\subsection{Turnbull's identity}
Let $M$ be an $n\times \pas{n\cdot p}$\bindestrich matrix, which we view as ``vector of columns $c_k$'':
% Thus we write
$$
M=\pas{c_1,c_2,\dots,c_{n\cdot p}}.
$$
Consider an arrangement of the ${n\cdot p}$ symbols $c_i$ in a $p\times n$\bindestrich tableau $T$.
Clearly, every such arrangement may be viewed as a \EM{permutation} $\sigma$ of the symbols
in the ``canonical arrangement''
\begin{equation}
\label{eq:canonical}
\begin{pmatrix}
c_1 & c_{p+1} & \cdots & c_{\pas{n-1}\cdot p +1} \\
c_2 & c_{p+2} & \cdots & c_{\pas{n-1}\cdot p +2} \\
\vdots & \vdots & \vdots & \vdots \\
c_p & c_{2\cdot p} & \cdots & c_{n\cdot p}
\end{pmatrix}
\end{equation}
We view $T$ as an ``encoding'' for the product of the minors given by $T$'s
rows, multiplied by the sign of the corresponding permutation $\sigma$. For example, for
$n=3$, $p=2$ we have
$$
\begin{pmatrix}
c_2 & c_3 & c_5 \\
c_1 & c_4 & c_6 \\
\end{pmatrix}
=-\det\of{c_2,c_3,c_5}\times\det\of{c_1,c_4,c_6}.
$$
By abuse of notation, we again call this signed product a \EM{tableau} and denote it by $T$.

Observe that a permutation of elements \EM{from the same row} in $T$ does not change
the value of the corresponding signed product of minors. In this sense, we consider
two tableaux differing only by such ``permutations inside rows'' as \EM{equivalent}:
When we consider sets of tableaux in the following, it goes without saying that these
tableaux are assumed to be \EM{pairwise non\bindestrich equivalent}.

Let $T$ be a $p\times n$ tableau, and let $A$ and $B$ be two disjoint subsets of the
entries of $T$. Imagine that the elements of $A$ are indicated by boxes, and the elements
of $B$ are indicated by circles. Then we will denote by $T^\square$ or $T^\bigcirc$
%% Referee:"respectively"
%respectively,
the sum over all (\EM{pairwise non\bindestrich equivalent}) tableaux which can be
obtained from $T$ by permuting the elements of $A$ or $B$, respectively.
For example, consider the following $2\times3$\bindestrich tableau with $A=\setof{a,b,e,f}$ and
$B=\emptyset$:
\begin{center}
\input graphics/exleclerc1.tex
\end{center}
%In the following, we will also consider only \EM{parts} of such sums, which correspond
%to permutations leaving certain subsets of elements in their original rows. Graphically,
%we shall indicate such ``fixed elements'' by stars. So, the above example with fixed
%subset $\setof{b}$ would give
%\begin{center}
%\input graphics/exleclerc2.tex
%\end{center}

In 1909, Turnbull stated the following generic identity \cite[Proposition 1.2.2]{leclerc:1993};
see also
% (according to \cite{leclerc:1993}, this identity can also be found in
\cite{turnbull:1909},
\cite[p.~48]{turnbull:1960} and \cite[V, p.51]{muir:1906}):
\begin{thm}[Turnbull's identity]
\label{thm:turnbull}
Let $T$ be a $p\times n$\bindestrich tableau, let $A$ be the subset of ``boxed'' entries
of $T$, and set $k\defeq\cardof{A}$.

Denote by $R$ the set of entries from the first row in $T$, and define
the subset of ``circled'' elements
$B\defeq R\setminus A$. 

If $k>n$, then $T^\square\equiv 0$.

