The simple idea for all the identities considered in the rest of this paper
can be stated in an \EM{abstract} manner as follows:
Assume some fixed \pointconf\ $\cnr{c}$.
%Consider an overlay of \nilp s, which corresponds to some \pointarr\ $C$.
%
\bit
\item
The set $S$ of \EM{all} overlays of \nilp s corresponding to $\cnr{c}$ and
some fixed admissible \pointarr\ $\cnr{r}$
corresponds to the set of \EM{all} terms in the product
$\schurf_{\lambda/\mu}\cdot\schurf_{\sigma/\tau}$, where the pair of shapes $\pas{\lambda/\mu,\sigma/\tau}$ corresponds to $\pas{\cnr{c},\cnr{r}}$.
\item 
The relation ``corresponds to the same \pointmatch\ as'' is an equivalence relation
on $S$. Stated otherwise, $S$ is \EM{partitioned} into \EM{matching classes} which are determined by the corresponding \pointmatch s $\cnr{m}_1, \cnr{m}_2,\dots$
of $\cnr{r}$:
$$S=S_{\cnr{m}_1}\cup S_{\cnr{m}_2}\cup \dots, \text{ where }S_{\cnr{m}_i}\cap S_{\cnr{m}_j} = \emptyset\text{ for } i\neq j.$$
%(%The left half of the left picture in
%\figref{fig:InvolStruct} visualizes this concept.)
\item Let $\cnr{m}_1$ be a fixed \pointmatch\ in $\cnr{r}$, and let $E\of{\cnr{r},\cnr{m}_1}=\setof{e_1,\dots,e_k}$ be
a fixed set of edges of $\cnr{m}_1$ in $\cnr{r}$: Then the \EM{recolouring} of the bicoloured trails $\setof{b_1,b_2,\dots,b_k}$ corresponding to $\setof{e_1,\dots,e_k}$ in $S_{\cnr{m}_1}$
% in some overlay $s\in S$
effectuates a \EM{weight\bindestrich preserving} involution
$$
S_{\cnr{m}_1} \leftrightarrow S^{\prime}_{\cnr{m}_1}\;,
$$
where  $S^\prime_{\cnr{m}_1}$ is the set of \EM{all} overlays of \nilp s corresponding to $\pas{\cnr{c},\cnr{r}^\prime}$, where $\cnr{r}^\prime$ is the \pointarr\ obtained from $\cnr{r}$ by reversing
the orientation of the points which are incident with the edges 
$\setof{e_1,\dots,e_k}$. Note that except for this reversing of orientations of \EM{points}, the \pointmatch\ $\cnr{m}_1$ appears \EM{unchanged} as an \pointmatch\ $\cnr{m}^\prime_1$
for $\cnr{r}^\prime$: We call this operation the \EM{reversing of the edges}
$E\of{\cnr{r},\cnr{m}_1}$. %$\setof{e_1,\dots,e_k}$ in $\cnr{m}$.
% \pointmatch\ $M$, so the \EM{same} \pointmatch\ $M$ will be admissible for another \pointarr\ $R^\prime$ which arises by such recolouring. This
%\pointarr\ $R^\prime$, together with $C$, determines \EM{another} set of overlays $S^\prime$, and the
%recolouring implements a \EM{weight\bindestrich preserving involution} between the \EM{matching classes}
%$S_{M}\leftrightarrow S^\prime_{M}$. (See the left picture in \figref{fig:InvolStruct} for an
%illustration of this idea.)
%
\eit
If we want to use this idea to obtain a Schur function identity where $\gfof{S}$
appears on the lefthand side and $\gfof{S^{\prime}}$ appears on the righthand side,
i.e.,
$$
\pas{\gfof{S}+\cdots}=\pas{\gfof{S^{\prime}}+\cdots},
$$
we have to consider \EM{all} triples $(\cnr{c},\cnr{r}^\prime,\cnr{m}^\prime)$,
	where $\cnr{m}^\prime$ is an \pointmatch\ for $\cnr{r}^\prime$, i.e.,
$$
\underbrace{\pas{\gfof{S_{m_1}}+\gfof{S_{m_2}}+\cdots}}_{\gfof{S}}+\cdots=
\underbrace{\pas{\gfof{S^{\prime}_{m^\prime_1}}+\gfof{S^{\prime}_{m^\prime_2}}+\cdots}}_{\gfof{S^\prime}}+\cdots,
$$
and have to iterate this involutive construction for \EM{all} pairs
$\pas{\cnr{r}^\star,\cnr{m}^\star}$
of \pointarr s/\pointmatch s
%$\cnr{m}^\star$
which arise by these operations in a \EM{consistent} way. This \EM{consistency}
is a condition for the set of edges to be recoloured, i.e.:
If the reversal of edges $E\of{\cnr{r}^\star,\cnr{m}^\star}$ leads to some
set of overlays corresponding to $\pas{\cnr{r}^{\prime\prime},\cnr{m}^{\prime\prime}}$,
then we must have
$$E\of{\cnr{r}^{\prime\prime},\cnr{m}^{\prime\prime}}=E\of{\cnr{r}^\star,\cnr{m}^\star}.$$

%we can assign a set of edges
%$E\of{\cnr{r}^\prime,\cnr{m}^\star}$ \EM{consistently}, i.e.: % such that
%If the reversal of edges $E\of{\cnr{r}^\prime,\cnr{m}^\star}$ leads to some
%set of overlays corresponding to $\pas{\cnr{r}^{\prime\prime},\cnr{m}^\star}$,
%then we have
%$E\of{\cnr{r}^{\prime\prime},\cnr{m}^\star}=E\of{\cnr{r}^\prime,\cnr{m}^\star}$.
%
%For \EM{every} triple $(\cnr{c},\cnr{r}^\prime,\cnr{m})$ obtained by such reversal
%	of edges, we have to consider \EM{all} triples $(\cnr{c},\cnr{r}^\prime,\cnr{m}^\star)$,
%	where $\cnr{m}^\star$ is a \pointmatch\ for $\cnr{r}^\prime$,
%
%Assume we can \EM{combine} such single \EM{involutions for matching classes}
%consistently, in the following sense:
%\bit
%\item For \EM{every} triple $(\cnr{c},\cnr{r}^\prime,\cnr{m})$ obtained by such reversal
%	of edges, we have to consider \EM{all} triples $(\cnr{c},\cnr{r}^\prime,\cnr{m}^\star)$,
%	where $\cnr{m}^\star$ is a \pointmatch\ for $\cnr{r}^\prime$,
%\item For \EM{all} \pointmatch s $\cnr{m}^\star$ %which must be considered
%in \EM{all} \pointarr s $\cnr{r}^\prime$ thus obtained, we can assign a set of edges
%$E\of{\cnr{r}^\prime,\cnr{m}^\star}$ \EM{consistently}, i.e.: % such that
%If the reversal of edges $E\of{\cnr{r}^\prime,\cnr{m}^\star}$ leads to some
%set of overlays corresponding to $\pas{\cnr{r}^{\prime\prime},\cnr{m}^\star}$,
%then we have
%$E\of{\cnr{r}^{\prime\prime},\cnr{m}^\star}=E\of{\cnr{r}^\prime,\cnr{m}^\star}$.
%\eit

We call such consistent assignment of edges (to be reversed) to pairs $\pas{\cnr{r},\cnr{m}}$
a \EM{(simple) recolouring scheme}.
Clearly, such recolouring scheme amounts to a \EM{weight preserving bijection}
$$
S_1\cup S_2\cup\dots\leftrightarrow S_1^\prime\cup S_2^\prime\cup \dots,
$$
where
%\bit
%\item
$\setof{S_1,S_2,\dots}$
% \item
and $\setof{S_1^\prime,S_2^\prime\dots}$
%\eit
are two families of sets of overlays. These two families correspond to two families
\bit
\item
$\setof{\pas{\lambda_1/\mu_1,\sigma_1/\tau_1}, \pas{\lambda_2/\mu_2,\sigma_2/\tau_2},\dots}$
\item
and $\setof{\pas{\lambda^\prime_1/\mu^\prime_1,\sigma^\prime_1/\tau_1},
\pas{\lambda^\prime_2/\mu^\prime_2,\sigma^\prime_2/\tau^\prime_2},\dots}$
\eit
of pairs of shapes, and the (\EM{simple}) recolouring scheme thus translates to an identity involving sums of products
of (skew) Schur functions
$$
\schurf_{\lambda_1/\mu_1}\cdot\schurf_{\sigma_1/\tau_1} +
\schurf_{\lambda_2/\mu_2}\cdot\schurf_{\sigma_2/\tau_2} + \cdots =
\schurf_{\lambda^\prime_1/\mu^\prime_1}\cdot\schurf_{\sigma^\prime_1/\tau^\prime_1} +
\schurf_{\lambda^\prime_2/\mu^\prime_2}\cdot\schurf_{\sigma^\prime_2/\tau^\prime_2} + \cdots.
$$
(\figref{fig:InvolStructSimple} visualizes this concept.)

