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\title{\textbf{A Simple Proof of an Identity Concerning Pfaffians of Skew Symmetric Matrices}}

\author{Andreas W. M. Dress\\
\textit{Fakult\"at f\"ur Mathematik, Universit\"at Bielefeld,}\\
\textit{Postfach 10 01 31, D-33501 Bielefeld, Germany}
\\[8pt]
\textsc{and}
\\[8pt]
Walter Wenzel\\
\textit{Labri, Universit\'e Bordeaux I, F-33405 Talence Cedex, France}}

\date{Published in: \textit{Advances in Mathematics} \textbf{112} (1995), pp.~120--134.}

\begin{document}

\maketitle

\noindent
Digitized with GPT-5.5 based on \href{https://doi.org/10.1006/aima.1995.1029}{a scan} from \href{https://www.elsevier.com/about/policies-and-standards/open-access-licenses/elsevier-user}{Elsevier Open Archive} \\(original scan: \href{https://doi.org/10.1006/aima.1995.1029}{doi:10.1006/aima.1995.1029}). \\
This is an unofficial re-edition of an article that appeared in an Elsevier publication. Elsevier has not endorsed this re-edition.\\
Proofread by Darij Grinberg 1 July 2026.

\begin{abstract}
In the present paper we prove an identity concerning Pfaffians similar to the well-known Grassmann--Pl\"ucker relations for determinants, using tools from multilinear algebra. More precisely, we shall derive the identity as a corollary to an equation concerning skew-symmetric bilinear forms and operators, acting on the exterior algebra $\Lambda(M)$ of an $R$-module $M$ over some fixed commutative ring $R$ with $1\in R$.
\end{abstract}

\section{Introduction}\label{sec:introduction}

The Pfaffian $\Pf(A)=\Pf_n(A)=\Pf_n^R(A)$ of a skew-symmetric\cdg{A \emph{skew-symmetric} matrix here means a square matrix $A$ that satisfies $A^T = -A$ and whose diagonal entries are $0$. Several sources refer to such matrices as \emph{alternating} instead.} $(n\times n)$-matrix
\begin{equation}
A=
\begin{pmatrix}
0 & a_{12} & a_{13} & \cdots & a_{1n}\\
-a_{12} & 0 & a_{23} & \cdots & a_{2n}\\
-a_{13} & -a_{23} & 0 & & \vdots\\
\vdots & \vdots & & \ddots & a_{n-1,n}\\
-a_{1n} & -a_{2n} & \cdots & -a_{n-1,n} & 0
\end{pmatrix}
\tag{1.1}\label{eq:1.1}
\end{equation}
with coefficients $a_{ij}$ ($1\leq i<j\leq n$) in some commutative ring $R$ has played an important r\^ole in many branches of mathematics, in particular in determinant theory, in symplectic geometry and in differential geometry (see for example \cite[Chapter~III]{A}; \cite[Kapitel~9]{K}; \cite{St}; \cite[Sect.~34]{DFN}; and \cite[Chap.~13, Sect.~3]{Sp}).

There are many ways to define the Pfaffian. A rather conceptual way is the following:

Let $\Skew_n:\mathbf{rings}\to\mathbf{sets}$ denote the covariant functor from the category $\mathbf{rings}$ of commutative rings into the category $\mathbf{sets}$ of all sets which associates to each commutative ring $R$ the set $\Skew_n(R)$ of all skew-symmetric $(n\times n)$-matrices over $R$ and let $\Forget:\mathbf{rings}\to\mathbf{sets}$ denote the forgetful functor from rings into sets which associates to a commutative ring $R$ the set $R$ itself. Then
\[
\Pf=\Pf_n=(\Pf_n^R)_{R\in\mathbf{rings}}
\]
can be defined as the unique natural transformation from $\Skew_n$ back into $\Forget$ such that
\begin{equation}
\bigl(\Pf_n^R(A)\bigr)^2=\det(A)
\tag{1.2}\label{eq:1.2}
\end{equation}
holds for every commutative ring $R$ and every $A\in\Skew_n(R)$, normalized in case $n\equiv0\mod 2$ such that for
\[
A:=A_0:=
\begin{pmatrix}
0 & I_{n/2}\\
-I_{n/2} & 0
\end{pmatrix}
\]
($I_m$ denoting the identity $(m\times m)$-matrix) one has\cdg{The formula \eqref{eq:1.3} has been corrected (the original had $1$ on the right hand side).}
\begin{equation}
\Pf(A_0):=(-1)^{(n/2)(n/2-1)/2}.
\tag{1.3}\label{eq:1.3}
\end{equation}
This means that the natural transformation $\Pf^2:\Skew_n\to\Forget$ which maps for every $R$ a matrix $A\in\Skew_n(R)$ onto $\bigl(\Pf_n^R(A)\bigr)^2$ coincides with the natural transformation ``det'' from $\Skew_n$ into $\Forget$, given by forming the determinant.

In case $n\equiv1\mod 2$ we have of course $\Pf(A)=0$ for all $A\in\Skew_n(R)$.

In case $n\equiv0\mod 2$ one can describe the Pfaffian equivalently as a polynomial
\[
\Pf(X_{12},X_{13},\ldots,X_{1n},X_{23},\ldots,X_{2n},\ldots,X_{n-1,n})
\in\ZZ[X_{12},X_{13},\ldots,X_{n-1,n}]
\]
such that for each commutative ring $R$ and each $A\in\Skew_n(R)$ of the form \eqref{eq:1.1} one has
\begin{equation}
\Pf(a_{12},a_{13},\ldots,a_{n-1,n})^2=\det(A),
\tag{1.2$'$}\label{eq:1.2prime}
\end{equation}
normalized such that\cdg{The formula \eqref{eq:1.3prime} has been corrected (the original had $1$ on the right hand side).}
\begin{equation}
\Pf(a_{12},a_{13},\ldots,a_{n-1,n}):=(-1)^{(n/2)(n/2-1)/2}
\quad\text{for}\quad
a_{ij}:=
\begin{cases}
1,&\text{for }j-i=\dfrac n2,\\
0,&\text{otherwise.}
\end{cases}
\qquad
\tag{1.3$'$}\label{eq:1.3prime}
\end{equation}

Much more constructively, on the other hand, one can define, for given $R$, even $n$, and $A\in\Skew_n(R)$ of the form \eqref{eq:1.1}, the value of $\Pf_n(A)$ in a recursive way by
\begin{equation}
\Pf_2\!\left(\begin{pmatrix}0&a_{12}\\-a_{12}&0\end{pmatrix}\right):=a_{12}
\qquad(a_{12}\in R)
\tag{1.4}\label{eq:1.4}
\end{equation}
for $n=2$ and by
\begin{equation}
\Pf_n(A):=\sum_{j=2}^{n}(-1)^j \cdot a_{1j} \cdot \Pf_{n-2}\bigl(A^{(j)}\bigr)
\qquad\text{for }n\geq4,
\tag{1.5}\label{eq:1.5}
\end{equation}
where $A^{(j)}$ denotes the $(n-2)\times(n-2)$-matrix one gets from $A$ by eliminating the first and the $j$th row and column\cdg{I would include the base case
\begin{equation}
\Pf_0\left(0\times 0\text{-matrix}\right) := 1
\tag{1.4$'$}\label{eq:1.4prime}
\end{equation}
for the sake of completeness. This way, \eqref{eq:1.4} becomes a particular case of \eqref{eq:1.5}.}.

