X^{q-t}\cdot w,\;w\in C_t(n).
\end{array} $$
\end{prop}

\begin{proof} To prove the first statement we show by induction with respect 
to\\ $m,\,0\leq m\leq  q$, that the identity
$$ \triangle^m\Big|_{C_q(n)} =m!\;H\,(q-m,q) \leqno{(1)\ldots} $$
holds. Of course, this is true for $m=0,1$. Assume this identity has been
proved already in case $m-1\geq 1$.

\smallskip

The following relation is well known in case $0\leq  s\leq  t\leq  q$
$$ H\,(s,t)\circ H\,(t,q)=\binom{q-s}{t-s}\;H\,(s,q)\leqno {(1)\ldots} $$
(see for example [2], Chapt 15, {\sc Lemma} 8.1). 

\smallskip
 
Therefore one has
$$ \begin{array}{ll}
& \triangle^m\Big|_{C_q(n)}=\triangle\Big|_{C_{q-m+1}(n)}\circ\triangle^{m-1}
\Big|_{C_q(n)}\\[4mm]
=& H\,(q-m,q-m+1)\cdot(m-1)!H\,(q-m+1,q)\\[2mm]
=& m\cdot(m-1)!\cdot H\,(q-m,q)=m!\,H\,(q-m,q).\quad -
\end{array} $$

To prove the second statement we show by induction with respect to $m\geq 0,$
\\$t+m\leq n$, that for $w\in C_t(n)$ the following relation holds
$$ \I X^m\cdot w=m!\;H\,(t,t+m)^T\,(w). $$
It is sufficient for our purpose to take $w$ as an element $X_{j_1}\cdot\ldots
\cdot X_{j_t}$ of the ``canonical'' basis of $C_t(n)$. The claim is evident in 
case $m=0$; moreover one has
$$ \I X w=\sum_{k\ne j_1,j_2,\ldots,j_t}^k X_{j_1} X_{j_2}\cdot\ldots\cdot 
X_{j_t}\cdot X_k=H\,(t,t+1)^T\,(w). $$
Assume that the statement has already been proved in case $m-1\geq 0$. By
transposing one gets from Eq (1)
$$ H\,(t,t+m)^T=m\cdot H\,(t+1,t+m)^T\circ H\,(t,t+1)^T, $$
Therefore according to the induction hypothesis (applied to $\I X w\in C_{t+1}
(n)$)
$$\begin{array}{ll} 
\I X^m w=\I X^{m-1}\cdot\I X w & =(m-1)!\;H\,(t+1,t+m)^T\circ H\,(t,t+1)^T 
(w)\\[2mm]
& =m!\;H\,(t,t+m)^T\,(w). 
\end{array} $$
\end{proof}

\bigskip

Suppose $w\in{\frak C}_\ast(n)$. Then we define the ``foundation'' of $w$ (in
signs Fund $(w)$) to be the product of all $X_j$ which appear in the basis
decomposition of $w$ with a coefficient unequal to zero.

\smallskip

For example one has with pairwise distinct $X_{j_1} ,X_{j_2},\ldots,
X_{j_{2t}}$
$$ \text{Fund }\Big((X_{j_1}-X_{j_2})(X_{j_3}-X_{j_4})\cdot\ldots\cdot(
X_{j_{2t-1}}-X_{j_{2t}})\Big) =\prod^{2t}_{k=1} X_{j_k}. $$
In a self--explaining manner we can treat the foundation also as a subset of
$\underline{\underline{n}}$.

\bigskip

\begin{prop} 

\begin{enumerate}
\item[i)] Assume $v,w\in{\frak C}_\ast(n)$ and Fund $(v)\,\cap$ Fund $(w)=
\emptyset$.\\ Then we have that
$$ \triangle\,(vw)=v\triangle\,(w)+w\triangle\,(v). $$
\item[ii)] Denote $\I Z=\sum\limits_{k=1}^pX_{j_k}$, the $X_{j_k}$ pairwise 
distinct. Assume $1\leq m\leq p$ (and put $\I Z^0=1$). Then we have that 
$$ \triangle\,(\I Z^m)=m\,(p-m+1)\I Z^{m-1}.$$
\end{enumerate}
\end{prop}

\bigskip

\begin{proof}

\smallskip

Ad i) Assume in the first instance $v=X_{i_1}\cdot\ldots\cdot X_{i_s},\;w=
X_{j_1}\cdot\ldots\cdot X_{j_t}$ are elements from a basis of
${\frak C}_\ast(n)$. According to hypothesis one has
$$ \bigl|\{i_1,\ldots,i_s,j_1,\ldots,j_t\}\bigr|=s+t. $$
Therefore
$$ \begin{array}{ll}
\triangle\,(vw)&=\sum\limits_{k=1}^s X_{i_1}\cdot\ldots\cdot\widehat X_{i_k}
\cdot\ldots\cdot X_{i_s}\cdot X_{j_1}\cdot\ldots\cdot X_{j_t}\\[5mm]
& +\sum\limits_{l=1}^t X_{i_1}\cdot\ldots\cdot X_{i_s}\cdot X_{j_1}\cdot\ldots
\cdot\widehat X_{j_l}\cdot\ldots\cdot X_{j_t}\\[5mm]
& =w\triangle\,(v)+v\triangle\,(w). 
\end{array} $$
The general case follows now from the law of distributivity.