%The crucial point is the ``glueing together'' of involutions
%for matching classes. Clearly, this amounts to a consistent \EM{rule} which identifies for every
%pair $\pas{\cnr{r},\cnr{m}}$ (where $\cnr{r}$ is a \pointarr\ and $\cnr{m}$ is a \pointmatch\ 
%in $\cnr{r}$) a \EM{family} $\setof{E_1,\dots,E_k}$ of sets of edges % (which correspond to bicoloured paths)
%to be \EM{reversed}, thus relating $\pas{\cnr{r},\cnr{m}}$ to $\pas{\cnr{r}_1^\prime,\cnr{m}},
%\dots,\pas{\cnr{r}_k^\prime,\cnr{m}}$, such that the iteration of this ``edge\bindestrich reversing
%procedure'' yields a ``bipartite substructure'', see \figref{fig:InvolStruct}.
%% (which corresponds to the recolouring of the bicoloured paths):
%We
%call such rule a \EM{recolouring scheme}.

% illustrate this concept by .

%Before summarizing these reasoning in elaborate abstract statements, we shall
%consider concrete examples.

\begin{figure}[h]
\caption{Illustration: The picture visualizes the situation where a (\EM{simple}) \EM{recolouring scheme}
amounts to a weight preserving bijection between overlays corresponding to \pointarr s
$\setof{\cnr{r}_1,\cnr{r}_2,\cnr{r}_3}$ and $\setof{\cnr{r}_1^\prime,\cnr{r}_2^\prime}$,
respectively. In the picture it is assumed that there are 6 \pointmatch s for these
\pointarr s, which are indicated by different shades of grey (white, light grey, grey and
black) and textures (slanted lines, crosshatch).
}
\label{fig:InvolStructSimple}
\begin{center}
\input graphics/InvolStructSimple
\end{center}
\end{figure}

%% Referee Punkt 7
% This sounds like general abstract nonsense, so
We shall give
some concrete examples to make this clear.

\subsection{Dodgson's condensation, revisited and generalized}
\label{sec:Dodgson}
Recall Dodgson's
condensation formula \eqref{eq:Dodgson}, or rather its Schur function equivalent \eqref{eq:DodgsonSchur}.
Given our above preparations,  \figref{fig:Dodgson} contains the
\EM{proof} of \eqref{eq:DodgsonSchur}! (A simple combinatorial proof of
Dodgson's condensation formula and its generalization, the Dodgson-Muir-identity, was given in \cite{Zeilberger:1997} and in \cite{Berliner-Brualdi:2008}, respectively.)

In order to see this, we consider once again an example: Let
$\lambda=\pas{\lambda_1,\dots,\lambda_6}=\pas{9, 7, 5, 3, 3, 1}$; for convenience, we
denote $\pas{\lambda_2,\dots,\lambda_5}$ by $\tau$. The product of Schur functions
corresponding to the lefthand side of Dodgson's formula
is 
$$\schurf_{\lambda}\cdot\schurf_{\tau} =
\schurf_{\lambda}\cdot\schurf_{\tau+\allconstpart{(-1)}/\allconstpart{(-1)}},$$
which by the \lgt\bindestrich interpretation appears as the generating function of overlays of
nonintersecting lattice paths. See \figref{fig:Dodgson2}, where the involutions
visualized in \figref{fig:Dodgson} are specialized to this concrete example.
These weight\bindestrich preserving involutions immediately imply \eqref{eq:DodgsonSchur},
and it is obvious that the argument is valid not only for our special example, but
in full generality, whence (by the reasoning following Corollary~\ref{cor:Dodgson})
we have proved \eqref{eq:DodgsonSchur} (and thus \eqref{eq:Dodgson}).
\hfill\qedsymbol

\begin{figure}[h]
\caption{Illustration: Graphical proof of Dodgson's condensation formula.
The picture
in the upper left corner
shows the three \pointarr s that may appear if the bicoloured trail starting in the
rightmost coloured upper point $s$ (drawn as black circle) is recoloured. Note that there is only
\EM{one} \pointmatch\ % (indicated by grey lines)
for the \pointarr s in the left half, % of the picture,
while there are \EM{two} \pointmatch s $\cnr{m}_1,\cnr{m}_2$ for the \pointarr\ in
the right half % of the picture
(the edges incident with $s$ in  $\cnr{m}_1$ and $\cnr{m}_2$ are indicated by grey
lines). The involution effectuated by the recolouring
scheme is indicated by black lines connecting the corresponding \pointarr s, each of
which represents a product of skew Schur functions. (The picture in the lower right corner
depicts this simple recolouring scheme in the ``generic'' manner of \figref{fig:InvolStructSimple}.)}
\label{fig:Dodgson}
\begin{center}
\input graphics/LemmaIllustrationSlide
\end{center}
\end{figure}
%\clearpage

For the proof of Dodgson's condensation formula, we used a simple
\EM{recolouring scheme}, which we may generalize as follows: 
%``Fix a single coloured point $s$, and in all \pointmatch s
%always reverse the edge starting in $s$''. Clearly,
%this idea can be generalized as follows: 
``Fix some nonempty subset $D$ of coloured points
of the same orientation (i.e., all points in $S$ are either all outwards oriented
or all inwards oriented). To \EM{all} \pointmatch s $\cnr{m}$ in some \pointarr\ 
$\cnr{r}$, assign $E\of{\cnr{r},\cnr{m}} = $ the set of edges incident
with some point in $D$''. We call this rule \EM{Dodgson's recolouring scheme}: It could
be restated sloppily as ``always reverse all edges incident with points of $D$''.

\begin{figure}[h]
\caption{Dodgson's condensation formula, by example.}
\label{fig:Dodgson2}
The pictures show
the positions of lower and upper points of the overlays corresponding to shapes
derived from
$$\lambda/\mu=\pas{9, 7, 5, 3, 3, 1}/\allconstpart{0}
\text{ and }
\sigma/\tau=\pas{6, 4, 2, 2,-1,-1,\dots}/\allconstpart{(-1)}.$$ 
(For the latter
we introduce the shorthand notation
$\sigma/\tau=\pas{6, 4, 2, 2,\periodic{(-1)}}/\allconstpart{(-1)}$.)
In the upper
picture,
% we chose shift parameters $t_1=0$ for the ``green''  shape $\lambda$ and $t_2=-1$;
% note that
all
coloured points are green, and the bicoloured trail $b$ starting in the rightmost upper
point $s=\pas{8,\n}$ \EM{must} have the rightmost lower point $\pas{-1,1}$ as its
other ending point.  Recolouring $b$ leads to the middle picture. Here, the bicoloured
trail starting in $s$ has \EM{two} possibilities: Its other ending point might be $\pas{-1,1}$
again,
but also the leftmost upper point $\pas{-5,\n}$ is possible. For the latter case, recolouring the
bicoloured trail connecting $s$ and $\pas{-5,\n}$ leads to the lower picture. Now, as in the
upper picture, the bicoloured trail starting in $s$ \EM{must} have $\pas{-5,\n}$ again as
its other ending point. By the above reasoning, this shows immediately the following instance
of
\eqref{eq:DodgsonSchur}:
$$
{\schurf_{\lambda/\mu}}\cdot{\schurf_{\sigma/\tau}}
+
{\schurf_{\lambda^{\prime\prime}/\mu^{\prime\prime}}}\cdot{\schurf_{\sigma^{\prime\prime}/\tau^{\prime\prime}}}
=
{\schurf_{\lambda^{\prime}/\mu^{\prime}}}\cdot{\schurf_{\sigma^{\prime}/\tau^{\prime}}}
$$
\begin{center}
\input graphics/Dodgson
\end{center}
\end{figure}
\clearpage