Of course, whatever definition one starts with, the point which needs proof is that both definitions are indeed equivalent, that is, that a natural transformation as described in the first definition exists%
\cdg{The uniqueness of this natural transformation follows from what was said above about polynomials:
Any natural transformation from $\Skew_n$ to $\Forget$ is uniquely determined by its value on the ``generic skew-symmetric matrix''
\[
X=
\begin{pmatrix}
0 & X_{12} & X_{13} & \cdots & X_{1n}\\
-X_{12} & 0 & X_{23} & \cdots & X_{2n}\\
-X_{13} & -X_{23} & 0 & & \vdots\\
\vdots & \vdots & & \ddots & X_{n-1,n}\\
-X_{1n} & -X_{2n} & \cdots & -X_{n-1,n} & 0
\end{pmatrix}
\]
over the polynomial ring $\ZZ[X_{12},X_{13},\ldots,X_{n-1,n}]$.
For the natural transformation $\Pf_n$, this value must be a polynomial $\Pf_n(X)$ that squares to $\det X$.
This uniquely determines this polynomial up to sign (if such a polynomial exists).
For $n \equiv 0 \mod 2$, the sign is then uniquely determined by the additional requirement \eqref{eq:1.3prime}.
For $n \equiv 1 \mod 2$, the sign does not matter, since a classical argument shows that $\det X = 0$ (in fact, $X^T = -X$ and thus $\det\left(X^T\right) = \det\left(-X\right) = -\det X$ because $n$ is odd, and therefore $-\det X = \det\left(X^T\right) = \det X$, so that $\det X = 0$), and the zero polynomial $0$ has only one square root, which of course is $0$.
Thus, in both cases, the polynomial $\Pf_n(X)$ is uniquely determined (if it exists); and this, in turn, determines the whole natural transformation $\Pf_n$ (since any skew-symmetric matrix $A$ is the image of $X$ under an appropriate ring homomorphism from $\ZZ[X_{12},X_{13},\ldots,X_{n-1,n}]$ to $R$).}
and can be constructed according to the second one. In this note we shall provide such a proof by starting with still another definition, based on exterior algebra concepts. The reason for doing this, though, is not so much to provide a rather nice and conceptual proof of the above-mentioned facts but rather to establish---in the course of doing this---a certain identity concerning the Pfaffian which can be viewed as a straightforward generalization of the famous Grassmann--Pl\"ucker identities.

To be more explicit, consider as above a skew-symmetric $(n\times n)$-matrix $A$ of the form \eqref{eq:1.1} with coefficients in some commutative ring $R$---with $n$ even or odd---and for each subset $I\subseteq\{1,\ldots,n\}$ of cardinality, say, $m$ let $A(I)$ denote the skew-symmetric $(m\times m)$-matrix one gets from $A$ by eliminating all rows and columns not indexed by indices from $I$. Moreover, put
\begin{equation}
p(I):=\Pf(A(I))
\qquad\text{for every }I\subseteq\{1,\ldots,n\},
\tag{1.6}\label{eq:1.6}
\end{equation}
where---as usual---$\Pf(A(\varnothing)):=1$. Then the following result holds:

\begin{theoremone}\refstepcounter{thmone}\label{thm:pfaffian-identity}
For any two subsets $I_1,I_2\subseteq\{1,\ldots,n\}$ of odd cardinality and elements $i_1,\ldots,i_t\in\{1,\ldots,n\}$ with $i_1<i_2<\cdots<i_t$ and
\[
\{i_1,i_2,\ldots,i_t\}=I_1\symdiff I_2:=(I_1\setminus I_2)\cup(I_2\setminus I_1)
\]
one has
\begin{equation}
\sum_{\tau=1}^{t}(-1)^\tau\cdot p(I_1\symdiff\{i_\tau\})\cdot p(I_2\symdiff\{i_\tau\})=0.
\tag{1.7}\label{eq:1.7}
\end{equation}
\end{theoremone}

Note that in the case in which $A$ is a skew-symmetric $(3m)\times(3m)$-matrix of the form
\begin{align*}
A
&=
\left(
\begin{array}{ccc@{\qquad}ccc}
 & & & a_{1,m+1} & \cdots & a_{1,3m}\\
 & 0 & & \vdots & & \vdots\\
 & & & a_{m,m+1} & \cdots & a_{m,3m}\\
-a_{1,m+1} & \cdots & -a_{m,m+1} & & &\\
\vdots & & \vdots & & 0 &\\
-a_{1,3m} & \cdots & -a_{m,3m} & & &
\end{array}
\right) ,
% \\
% &=
% \begin{pmatrix}
% 0_{m\times m} & B\\
% -B^{\mathsf T} & 0_{2m\times2m}
% \end{pmatrix},
\end{align*}
% where
% \[
% B=
% \begin{pmatrix}
% a_{1,m+1}&\cdots&a_{1,3m}\\
% \vdots&&\vdots\\
% a_{m,m+1}&\cdots&a_{m,3m}
% \end{pmatrix},
% \]
$I_1$ equals $\{1,\ldots,2m+1\}$, and $I_2$ equals $\{1,\ldots,m,2m+2,\ldots,3m\}$, the above identity indeed yields the Grassmann--Pl\"ucker identities\cdg{This conclusion rests on the known fact that every $m\times m$-matrix $B$ satisfies
\[
\Pf_{2m} \left( \begin{pmatrix} 0 & B \\ -B^T & 0 \end{pmatrix} \right) = (-1)^{m(m-1)/2} \det B .
\]
} (for more details see \cite[Sect.~7]{W4}).

The above identity was discovered by the second named author in \cite[Proposition~2.3]{W1} while trying to understand the geometric algebra associated with $\Delta$-matroids (cf.~\cite{B1,B2,BDH,DH,W2,W3,W4}). Though the identity looks very ``classical,'' we have sofar been unable to find any reference to it in the existing literature. While the original proof given in \cite{W1} and based on the recursive definition of the Pfaffian outlined above is rather computational, the new proof presented here is more conceptual and, we hope, transparent---at least, once one is willing to accept\cdg{In the original, ``except''.} exterior algebra concepts as something of that sort.

The outline of the paper is as follows:

In Section~\ref{sec:some-exterior-algebra} we present the basic exterior algebra machinery we shall need later on. In Section~\ref{sec:exterior-algebra-skew-bilinear} we associate with any skew-symmetric bilinear form $\beta:M\times M\to R$, defined on some $R$-module $M$ over a commutative ring $R$, a canonical $R$-linear map, also called $\beta$, from the exterior algebra $\Lambda(M)$ of $M$ back into $R$ such that $\beta(m\wedge m')=\beta(m,m')$ for all $m,m'\in M$, and we establish

\begin{theoremtwo}
For all $m_1,\ldots,m_n\in M$ one has
\[
\beta(m_1\wedge\cdots\wedge m_n)
=\Pf\bigl((\beta(m_i,m_j))_{i,j=1,\ldots,n}\bigr).
\]
\end{theoremtwo}

Here we assume the Pfaffian to be defined recursively according to \eqref{eq:1.4} and \eqref{eq:1.5}.

In Section~\ref{sec:pfaffian-identity} we prove a simple identity between the values of $\beta$, namely

\begin{theoremthree}
For all $m_1,\ldots,m_k,n_1,\ldots,n_l\in M$ one has
\begin{align*}
&\sum_{i=1}^{k}(-1)^i\cdot \beta(m_1\wedge\cdots\wedge m_{i-1}\wedge m_{i+1}\wedge\cdots\wedge m_k)
  \cdot\beta(m_i\wedge n_1\wedge\cdots\wedge n_l)\\
&\quad=-\sum_{j=1}^{l}(-1)^j\cdot\beta(n_1\wedge\cdots\wedge n_{j-1}\wedge n_{j+1}\wedge\cdots\wedge n_l)
  \cdot\beta(n_j\wedge m_1\wedge\cdots\wedge m_k).
\end{align*}
\end{theoremthree}

Moreover, we will show that Theorem~\ref{thm:pfaffian-identity} is a simple consequence of Theorem~\ref{thm:beta-identity}, thus establishing the primary goal of this note. Even so, we go on in Section~\ref{sec:pfaffian-determinant} to show as promised how the exterior algebra machinery we have set up before can also be used to give a rather conceptual proof of the identity \eqref{eq:1.2}, where again we assume that the Pfaffian is defined recursively according to \eqref{eq:1.4} and \eqref{eq:1.5}.