\medskip

Ad ii) Without loss of generality assume $\I Z=\sum\limits_{j=1}^p X_j$. Then 
we have
$$ \I Z^k=\left\{ \begin{array}{cl}
k!\sum\limits_{1\leq j_1<\ldots<j_k\leq  p} X_{j_1}\cdot\ldots\cdot
X_{j_k}, & 1\leq k\leq  p,\\[4mm]
0, & k>p.
\end{array} \right.\leqno {(2)\ldots} $$

In this sum a term $X_{i_1}\cdot\ldots\cdot X_{i_{m-1}},\;(1\leq i_1<\ldots<
i_{m-1}\leq p)$ occurs exactly in the terms $\triangle\,(X_{i_1}
\cdot\ldots\cdot X_{i_{m-1}}\cdot X_t),\; t\in\underline{\underline{p}}
\backslash\{i_1,\ldots,i_{m-1}\}$ with factor 1; therefore it occurs in 
$\triangle\,(\I Z^m)$ with factor $m!(p-m+1)=m\,(p-m+1)(m-1)$!. The conclusion
follows now from equation (2). 
\end{proof}

\bigskip

We remark that $\triangle$ is no derivation of ${\frak C}_\ast(n)$.

\medskip

Let ${\frak m}:=(X_1,\ldots,X_n)$ denote the maximal ideal of ${\frak C}_\ast
(n)$. Then it holds that\\ ${\frak m}^{n+1}=0$. 
If ${\frak p}$ is a prime ideal of ${\frak C}_\ast(n)$, then from 
${\frak m}^{n+1}\subseteq{\frak p}$ one concludes ${\frak m}={\frak p}$.
Therefore ${\frak C}_\ast(n)$ is an Artinian local ring. Let $\widetilde
\triangle$ denote a derivation of ${\frak C}_\ast(n)$ (into itself). Then it 
must hold for all $j\in\underline{\underline{n}}$ that
$$ 0=\widetilde\triangle\,(X_j^2)=2X_j\widetilde\triangle\,(X_j), $$
therefore $\widetilde\triangle\,(X_j)\in{\frak m}$. -- So $\widetilde
\triangle$ maps ${\frak m}$ (and ${\frak C}_\ast(n)$, too) into ${\frak m}$ 
(compare with [4], \S 1, Exercise 4).

\bigskip

Now we extend Prop.2, ii):

\medskip

\begin{prop} Assume $0\leq s\leq n-1,\;\alpha\in\I N,\;\alpha+s\leq n$ and 
$w\in C_s(n)$. Then the following identity holds: 
$$ \triangle\,(\I X^\alpha\cdot w)=\I X^\alpha\cdot\triangle\,(w)+\alpha\,
(n-\alpha-2s+1)\I X^{\alpha-1}\cdot w. $$
\end{prop}

\begin{proof}
In case $s=0$ the statement is true according to Prop. 2, ii). Assume now
$s\geq 1$ and take
$$ w=X_{j_1}\cdot\ldots\cdot X_{j_s} $$
as a basis element of $C_s(n)$. One defines
$$ \I Y=\sum_{k=1}^s X_{j_k},\;\I Z=\I X-\I Y. $$
Since $w$ is a word in {\it all} $X_{j_1},\ldots,X_{j_s}$ one has $\I Y\cdot 
w=0$. Assume $t\geq 0$. Then the following holds
$$ \I X^tw=(\I Y+\I Z)^t\cdot w=\sum_{r=0}^t\binom{t}{r}\I Z^r \I Y^{t-r}\cdot
w=\I Z^tw, $$
furthermore according to the definition of $\I Y$ and $\I Z\;$ Fund $(w)\,
\cap$ Fund $(\I Z^t)=\emptyset$; from Prop. 2 one concludes
$$ \left\{ \begin{array}{ll}
\triangle\,(\I X^\alpha\cdot w) &=\triangle\,(\I Z^\alpha\cdot w)=\I Z^\alpha
\cdot\triangle\,(w)+w\cdot\triangle\,(\I Z^\alpha)\\[1mm]
& = \I Z^\alpha\cdot\triangle\,(w)+\alpha\,(n-s-\alpha+1)\,\I Z^{\alpha-1}w\\[1mm]
& = \I Z^\alpha\cdot\triangle\,(w)+\alpha\,(n-s-\alpha+1)\,\I X^{\alpha-1}w.
\end{array} \right. \leqno {(3)\ldots} $$
Now we use
$$ \I Y^t\cdot\triangle\,(w)=\left\{ \begin{array}{cl}
0, & t>1,\\
s\cdot w, &t=1. 
\end{array} \right. $$
Therefore
$$ \begin{array}{ll}
\I X^\alpha\triangle\,(w) & =\left(\sum\limits_{r=0}^\alpha\binom{\alpha}{r}
\I Y^r\I Z^{\alpha-r}\right)\cdot\triangle\,(w)=(\alpha\I Y\cdot
\I Z^{\alpha-1}+\I Z^\alpha)\cdot\triangle\,(w)\\[2mm]
& =\alpha s\I Z^{\alpha-1}\,w+\I Z^\alpha\cdot\triangle\,(w)=\alpha\cdot s\cdot
\I X^{\alpha-1}w+\I Z^\alpha\cdot\triangle\,(w).
\end{array} $$