%This construction is described in detail in \cite{fulmek:2001}. Here, we simply refer to
%the left picture of \figref{fig:bicolouredPaths}, where the white paths are indicated by dashed
%lines, and all bicoloured paths are indicated by thick grey lines.
%
%The following observations are immediate:
%\begin{obs}[{\bf Bicoloured paths always exist}]
%\label{obs:arbitrary-point}
%For \EM{every} coloured point $q$, there exists a bicoloured path starting at $q$.
%\end{obs}
%
%\begin{obs}[{\bf Bicoloured paths connect points of different radial orientation}]
%\label{obs:different-orientation}
%The bicoloured paths thus constructed never connect points of the same radial orientation
%(i.e., two points oriented both towards or both away from the center).
%\end{obs}
%
%In the lower picture of \figref{fig:twoSkewShapes}, a possible pattern of ``connections
%by bicoloured paths'' is indicated by dashed lines.
%
%The following observation is easy to see:
%\begin{obs}[{\bf Bicoloured path connect points of different parity}]
%\label{obs:different-parity}
%Two different bicoloured paths may have lattice points in common (they may intersect),
%but they can never \EM{cross}.
%%
%If we assume some consecutive numbering of the coloured points in their circular
%orientation (see the lower picture of \figref{fig:twoSkewShapes}), then this non\bindestrich crossing condition implies that there can never be a bicoloured path connecting two points with numbers of the same parity.
%\end{obs}
%
%The non\bindestrich crossing condition means that
%if all such connections were drawn as straight lines connecting points on the circle, then
%no two such lines can intersect. (In \figref{fig:twoSkewShapes}, not all connections are
%drawn as straight lines for graphical reasons.) 
%
%% \section{Recolouring bicoloured paths}
%
%Consider some bicoloured path $b$ in the overlay of nonintersecting lattice paths $\pas{P_1,P_2}$: 
%Changing colours (black to white and vice versa)
%\bit
%\item of both upper points of $b$
%\item and of all arcs of $b$
%\eit
%gives a new overlay of nonintersecting lattice paths  $\pas{P_1^\prime,P_2^\prime}$
%(with different lower/upper points). It is easy to see that we have for this
%\EM{recolouring} of a bicoloured path:
%\begin{obs}[{\bf Recolouring bicoloured paths is a weight preserving involution}]
%\label{obs:weight-preserving-involution}
%The recolouring of a bicoloured path $b$ in an overlay  of nonintersecting lattice
%paths $\pas{P_1,P_2}$ is an \EM{involutive operation} (i.e., if we obtain
%the overlay $\pas{P_1^\prime,P_2^\prime}$ by recolouring $b$ in $\pas{P_1,P_2}$, then
%recolouring $b$ \EM{again} in $\pas{P_1^\prime,P_2^\prime}$ yields the original
%$\pas{P_1,P_2}$), which \EM{preserves % (the product of)
%the respective weights}, i.e.,
%$$
%\weight\of{P_1}\cdot\weight\of{P_2}=\weight\of{P_1^\prime}\cdot\weight\of{P_2^\prime}.
%$$
%\end{obs}
%
%%%%%%%%%%%%%%%%%%%%%%%%%%
%\begin{figure}
%\caption{The two (necessarily different, by
%Observation~\ref{obs:different-orientation}!)
%bicoloured paths in \figref{fig:twoSkewShapes} which start at the white upper point
%$\pas{15,N}$ and at the the black lower point $\pas{5,1}$ may have their respective other
%upper points in $\pas{10, 1}$ and $\pas{-8,1}$. In the picture below, the corresponding points are marked by
%white rectangles, the bicoloured paths are indicated by arrows. The picture shows the situation \EM{after} recolouring these paths.
%}
%\label{fig:twoSkewShapesRecoloured}
%\input graphics/twoSkewShapesRecoloured
%\end{figure}
%
%Return to the example illustrated in \figref{fig:twoSkewShapes} and
%consider the white upper point $\pas{15,N}$
%and the black lower point $\pas{5,1}$ there. Note that both of these points are marked
%with a triangle pointing \EM{downward}. The bicoloured paths ending at these points must have
%their other upper points marked with a triangle pointing \EM{upward}. One possible choice of these other
%ending points is depicted in \figref{fig:twoSkewShapesRecoloured}: The corresponding ending
%points are marked by
%white rectangles, the bicoloured paths are indicated by arrows. The picture shows the situation \EM{after} recolouring these paths.
%
%The skew shape corresponding to the white points in \figref{fig:twoSkewShapesRecoloured} is
%$\lambda^\prime/\mu^\prime$, where
%\begin{align}
%\lambda^\prime &=\pas{13, 13, 11, 11, 9, 9, 8, 8, 7, 5, 3},\notag\\
%\mu^\prime&= \pas{9, 9, 7, 7, 7, 6, 5, 5, 5, 4, 0}.\label{eq:lambdaprime-muprime}
%\end{align}
%The skew shape corresponding to the black points in \figref{fig:twoSkewShapesRecoloured}
%is  $\sigma^\prime/\tau^\prime$, where
%\begin{align}
%\sigma^\prime &= \pas{15, 14, 14, 12, 12, 11, 11, 11, 9, 8, 7, 7, 5},\notag\\
%\tau^\prime &= \pas{10, 10, 10, 7, 7, 6, 5, 5, 5, 4, 2, 2, 0}.\label{eq:sigmaprime-tauprime}
%\end{align}
%For both skew shapes, the lower and ending points are shifted by $1$,
%so, for instance, the lower point of the first path corresponding to
%$\sigma^\prime/\tau^\prime$ is $\pas{\tau^\prime_1-1+1,1}=\pas{10,1}$ and the
%ending point of the last (eleventh) path corresponding to $\lambda^\prime/\mu^\prime$
%is $\pas{\lambda_{11}-11+1,N}=\pas{-7,N}$.
%
%
%The pictures in \figref{fig:twoSkewShapesRecoloured} contain 
%redundant information: All uncoloured points are ``doubled'', and the colour (black or white)
%of the coloured points is determined uniquely
%by their circular orientation. So we may encode the information in a more terse
%way, namely as
%\bit
%\item the \EM{\pointconf} (see the upper picture in \figref{fig:GenericConfiguration}),
%\item and the \EM{\pointarr} (see the lower picture in \figref{fig:GenericConfiguration}).
%\eit
%%%%%%%%%%%%%%%%%%%%%%%%%%
%\begin{figure}
%\caption{The information of \figref{fig:twoSkewShapesRecoloured} can be encoded in a more
%terse way, namely as the \pointconf\ (shown in the upper picture) and the \pointarr\ 
%(shown in the lower picture): The grey points in the upper picture are ``doubled''
%points (coloured black \EM{and} white), and the colour of the white points in the upper
%picture is determined by the \EM{orientation} of the corresponding points in the lower
%picture (points in the upper half which are inwardly orientated and points in the lower
%half which are outwardly oriented should be coloured black; all other points
%should be coloured white).
%}
%\label{fig:GenericConfiguration}
%\input graphics/GenericConfiguration
%\end{figure}


The following Lemma and its proof are a reformulation of \cite[Lemma~15]{fulmek:2001}):
\begin{lem}
\label{lem:DodgsonScheme}
Except for degenerate cases, Dodgson's recolouring scheme always yields a Schur function identity.
\end{lem}
\begin{proof}
Clearly, Dodson's recolouring scheme consistently determines the set of edges to be
reversed for every \pointmatch.
The following reasoning is merely a clarification of the statement:

Let $\lambda/\mu$ and $\sigma/\tau$ be two skew shapes and consider the \pointconf\ $\cnr{c}$ which corresponds to $\pas{\lambda/\mu,\sigma/\tau}$. Without loss of generality, let $D$
be a nonempty subset of \EM{inwards} oriented points in $\cnr{c}$.

Let $V$ be the set of \EM{all} admissible \pointarr s for $\cnr{c}$ where \EM{all} points
of $D$ have the \EM{same} orientation.
Consider the graph $G$ with vertex set $V$, where two vertices
$v_1$, $v_2$ are connected by an edge if and only if there exists a \pointmatch\ $\cnr{m}$
for $v_1$ such that $v_2$ is obtained by applying Dodgson's recolouring scheme (i.e.,
reverse
all edges in $\cnr{m}$ which are incident with a point of $D$).
Clearly, this graph $G$ is \EM{bipartite}: 
$V=I\cup O$, where $I$ is the subset of $V$ with all points in $D$ oriented \EM{inwards}
and $O$ is the subset of $V$ with all points in $D$ oriented \EM{outwards}, and there
is no edge connecting two vertices of $I$ or two vertices of $O$ (see again \figref{fig:Dodgson},
where the bipartite structure is indicated by a dashed line separating the left and right half of the
picture).