\section{Some Exterior Algebra}\label{sec:some-exterior-algebra}

As above, let $R$ denote a commutative ring and let $M$ be a left $R$-module. Adopting standard notations we denote by $\Lambda(M)$ the \emph{exterior algebra} of $M$, which we assume to contain $R$ as a subring and $M$ as an $R$-submodule so that---with ``$\wedge$'' denoting the multiplication in the exterior algebra---$\Lambda(M)$ can be considered as a \emph{graded $R$-algebra}
\begin{equation}
\Lambda(M)=\bigoplus_{k=0}^{\infty}\Lambda_k(M)
=\Lambda_0(M)\oplus\Lambda_1(M)\oplus\cdots\oplus\Lambda_k(M)\oplus\cdots
\tag{2.1}\label{eq:2.1}
\end{equation}
with $\Lambda_0(M)=R$, $\Lambda_1(M)=M$, and $\Lambda_k(M)$ the $R$-submodule of $\Lambda(M)$ generated\cdg{Removed the comma before ``generated'' to make this less ambiguous.} by all products of the form $m_1\wedge\cdots\wedge m_k$ with $m_1,\ldots,m_k\in M$. As usual, we also put
\begin{equation}
\Lambda_+(M):=\bigoplus_{k\equiv0\bmod 2}\Lambda_k(M)
\tag{2.2}\label{eq:2.2}
\end{equation}
and
\begin{equation}
\Lambda_-(M):=\bigoplus_{k\equiv1\bmod 2}\Lambda_k(M),
\tag{2.3}\label{eq:2.3}
\end{equation}
so that
\begin{equation}
\Lambda(M)=\Lambda_+(M)\oplus\Lambda_-(M)
\tag{2.4}\label{eq:2.4}
\end{equation}
can also be viewed as a bigraded algebra\cdg{The word ``bigraded'' means ``$\mathbb{Z}_2$-graded'' in this paper.}.

For $x\in\Lambda(M)$ and $k\in\NN_0$ we denote by $x^{(k)}\in\Lambda_k(M)$ its component in $\Lambda_k(M)$, that is, the unique element in $\Lambda_k(M)$ such that $x-x^{(k)}\in\bigoplus_{i\ne k}\Lambda_i(M)$. Hence, $x^{(k)}=0$ for almost all $k\in\NN_0$ and $x=\bigoplus_{k=0}^{\infty}x^{(k)}$. Similarly, we put
\begin{align*}
x^+ &:= \bigoplus_{k\equiv0\bmod2}x^{(k)}\in\Lambda_+(M),
\\
x^- &:= \bigoplus_{k\equiv1\bmod2}x^{(k)}\in\Lambda_-(M),
\end{align*}
and $\bar x:=x^+-x^-$, so that $x=x^+\oplus x^-$ is the decomposition of $x$ relative to the bigraded structure \eqref{eq:2.4} of $\Lambda(M)$, and one has $\bar{\bar x}=x$ as well as
\[
\overline{x\wedge y}=\bar x\wedge\bar y
\qquad\text{for all }x,y\in\Lambda(M).
\]

Recall that the $R$-multilinear map
\[
\lambda_k:M^k\to\Lambda_k(M):(m_1,\ldots,m_k)\mapsto m_1\wedge\cdots\wedge m_k
\]
satisfies $\lambda_k(m_1,\ldots,m_k)=0$ whenever $m_i=m_{i+1}$ for some $i\in\{1,\ldots,k-1\}$, and that for every other $R$-multilinear map $\lambda:M^k\to N$ from $M^k$ into some $R$-module $N$ with $\lambda(m_1,\ldots,m_k)=0$ whenever $m_i=m_{i+1}$ for some $i\in\{1,\ldots,k-1\}$ there exists a unique \emph{induced} $R$-linear map $\widehat{\lambda}:\Lambda_k(M)\to N$ such that the diagram
\[
\begin{tikzcd}[column sep=huge,row sep=large]
& M^k \arrow[dl,"\lambda_k"'] \arrow[dr,"\lambda"] & \\
\Lambda_k(M) \arrow[rr,dashed,"\widehat{\lambda}"'] && N
\end{tikzcd}
\]
commutes.
Equivalently, given an $R$-bilinear map $\lambda:M\times\Lambda_{k-1}(M)\to N$, there exists a unique induced $R$-linear map $\widehat{\lambda}:\Lambda_k(M)\to N$ with $\lambda(m,x)=\widehat{\lambda}(m\wedge x)$ for all $m\in M$ and $x\in\Lambda_{k-1}(M)$ if and only if $\lambda(m,m\wedge x)=0$ for all $m\in M$ and $x\in\Lambda_{k-2}(M)$.\ \ \ \ \cdg{Indeed, the ``only if'' part is clear (since $m \wedge (m\wedge x) = 0$), whereas the ``if'' part is an easy consequence of the previous sentence: If $\lambda(m,m\wedge x)=0$ for all $m\in M$ and $x\in\Lambda_{k-2}(M)$, then the $R$-multilinear map $\lambda' : M^k \to N$ sending each $(m_1, \ldots, m_k) \in M^k$ to $\lambda(m_1, m_2 \wedge \cdots \wedge m_k) \in N$ satisfies the property that $\lambda'(m_1,\ldots,m_k)=0$ whenever $m_i=m_{i+1}$ for some $i\in\{1,\ldots,k-1\}$ (this is clear for $i > 1$, and follows from the $\lambda(m,m\wedge x)=0$ assumption when $i = 1$), and thus induces a unique $R$-linear map $\widehat{\lambda'} : \Lambda_k(M) \to N$; this latter map $\widehat{\lambda'}$ is precisely the $\widehat{\lambda}$ that we are looking for.}
This observation is the key for many recursive constructions of maps defined on $\Lambda(M)$: For instance, if $\varphi:M\to\Lambda(M)$ is an $R$-linear map, then there exists a unique $R$-linear extension $\widehat{\varphi}:\Lambda(M)\to\Lambda(M)$ satisfying $\widehat{\varphi}(R)=\{0\}$ and
\begin{equation}
\widehat{\varphi}(m\wedge x)=\varphi(m)\wedge x-m\wedge\widehat{\varphi}(x)
\quad\text{for all }m\in M\text{ and }x\in\Lambda(M)
\tag{2.5}\label{eq:2.5}
\end{equation}
if and only if
\begin{equation}
\varphi(m)\wedge m=m\wedge\varphi(m)
\quad\text{for all }m\in M.
\tag{2.6}\label{eq:2.6}
\end{equation}
Indeed, putting $x:=m$ in \eqref{eq:2.5} yields that \eqref{eq:2.6} must hold if $\varphi$ can be extended to some $\widehat{\varphi}:\Lambda(M)\to\Lambda(M)$ such that \eqref{eq:2.5} is satisfied. Vice versa, if \eqref{eq:2.6} holds then $\widehat{\varphi}$ can be constructed recursively on each $\Lambda_k(M)$, $k\geq2$, so that \eqref{eq:2.5} holds, because, independent of the actual value of $\widehat{\varphi}(x)$, \eqref{eq:2.6} implies
\[
\varphi(m)\wedge(m\wedge x)-m\wedge\bigl(\varphi(m)\wedge x-m\wedge\widehat{\varphi}(x)\bigr)=0
\]
for all $m\in M$ and $x\in\Lambda(M)$.