Replacing the term $\I Z^\alpha\triangle\,(w)$ in Eq. (3) yields
$$ \triangle\,(\I X^\alpha w)=\I X^\alpha\cdot\triangle\,(w)+\alpha\,(n-2s-\alpha+1)
\,\I X^{\alpha-1}\cdot w, $$
as claimed. --
\end{proof}
 
\bigskip

Prop. 3 is a special case of

\medskip

\begin{prop} 
The same assumptions about $s,\alpha$ are in force as in Prop. 3. 
Assume $r\in\I R$ and $k\in\I N$ and define
$$ [r]_k=r\,(r-1)\cdot(r-2)\cdot\ldots\cdot (r-k+1),\,[r]_0=1. $$
Let $\beta$ be a non--negative integer and assume $0\leq\beta\leq\alpha$. Then 
the following identity holds
$$ \triangle^\beta(\I X^\alpha\cdot w)=\sum_{k=0}^\beta\binom{\beta}{k}
[\alpha]_k[n-\alpha-2s+\beta]_k\cdot\I X^{\alpha-k}\cdot
\triangle^{\beta-k}(w). $$
\end{prop}

\begin{proof}
There is nothing to prove in case $\beta=0$; in case $\beta=1$ the claim boils
down to Prop. 3. If $0\leq  j\leq \beta$ we put
$$ [\beta,j]=\I X^{\alpha-j}\cdot\triangle^{\beta-j}(w) $$
and prove first of all by induction with respect to $\beta\geq 1$, that the
identity
$$ \triangle^\beta(\I X^\alpha w)=\sum_{j=0}^\beta c\,(\beta,j)\cdot[\beta,j],\quad
c\,(\beta,j)\in\I Q, \leqno {(4)\ldots} $$
holds; furthermore this will yield recursion formulas for the coefficients
$c\,(\beta,j)$. In case $\beta=1$ one has according to Prop. 3
$$ c\,(1,0)=1,\;c\,(1,1)=\alpha\,(n-\alpha-2s+1)=:\lambda\,(\alpha,s). $$
Since $\triangle^{k-j} (w)\in C_{s-k+j}(n)$, Prop. 3 now yields
$$ \begin{array}{c}
\triangle\,\bigl([k,j]\bigr)=\triangle\bigl(\I X^{\alpha-j}\cdot
\triangle^{k-j}(w)\bigr)=\\[2mm]
=\I X^{\alpha-j}\cdot\triangle^{k-j+1}(w)+(\alpha-j)\cdot\bigl(n-(\alpha-j)-
2(s-k+j)+1\bigr)\cdot\I X^{\alpha-j-1}\cdot\triangle^{k-j}(w)=\\[2mm]
=\I X^{\alpha-j}\cdot\triangle^{k-j+1}(w)+\lambda\,(\alpha-j,s-k+j)\cdot
\I X^{\alpha-j-1}\cdot\triangle^{k-j}(w),
\end{array}$$
that is
$$ \triangle\,\bigl([k,j]\bigr)=[k+1,j]+\lambda(\alpha-j,s-k+j)[k+1,j+1]. $$
According to the induction hypothesis we obtain
$$ \begin{array}{rl}
\triangle^{\beta+1}(\I X^\alpha w) & =\triangle\,\bigl(\triangle^\beta(\I X^\alpha
\cdot w)\bigr)=\\[3mm]
c\,(\beta,0)[\beta+1,0] & +\sum\limits_{j=1}^\beta \bigl\{c\,(\beta,j)+c\,
(\beta,j-1)\cdot\lambda\,(\alpha-j+1,s-\beta+j-1)\bigr\}\cdot[\beta+1,j]\\[5mm]
& +\lambda\,(\alpha-\beta,s)\cdot c\,(\beta,\beta)[\beta+1,\beta+1],
\end{array} $$
which proves Eq. (4); at the same time we have proved the following recursion
formulas
$$ c\,(\beta+1,0)=c\,(\beta,0),\;c\,(\beta+1,\beta+1)=\lambda
\,(\alpha-\beta,s)\,c\,(\beta,\beta),\leqno{(5.1)\;\ldots} $$

$$ c\,(\beta+1,j)=c\,(\beta,j)+\lambda\,(\alpha-j+1,s-\beta+
j-1)\cdot c\,(\beta,j-1),(1\leq j\leq \beta). \leqno{(5.2)\;\ldots} $$

\bigskip

Eq. (5.1) yields immediately
$$ \left\{ \begin{array}{lll}
c\,(\beta,0)=1 & &\\
& & (0\leq \beta\leq \alpha).\\
c\,(\beta,\beta)=[\alpha]_\beta[n-\alpha-2s+\beta]_\beta & &
\end{array} \right. \leqno {(6)\ldots} $$