Assume that graph $G$ has a connected component $Z$ with at least $2$ vertices
(if there is no such component, we call this a \EM{degenerate case}). Then we
have the following identity for skew Schur functions:
\begin{equation}
\label{eq:lemma-general}
\sum_{\pas{\lambda/\mu,\;\sigma/\tau}\in Z_I}
	\schurf_{\lambda/\mu}\cdot\schurf_{\sigma/\tau}
=
\sum_{\pas{\lambda^\prime/\mu^\prime,\;\sigma^\prime/\tau^\prime}\in Z_O}
	\schurf_{\lambda^\prime/\mu^\prime}\cdot\schurf_{\sigma^\prime/\tau^\prime},
\end{equation}
where $Z_O$ and $Z_I$
denote the sets of pairs of skew shapes corresponding to $\pas{\cnr{c},x}$ for
$x\in O$ and 
$x\in I$, respectively.
\end{proof}

\begin{figure}[h]
\caption{Dodgson's recolouring scheme, where the set $D$ consists only of the
rightmost coloured upper point (drawn as black circle), applied to 3 coloured upper/lower points.
%: Recolouring $s$ when it is green will introduce 
%a new lower red point or remove an upper red point, and when the point is red, recolouring
%will remove a lower red point or introduce a new upper red point.
}
\label{fig:Dodgson3}
\begin{center}
\input graphics/Dodgson3
\end{center}
\vspace*{-2ex}
\end{figure}

%%%%%%%%%%%%%%%%%%%%%%%%%%

We illustrate Lemma~\ref{lem:DodgsonScheme} by the following generalization of
Dodgson's condensation, which might be new. Consider the \pointarr\ with
$k$ upper and $k$ lower points ($k> 1$) which are all green, and always recolour the bicoloured
trail starting in the rightmost upper point ($k=2$ corresponds to Dodgson's condensation,
see \figref{fig:Dodgson}; the cases $k=3$ and $k=4$ are depicted in Figures~\ref{fig:Dodgson3}
and \ref{fig:Dodgson4}, respectively).
\begin{thm}
\label{thm:maybenew}
Let $a$ be an $\pas{m+k}\times\pas{m+k}$\bindestrich matrix, and let $1\leq i_1<i_2<\cdots<i_k\leq m+k$ and
$1\leq j_1<j_2<\cdots<j_k\leq m+k$ be (the indices of) $k$ fixed rows and $k$ fixed columns,
respectively, of $a$. Denote the sets of these (indices of) rows and columns by $R$ and $C$,
respectively.

Let $E\defeq\setof{j_2,j_4,\dots,j_{2\cdot\floor{k/2}}}$ and $O\defeq\setof{i_1,i_3,\dots,i_{2\cdot\ceil{k/2}-1}}$ be the sets of \EM{odd} fixed row indices and of \EM{even} fixed column indices,
respectively. Then we have:
\begin{equation}
\label{eq:maybenew}
\sum_{\substack{S\subseteq E,\\ T\subseteq O,\\ \cardof{S}=\cardof{T}}}
\detof{\deleterowcols{a}{T}{S}}\cdot\detof{\deleterowcols{a}{R\setminus T}{C\setminus S}}
=
\sum_{\substack{S\subseteq E,\\ T\subseteq O,\\ \cardof{S}=\cardof{T}-1}}
\detof{\deleterowcols{a}{T}{S\cup\setof{j_1}}}\cdot\detof{\deleterowcols{a}{R\setminus T}{C\setminus \pas{S\cup\setof{j_1}}}}.
\end{equation}
%I.e., we form square minors/cominors from the columns/rows
%$\setof{1,2,4,6,\dots}$/$\setof{1,3,5,7,\dots}$: All the minors containing column 1 go to
%$E$, and all others go to $O$. Then we have a determinantal identity.
\end{thm}

\begin{proof}
We shall prove the Schur function identity equivalent to \eqref{eq:maybenew}: Note that
the corresponding \pointconf\ shows $m+k$ lower points, of which $k$ are coloured, and
$m+k$ upper points, of which $k$ are coloured. As always, we concentrate on the coloured
points: Denote the upper coloured points by ${t_1,t_2,\dots,t_k}$,
and the lower coloured points by ${s_1,s_2,\dots,s_k}$
(counted from the right, as always).
Consider the \pointarr\ where \EM{all} coloured points are \EM{green} (see the uppermost
configuration in the left parts of
Figures~\ref{fig:Dodgson3} and \ref{fig:Dodgson4}): This \pointarr\ corresponds to the summand
for $S=T=\emptyset$ in the left\bindestrich hand side of \eqref{eq:maybenew}.
Clearly, the other end of a bicoloured trail starting in the rightmost upper point $t_1$
must either be in the set $O\defeq\setof{s_1,s_3,\dots,s_{2\cdot\ceil{k/2}-1}}$ or in
the set $E\defeq\setof{t_2,t_4,\dots,t_{2\cdot\floor{k/2}}}$. Now observe that it is
possible to recolour points $s_1, t_2, s_3, t_4, \dots$ (in this order;
these recolouring steps are indicated by thick grey lines in Figures~\ref{fig:Dodgson3}
and \ref{fig:Dodgson4}) until
we obtain the \pointarr\ where \EM{all} points in $E\cup O$ are coloured red. (Note that
in this \pointarr, $t_1$ is green if $k$ is even, otherwise $t_1$ is red.)
Now it is easy to see that for an arbitrary choice of subsets $S$, $T$,
where $S\subseteq E\cup\setof{t_1}$, $T\subseteq O$ and $\cardof{S}=\cardof{T}$,
by the recolouring of red upper and lower points in the appropriate order, we can obtain the
situation where the set of red points is precisely the union $S\cup T$: This translates to the assertion.
\end{proof}

As an example, we state the determinantal identity corresponding to
\figref{fig:Dodgson4} ($k=4$) for the special case where $a$ is a
$4\times4$\bindestrich matrix (i.e., $m=0$):
\begin{multline*}
\detof{a} = 
\detprodcminors{a}{1}{1} + \detprodcminors{a}{1,3}{1,2} + \detprodcminors{a}{1,3}{1,4} +
\detprodcminors{a}{3}{1} \\ -
\detprodcminors{a}{1}{2} - \detprodcminors{a}{1,3}{2,4} -
\detprodcminors{a}{3}{2} \\ - \detprodcminors{a}{1}{4} - \detprodcminors{a}{3}{4}.
\end{multline*}

\begin{figure}[h]
\caption{Dodgson's recolouring scheme, where the set $D$ consists only of the
rightmost coloured upper point (drawn as black circle), applied to 4 coloured upper/lower points.
}
\label{fig:Dodgson4}
\begin{center}
\input graphics/Dodgson4
\end{center}
\end{figure}

\subsection{(Generalized) PlŸcker relations}
\label{sec:Pluecker}

There is a particularly simple special case of Lemma~\ref{lem:DodgsonScheme}
(this is a reformulation of
\cite[Lemma~16]{fulmek:2001}):
\begin{cor}
\label{cor:fulmek}
Let $\cnr{r}$ be the \pointarr\ for the
pair of shapes $\pas{\lambda/\mu,\;\sigma/\tau}$, and assume that the orientation of
the points in $\cnr{r}$  is \EM{alternating} (with respect to their circular arrangement).
%Let $S$
%be some fixed subset of points of $C$, where all points of $S$ have the same orientation
Consider the set of all \pointarr s which arise by applying Dodgson's recolouring scheme
(for some fixed nonempty set of points of the same orientation) to $\cnr{r}$ and denote the corresponding
set of pairs of
skew shapes by $Q$.

Then we have:
\begin{equation}
\label{eq:lemma-special}
\schurf_{\lambda/\mu}\cdot\schurf_{\sigma/\tau}
=
\sum_{\pas{\lambda^\prime/\mu^\prime,\;\sigma^\prime/\tau^\prime}\in Q}
	\schurf_{\lambda^\prime/\mu^\prime}\cdot\schurf_{\sigma^\prime/\tau^\prime}.
\end{equation}
\end{cor}
\begin{proof}
Consider the \pointarr\ $\cnr{r}^\prime$ corresponding to some Schur function product
$\schurf_{\lambda^\prime/\mu^\prime}\cdot\schurf_{\sigma^\prime/\tau^\prime}$
from the right hand side of \eqref{eq:lemma-special}: By the
combination of Observations \ref{obs:different-orientation} and \ref{obs:different-parity},
it is clear that
re\bindestrich applying Dodgson's recolouring scheme \EM{always yields} the \pointarr\ $\cnr{r}$
corresponding
to the Schur function product $\schurf_{\lambda/\mu}\cdot\schurf_{\sigma/\tau}$.
\end{proof}