Note that \eqref{eq:2.5} implies
\begin{align}
\widehat{\varphi}(m_1\wedge\cdots\wedge m_k)
&=\sum_{i=1}^{k}(-1)^{i-1}m_1\wedge\cdots\wedge m_{i-1}\wedge\varphi(m_i)
  \wedge m_{i+1}\wedge\cdots\wedge m_k\notag\\
&=\sum_{i=1}^{k}\overline{m_1\wedge\cdots\wedge m_{i-1}}\wedge\varphi(m_i)
  \wedge m_{i+1}\wedge\cdots\wedge m_k
\tag{2.7}\label{eq:2.7}
\end{align}
for all $m_1,\ldots,m_k\in M$ and therefore also
\begin{equation}
\widehat{\varphi}(x\wedge y)=\widehat{\varphi}(x)\wedge y+\bar x\wedge\widehat{\varphi}(y)
\quad\text{for all }x,y\in\Lambda(M).
\tag{2.8}\label{eq:2.8}
\end{equation}
\cdg{A linear map $f : \Lambda(M) \to \Lambda(M)$ satisfying $f(x \wedge y) = f(x) \wedge y + \bar x \wedge f(y)$ for all $x, y \in \Lambda(M)$ is sometimes called a \emph{superderivation} of $\Lambda(M)$. Thus, \eqref{eq:2.8} shows that $\widehat{\varphi}$ is a superderivation of $\Lambda(M)$.}

Note also that for any two elements $r_1,r_2\in R$ and $R$-linear maps
\[
\varphi_1,\varphi_2:M\to\Lambda(M)
\]
satisfying \eqref{eq:2.6}, $R$-linearity of the formulas \eqref{eq:2.5} and \eqref{eq:2.6} ensures also that $\varphi:=r_1\cdot \varphi_1+r_2\cdot \varphi_2:M\to\Lambda(M)$ satisfies \eqref{eq:2.6} and that
\begin{equation}
\widehat{\varphi}=r_1\cdot \widehat{\varphi_1}+r_2\cdot \widehat{\varphi_2}.
\tag{2.9}\label{eq:2.9}
\end{equation}

Finally, note that \eqref{eq:2.6} holds in particular for all $R$-linear maps $\varphi:M\to R\subseteq\Lambda(M)$, so definitely all these maps can be extended as described above. For such maps $\varphi$ we evidently have
\begin{equation}
\widehat{\varphi}(\Lambda_k(M))\subseteq\Lambda_{k-1}(M)
\quad\text{for all }k\in\NN.
\tag{2.10}\label{eq:2.10}
\end{equation}
They also satisfy
\begin{equation}
\widehat{\varphi}(\varphi(m))=0
\qquad\text{and}\qquad
\overline{\varphi(m)}=\varphi(m)
\quad\text{for each }m\in M.
\tag{2.11}\label{eq:2.11}
\end{equation}
These conditions in turn imply\cdg{This is all under the condition that $\varphi : M \to R$.}
\begin{equation}
\widehat{\varphi}(\widehat{\varphi}(x))=0
\quad\text{for all }x\in\Lambda(M),
\tag{2.12}\label{eq:2.12}
\end{equation}
\cdg{Thus, $\widehat{\varphi}$ can be used to define a chain complex
\[
\cdots \overset{\widehat{\varphi}}{\longrightarrow} \Lambda_2(M) \overset{\widehat{\varphi}}{\longrightarrow} \Lambda_1(M) \overset{\widehat{\varphi}}{\longrightarrow} \Lambda_0(M) \longrightarrow 0.
\]
}
as follows directly by induction on
\[
\min\bigl\{k\in\NN_0\mid x^{(\ell)}=0\text{ for all }\ell>k\bigr\}
\]
\cdg{Typo corrected in the above display, and $\min(...)$ replaced by $\min\{...\}$.} in view of the formula
\begin{align}
\widehat{\varphi}(\widehat{\varphi}(m\wedge x))
&=\widehat{\varphi}\bigl(\varphi(m)\wedge x-m\wedge\widehat{\varphi}(x)\bigr)\notag\\
&=\widehat{\varphi}(\varphi(m)\wedge x)-\widehat{\varphi}(m\wedge\widehat{\varphi}(x))\notag\\
&=\widehat{\varphi}(\varphi(m))\wedge x+\overline{\varphi(m)}\wedge\widehat{\varphi}(x) \notag\\
& \qquad
  -\varphi(m)\wedge\widehat{\varphi}(x)+m\wedge\widehat{\varphi}(\widehat{\varphi}(x))
\tag{2.13}\label{eq:2.13}
\end{align}
\cdg{Corrected the sign before the $m\wedge\widehat{\varphi}(\widehat{\varphi}(x))$ term (it was a $-$ in the original).}
for all $m\in M$ and $x\in\Lambda(M)$.

Hence, for $R$-linear maps\cdg{Added ``$R$-linear maps''.} $\varphi_1,\varphi_2:M\to R$ and $\varphi:=\varphi_1+\varphi_2$ we have
\begin{align}
0&=\widehat{\varphi}\circ\widehat{\varphi}
=(\widehat{\varphi_1}+\widehat{\varphi_2})\circ(\widehat{\varphi_1}+\widehat{\varphi_2})\notag\\
&=\widehat{\varphi_1}\circ\widehat{\varphi_1}+\widehat{\varphi_1}\circ\widehat{\varphi_2}
 +\widehat{\varphi_2}\circ\widehat{\varphi_1}+\widehat{\varphi_2}\circ\widehat{\varphi_2}\notag\\
&=\widehat{\varphi_1}\circ\widehat{\varphi_2}+\widehat{\varphi_2}\circ\widehat{\varphi_1}
\tag{2.14}\label{eq:2.14}
\end{align}
and therefore
\begin{equation}
\widehat{\varphi_1}\circ\widehat{\varphi_2}=-\widehat{\varphi_2}\circ\widehat{\varphi_1}.
\tag{2.15}\label{eq:2.15}
\end{equation}

Similarly, \eqref{eq:2.7} implies by induction relative to $k$ that for $m_1,\ldots,m_k\in M$ and for $R$-linear maps\cdg{Added ``for $R$-linear maps''.} $\varphi_1,\ldots,\varphi_k:M\to R$ one has
\begin{equation}
\widehat{\varphi_k}\circ\cdots\circ\widehat{\varphi_1}(m_1\wedge\cdots\wedge m_k)
=\det\bigl((\varphi_i(m_j))_{i,j=1,\ldots,k}\bigr)
\tag{2.16}\label{eq:2.16}
\end{equation}
and therefore, using \eqref{eq:2.15},
\begin{equation}
\widehat{\varphi_1}\circ\cdots\circ\widehat{\varphi_k}(m_1\wedge\cdots\wedge m_k)
=(-1)^{k(k-1)/2}\det\bigl((\varphi_i(m_j))_{i,j=1,\ldots,k}\bigr).
\tag{2.17}\label{eq:2.17}
\end{equation}

\section{Exterior Algebra and Skew-Symmetric Bilinear Forms}\label{sec:exterior-algebra-skew-bilinear}