Now we claim that the following holds in case $\beta\geq j$
$$ c\,(\beta,j)=\binom{\beta}{j}[\alpha]_j[n-\alpha-2s+\beta]_j\leqno{
(7)\ldots} $$

which we prove by induction with respect to the pairs $(\beta,j),\;\beta\geq 
j$. The claim is true for pairs $(\beta,0)$ according to Eq. (6).
Assume that it has already been proved that in case $j\geq 1$ the claim is
true for all pairs $(\gamma,j-1),\;\gamma\geq j-1$. We rewrite a term in the 
recursion formula (5.2)
$$ \lambda\,(\alpha-j+1,s-\beta+j-1)=(\alpha-j+1)(n-\alpha-2s+2\beta-j+2). $$

Now we proceed by induction with respect to $\beta,\beta\geq j$. In case 
$\beta=j$ the statement is true according to Eq. (6). Assume that the
statement has already be proved for some $\beta\geq j$. Then we have according
to Eq. (5.2)
$$ \begin{array}{ll}
c\,(\beta+1,j)= &\binom{\beta}{j}[\alpha]_j[n-\alpha-2s+j]_j\;+\\[2mm]
& +(n-\alpha-2s+2\beta-j+2)\cdot\binom{\beta}{j-1}[\alpha]_j
[n-\alpha-2s+\beta]_{j-1}. 
\end{array} $$

Using the identities $\binom{\beta}{j}=\binom{\beta}{j-1}\cdot
\displaystyle{\frac{\beta-j+1}{j}}$ and $\binom{\beta}{j}+\binom{\beta}{j-1}
=\binom{\beta+1}{j}$ we conclude
$$ \begin{array}{rcl}
c\,(\beta+1,j) & = & [\alpha]_j[n-\alpha+2s+\beta]_{j-1}\cdot\left\{
\binom{\beta+1}{j}(n-\alpha-2s)\right.+\binom{\beta}{j}(\beta-j+1)+\\[4mm]
& & \left. +\binom{\beta}{j-1}(2\beta-j+2)\right\}\\[4mm]
& = & [\alpha]_j[n-\alpha-2s+\beta]_{j-1}\cdot\binom{\beta+1}{j}
(n-\alpha-2s +\beta+1)\\[4mm]
& = &\binom{\beta+1}{j}[\alpha]_j[n-\alpha-2s+(\beta+1)]_j
\end{array} $$

This proves Eq. (7). --
\end{proof}

\bigskip

{\bf 3.} Assume now $0\leq t\leq q\leq n$ and $\binom{n}{t}\leq\binom{n}{q}$. 
We denote $\nabla:=\triangle^{q-t}:$\\ 
$C_q(n)\longrightarrow C_t(n)$ and fix some 
$w\in C_t(n)$. Our aim is to construct explicitely a primage $u\in C_q(n)$
of $w$ with respect to $\nabla$. Therefore we make the following ansatz (we
will see later that this method will work): If $0\leq j\leq t$ we denote
$$ U_j:=\I X^{q-t+j}\cdot\triangle^j(w) $$
and put
$$ u=\sum_{j=0}^t x_j U_j $$
where the $x_j\in\I Q$ have to be determined. To compute $\nabla U_j$ apply
Prop. 4 (replace here $w$ by $\triangle^j(w)\in C_{t-j}(n)$) and obtain 
with the convention
$$ c\,(k,j) :=\binom{q-t}{k}[q-t+j]_k\cdot[n-2t+j]_k, \quad
(0\leq j\leq t,\;0\leq k\leq q-t) \leqno {(8)\ldots} $$
$$\nabla U_j=\sum_{k=0}^{q-t}c\,(k,j)\cdot\I X^{q-t+j-k}\cdot
\triangle^{q-t+j-k}(w)\in C_t(n), $$
therefore
$$ \nabla u=\sum_{k=0}^{q-t}\sum_{j=0}^t x_j c\,(k,j)\cdot\I X^{q-t+j-k}\cdot
\triangle^{q-t+j-k}(w).\leqno{(9)\ldots} $$
We order the right hand side of this equation with respect to the terms  
$$ V_m:=\I X^m\cdot\triangle^m(w),\;0\leq m\leq t, $$
by defining
$$ V_{k,j}=\I X^{q-t+j-k}\cdot\triangle^{q-t+j-k}(w).$$

Then it holds that $V_{k,j}=0$ if $q-t+j-k>t$ and
$$ V_{k,j}=V_m $$
exactly if $k=(q-t)-l,\,j=m-l,\,0\leq l\leq \min\{m,q-t\}$.
Therefore Eq. (9) yields now
$$ \nabla u=\sum_{m=0}^t d_m V_m $$
with
$$ d_m=\sum_{l=0}^{\min\{m,q-t\}} c\,\bigl((q-t)-l,\,m-l\bigr)\cdot x_{m-l}.
\leqno{(10)\ldots}  $$