We will use this Corollary for a proof of the PlŸcker relations (also known as
GrassmannÐPl\"ucker syzygies, see \cite{sturmfels:1993},
% or \cite[Chapter 3, Section 9, formula II]{turnbull:1960},
or as Sylvester's Theorem, see \cite[section 137]{muir:1933}).
In addition to notation $\range{m}\defeq\setof{1,2,\dots,m}$ we introduce the notation
$$\shiftrange{m}{n}\defeq\setof{n+1,n+2,\dots,n+m}.$$
Moreover, for finite ordered
sets $X\subseteq S$ and $Y$ with $Y\cap S=\emptyset$, $\cardof{Y}=\cardof{X}$, we introduce the
notation $\shuffle{S}{X}{Y}$ for the set $S$ where the elements of $X$ are replaced
by the elements of $Y$ \EM{in the same order}, i.e., if ordered sets $X$ and $Y$ are
given as $X=\pas{x_1,x_2,\dots}$ and $Y=\pas{y_1,y_2,\dots}$, respectively, and $S$ is given as
$$S=\pas{s_1,\dots,s_{(k_1-1)},x_1,s_{(k_1+1)},\dots,s_{(k_2-1)},x_2,s_{(k_2+1)},\dots},$$
then ordered set $\shuffle{S}{X}{Y}$ is given as
$$\shuffle{S}{X}{Y}=\pas{s_1,\dots,s_{(k_1-1)},y_1,s_{(k_1+1)},\dots,s_{(k_2-1)},y_2,s_{(k_2+1)},\dots}.$$
\begin{thm}[PlŸcker relations]
\label{thm:Pluecker}
Let $a=\pas{a_{i,j}}_{(i,j)\in\range{m}\times\range{2m}}$ be a $m\times 2\cdot m$ matrix.
Consider some fixed % list $1\leq r_1<r_2<\cdots<r_k\leq m$
set $R\subseteq\range{m}$.
Then we have
\begin{equation}
\label{eq:pluecker}
\detof{\minor{a}{\range{m}}{\range{m}}}\cdot\detof{\minor{a}{\range{m}}{\shiftrange{m}{m}}}=
\sum_{\substack{S\subseteq\shiftrange{m}{m},\\\cardof{S}=\cardof{R}}}
\detof{\minor{a}{\range{m}}{\shuffle{\range{m}}{R}{S}}}\cdot
\detof{\minor{a}{\range{m}}{\shuffle{\pas{\range{2m}\setminus\range{m}}}{S}{R}}}.
\end{equation}
\end{thm}
\begin{proof}
We shall prove a Schur function identity which is equivalent to \eqref{eq:pluecker}: Consider
$$
\lambda=\pas{(2\cdot m)\cdot(m-1),(2\cdot m-2)\cdot(m-1),\dots,4\cdot(n-1),2\cdot(n-1)}
$$
and
$$
\sigma=\pas{(2\cdot m-1)\cdot(m-1),(2\cdot m-3)\cdot(m-1),\dots,3\cdot(n-1),(n-1)}
$$
and assume that matrix $\minor{a}{\range{m}}{\range{m}}=\jtmat{\lambda}$ and
matrix $\minor{a}{\shiftrange{m}{m}}{\range{m}}=\jtmat{\sigma}$. It is clear that
the \pointarr\ corresponding to the pair of shapes
$\pas{\lambda/\allconstpart{0},\sigma/\allconstpart{0}}$ has only \EM{upper} coloured points
which alternate in orientation, see the configurations in the left part of \figref{fig:Pluecker}. Corollary~\ref{cor:fulmek} immediately translates to (the Schur function
equivalent of) \eqref{eq:pluecker}; see \figref{fig:Pluecker} for an illustration.
\end{proof}

\begin{figure}[h]
\caption{Illustration of the Pl\"ucker relation \eqref{eq:pluecker} for $m=3$: The left
part shows \EM{three copies} of the \pointarr\ considered in the proof of
Theorem~\ref{thm:Pluecker} with different \pointmatch s; the right part shows the
effect of recolouring the fixed points which are indicated by black circles. Note that
there are \EM{five} possible \pointmatch s in this situation: For the upper and lower
pictures, there are \EM{two} \pointmatch s which yield the same recolouring; these
two \pointmatch s are shown in the left and right halves of the pictures, respectively.
In the middle picture, there is only one \pointmatch\ (which is the same in the left half
and in the right
half).}
\label{fig:Pluecker}
\begin{center}
\input graphics/Pluecker
\end{center}
\end{figure}

Note that the proof (which basically is \EM{contained} in \figref{fig:Pluecker}!)
implies an obvious generalization: Of course, there might be \EM{uncoloured} points
in the \pointconf s associated to the \pointarr s shown in \figref{fig:Pluecker}!
Such uncoloured points amount to a slightly more general % Schur function identity
% or to a
Pl\"ucker\bindestrich like identity involving a \EM{common minor} in the products of determinants
(similar to Theorem~\ref{thm:maybenew}).
(In \cite[(3.3)]{gurevichPyatovSaponov:2009}, this identity is stated by describing
the situation with certain operations of Ferrers diagrams and proved by using
the Pl\"ucker relations; the connection of this identity to the more general
%% Referee Punkt 8
statements presented here was already explained \EM{ad hoc} in \cite{fulmek:2009}.)
\begin{thm}
\label{them:plueckergeneral}
Let $a$ be an $\pas{m+k}\times\pas{m+2\cdot k}$\bindestrich matrix, and let
% $1\leq i_1<i_2<\cdots<i_k\leq m+k$ and
$1\leq j_1<j_2<\cdots<j_{2k}\leq m+2\cdot k$ be (the indices of) % $k$ fixed rows and
$2\cdot k$ fixed columns of $a$, and set $A\defeq\setof{j_1,\dots,j_k}$ and 
$O\defeq\setof{j_{k+1},\dots,j_{2\cdot k}}$. Let $A^\prime\defeq\range{m+2\cdot k}\setminus A$
and $O^\prime\defeq\range{m+2\cdot k}\setminus O$.
% Let 
%respectively, of $a$. Denote the set of these (indices of) rows and columns by $R$ and $C$,
%respectively.
Consider some fixed % list $1\leq r_1<r_2<\cdots<r_k\leq m$
set $R\subseteq A$.
Then we have:
\begin{equation}
\label{eq:plueckergeneral}
\detof{\minor{a}{\range{m+k}}{A^\prime}}
\detof{\minor{a}{\range{m+k}}{O^\prime}}=
\sum_{\substack{S\subseteq O,\\\cardof{S}=\cardof{R}}}
\detof{\minor{a}{\range{m+k}}{\shuffle{A^\prime}{R}{S}}}\cdot
\detof{\minor{a}{\range{m+k}}{\shuffle{O^\prime}{S}{R}}}.
\end{equation}
\end{thm}

\subsection{Generalized recolouring scheme}
For Laplace's Theorem (stated here as Theorem~\ref{thm:Laplace}), we need a slight
generalization of \EM{simple} recolouring schemes. Observe that
%% Referee Punkt 8
it is not necessary for our purposes that a recolouring
scheme constitutes a \EM{bijective mapping} between matching classes (as visualized in
\figref{fig:InvolStructSimple}); it suffices that it establishes a ``bipartite
structure''in the following sense:
\bit
\item Let $\cnr{m}$ be a fixed \pointmatch\ for $\pas{\cnr{c},\cnr{r}}$ (where
$\cnr{r}$ is a \pointarr\ for \pointconf\ $\cnr{c}$), and let
$$
{\mathcal E}\of{\cnr{r},\cnr{m}}=
\setof{E_1\of{\cnr{r},\cnr{m}},\dots,E_k\of{\cnr{r},\cnr{m}}}$$
be
a fixed \EM{family} of sets of edges of $\cnr{m}$ in $\cnr{r}$: Then the reversing
of edges in $E_1, \dots, E_k$ amounts to a $k$\bindestrich valued relation\begin{align*}
S_{\cnr{m}} &\leftrightarrow S^{1}_{\cnr{m}}\text{ (weight\bindestrich preserving bijection),}\\
\dots \\
S_{\cnr{m}} &\leftrightarrow S^{k}_{\cnr{m}}\text{ (weight\bindestrich preserving bijection),}
\end{align*}
where 
\bit
\item $S_{\cnr{m}}$ is the set of all overlays of \nilp s corresponding to
$\pas{\cnr{c},\cnr{r},\cnr{m}}$,
\item $S_{\cnr{m}}^i$ is the set of \EM{all} overlays of \nilp s corresponding to
$\pas{\cnr{c},\cnr{r}_i^\prime,\cnr{m}}$, where $\cnr{r}_i^\prime$ is the \pointarr\ obtained from $\cnr{r}$ by reversing the edges $E_i$.
\eit
\item Assume that we can assign to \EM{each} pair $\pas{\cnr{r},\cnr{m}}$ a family of
sets of edges \EM{consistently}, i.e., such that
%\bit
%\item
${\mathcal E}\of{\cnr{r},\cnr{m}}\cap{\mathcal E}\of{\cnr{r}^\prime,\cnr{m}}\supseteq\setof{E}$ 
	%contains precisely one element %(set of edges) $E$
	if $\cnr{r}^\prime$ is obtained by
	$\cnr{r}$ by reversing the edges $E$.
% \item If 
\item Assume further that the graph $G$
\bit
\item with vertex set equal to \EM{all} matching classes $S^{\prime}_{\cnr{m}}$,
	%(corresponding to fixed \nop\ $\cnr{m}$),
\item where the edges are given by the weight\bindestrich preserving bijections
	$S^{\prime}_{\cnr{m}} \leftrightarrow S^{\prime\prime}_{\cnr{m}}$
	just described
\eit
is \EM{bipartite} such that for every connected component of $G$ the bipartition classes
are of equal size.
\eit
Then we call such assignment of families of edges a (generalized) recolouring scheme.
It is clear, that a recolouring scheme corresponds to a Schur function identity.
This idea is visualized in \figref{fig:InvolStruct}.