If $k\leq n$, we may construct another tableau $T_B$ by exchanging \EM{all} the elements
of $A\setminus R$ with elements of $B$. Set $m\defeq\cardof{A\setminus R}$.
%, and let $B$ be the subset of ``circled'' entries
%of $T_B$.
Then we have
\begin{equation}
\label{eq:idturnbull}
T^\square\equiv\pas{-1}^{m}\cdot T_B^\bigcirc.
\end{equation}
\end{thm}

%Assuming that all marked (i.e., boxed or circled) elements appear in the
%\EM{first} columns in each row, a graphical representation (of a generic example) of Turnbull's
%identity is
%\begin{equation}
%\input graphics/tbleft.tex
%=
%(-1)^5\cdot
%\input graphics/tbright.tex
%.
%\end{equation}
%\vskip0.5cm
%
Of course, we want to present a proof of \eqref{eq:idturnbull} in terms of
recolouring bicoloured trails. So from now on,
%In the following, 
we assume that for every tableaux
$T^\prime$, which is obtained by permuting entries of the ``original'' tableau $T$,
the entries in each row of $T^\prime$ appear in the lexicographic order determined by $T$
(see \eqref{eq:canonical}):
\begin{equation}
\label{eq:lexorder}
T_{i,j} < T_{k,l}:\!\iff j<l \text{ or } \pas{j=l \text{ and } i < k}.
\end{equation}
We may \EM{identify} the $k$\bindestrich th entry $T_{i,j}$ (in this order) of $T^\prime$ with
the $k$\bindestrich th (coloured) endpoint $e$ of lattice paths (ordered from right to left), where
the \EM{colour} of $e$ is determined by the row $l$ in which $T_{i,j}$ appears in $T^\prime$.

We illustrate this by the following example for Turnbull's identity: 
\begin{equation}
\label{eq:exturnbull}
\input graphics/exturnbull-final.tex
\end{equation}
I.e., we consider a $3\times 4$\bindestrich tableau $\tableau$ with $A=\setof{a,b,c,d}$ and
$B=\setof{\alpha,\beta,\gamma}$ (we do not introduce notations for the remaining
entries of $\tableau$ since they are never involved in any permutation of
entries): \figref{fig:exturnbull} shows the tableaux involved in this identity.


From the pictures in \figref{fig:exturnbull} it is not obvious how to 
view Turnbull's identity \EM{directly} as a recolouring scheme: In fact,
our proof takes the following detour.
% In order to prove Turnbull's identity, we start with the following Lemma:
\begin{lem}
\label{lem:turnbull}
Let $T$ be a $p\times n$\bindestrich tableau, let $A\neq\emptyset$ be the subset of ``boxed'' entries
of $T$, and set $k\defeq\cardof{A}$.

Denote by $R$ the set of entries from the first row in $T$, and define
the subset of ``circled'' elements
$B\defeq R\setminus A$. Set $m\defeq\cardof{A\setminus R}$.

%Let $T$ be a $\pas{p\times n}$\bindestrich tableau. Denote by $R$ the set of entries from
%the first row of $T$. Let $A$ be some arbitrary subset of entries
%of ``boxed'' elements with $\cardof{A}=k>0$, and let $B=R\setminus A$ be the
%subset of ``circled'' elements. Denote by $m$ the number of boxed elements % which are
%\EM{not} in the first row of $T$: $m=\cardof{A\setminus R}$. % Assume that $m>0$.

Consider the set %/generating function
$T^{A\cup B}$ of tableaux which are
obtained from $T$ by permuting marked (boxed or circled) elements, i.e., elements
from $A\cup B$.
Denote by $\tbsum{k}$ the % generating function
set of all tableaux
in $T^{A\cup B}$
% which are obtained by permuting the marked (i.e., boxed or circled) entries of $T$ and
% where the sets of circled entries in rows $<1$ is precisely $M$, and
with
\bit
\item exactly $k$ circled entries
\item and exactly $(m-k)$ boxed entries
\eit
in the rows $>1$.
Then we have the following $m$ identities:
\begin{equation}
\label{eq:tubueq}
\sum_{j=0}^{m}\binom{j}{k}\gfof{\tbsum{k}}=0\text{ for } 0\leq k < m.
\end{equation}
\end{lem}