Continuing with our notations let us now assume that in addition to $R$ and $M$ we are given a skew-symmetric $R$-bilinear form\cdg{A \emph{skew-symmetric $R$-bilinear form} (also known as an \emph{alternating $R$-bilinear form}) means an $R$-bilinear form $\beta:M\times M\to R$ that satisfies $\beta(m,m) = 0$ for each $m \in M$. This automatically implies $\beta(m,n) = -\beta(n,m)$ for all $m,n \in M$. Each skew-symmetric $R$-bilinear form $\beta:M\times M\to R$ can be written as $\beta'\circ\lambda_2$ for a unique $R$-linear map $\beta':\Lambda_2(M) \to R$. By abuse of notation, we denote the latter $\beta'$ by $\beta$ as well.} $\beta:M\times M\to R$ or---equivalently---an $R$-linear map $\beta:\Lambda_2(M)\to R$. As $\beta(m,\cdot):M\to R\subseteq\Lambda(M):m'\mapsto\beta(m,m')$ is $R$-linear, we can apply Section~\ref{sec:some-exterior-algebra} to extend this map to an $R$-linear map $\widehat\beta(m,\cdot):\Lambda(M)\to\Lambda(M):x\mapsto\widehat\beta(m,x)$ satisfying\cdg{In the notations of Section~\ref{sec:some-exterior-algebra}, this map $\widehat\beta(m,\cdot)$ is $\widehat\varphi$, where $\varphi$ is the $R$-linear map $\beta(m,\cdot)$.}
\begin{equation}
\widehat\beta(m,m'\wedge x)=\beta(m,m')\cdot x-m'\wedge\widehat\beta(m,x)
\tag{3.1}\label{eq:3.1}
\end{equation}
for all $m,m'\in M$, $x\in\Lambda(M)$, as well as\cdg{Specifically: the equalities \eqref{eq:3.2}, \eqref{eq:3.3}, \eqref{eq:3.4} and \eqref{eq:3.5} follow from \eqref{eq:2.15}, \eqref{eq:2.12}, \eqref{eq:2.10}, \eqref{eq:2.9}, respectively.}
\begin{equation}
\widehat\beta(m,\widehat{\varphi}(x))=-\widehat{\varphi}(\widehat\beta(m,x))
\tag{3.2}\label{eq:3.2}
\end{equation}
\cdg{Typo corrected ($\beta(m,x)$ replaced by $\widehat\beta(m,x)$).} for all $m\in M$, $R$-linear maps $\varphi:M\to R$, and $x\in\Lambda(M)$,
\begin{equation}
\widehat\beta(m,\widehat\beta(m,x))=0,
\tag{3.3}\label{eq:3.3}
\end{equation}
\begin{equation}
\widehat\beta(m,\Lambda_k(M))\subseteq\Lambda_{k-1}(M),
\tag{3.4}\label{eq:3.4}
\end{equation}
and
\begin{align}
\widehat\beta(r_1\cdot m_1+r_2\cdot m_2,x)
&=r_1\cdot \widehat\beta(m_1,x)+r_2\cdot \widehat\beta(m_2,x)
\tag{3.5}\label{eq:3.5}
\\
&\qquad\text{for all }m,m_1,m_2\in M,\ x\in\Lambda(M),\ r_1,r_2\in R.
\notag
\end{align}

In consequence, we can extend $\beta:\Lambda_2(M)\to R$, considered as an $R$-linear form defined on $\Lambda_2(M)$, recursively to a uniquely determined $R$-linear form $\beta:\Lambda(M)\to R$ satisfying $\beta(r)=r$ for all $r\in R$ as well as
\begin{equation}
\beta(m\wedge x)=\beta(\widehat\beta(m,x))
\quad\text{for all }m\in M\text{ and }x\in\Lambda(M)
\tag{3.6}\label{eq:3.6}
\end{equation}
in view of the formula\cdg{Let me explain the proof of the well-definedness of this form $\beta:\Lambda(M) \to R$ in a bit more detail:
\par
First, we define an $R$-linear form $\beta_k : \Lambda_k(M) \to R$ for each $k \geq 0$ by recursion on $k$, starting with $\beta_0 = \id$ and $\beta_1 = 0$. For each $k > 1$, if we assume that $\beta_{k-2} : \Lambda_{k-2}(M) \to R$ has already been defined, then we define $\beta_k : \Lambda_k(M) \to R$ by
\begin{align}
\beta_k(m\wedge x)=\beta_{k-2}(\widehat\beta(m,x))
\quad\text{for all }m\in M\text{ and }x\in\Lambda_{k-1}(M);
\tag{3.6$'$}
\label{eq:3.6prime}
\end{align}
this is well-defined for the following reason:
Recall from Section~\ref{sec:some-exterior-algebra} that
given an $R$-bilinear map $\lambda:M\times\Lambda_{k-1}(M)\to N$, there exists a unique induced $R$-linear map $\widehat{\lambda}:\Lambda_k(M)\to N$ with $\lambda(m,x)=\widehat{\lambda}(m\wedge x)$ for all $m\in M$ and $x\in\Lambda_{k-1}(M)$ if and only if $\lambda(m,m\wedge x)=0$ for all $m\in M$ and $x\in\Lambda_{k-2}(M)$.
We apply this construction to $N = R$ and to the $R$-bilinear map $\lambda:M\times\Lambda_{k-1}(M)\to R$ given by $\lambda(m, x) = \beta_{k-2}(\widehat{\beta}(m,x))$ for all $m \in M$ and $x \in \Lambda_{k-1}(M)$, which is well-defined by \eqref{eq:3.4}, and which satisfies $\lambda(m,m\wedge x)=0$ for all $m\in M$ and $x\in\Lambda_{k-2}(M)$ because
\begin{align*}
\lambda(m,m\wedge x)
&= \beta_{k-2}(\widehat{\beta}(m, m \wedge x))
%= \beta_{k-2}(\beta(m,m)\wedge x-m\wedge\widehat\beta(m,x)) \\
= \beta_{k-2}(-m\wedge\widehat\beta(m,x)) \\
&\qquad \qquad \left( %\begin{array}{c}
\text{since \eqref{eq:3.1} yields }
\widehat{\beta}(m, m \wedge x)
= \beta(m, m) \cdot x - m \wedge \widehat{\beta}(m, x)
= - m \wedge \widehat{\beta}(m, x)
%\end{array}
\right) \\
&= -\beta_{k-2}(m \wedge \widehat\beta(m,x))
= -\beta_{k-4}(\widehat\beta(m, \widehat\beta(m,x)))
\\
&\qquad \qquad \left(
\text{by \eqref{eq:3.6prime} for $k-2$ instead of $k$ (which holds by the inductive hypothesis)}\right) \\
&= -\beta_{k-4}(0) \qquad \left(\text{by \eqref{eq:3.3}}\right)
\\
&= 0
\end{align*}
(here we have assumed $k>3$ in the third-to-last equality sign; the case $k\leq 3$ can be done by hand).
Thus, we obtain a unique induced $R$-linear map $\widehat{\lambda}:\Lambda_k(M)\to R$ with $\widehat{\lambda}(m\wedge x) = \lambda(m,x) = \beta_{k-2}(\widehat{\beta}(m,x))$ for all $m\in M$ and $x\in\Lambda_{k-1}(M)$.
We denote this map $\widehat{\lambda}$ by $\beta_k$; then the equality just stated is \eqref{eq:3.6prime}, and so the recursive definition of $\beta_k$ is complete.
\par
Now, the $R$-linear form $\beta : \Lambda(M) \to R$ is the sum of the forms $\beta_k : \Lambda_k(M) \to R$ over all $k \geq 0$ (that is, it sends each $x \in \Lambda_k(M)$ to $\beta_k(x)$ for each $k \geq 0$).
By linearity, its property \eqref{eq:3.6} follows from \eqref{eq:3.6prime}.
}
\begin{equation}
\widehat\beta(m,m\wedge x)=\beta(m,m)\wedge x-m\wedge\widehat\beta(m,x)
=-m\wedge\widehat\beta(m,x)
\tag{3.7}\label{eq:3.7}
\end{equation}
which together with \eqref{eq:3.3} and \eqref{eq:3.4} implies that in the case $\beta$ has been defined on $\bigoplus_{i\leq k}\Lambda_i(M)$ such that\cdg{``$\bigoplus_{i\leq k}\Lambda_k(M)$'' corrected to ``$\bigoplus_{i\leq k}\Lambda_i(M)$''.} \eqref{eq:3.6} holds, the resulting map
\[
M\times\Lambda_k(M)\to R:(m,x)\mapsto\beta(\widehat\beta(m,x))
\]
satisfies
\begin{align*}
\beta(\widehat\beta(m,m\wedge x))
&=\beta(-m\wedge\widehat\beta(m,x))
=-\beta(m\wedge\widehat\beta(m,x))\\
&=-\beta(\widehat\beta(m,\widehat\beta(m,x)))=0
\end{align*}
for all $m\in M$ and $x\in\Lambda_{k-1}(M)$\ \ \ \ \cdg{In the original, this was ``$x\in\Lambda_k(M)$'', which would not fit with the domain of the map.}.