%The crucial point is the ``glueing together'' of involutions
%for matching classes. Clearly, this amounts to a consistent \EM{rule} which identifies for every
%pair $\pas{\cnr{r},\cnr{m}}$ (where $\cnr{r}$ is a \pointarr\ and $\cnr{m}$ is a \pointmatch\ 
%in $\cnr{r}$) a \EM{family} $\setof{E_1,\dots,E_k}$ of sets of edges % (which correspond to bicoloured paths)
%to be \EM{reversed}, thus relating $\pas{\cnr{r},\cnr{m}}$ to $\pas{\cnr{r}_1^\prime,\cnr{m}},
%\dots,\pas{\cnr{r}_k^\prime,\cnr{m}}$, such that the iteration of this ``edge\bindestrich reversing
%procedure'' yields a ``bipartite substructure'', see \figref{fig:InvolStruct}.
%% (which corresponds to the recolouring of the bicoloured paths):
%We
%call such rule a \EM{recolouring scheme}.

\begin{figure}[h]
\caption{Illustration: The picture visualizes the situation where a \EM{recolouring scheme}
amounts to a weight preserving bijection between overlays corresponding to \pointarr s
$\setof{\cnr{r}_1,\cnr{r}_2,\cnr{r}_3}$ and $\setof{\cnr{r}_1^\prime,\cnr{r}_2^\prime}$,
respectively. In the picture it is assumed that there are only 4 \pointmatch s for these
\pointarr s, which are indicated by different shades of grey (white, light grey, grey and
black).
The crucial point is that the recolouring scheme assigns every \pointmatch\ 
to a ``bipartite substructure'' with bipartition classes \EM{of the same size} (the edges of these
``bipartite substructures'' are shown as thin black lines). The simplest case of such ``bipartite substructure'' appears for the black \pointmatch\ 
$\cnr{m}_{\text{black}}$: There is a \EM{mapping} taking $\pas{\cnr{r}_3,\cnr{m_{\text{black}}}}$
to $\pas{\cnr{r}_2^\prime,\cnr{m_{\text{black}}}}$
and vice versa. However, this is not the only possibility: Note that, for example, every white \pointmatch\ 
$\cnr{m}_{\text{white}}$ is related to \EM{two} instances of $\cnr{m}_{\text{white}}$:
The corresponding ``bipartite substructure'' has bipartitions classes 
$\setof{\pas{\cnr{r}_1,\cnr{m_{\text{white}}}},\pas{\cnr{r}_3,\cnr{m_{\text{white}}}}}$ and
$\setof{\pas{\cnr{r}_1^\prime,\cnr{m_{\text{white}}}},\pas{\cnr{r}_2^\prime,\cnr{m_{\text{white}}}}}$ (\EM{both of size $2$)}.}
\label{fig:InvolStruct}
\begin{center}
\input graphics/InvolStruct
\end{center}
\end{figure}

\subsubsection{Laplace expansion}
\label{sec:Laplace}
Assume we are given some \pointarr\ $\cnr{r}$ with at least $m$ lower points $s_1,\dots,s_m$.
%and $m>k$ upper points.
%all of which are green.
Consider the following recolouring scheme:
%Always recolour the trail
%starting 
%\bit
%\item in the rightmost \EM{upper} point $s$ which is \EM{inwards} oriented, if there is one,
%\item else in the $j$\bindestrich th \EM{lower} point $s$. % which is \EM{outwards} oriented.
%\eit
Fix lower points $S\defeq\setof{s_{i_1},\dots,s_{i_k}}$ in $\cnr{r}$ ($k\leq m$). For every
\pointmatch\ $\cnr{m}$, reverse  
\bit
\item \EM{one} of the edges of $\cnr{m}$ connecting \EM{two consecutive} coloured upper
	points (we call such edge an upper \EM{handle}), if there is one,
\item else \EM{all} the edges of $\cnr{m}$ starting at a point of $S$.
\eit
(I.e., the family ${\mathcal E}\of{\cnr{r},\cnr{m}}$ of edges assigned to $\pas{\cnr{r},\cnr{m}}$ is either the family
of sets (singletons), where each set contains an upper handle, or it consists of the
single set of edges which are incident with points in $S$.) A moment's thought shows that this
is a recolouring scheme in the sense just described:
We call this the \EM{Laplace recolouring scheme}.

 Applying this scheme for $S=\setof{j}$ (i.e., $k=1$)
to the
\pointconf\ with $m$ upper and $m$ lower points, $m>1$, all of which are coloured green,
% of recolouring the bicoloured trail starting in $s$
implies the Laplace expansion of the $m\times m$\bindestrich determinant by its $j$\bindestrich th column
\begin{equation}
\label{eq:Laplace}
\detof{a} = \sum_{i=1}^m (-1)^{i-j}\cdot\detof{a_{i,j}}\cdot\detof{\deleterowcols{a}{i}{j}}.
\end{equation}
As for Dodgson's condensation, \figref{fig:Laplace} contains the \EM{proof}: The concrete
example presented there illustrates the case $m=4$, $j=1$, and generalizes to a general Schur function identity, which
implies \eqref{eq:Laplace}. (Observe that this gives, in fact, a \EM{simple} recolouring
scheme, since there can be at most \EM{one} upper handle in this case.)

\begin{figure}[!h]
\caption{The proof of the Laplace expansion of the $4\times 4$\bindestrich determinant.}
\label{fig:Laplace}
The proof is contained
in the pictures below. The upper point $s$ of the bicoloured trail to be recoloured is
indicated by a coloured circle,
and the possible connections by bicoloured trails are indicated by grey arcs. Note that
for some partition $\lambda=\pas{\lambda_1,\dots,\lambda_4}$ ($m=4$), %and $\mu=\constpart{1}{4}$,
the involutions
(indicated in the pictures by thick black lines) imply that the generating function of the
left half,
$$
%\schurf_{\addtoparts{\lambda}{1}/{\constpart{1}{m}}}\cdot\schurf_{\pas0} +
\schurf_{\lambda}\cdot\schurf_{\pas0} +
\schurf_{\deletepart{\addtoparts{\lambda}{1}}{2}}\cdot\schurf_{\pas{\lambda_2-1}}+
\schurf_{\deletepart{\addtoparts{\lambda}{1}}{4}}\cdot\schurf_{\pas{\lambda_4-3}},
$$
is equal to the generating function of the right half,
$$
\schurf_{\deletepart{\addtoparts{\lambda}{1}}{1}}\cdot\schurf_{\pas{\lambda_1-1}}+
\schurf_{\deletepart{\addtoparts{\lambda}{1}}{3}}\cdot\schurf_{\pas{\lambda_3-2}}.
$$
By the connections between minors of $\jtmat{\sigma/\tau}$ and operations on partitions
$\sigma$ and $\tau$ (which were explained before Corollary~\ref{cor:Dodgson}), this
means $\detof{\jtmat\lambda}=
\detof{\deleterowcols{\jtmat\lambda}{1}{1}}\cdot\pas{\jtmat\lambda}_{1,1}-
\detof{\deleterowcols{\jtmat\lambda}{2}{1}}\cdot\pas{\jtmat\lambda}_{2,1}+
\detof{\deleterowcols{\jtmat\lambda}{3}{1}}\cdot\pas{\jtmat\lambda}_{3,1}-
\detof{\deleterowcols{\jtmat\lambda}{4}{1}}\cdot\pas{\jtmat\lambda}_{4,1},
$
which implies the Laplace expansion by the first column.
\begin{center}
\input graphics/Laplace
\end{center}
\end{figure}

\subsubsection{Laplace's Theorem}

For the generalization of Laplace's expansion (see \cite[section 93]{muir:1933}),
we introduce the following notation: Let $X=\setof{x_1,\dots,x_m}\subset\Z$ be an ordered
set, and let $S=\setof{x_{i_1},\dots,x_{i_k}}$, $k\leq m$, be a subset. In this
situation, we define
$$\sumset{S}{X}\defeq\sum_{j=1}^k i_j.$$
\begin{thm}[Laplace's Theorem]
\label{thm:Laplace}
Let $a=\pas{a_{i,j}}_{(i,j)\in\range{m}\times\range{m}}$ be an $m\times m$\bindestrich matrix.
Consider some fixed set $I\subseteq\range{m}$.
Then we have
\begin{equation}
\label{eq:laplace-general}
\detof{a} = %\pas{-1}^{\setsum{\range{m}}}\cdot
\sum_{\substack{J\subseteq\range{m},\\\cardof{J}=\cardof{I}}}
\pas{-1}^{\sumset{I}{\range{m}}+\sumset{J}{\range{m}}}\cdot
\detprodcminors{a}{I}{J}.
\end{equation}
\end{thm}

\begin{figure}[h]
\caption{Illustration of Laplace's Theorem, for $m=5$ and $I=\setof{2,4}$: The lower points
corresponding to $I$ are indicated by black circles. The \pointmatch\ shown in the pictures
has two upper handles
(edges connecting neighbouring upper points); handles that have been ``reversed'' (i.e., their
endpoints were recoloured) are drawn in black. The important point is that
the two bipartition classes (corresponding to
the left and the right half of the picture) are of the same size; see also \figref{fig:InvolStruct} where
this is situation appears in a more ``abstract'' way.}
\label{fig:Laplace2}
\begin{center}
\input graphics/Laplace2
\end{center}
\end{figure}