\begin{figure}
\caption{Graphical representations of the tableaux corresponding to
% the left\bindestrich hand side of
\eqref{eq:exturnbull}. In the picture, the tableaux are shown together with
the corresponding coloured endpoints: Endpoints corresponding to the set $A$ are boxed,
endpoints corresponding to the set $B$ are circled. The colour of the endpoints indicates
the row they belong to: Black endpoints belong to the first row, red endpoints belong to
the second row, and green endpoints (shown as dotted points in the pictures)
belong to the third row. The tableaux with positive
sign appear on the left side, and the tableaux with negative sign appear on the right side.}
\label{fig:exturnbull}
\begin{center}
The tableaux corresponding to the left\bindestrich hand side of \eqref{eq:exturnbull}:\par
\input graphics/exturnbullAv2.tex
\par The tableaux corresponding to the  right\bindestrich hand side of \eqref{eq:exturnbull}:\par
\input graphics/exturnbullBv2.tex
\end{center}
\end{figure}

\begin{proof}
View the determinant corresponding to the $i$\bindestrich th row of % some $p\times n$\bindestrich tableau
$T$
as the generating function of nonintersecting lattice paths with starting points
$$
\pas{-1,1},\pas{-2,1},\cdots,\pas{-n+1,1},\dots,\pas{-n,1}
$$
and with ending points
$$
\pas{n\cdot p - i,N},
\pas{\pas{n-1}\cdot p - i,N},
\cdots,
\pas{\pas{2\cdot p} - i,N},
\pas{\pas{1\cdot p} - i,N}.
$$

Consider the set %/generating function
$T^{A\cup B}$ of all tableaux which are
obtained from $T$ by rearranging the marked (boxed or circled) elements
from $A\cup B$. Assume that the order \eqref{eq:lexorder} for the entries of $T$
is inherited for every row in every $T^\prime\in T^{A\cup B}$:
Then each tableau $T^\prime\in T^{A\cup B}$ can be described by the \EM{same} starting and ending points, where only the colours change: For each such $T^\prime$, assign colour $c_i$ to the ending points whose
corresponding entry appears in row $i$ in $T^\prime$; denote these points by $c_i\of{T^\prime}$.
%the the colours of the ending points are determined
%by the row in which $e$ appears in $T^\prime$:
%Imagine that the ending points corresponding to the $i$\bindestrich th row have colour $c_i$:
%Assume $c_1$ is \EM{green}, and call colours $c_2,\dots,c_p$ the \EM{other colours}.
% where the colour of the ending point $e$ is given as
%determined by the row in which the
%corresponding entry (of $T$) appears in $T^\prime$:
%$$
%\text{colour of }e\text{ in } T^\prime = c_i,
%$$
%where $i$ is the number of the row where $e$ appears in $T^\prime$.
%
% it appears in in the
%tableau $\tableau^\prime$.
%
% overlay the $p$-fold product of determinants corresponding to a 
%and alternating ending points;
%assume the ending points corresponding to the first row are green, and the ending points
%corresponding to the second row are red.

For some arbitrary subset $F\subseteq B$ of \EM{circled} entries,
denote by $\tbsum{F}$ the % generating function
family of all tableaux
in $T^{A\cup B}$
where the set of circled entries in rows $>1$ \EM{is precisely} $F$, and denote by
$\tbsub{F}$ the % generating function
family of all tableaux
in $T^{A\cup B}$
where the set of circled entries in rows $>1$ \EM{contains} $F$.

%For fix an arbitrary subset $F\subseteq B$ of \EM{circled} entries with $0\leq\cardof{F}=i<m$ and 
%consider the family $T^{A\cup B}_F\subseteq T^{A\cup B}$ of all tableaux where the entries of $F$ only appear in rows $>1$.
%Stated otherwise: Denote by $c_1\of{\tableau^\prime}$
%the set of elements of colour $c_1$ in $\tableau^\prime$, then
%$F\subseteq B\setminus c_1\of{\tableau^\prime}$ iff $T^\prime\in T^{A\cup B}_F$.
%% (i.e., the ending points corresponding to $F$ are \EM{not} green).