Note that \eqref{eq:3.6} implies that $\beta(m)=0$ for all $m\in M$, and thus $\beta(\Lambda_-(M))=0$, because, using \eqref{eq:3.4}, it implies easily by induction on $k$ that $\beta(x)=0$ for all $x$ in $\Lambda_k(M)$, $k\equiv1\mod2$. Thus $\beta$ ``lives'' essentially on $\Lambda_+(M)$.

Obviously, for $m_1,m_2,\ldots,m_n\in M$ one has
\begin{equation}
\beta(m_1\wedge m_2)=\beta(m_1,m_2)
\tag{3.8}\label{eq:3.8}
\end{equation}
and
\begin{align}
\beta(m_1\wedge\cdots\wedge m_n)
&=\beta(\widehat\beta(m_1,m_2\wedge\cdots\wedge m_n))\notag\\
&=\beta\left(\sum_{j=2}^{n}(-1)^j\cdot \beta(m_1,m_j)\cdot
  m_2\wedge\cdots\wedge m_{j-1}\wedge m_{j+1}\wedge\cdots\wedge m_n\right)\notag\\
&=\sum_{j=2}^{n}(-1)^j\cdot \beta(m_1,m_j)\cdot
  \beta(m_2\wedge\cdots\wedge m_{j-1}\wedge m_{j+1}\wedge\cdots\wedge m_n)
\qquad
\tag{3.9}\label{eq:3.9}
\end{align}
for $n \geq 2$,\ \ \ \ \cdg{Added ``for $n \geq 2$''.}
so if the Pfaffian is defined recursively according to \eqref{eq:1.4} and \eqref{eq:1.5}, we indeed have

\begin{theoremtwo}\refstepcounter{thmtwo}\label{thm:beta-pfaffian}
For all $m_1,\ldots,m_n\in M$ one has
\begin{equation}
\beta(m_1\wedge\cdots\wedge m_n)
=\Pf_n^R\bigl((\beta(m_i,m_j))_{i,j=1,\ldots,n}\bigr).
\tag{3.10}\label{eq:3.10}
\end{equation}
\end{theoremtwo}

\section{The Pfaffian Identity}\label{sec:pfaffian-identity}

Continuing with our notations we now prove

\begin{theoremthree}\refstepcounter{thmthree}\label{thm:beta-identity}
With $R$, $M$, and $\beta:\Lambda_2(M)\to R$ as above we have
\begin{align*}
&\sum_{i=1}^{k}(-1)^i\cdot \beta(m_1\wedge\cdots\wedge m_{i-1}\wedge m_{i+1}\wedge\cdots\wedge m_k)
\cdot\beta(m_i\wedge n_1\wedge\cdots\wedge n_l)\\
&\quad=-\sum_{j=1}^{l}(-1)^j\cdot\beta(n_1\wedge\cdots\wedge n_{j-1}\wedge n_{j+1}\wedge\cdots\wedge n_l)
\cdot\beta(n_j\wedge m_1\wedge\cdots\wedge m_k)
\end{align*}
for all $m_1,\ldots,m_k,n_1,\ldots,n_l\in M$.
\end{theoremthree}

\begin{proof}
This formula follows immediately from the equations
\begin{align*}
\beta(m_i\wedge n_1\wedge\cdots\wedge n_l)
&=\beta(\widehat\beta(m_i,n_1\wedge\cdots\wedge n_l))\\
&=\beta\left(\sum_{j=1}^{l}(-1)^{j-1}\cdot \beta(m_i,n_j)\cdot
 n_1\wedge\cdots\wedge n_{j-1}\wedge n_{j+1}\wedge\cdots\wedge n_l\right)\\
&=\sum_{j=1}^{l}(-1)^{j-1}\cdot \beta(m_i,n_j)\cdot
 \beta(n_1\wedge\cdots\wedge n_{j-1}\wedge n_{j+1}\wedge\cdots\wedge n_l)
\end{align*}
for $1\leq i\leq k$ and
\begin{align*}
\beta(n_j\wedge m_1\wedge\cdots\wedge m_k)
=\sum_{i=1}^{k}(-1)^{i-1}\cdot \beta(n_j,m_i)\cdot
\beta(m_1\wedge\cdots\wedge m_{i-1}\wedge m_{i+1}\wedge\cdots\wedge m_k)
\end{align*}
for $1\leq j\leq l$.
\end{proof}

Now assume that $x_1,\ldots,x_n\in M$ and put
\[
p(I):=\beta(x_{a_1}\wedge\cdots\wedge x_{a_k})
\]
for any subset $I=\{a_1,\ldots,a_k\}\subseteq\{1,\ldots,n\}$ with $a_1<a_2<\cdots<a_k$\ \ \ \ \cdg{In the original: ``for $I=\{a_1,\ldots,a_k\}\subseteq\{1,\ldots,n\}$ and $a_1<a_2<\cdots<a_k$''; edited for additional clarity.}.

Next assume that $I_1,I_2$ are two arbitrary subsets of the index set $I_0:=\{1,\ldots,n\}$. Write $I_1=\{a_1,\ldots,a_k\}$, $I_2=\{b_1,\ldots,b_l\}$, where $a_1<\cdots<a_k$ and $b_1<\cdots<b_l$. Put $m_i:=x_{a_i}$ for $i=1,\ldots,k$ and $n_j:=x_{b_j}$ for $j=1,\ldots,l$. Then Theorem~\ref{thm:beta-identity} implies
\begin{align*}
&\sum_{i=1}^{k}(-1)^i\cdot \beta(x_{a_1}\wedge\cdots\wedge x_{a_{i-1}}\wedge x_{a_{i+1}}\wedge\cdots\wedge x_{a_k})
\cdot\beta(x_{a_i}\wedge x_{b_1}\wedge\cdots\wedge x_{b_l})\\
&\quad+\sum_{j=1}^{l}(-1)^j\cdot \beta(x_{b_1}\wedge\cdots\wedge x_{b_{j-1}}\wedge x_{b_{j+1}}\wedge\cdots\wedge x_{b_l})
\cdot\beta(x_{b_j}\wedge x_{a_1}\wedge\cdots\wedge x_{a_k})=0.
\end{align*}

Note that for all $i=1,\ldots,k$ one has
\[
\beta(x_{a_i}\wedge x_{b_1}\wedge\cdots\wedge x_{b_l})
=
\begin{cases}
0,&\text{if }a_i\in I_2,\\
(-1)^{\#\{j\mid b_j<a_i\}}\cdot p(I_2\symdiff\{a_i\}),&\text{otherwise,}
\end{cases}
\]
and for all $j=1,\ldots,l$ one has similarly
\[
\beta(x_{b_j}\wedge x_{a_1}\wedge\cdots\wedge x_{a_k})
=
\begin{cases}
0,&\text{if }b_j\in I_1,\\
(-1)^{\#\{i\mid a_i<b_j\}}\cdot p(I_1\symdiff\{b_j\}),&\text{otherwise.}
\end{cases}
\]