\bigskip

We observe that $V_0=w$. -- So $u$ is in fact a preimage, if the system of 
linear equations in the unknowns $x_0,\ldots,x_t$ 
$$ \begin{array}{c}
d_0=1,\\[2mm]
d_m=0,1\leq m\leq t,
\end{array} $$
is solvable. This will be the case if all $c\,(q-t,m)$ don't vanish.
In fact we have in the trivial case $q-t=0$
$$ c\,(0,m)=1,\; 0\leq m\leq t, $$
(and the system has the solution $x_0=1,\; x_1=x_2=\ldots=x_t=0$). In
case $q-t>0$ it is sufficient (see Eq. (8)) to show that the
$$ [n-2t+m]_{q-t} $$
don't vanish. Now the smallest factor $f_m$ in the above mentioned falling
factorial is
$$ f_m=n-2t+m-(q-t)+1=n-(q+t)+m+1. $$
Assume first $\binom{n}{t}=\binom{n}{q}$. Then it holds that $t+q=n$, 
therefore $f_m=m+1$. In the second case the condition $\binom{n}{k}<
\binom{n}{q}$ is equivalent to $q+t+1\leq n$, therefore $f_m\geq m+2$.

\smallskip

So our ansatz has worked and we have proved 

\medskip

\begin{theor} Assume $0\leq t\leq q\leq n$ and $\binom{n}{t}\leq\binom{n}{q}$. 
Then the mapping
$$ \nabla:=\triangle^{q-t}:C_q(n)\longrightarrow C_t(n) $$
is surjektiv. There exist $x_0,x_1,\ldots,x_t\in\I Q$ such that the mapping
$$ \nabla^{[-1]}:=\sum_{j=0}^t x_j\cdot\I X^{q-t+j}\cdot\triangle^j
\Big|_{C_t(n)} $$
is a right inverse with respect to $\nabla$ (in case $\binom{n}{t}=
\binom{n}{q}\quad\nabla^{[-1]}$ is the inverse of $\nabla$).

\medskip

If one denotes in case $0\leq k\leq q-t,\,0\leq j\leq t$,
$$ c\,(k,j):=\binom{q-t}{k}\cdot[q-t+j]_k\cdot[n-2t+j]_k, $$
then one can choose the $x_0,\ldots,x_t$ as the (existing) solution of the 
system of linear equations
$$ \begin{array}{c}
c\,(q-t,0)\cdot x_0=1,\\
\sum\limits_{l=0}^{\min\{m,q-t\}}c\,\bigl((q-t)-l,m-l\bigr)\cdot x_{m-l}=0,\;
1\leq m\leq t. 
\end{array} $$
\end{theor}

\bigskip

\begin{cor} 
If one defines
$$ y_j=(q-t)!j!(q-t+j)!\cdot x_j,\;0\leq j\leq t, $$
then the mapping $K\,(q,t): C_t(n)\longrightarrow C_q(n)$ defined by
$$ K\,(q,t):=\sum_{j=0}^t y_j\cdot H\,(t-j,q)^T\circ H\,(t-j,t) $$
is a right inverse of $H\,(t,q)$.
\end{cor}

\bigskip

{\it Proof} (of the corollary). According to Prop. 1 one has
$$ \nabla=(q-t)!\cdot H\,(t,q) $$
and
$$ \I X^{q-t+j}\cdot\triangle^j\Big|_{C_t(n)}=(q-t+j)!j!\cdot H\,(t-j,q)^T
\circ H\,(t-j,t). $$\\[-1.2cm]
\hspace*{\fill}$\square$

\bigskip

Now we exploit the the theorem and the corollary. The following result is
properly spoken another corollary; however, we state it as

\bigskip

\begin{theor}
Assume $0\leq t< q\leq n$ and $t+q=n$. Then the following identity holds:
$$ H\,(t,q)^{-1}=\sum_{j=0}^t(-1)^j\frac{q-t}{q-t+j}\,H\,(t-j,q)^T\circ H\,
(t-j,t). $$
\end{theor}

{\it Proof}. We introduce a new parameter $p=q-t$. Then we obtain $n-2t=p$
and 
$$ c\,(k,j)=\binom{p}{k}\cdot\bigl([p+j]_k\bigr)^2. $$

We switch now to new indeterminates $\alpha_j$ in the system of linear 
equations of the Theorem by defining
$$ x_j=(-1)^j\frac{\alpha_j}{((p+j)!)^2},\;\alpha_j\in\I Q. $$
An elementary computation which we omit yields
$$ \alpha_0=1 $$
and the following recursion formula
$$ \alpha_j=\sum_{l=1}^{\min\{p,j\}} (-1)^{l+1}\binom{p}{l}\alpha_{j-l},\;
1\leq j\leq t. $$
We compute the $\alpha_j$ by elementary difference calculus; let us define
therefore functions
$$ f_p:\I N_0\longrightarrow \I Q $$
by the following conditions
$$ \left\{ \begin{array}{c}
f_p(0)=1,\\[4mm]
f_p(j)=\sum\limits_{l=1}^{\min\{p,j\}}(-1)^{l+1}\displaystyle{\binom{p}{l}}
f_p(j-l),\;1\leq  j.
\end{array} \right. \leqno{(11)\ldots} $$