We claim that for an arbitrary subset $F\subseteq B$ with
$0\leq\cardof{F}=k<m$ there holds:
$$
\gfof{\tbsub{F}}\equiv 0.
$$
This is easily established by observing that the recolouring scheme
\begin{quote}
Choose a marked ending point $e\in\pas{A\cup B\setminus c_1\of{\tableau^\prime}}\setminus F$;
assume $e$ has colour $c_i$, $i>1$.
Recolour the bicoloured trail starting in $e$ in the overlay of lattice paths of colours
$c_1$ and $c_i$.
\end{quote}
yields a weight\bindestrich preserving and sign\bindestrich reversing ``involutive bipartite structure'' on
$T^{A\cup B}_F$: Note that there \EM{is} such marked ending point $e$ by assumption, since
$$
\cardof{F} < m = \cardof{A\cup B\setminus c_1\of{\tableau^\prime}}.
$$

%%denote the sum of (pairwise non\bindestrich equivalent)
%%tableaux (obtained by permuting the entries of $T$),
%\bit
%\item for which \EM{all} elements in the first row are marked (boxed or circled)
%\item and for which all the elements of $C$ appear in the \EM{first} row
%\eit
%by $\tbbsum{C}$. (Clearly, $T$ itself meets these conditions.)
%
%We claim that the recolouring rule
%\begin{quote}
%``Choose a bicoloured path $b$ connecting a \EM{marked (i.e., boxed or coloured)} red point
%with a green point which is \EM{not} in $C$
%and recolour $b$''
%\end{quote}
%yields a weight\bindestrich preserving and sign\bindestrich reversing involution for the overlays of
%non\bindestrich intersecting lattice paths corresponding to these tableaux. Note that \EM{at least} one
%such a path does always exist, since the number of marked red points is $m>\cardof{C}=i$ by assumption; if there are more, then {\red the bipartite structure is multi\bindestrich valued.}
%
%Thus we have $\tbbsum{C}\equiv 0$.

Thus we obtain by summing and elementary counting
$$
0 = 
\sum_{\substack{F\subseteq B,\\ \cardof{F}=k}}\gfof{\tbsub{F}} =
\sum_{\substack{M\subseteq B,\\ \cardof{M}=j\geq k}}\binom{j}{k}\gfof{\tbsum{M}} = 
\sum_{j=0}^{m}\binom{j}{k}\gfof{\tbsum{j}}.
$$
\end{proof}

\begin{cor}
\label{cor:turnbull}
Under the assumptions and with the definitions from Lemma~\ref{lem:turnbull} we have
\begin{equation*}
\tbsum{0} = \pas{-1}^{m}\tbsum{m},
\end{equation*}
which amounts to Turnbull's identity.
\end{cor}
\begin{proof}
The alternating sum of equations \eqref{eq:tubueq} gives
\begin{align*}
0 &= \sum_{i=0}^{m-1}\pas{-1}^i\sum_{j=0}^{m}\binom{j}{i}\tbsum{j} \\
  &= \sum_{j=0}^{m}\tbsum{j}\pas{\sum_{i=0}^{m-1}\pas{-1}^i\binom{j}{i}}\\
  &= \tbsum{0}-\pas{-1}^{m}\tbsum{m}.
\end{align*}
(Note that $\tbsum{m}=0$ if $k>n$.)
\end{proof}
%\begin{figure}
%\caption{Graphical representations of the tableaux corresponding to the right\bindestrich hand
%side of \eqref{eq:exturnbull}. In the picture, the tableaux are shown together with
%the corresponding coloured endpoints: Endpoints corresponding to the set $A$ are boxed,
%endpoints corresponding to the set $B$ are circled. The colour of the endpoints indicates
%the row they belong to: Black endpoints belong to the first row, red endpoints belong to
%the second row, and green endpoints belong to the third row. The tableaux with positive
%sign appear on the left, and the tableaux with negative sign appear on the right.} 
%\begin{center}
%\input graphics/exturnbullB.tex
%\end{center}
%\end{figure}
%
\endinput