Finally, note that with $I_1\symdiff I_2=\{c_1,\ldots,c_t\}$, $c_1<c_2<\cdots<c_t$, and $a_i=c_\tau\in I_1\setminus I_2$ for some $i\in\{1,\ldots,k\}$ and $\tau\in\{1,\ldots,t\}$ one has
\[
(-1)^i\cdot(-1)^{\#\{j\mid b_j<a_i\}}=(-1)^\tau,
\]
because
\begin{align*}
\tau
&=\#\{c\in I_1\symdiff I_2\mid c\leq a_i\}\\
&=\#\{c\in I_1\setminus I_2\mid c\leq a_i\}
 +\#\{c\in I_2\setminus I_1\mid c\leq a_i\}\\
&\equiv\#\{c\in I_1\setminus I_2\mid c\leq a_i\}
 +\#\{c\in I_1\cap I_2\mid c\leq a_i\}\\
&\qquad+\#\{c\in I_2\setminus I_1\mid c\leq a_i\}
 +\#\{c\in I_2\cap I_1\mid c\leq a_i\}\\
&=\#\{c\in I_1\mid c\leq a_i\}+\#\{c\in I_2\mid c\leq a_i\}\\
&=i+\#\{j\mid b_j<a_i\}\mod 2.
\end{align*}
Similarly, for $b_j=c_\tau\in I_2\setminus I_1$ for some $j\in\{1,\ldots,l\}$ and $\tau\in\{1,\ldots,t\}$ one has
\[
(-1)^j\cdot(-1)^{\#\{i\mid a_i<b_j\}}=(-1)^\tau.
\]
Hence, Theorem~\ref{thm:beta-identity} indeed yields that
\[
\sum_{\tau=1}^{t}(-1)^\tau\cdot p(I_1\symdiff\{c_\tau\})\cdot p(I_2\symdiff\{c_\tau\})=0.
\]
\cdg{The ``$c_\tau$''s in the preceding equality were called ``$i_\tau$''s in the original; fixed for consistency with the equality $I_1\symdiff I_2=\{c_1,\ldots,c_t\}$.}
As this is the identity \eqref{eq:1.7} for $A:=(\beta(x_i,x_j))_{i,j=1,\ldots,n}$ and as any skew-symmetric $(n\times n)$-matrix $A\in\Skew_n(R)$ is of this form for some $M$, $\beta$, and $x_1,\ldots,x_n\in M$, Theorem~\ref{thm:pfaffian-identity} follows.

\section{The Pfaffian and the Determinant}\label{sec:pfaffian-determinant}

In this section we use our exterior algebra formalism to prove

\begin{theoremfour}\refstepcounter{thmfour}\label{thm:pfaffian-square-det}
For $R$, $M$, $\beta$ as above and $m_1,\ldots,m_n\in M$ one has
\begin{equation}
\beta(m_1\wedge\cdots\wedge m_n)^2
=\det\bigl((\beta(m_i,m_j))_{i,j=1,\ldots,n}\bigr).
\tag{5.1}\label{eq:5.1}
\end{equation}
\end{theoremfour}

Note that Theorem~\ref{thm:beta-pfaffian} and Theorem~\ref{thm:pfaffian-square-det} yield the classical relation \eqref{eq:1.2}.

To prove Theorem~\ref{thm:pfaffian-square-det} it is convenient to define an $R$-linear map $I_\beta:\Lambda(M)\to\Lambda(M)$ recursively by
\begin{equation}
I_\beta(r):=r\qquad\text{for }r\in R,
\tag{5.2a}\label{eq:5.2a}
\end{equation}
\begin{equation}
I_\beta(m\wedge x):=m\wedge I_\beta(x)+\widehat\beta(m,I_\beta(x))
\qquad\text{for }m\in M,\quad x\in\Lambda(M).
\tag{5.2b}\label{eq:5.2b}
\end{equation}
Since \eqref{eq:3.1} and \eqref{eq:3.3} imply
\begin{align*}
&m\wedge\bigl(m\wedge I_\beta(x)+\widehat\beta(m,I_\beta(x))\bigr)
 +\widehat\beta\bigl(m,m\wedge I_\beta(x)+\widehat\beta(m,I_\beta(x))\bigr)\\
&\quad=m\wedge\widehat\beta(m,I_\beta(x))+\beta(m,m)\cdot I_\beta(x)
 -m\wedge\widehat\beta(m,I_\beta(x))
 +\widehat\beta(m,\widehat\beta(m,I_\beta(x)))=0
\end{align*}
---independent of the actual value of $I_\beta(x)$---the map $I_\beta$ is indeed well defined.

In order to study $I_\beta$ we first observe that for every $R$-linear\cdg{Added ``$R$-linear''.} $\varphi:M\to R$ and its canonical extension $\widehat{\varphi}:\Lambda(M)\to\Lambda(M)$ one has
\begin{equation}
I_\beta(\widehat{\varphi}(x))=\widehat{\varphi}(I_\beta(x))
\qquad\text{for all }x\in\Lambda(M).
\tag{5.3}\label{eq:5.3}
\end{equation}
\cdg{Removed ``$m\in M$ and''.}
If $x\in R$, then both sides in \eqref{eq:5.3} vanish; thus by induction it suffices to observe that by \eqref{eq:2.5}, \eqref{eq:5.2b}, and \eqref{eq:3.2} the assumption $I_\beta(\widehat{\varphi}(x))=\widehat{\varphi}(I_\beta(x))$ implies
\begin{align*}
I_\beta(\widehat{\varphi}(m\wedge x))
&=I_\beta(\varphi(m)\cdot x-m\wedge\widehat{\varphi}(x))\\
&=\varphi(m)\cdot I_\beta(x)-m\wedge I_\beta(\widehat{\varphi}(x))
 -\widehat\beta(m,I_\beta(\widehat{\varphi}(x)))\\
&=\varphi(m)\cdot I_\beta(x)-m\wedge\widehat{\varphi}(I_\beta(x))
 -\widehat\beta(m,\widehat{\varphi}(I_\beta(x)))\\
&=\widehat{\varphi}(m\wedge I_\beta(x))+\widehat{\varphi}(\widehat\beta(m,I_\beta(x)))\\
&=\widehat{\varphi}(I_\beta(m\wedge x)).
\end{align*}

Next we prove that with $P_0:\Lambda(M)\twoheadrightarrow\Lambda_0(M):x\mapsto x^{(0)}$ denoting the canonical projection from $\Lambda(M)$ onto $\Lambda_0(M)=R$ one has
\begin{equation}
P_0(I_\beta(x))=\beta(x)
\qquad\text{for all }x\in\Lambda(M).
\tag{5.4}\label{eq:5.4}
\end{equation}
For $x\in R$ we have indeed $P_0(I_\beta(x))=x=\beta(x)$. By linearity we may now assume that $x=m_1\wedge\cdots\wedge m_k$ for some $m_1,\ldots,m_k\in M$ and that \eqref{eq:5.4} is true for all $x'\in\bigoplus_{i=0}^{k-1}\Lambda_i(M)$. Then we get, by \eqref{eq:5.2b}, \eqref{eq:5.3}\cdg{Specifically, \eqref{eq:5.3} is being applied to $\varphi = \beta(m_1,\cdot)$ to obtain $I_\beta(\widehat{\beta}(m_1,x)) = \widehat{\beta}(m_1,I_\beta(x))$ for all $x \in \Lambda(M)$.}, \eqref{eq:3.4}, and \eqref{eq:3.6},
\begin{align*}
P_0(I_\beta(m_1\wedge\cdots\wedge m_k))
&=P_0(\widehat\beta(m_1,I_\beta(m_2\wedge\cdots\wedge m_k)))\\
&=P_0(I_\beta(\widehat\beta(m_1,m_2\wedge\cdots\wedge m_k)))\\
&=\beta(\widehat\beta(m_1,m_2\wedge\cdots\wedge m_k))\\
&=\beta(m_1\wedge\cdots\wedge m_k),
\end{align*}
so \eqref{eq:5.4} follows by induction.