Then we claim
$$ f_p(j)=\binom{j+p-1}{p-1}.\leqno {(12)\ldots} $$

\bigskip

In order to prove Eq. (12) we need the following

\medskip

\begin{lem} 
Assume $p\geq 1$ and $0\leq l\leq p$. Then the following identity holds 
$$ \binom{p}{l}-\binom{p}{l-1}+\binom{p}{l-2}\mp\ldots+(-1)^l\binom{p}{0}=
\binom{p-1}{l}. $$
\end{lem}

\bigskip

{\it Proof} (of the lemma): Denote the left hand side of this identity by 
$S_l(p)$. Then we have
$$ S_l(p)=\binom{p}{l}-S_{l-1}(p). $$

Now we proceed by induction with respect to $l.\hspace{\fill}\square$ 

\bigskip

Denote by $\triangle f_p$ the first difference series of $f_p,$ that is
$$ (\triangle f_p)(j)=f_p(j+1)-f_p(j),\,j\in\I N_0. $$

Then we have by placing $f_p(k)=\alpha_k$ and $h=\min\{p,j\}$, in case 
$j\geq  1$
$$ \begin{array}{ll}
\alpha_{j+1}-\alpha_j & =\left[\binom{p}{1}-\binom{p}{0}\right]\cdot(\alpha_j 
-\alpha_{j-1})-\left[\binom{p}{2}-\binom{p}{1}+\binom{p}{0}\right](
\alpha_{j-1}-\alpha_{j-2})\pm\ldots\\[4mm]
& +(-1)^{h+1}\left[\binom{p}{h}-\binom{p}{h-1}+\binom{p}{h-2}\mp\ldots+
(-1)^h\binom{p}{0}\right]\cdot(\alpha_{j-h+1}-\alpha_{j-h}).
\end{array} $$

According to the lemma this rewrites to
$$ (\triangle f_p)(j)=\sum_{l=1}^{\min\{j,p\}} (-1)^{l+1}\binom{p-1}{l}
(\triangle f_p)(j-l),\; j\geq 1. $$

Furthermore we have
$$ (\triangle f_p)(0)=p-1=f_{p-1}(1),\; p\geq 2.$$

This yields in case $p\geq 2$
$$ (\triangle f_p)(j)=f_{p-1}(j+1),\; j\in\I N_0,\leqno{(13)\ldots} $$
since $\triangle f_p$ satisfies the same recursion formula (see Eq. (12)) as 
$f_{p-1}$ does.

\medskip

Now Eq. (12) is certainly true if $p=1$. Assume our claim is true if
$p-1\geq 1$. Then according to Eq. (13) we have
$$ (\triangle f_p)(j)=\binom{j+p-1}{p-2},\;j\in\I N_0. $$

Therefore the function $F:\I N_0\to\I Q$ defined by
$$ F\,(j)=\binom{j+p-1}{p-1} $$
is a discrete indefinite integral of $\triangle f_p$; that is $\triangle F=
\triangle f_p$. It follows that
$$ F=f_p+\mbox{ const}, $$
but $F\,(0)=f_p(0)$, so we have $F=f_p$. Therefore the claim Eq. (12) is 
proved.

\medskip

Now we use the notations introduced in the corollary of Theorem 1 and obtain
$$ y_j=(-1)^j\;\frac{p!\,j!\,(p+j)!}{((p+j)!)^2}\;\binom{j+p-1}{p-1}=(-1)^j
\frac{p}{p+j}=(-1)^j\frac{q-t}{q-t+j}. $$

The statement of Theorem 2 now follows from the corollary to Theorem 1. 
\hspace{\fill}$\square$

\bigskip

To exploit Theorem 1 and the corollary in the case $\binom{n}{t}\leq
\binom{n}{q}$, we restrict ourselves to the condition $q-t=1$. We then have
$q\leq \frac{n}{2}$ if $n$ is even and $q\leq \lfloor\frac{n}{2}\rfloor+1$ if
$n$ is odd.

\smallskip

Here we have under the assumption $0\leq  j\leq t = q-1$
$$\begin{array}{lcl}
c\,(0,j) &=& 1,\\[1mm]
c\,(1,j) &=& (j+1)(n-2q+j+2).
\end{array} $$

We obtain the following system of linear equations in the unknowns $x_0,
\ldots,x_t$ 
$$ \begin{array}{rcl}
c\,(1,0)\,x_0 &=& 1,\\[1mm]
c\,(1,j)\,x_j+x_{j-1} &=& 0,\;1\leq j\leq t,
\end{array} $$

which has the solution

$$ x_j=\frac{(-1)^j}{(j+1)![n-2q+j+2]_{j+1}},\; 0\leq j\leq t. $$

\bigskip

Again with the notations of the corollary to Theorem 1 we obtain

$$ y_j=j!\,(j+1)!\,x_j=\frac{(-1)^j}{j+1}\cdot\binom{n-2q+j+2}{j+1}^{-1}. $$

Now we change the indices from $j+1$ to $j$ and obtain the statement of 
example\quad $\bullet\bullet$~.