%\begin{figure}
%\caption{Illustration of the ``bipartite structure'' occurring for the generalized
%Laplace recolouring scheme: The picture illustrates that some \pointmatch\ 
%$M$ of some \pointarr\ $R_1$ can be related to \EM{more than} one \pointarr s (in the
%picture, these are $R_1^\prime$ and $R_2^\prime$). If for the corresponding
%bipartite structure connecting objects $\pas{R_1,M},\pas{R_2,M},\dots,\pas{R_i,M}$ and $\pas{R_1^\prime,M},\pas{R_2^\prime,M},\dots,\pas{R_j^\prime,M}$ we have $i=j$ \EM{for all} \pointarr s
%$M$, then clearly we obtain a Schur function identity: It is easy to see that this is the
%case for the generalized
%Laplace recolouring scheme.}
%\label{fig:InvolStruct2}
%\input graphics/InvolStruct2
%\end{figure}

\begin{proof} As always, we consider the equivalent Schur function identity:
The \pointarr\ corresponding to the left\bindestrich hand side of \eqref{eq:laplace-general}
consists of
\bit
\item $m$ upper points, which are outward oriented,
\item and $m$ lower points, which are inward oriented
\eit
(stated otherwise: All $2\cdot m$ coloured points are green). We may assume $0<\cardof{I}<m$ (otherwise, there is nothing to prove),
i.e., $I$ corresponds to $k$ lower points $s_{i_1},\dots,s_{i_k}$, $0<k<m$.

Now we apply the Laplace recolouring scheme.
Note that a \pointmatch\ $\cnr{m}$ \EM{always} contains an upper handle if it contains an edge connecting two
upper points.
If there is more than one upper handle, the recolouring scheme is \EM{not} simple, i.e., it
does \EM{not}
describe a \EM{mapping}, but a \EM{multi\bindestrich valued relation} on pairs $\pas{\cnr{r},\cnr{m}}$, where
$\cnr{r}$ is a \pointarr\ and $\cnr{m}$ is a \pointmatch\ in $\cnr{r}$: It is easy
to see that nevertheless a \EM{bipartite substructure} occurs (see \figref{fig:InvolStruct}): Starting with some arbitrary but fixed pair
$\pas{\cnr{r},\cnr{m}}$, by the Laplace recolouring scheme all the pairs $\pas{\cnr{r}^{\prime\prime},\cnr{m}}$ which
are obtained by reversing an \EM{even} number of upper handles, end up in the same bipartition
class as $\pas{\cnr{r},\cnr{m}}$: Clearly, these are of the same number as all the pairs $\pas{\cnr{r}^{\prime},\cnr{m}}$ which
are obtained by reversing an \EM{odd} number of upper handles (constituting the other bipartition
class). See \figref{fig:Laplace2} for an
illustration.
\end{proof}

Note that the same proof also works for \pointconf s which contain \EM{uncoloured}
points: This amounts to a slightly more general statement (see \cite[section 148]{muir:1933})
\begin{thm}
\label{thm:muir148}
Let $a$ be an $\pas{m+k}\times\pas{m+ k}$\bindestrich matrix, and let
$1\leq i_1<i_2<\cdots<i_m\leq m+k$ and
$1\leq j_1<j_2<\cdots<j_m\leq m+k$ be (the indices of) $k$ fixed rows and
$k$ fixed columns of $a$. Denote the set of these (indices of) rows and columns by $R$ and $C$,
respectively.
%and set $A\defeq\setof{j_1,\dots,j_k}$ and 
%$O\defeq\setof{j_{k+1},\dots,j_{2\cdot k}}$. Let $A^\prime\defeq\range{m+2\cdot k}\setminus A$
%and $O^\prime\defeq\range{m+2\cdot k}\setminus O$.
% Let 
%respectively, of $a$. 
Consider some fixed % list $1\leq r_1<r_2<\cdots<r_k\leq m$
set $I\subseteq R$.
Then we have:
\begin{equation}
\label{eq:muir148}
\detof{a}\cdot\detof{\deleterowcols{a}{R}{C}} = %\pas{-1}^{\setsum{\range{m}}}\cdot
\sum_{\substack{J\subseteq C,\\\cardof{J}=\cardof{I}}}
\pas{-1}^{\sumset{I}{R}+\sumset{J}{C}}\cdot
\detspecminor{a}{\overline{R\setminus I}}{\overline{C\setminus J}}
\cdot
\detspecminor{a}{\overline{I}}{\overline{J}}
.
\end{equation}
\end{thm}

% \alert{Danilov et.al.: \cite{Danilov:2008,Danilov:2010,Danilov:2010a}; setsum redefinieren!!!}