Now let $\tbet:M\times M\to R$ denote the skew-symmetric $R$-bilinear form given by
\[
\tbet(m_1,m_2):=\beta(m_2,m_1)=-\beta(m_1,m_2)
\qquad\text{for }m_1,m_2\in M.
\]
The induced $R$-linear form $\tbet:\Lambda(M)\to R$ satisfies $\tbet(x)=0$ for all $x\in\Lambda_-(M)$, and \eqref{eq:3.9} implies by induction for $k\equiv0\mod2$ and all $m_1,\ldots,m_k\in M$:
\begin{align}
\tbet(m_1\wedge\cdots\wedge m_k)
&=(-1)^{k\cdot(k-1)/2}\cdot \beta(m_1\wedge\cdots\wedge m_k) \notag\\
&=
\begin{cases}
\beta(m_1\wedge\cdots\wedge m_k),&\text{for }k\equiv0\mod4,\\
-\beta(m_1\wedge\cdots\wedge m_k),&\text{for }k\equiv2\mod4.
\end{cases}
\tag{5.5}\label{eq:5.5}
\end{align}

Theorem~\ref{thm:pfaffian-square-det} now follows directly from \eqref{eq:5.4} and \eqref{eq:5.5}\cdg{... and from the fact that the projection $P_0$ is a ring homomorphism} once we have proved that for all $k\in\NN$ and $m_1,\ldots,m_k\in M$ the following much stronger identity holds:

\begin{theoremfourprime}\refstepcounter{thmfourprime}\label{thm:stronger-identity}
For $R$, $M$, $\beta$ as above and $m_1,\ldots,m_k\in M$ one has
\[
I_\beta(m_1\wedge\cdots\wedge m_k)\wedge I_{\tbet}(m_1\wedge\cdots\wedge m_k)
=(-1)^{k\cdot(k-1)/2}\cdot \det\bigl((\beta(m_i,m_j))_{i,j=1,\ldots,k}\bigr).
\]
\end{theoremfourprime}

\begin{proof}
Note first that \eqref{eq:2.8} and \eqref{eq:2.9} imply
\begin{equation}
\widehat\beta(m,x\wedge y)=\widehat\beta(m,x)\wedge y+\bar x\wedge\widehat\beta(m,y)
\tag{5.6}\label{eq:5.6}
\end{equation}
and
\begin{equation}
\widehat{\,\tbet\,}(m,x)=-\widehat\beta(m,x)
% Added \,'s around \tbet to make the scope of the \widehat clearer in the output.
\tag{5.7}\label{eq:5.7}
\end{equation}
for all $m\in M$ and $x,y\in\Lambda(M)$. Now assume that $m\in M$ and $y\in\Lambda(M)$ satisfy $m\wedge y=0$. Then $I_{\tbet}(m \wedge y) = 0$, so that \eqref{eq:5.2b} and \eqref{eq:5.7} yield\cdg{Added some more details to this sentence.}
\[
m\wedge I_{\tbet}(y)=-\widehat{\tbet}(m,I_{\tbet}(y))=\widehat\beta(m,I_{\tbet}(y)).
\]
Hence, for any $x\in\Lambda(M)$ we have\cdg{The second equality sign in the following computation relies on the (easy) fact that every $m \in M$ and $z \in \Lambda(M)$ satisfy $m \wedge z = \overline z \wedge m$.}
\begin{align*}
I_\beta(m\wedge x)\wedge I_{\tbet}(y)
&=\bigl(m\wedge I_\beta(x)+\widehat\beta(m,I_\beta(x))\bigr)\wedge I_{\tbet}(y)\\
&=\widehat\beta(m,I_\beta(x))\wedge I_{\tbet}(y)
 +\overline{I_\beta(x)}\wedge m\wedge I_{\tbet}(y)\\
&=\widehat\beta(m,I_\beta(x))\wedge I_{\tbet}(y)
 +\overline{I_\beta(x)}\wedge\widehat\beta(m,I_{\tbet}(y))
\end{align*}
and thus by \eqref{eq:5.6}\cdg{We are still assuming $m \wedge y = 0$ here.}
\begin{equation}
I_\beta(m\wedge x)\wedge I_{\tbet}(y)
=\widehat\beta\bigl(m,I_\beta(x)\wedge I_{\tbet}(y)\bigr).
\tag{5.8}\label{eq:5.8}
\end{equation}
It follows by induction that for $m_1,\ldots,m_k\in M$, $y\in\Lambda(M)$, and $\varphi_i:=\beta(m_i,\cdot)$ ($i=1,\ldots,k$)\ \ \cdg{In the original, ``$\varphi_i:=\beta(m_i,\cdot)$'' was ``$\varphi_i:=\beta(m_i,-)$''. The change was made to ensure consistency with the ``$\widehat\beta(m,\cdot)$'' earlier on.} one has\cdg{In the following computation, \eqref{eq:5.8} is applied first to $m_1$, $m_2 \wedge \cdots \wedge m_k$ and $m_1\wedge\cdots\wedge m_k\wedge y$ instead of $m$, $x$ and $y$; then to $m_2$, $m_3 \wedge \cdots \wedge m_k$ and $m_1\wedge\cdots\wedge m_k\wedge y$ instead of $m$, $x$ and $y$; then to $m_3$, $m_4 \wedge \cdots \wedge m_k$ and $m_1\wedge\cdots\wedge m_k\wedge y$ instead of $m$, $x$ and $y$; and so on. Each time, we are using $m_i \wedge (m_1\wedge\cdots\wedge m_k\wedge y) = 0$.}
\begin{align*}
&I_\beta(m_1\wedge\cdots\wedge m_k)\wedge I_{\tbet}(m_1\wedge\cdots\wedge m_k\wedge y)\\
&\quad=\widehat\beta\bigl(m_1,I_\beta(m_2\wedge\cdots\wedge m_k)
 \wedge I_{\tbet}(m_1\wedge\cdots\wedge m_k\wedge y)\bigr)\\
&\quad=\widehat{\varphi_1}\bigl(I_\beta(m_2\wedge\cdots\wedge m_k)
 \wedge I_{\tbet}(m_1\wedge\cdots\wedge m_k\wedge y)\bigr)\\
&\quad=\cdots=\widehat{\varphi_1}\circ\cdots\circ\widehat{\varphi_k}
 \bigl(I_{\tbet}(m_1\wedge\cdots\wedge m_k\wedge y)\bigr)\\
&\quad=I_{\tbet}\bigl(\widehat{\varphi_1}\circ\cdots\circ\widehat{\varphi_k}
 (m_1\wedge\cdots\wedge m_k\wedge y)\bigr)
\end{align*}
\cdg{The last equality sign here used \eqref{eq:5.3}.}
and therefore, putting $y:=1$ and using \eqref{eq:2.17},
\begin{align*}
&I_\beta(m_1\wedge\cdots\wedge m_k)\wedge I_{\tbet}(m_1\wedge\cdots\wedge m_k)\\
&\quad=I_{\tbet}\bigl(\widehat{\varphi_1}\circ\cdots\circ\widehat{\varphi_k}(m_1\wedge\cdots\wedge m_k)\bigr)\\
&\quad=I_{\tbet}\left((-1)^{k(k-1)/2}\det\bigl((\varphi_i(m_j))_{i,j=1,\ldots,k}\bigr)\right)\\
&\quad=(-1)^{k(k-1)/2}\det\bigl((\beta(m_i,m_j))_{i,j=1,\ldots,k}\bigr).
\qedhere
\end{align*}
\end{proof}

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\end{document}