\bigskip

Finally let us evaluate Theorem 2 in case $n=5,\,q=3,\,t=2$. With the
notation\\ $\{j_1,j_2,j_3,j_4,j_5\}=\underline{\underline 5}$ we obtain

$$ H\,(2,3)^{-1}(X_{j_1}\cdot X_{j_2})=$$
$$ \begin{array}{ll}
\displaystyle{\frac{1}{6}}\Big[2X_{j_1}X_{j_2}X_{j_3}&-X_{j_2}X_{j_3}X_{j_4}
+2X_{j_3}X_{j_4}X_{j_5}-X_{j_1}X_{j_4}X_{j_5}+2X_{j_1}X_{j_2}X_{j_4}-X_{j_1}X_{j_3}
X_{j_5}+\\[2mm]
& +2X_{j_1}X_{j_2}X_{j_5}-(X_{j_1}X_{j_3}X_{j_4}
+X_{j_2}X_{j_3}X_{j_5}+X_{j_2}X_{j_4}X_{j_5})\Big],
\end{array} $$

\medskip

and conclude that the matrix $H\,(2,3)^{-1}$ is ``non--sparse''.

\vspace*{1cm}

{\bf 4.} Finally we demonstrate the usefulness (as we hope) of the algebra 
${\frak C}_\ast(n)$ by giving new proofs or extending, respectively, some
results in [2].

\medskip

\begin{itemize}
\item[$\bullet\bullet\bullet$] (l.c., 3.3) {\it Let} $v\in\mbox{Ker }\triangle,
\,w\in C_t(n)$, {\it such that} Fund $(v)\,\cap$ Fund $(w)=\emptyset$. {\it 
Then it holds that} $\triangle^{t+1}(v\cdot w)=0$.
\end{itemize}

\medskip

This is an immidediate consequence of the property of $\triangle$ to be a
``quasi--derivation'' (see Prop. 2).

\vspace*{8mm}

Secondly, we exhibit a system of generators of $\text{Ker } H\,(t,q)$ provided
$\binom{n}{t}<\binom{n}{q}$.

\medskip

We denote $K_q(n)=\mbox{Ker }\triangle\Big|_{C_q(n)}$. Then obviosly one has 
$K_q(n-1)\subset K_q(n)$. To determine remaining candidates $u$ of $K_q(n)$ 
we make the following ansatz
$$ u=vX_n+w,\,v\in C_{q-1}(n-1),\, w\in C_q(n-1). \leqno{(14)\ldots}. $$

\bigskip

Since Fund $(v)\,\cap$ Fund $(X_n)=\emptyset$, Prop. 2, i) yields 
$$ 0=\triangle u=(\triangle v)\cdot X_n+(v+\triangle w), $$
therefore
$$ \triangle v=0,\; v+\triangle w=0. \leqno{(15)\ldots} $$

Assume now that is has been proved by induction with respect to $s$ that 
$K_t(s)$,\\$t\leq \lfloor\frac{s}{2}\rfloor,s\leq n-1$, is generated by 
elements of the shape
$$ (X_{j_1}-X_{j_2})(X_{j_3}-X_{j_4})\cdot\ldots\cdot(X_{j_{2t-1}}-
X_{j_{2t}})$$
with pairwise distinct $X_{j_k}$. Therefore we may assume (in Eq. (14), (15))
that
$$ v=(X_{j_1}-X_{j_2})\cdot\ldots\cdot\Big(X_{j_{2(q-1)-1}}-X_{j_{2(q-1)}}
\Big). $$
There are two cases to be considered. In the first case assume $2\,(q-1)=n-1$; 
then we have $n$ odd, $q=\lfloor\frac{n}{2}\rfloor+1$. According to $\bullet
\quad K_q(n)=0$ and then there is no more to prove.

\medskip

In the second case assume $2\,(q-1)<n-1$. Then there exists $X_j,j\leq  n-1$ 
such that Fund $(v)\cap\{X_j\}=\emptyset$. We conclude that
$$ w:=-v\cdot X_j  $$
solves the second equation in Eq. (15); furthermore one has $u=v\,(X_n-X_j)$.

\medskip

This proves

\medskip
$\bullet\bullet\bullet\,\bullet$\, (contained in l.c., 4.2)\, {\it Assume} 
$q\leq\lfloor\frac{n}{2}\rfloor$. {\it Then it holds that} Ker
$H\,(q-1,q)\ne 0$\\
\hspace*{12mm} {\it and this kernel is generated by elements of the shape}
$$ (X_{j_1}-X_{j_2})(X_{j_3}-X_{j_4})\cdot\ldots\cdot(X_{j_{2q-1}}-
X_{j_{2q}}). $$

\bigskip

In as subsequent paper (in which we use the algebra ${\frak C}_\ast(n)$
to exhibit explicitely eigenspace decompositions of the matrix $H\,(t,q)^T
\circ H\,(t,q)$) we will show

\bigskip

\begin{theor} 
Assume $\binom{n}{t}<\binom{n}{q}$. Then it holds that
$$ {\rm Ker }\,H\,(t,q)=\bigoplus_{s=t+1}^{\min\{q,n-q\}} H\,(s,t)^T\,\Big(
{\rm Ker }\,H\,(s-1,s)\Big). $$
\end{theor}

This result provides together with $\,\bullet\bullet\bullet\,\bullet\,$ 
systems of generators of $\text{Ker } H\,(t,q)$ which are in general distinct
from those which are exhibited in [2].