%\subsection{Summary} The examples presented in sections~\ref{sec:Dodgson} and
%\ref{sec:Laplace} illustrate the following general concept. Assume we have some overlay
%of nonintersecting lattice trails corresponding to skew shapes $\lambda/\mu$ and $\sigma/\tau$
%with shift parameters $0$ and $t\in\Z$, respectively. As above, we ``separate'' this
%overlay in the \pointconf\ and the \pointarr. Consider the set $V$ of \EM{all} admissible \pointarr s for this \pointconf, and assume further that we have a \EM{rule} which 
%determines \EM{uniquely} a non\bindestrich empty set $S$ of bicoloured trail for each $v\in V$.
%(For Dodgson's
%condensation, this rule was ``Choose the bicoloured trail starting in the rightmost upper
%point of the overlay''.) Then the operation of recolouring \EM{all} paths in this set
%$S$ is a \EM{fixed\bindestrich point free} and \EM{weight\bindestrich preserving} involution. We call
%such operation a \EM{recolouring scheme}.
%
%%which , we see: If we succeed in devising a scheme $\involution{r}$ of
%%recolouring bicoloured paths,
%%\bit
%%\item which is \EM{well\bindestrich defined} (i.e., it is clear in all cases \EM{which} paths are
%%	to be recoloured), 
%%\item and which implements a \EM{involution},
%%\eit
%%then $\involution{r}$ is weight\bindestrich preserving and we obtain immediately a \EM{generic}
%%identity for skew Schur functions, which is equivalent to the corresponding determinant
%%identity.
% 
%Now we may summmarize our previous considerations as follows (this is a reformulation of
%\cite[Lemma~15]{fulmek:2001}):
%%\begin{dfn}
%%Pair of skew shape with relative shift $t$ may be encoded in
%%configuration of lower/upper points
%%AND
%%admissible circular arrangement of oriented points.
%
%%Admissible matchings of oriented edges derived from bicoloured paths: Changing orientation
%%of edges in such matchings amounts to recolouring the corresponding bicoloured paths.
%%\end{dfn}
%\begin{thm}
%\label{thm:fulmek}
%Let $\lambda/\mu$ and $\sigma/\tau$ be two skew shapes, and let $t$ be an arbitrary
%integer. Consider the \pointconf\ $C$ which corresponds to % the lower/upper points
%%\bit
%%\item
%$\lambda/\mu$ with shift $0$
%%\item
%and $\sigma/\tau$ with shift $t$.
%%\eit
%Consider the set $V$ of \EM{all} admissible \pointarr s corresponding to $C$ %\pointconf\
%and assume we have a recolouring scheme $\involution{r}$.
%%In the corresponding \pointarr,
%%choose a nonempty subset $S$ (arbitrary, but fixed) of the points oriented towards
%%the center.
%
%Consider the graph $G$ with vertex set $V$, where two vertices
%$v_1$, $v_2$ are connected by an edge if and only if there are overlays of lattice
%paths $\pas{P_1,P_2}$ and $\pas{P^\prime_1,P^\prime_2}$ connecting the lower/upper points corresponding to $\pas{C,v_1}$
%and $\pas{C,v_2}$, respectively, such that $\pas{P^\prime_1,P^\prime_2}$
%is obtained from $\pas{P_1,P_2}$ by applying $\involution{r}$. 
%% Then $\involution{r}$ is a fixed\bindestrich point free involution.
%
%If the graph $G$ is \EM{bipartite},
%i.e.,
%$V=E\cup O$ with $E\cap O=\emptyset$,
%such that there is no edge connecting two vertices of $E$ or two vertices of $O$, then
%we have the following identity for skew Schur functions:
%\begin{equation}
%\label{eq:lemma-general}
%\sum_{\pas{\lambda/\mu,\;\sigma/\tau}\in Z_E}
%	\schurf_{\lambda/\mu}\cdot\schurf_{\sigma/\tau}
%=
%\sum_{\pas{\lambda^\prime/\mu^\prime,\;\sigma^\prime/\tau^\prime}\in Z_O}
%	\schurf_{\lambda^\prime/\mu^\prime}\cdot\schurf_{\sigma^\prime/\tau^\prime},
%\end{equation}
%where $Z$ is an arbitrary
%connected component of $G$ with at least $2$ vertices, and where $Z_O$ and $Z_E$
%denote the sets of pairs of skew shapes corresponding to $\pas{C,x}$ for
%$x\in O$ and 
%$x\in E$, respectively.
%\end{thm}
%
%The condition that the graph $G$ described in Theorem~\ref{thm:fulmek} is \EM{bipartite}
% clearly is satisfied in our preceding examples:
%\bit
%\item In the case of the ``Dodgson's recolouring scheme'' (always recolour the bicoloured path ending in the rightmost upper point $s$), the bipartition is determined by the \EM{colour} of
%$s$ (i.e., $O$ is the set where $s$ is red, and $e$ is the set where $s$ is green),
%\item In the case of the ``Laplace recolouring scheme'' (always recolour the bicoloured path
%ending in $s$, where $s$ is the rightmost red upper point if there is one, else $s$ is the rightmost upper point), the bipartition is determined by the \EM{parity} of the number
%of the red upper point, counted from the right; this number is defined as zero if there is
%no red upper point.
%\eit
%
%
%%\begin{figure}
%%\caption{Illustration of the theorem.}
%%\input graphics/TheoremIllustrationSlide
%%\end{figure}
%
%
%\begin{ex}
%To illustrate that Theorem~\ref{thm:fulmek} states a more general identity than Dodgson's condensation and
%Laplace's expansion, we start with the same configuration as for
%our proof of Laplace's expansion, but apply \EM{Dodgson's} recolouring scheme, see
%\figref{fig:LemmaIllustration}: By Observation~\ref{obs:deletepointminors}, the pictures
%there translate immediately to the following expansion of $\detof{a}$ as a signed sum of complementary minors, where
%$a=\jtmat{\lambda/\mu}$:
%\begin{multline*}
%\detof{a} = 
%\prodcminors{a}{1}{1} + \prodcminors{a}{1,3}{1,2} + \prodcminors{a}{1,3}{1,4} +
%\prodcminors{a}{3}{1} - \\
%\prodcminors{a}{1}{2} - \prodcminors{a}{1,3}{2,4} -
%\prodcminors{a}{3}{2} - \prodcminors{a}{1}{4} - \prodcminors{a}{3}{4}.
%\end{multline*}
%%\begin{multline*}
%%\schurf_{\pas{8, 8, 5}/\pas{1}}\cdot\schurf_{\pas{7, 6, 5, 5, 5}/\pas{1, 1, 1}} +
%%\schurf_{\pas{7, 7, 7, 5}/\pas{1, 1}}\cdot\schurf_{\pas{8, 5, 5, 5}/\pas{1, 1}} + \\
%%\schurf_{\pas{7, 7, 4, 4}/\pas{1, 1}}\cdot\schurf_{\pas{8, 7, 6, 6}/\pas{1, 1}}+
%%\schurf_{\pas{8, 8, 5}/\pas{2, 2}}\cdot\schurf_{\pas{7, 6, 5, 5, 5}/\pas{\;}}= \\
%%\schurf_{\pas{8, 7, 5}/\pas{1}}\cdot\schurf_{\pas{7, 7, 5, 5, 5}/\pas{1, 1, 1}}+
%%\schurf_{\pas{8, 4, 4}/\pas{1}}\cdot\schurf_{\pas{7, 7, 7, 6, 6}/\pas{1, 1, 1}}+\\
%%\schurf_{\pas{9, 5}/\pas{2}}\cdot\schurf_{\pas{6, 6, 6, 5, 5, 5}/\pas{\;}}+
%%\schurf_{\pas{8, 7, 5}/\pas{2, 2}}\cdot\schurf_{\pas{7, 7, 5, 5, 5}/\pas{\;}}+\\
%%\schurf_{\pas{7, 6, 4, 4}/\pas{1, 1}}\cdot\schurf_{\pas{8, 8, 6, 6}/\pas{1, 1}}+
%%\schurf_{\pas{8, 4, 4}/\pas{2, 2}}\cdot\schurf_{\pas{7, 7, 7, 6, 6}/\pas{\;}}
%%\end{multline*}
%\end{ex}
%
%\alert{Dodgson\bindestrich type scheme leads to new determinant identity!?}
%
%Beispiel Schurfunktionen\bindestrich Identit\"at:
%\begin{multline}
%\schurf_ {\pas{\lambda_1+2}}\cdot\schurf_{\pas{\lambda_2, \lambda_3,
%\lambda_4}/\pas{1, 1}}+ \\
%\schurf_{\pas{\lambda_1+1,
%\lambda_2+1}/\pas{1}}\cdot\schurf_{\pas{\lambda_3, \lambda_4}/\pas{1}}+ \\
%\schurf_{\pas{\lambda_1+1, \
%\lambda_4-1}/\pas{1}}\cdot\schurf_{\pas{\lambda_2+1,
%\lambda_3+1}/\pas{1}}+ \\
%\schurf_{\pas{\lambda_1}}\cdot\schurf_{\pas{\lambda_2, \lambda_3,
%\lambda_4}}= \\
%\schurf_{\pas{\lambda_2+1}}\cdot\schurf_{\pas{\lambda_1+1, \lambda_3,
%\lambda_4}/\pas{1, 1}}+ \\
%\schurf_{\pas{\lambda_4-1}}\cdot\schurf_{\pas{\lambda_1+1,
%\lambda_2+1, \lambda_3+1}/\pas{1, 1}}+
%\schurf_{\pas{\;}}\cdot\schurf_{\pas{\lambda_1, \lambda_2, \lambda_3,
%\lambda_4}}+ \\
%\schurf_{\pas{\lambda_2-1}}\cdot\schurf_{\pas{\lambda_1+1,
%\lambda_3, \lambda_4}}+ \\
%\schurf_{\pas{\lambda_2,
%\lambda_4-1}/\pas{1}}\cdot\schurf_{\pas{\lambda_1+2,
%\lambda_3+1}/\pas{1}}+ \\
%\schurf_{\pas{\lambda_4-3}}\cdot\schurf_{\pas{\lambda_1+1,
%\lambda_2+1, \lambda_3+1}}
%\end{multline}
%
%€quivalent zu folgender generischen Determinanten\bindestrich Identit\"at (komplementŠre Minoren
%in einer $4\times4$-Matrix):
%\begin{multline}
%\detminor{1}{3}\cdot\detminor{2, 3, 4}{1, 2, 4}+
%\detminor{1, 2}{1, 3}\cdot\detminor{3, 4}{2, 4}+\\
%\detminor{1, 4}{1, 3}\cdot\detminor{2, 3}{2, 4}+
%\detminor{1}{1}\cdot\detminor{2, 3, 4}{2, 3, 4}
%=\\
%\detminor{2}{3}\cdot\detminor{1, 3, 4}{1, 2, 4}+
%\detminor{4}{3}\cdot\detminor{1, 2, 3}{1, 2, 4}+\\
%1\cdot\detminor{1, 2, 3, 4}{1, 2, 3, 4}+
%\detminor{2}{1}\cdot\detminor{1, 3, 4}{2, 3, 4}+\\
%\detminor{2, 4}{1, 3}\cdot\detminor{1, 3}{2, 4}+
%\detminor{4}{1}\cdot\detminor{1, 2, 3}{2, 3, 4}
%\end{multline}
%
%Ebenso: Pl\"ucker\bindestrich Relations (Anfangspunkte identisch, Endpunkte verschieden\dots)
%und Dodgson's Condensation (Einfaches Beispiel aus dem Vortrag).
%
%Und was ist mit Laplace\bindestrich expansion!? Regel: FŠrbe immer den rechtesten schwarzen Endpunkt um, wenn es einen gibt, sonst den rechtesten wei§en Endpunkt. "UmfŠrbere\-geln" dŸrfen auch
%"konfigurationsabhŠngig" sein : wichtig ist aber wohl, da§ der Graph bipartit bleibt!!
%
%Allgemeiner Laplace: Fixiere $m$ Anfangspunkte; Regel: FŠrbe alle \EM{wei§en} fixierten
%Anfangspunkte \EM{und} alle schwarzen Endpunkte um.
%
%PlŸcker relations = Sylvester's Theorem, see \cite[section 137]{muir:1933}!?
%
%% Allgemeine Laplace\bindestrich Entwicklung, see \cite[section 93]{muir:1933}!?
%
%Cauchy\bindestrich Entwicklung \cite[section 110]{muir:1933} mit Minoren und komplementŠren Minoren
%(evtl. Vorzeichen falsch!?):
%
%$$
%\detwhole=\detminor{}{}\cdot\detcominor{}{}=\detminor{r}{s}\cdot\detcominor{r}{s}+
%	\sum_{i\neq s,j\neq r}\detminor{r}{i}\cdot\detcominor{j}{s}.
%$$

\input turnbull