\bigskip

With the aid of Theorem 3 (and Prop. 4) one easily proves

\medskip

$\bullet\bullet\bullet\bullet\bullet$\, (l.c., 4.3)\, {\it Assume $0\leq t<q<n
-t$. Then $H\,(t+1,q)$ maps the  kernel of\linebreak[4] 
\hspace*{16mm} $H\,(t,q)$ onto the kernel of $H\,(t,t+1)$.}

\bigskip

Finally we exhibit in a special case a system of generators of
$\mbox{Ker } H\,(q-1,q)$ which is different from the ''canonical'' one 
constructed above as follows:

\medskip

Assume $n=7,\,q=3$. Any enumeration $\sigma$ of the points of the projective
plane $\I P$ consisting of 7 points and 7 lines yields the family 
${\frak G}_\sigma\subset\binom{\underline{\underline 7}}{3}$ consisting of the
lines of $\I P$. We claim that the elements
$$ u_\sigma:=4\cdot\sum_{M\in{\frak G}_\sigma}[M]-
\sum_{M\not\in{\frak G}_\sigma}^{|M|=3}[M] $$
also generate $\text{Ker } H\,(2,3)$.

\medskip

First, we sketch a proof that the $u_\sigma$ indeed are contained in the 
kernel of $H\,(2,3)$ as follows: Define for $q\in\underline{\underline n}$
$$ H_q:=H\,(q-1,q)^T\circ H\,(q-1,q):C_q\,(n)\longrightarrow C_q\,(n).$$
Then it holds, if $[M]$ is an element of the canonical basis of $C_q(n)$,
$$ H_q\bigl([M]\bigr)=q\cdot[M]+\sum_{|M\cap M'|=q-1}^{|M'|=q}[M']. \leqno{
(16)\ldots} $$
Now it can be seen easily that
$$ w_q:=\sum_{|M|=q}^M [M] $$
is an eigenvector of $H_q$ with eigenvalue $q\,(n-q+1)$. In addition we need
the well known result (see for instance [1], Chapt. II, 2.5 Lemma), that
$$ \text{rang }H\,(t,q)=\text{rang }H\,(t,q)^T\circ H\,(t,q)\;\Bigl(=
\text{rang } H\,(t,q)\circ H\,(t,q)^T\Bigr), $$
which yields
$$ \text{Ker } H\,(t,q)=\text{Ker }\,\Bigl(H\,(t,q)^T\circ H\,(t,q)\Bigr). $$

\medskip

After this digression suppose now again $n=7,\, q=3,\, t=2$. Since two 
different projective lines intersect in one point we have $|M_1\cap M_2|=1$
provided $M_1,M_2\in{\frak G}_\sigma$ and $M_1\ne M_2$. This in turn yields
according to Eq. (16):
$$ H_3\left(\sum_{M\in{\frak G}_\sigma}[M]\right)=3\cdot\sum_{M\in{\frak G}_\sigma}
[M]+\sum_{M'\not\in{\frak G}_\sigma}^{|M'|=3}\lambda_{M'}[M']. $$
The coefficients $\lambda_{M'}\in\I Q$ are determined as follows: Any 
$M'\not\in{\frak G}_\sigma$ consists of three non--collinear points; therefore
for given $M'$ there are exactly three $M\in{\frak G}_\sigma$ such that 
$|M'\cap M|=2$. We conclude $\lambda_{M'}=3$, in turn
$$ H_3\left(\sum_{M\in{\frak G}_\sigma}[M]\right)=3\cdot w_3=H_3\left(
\frac{1}{5}w_3\right),$$
that is 
$$\sum_{M\in{\frak G}_\sigma}[M]-\frac{1}{5} w_3\in\mbox{Ker }\Bigl(H\,(2,3)^T
\circ H\,(2,3)\Bigr)=\mbox{Ker } H\,(2,3). - $$

Secondly, assume that $v$ is an element of the ''canonical'' system of
generators of $\mbox{Ker } H\,(2,3)$, say
$$ v=(X_1-X_2)(X_3-X_4)(X_5-X_6). $$
Suppose the enumerations $\tau,\tau'$ of the points of $\I P$ are given by

%\bigskip

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%\end
\end{picture}
\end{center} 

Then a lengthy but elementary computation which we omit yields
$$ v=\frac{1}{5}\,(u_\tau-u_{\tau'}). $$
We conclude that the $u_\sigma$ generate $\text{Ker } H\,(2,3)$, too.

\vspace*{15mm}

{\bf Acknowledgement}

\smallskip

The author is indebted to the referee for helpful comments.

%\pagebreak

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\bibitem{2}
{\sc C.D. Godsil}: {\it Algebraic Combinatorics}.
New York / London 1993\\ (Chapman \& Hall)
\bibitem{3}
{\sc J.B. Graver} und {\sc W.B. Jurkat}: {\it The
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\bibitem{4}
{\sc E. Kunz}: {\it K\"ahler Differentials}. 
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\end{thebibliography}

%\vspace{\fill}

%Helmut Kr\"amer\\
%Mathematisches Seminar\\ 
%der Universit\"at Hamburg\\
%Bundesstra\ss e 55\\
%D--20146 Hamburg


\end{document}








