\documentclass[reqno]{amsart} 
\begin{document}
\setcounter{page}{121} 
{\noindent\small Mathematical Physics and Quantum Field Theory,\newline 
 Electron. J. Diff. Eqns., Conf. 04, 2000, pp. 121--145.\newline 
http://ejde.math.swt.edu  or http://ejde.math.unt.edu 
\newline ftp ejde.math.swt.edu \quad ejde.math.unt.edu (login: ftp)}
\thanks{\copyright 2000 Southwest Texas State University  and 
University of North Texas.} \vspace{1cm} 

\title[ Feynman's problem ]
{A partial solution for Feynman's problem:\\
 A new derivation of the Weyl equation} 
\author[Atsushi Inoue]
{Atsushi Inoue} 

\address {Atsushi Inoue \hfill\break\indent
Department of Mathematics, Tokyo Institute of Technology,  
\hfill\break\indent
2-12-1, Oh-okayama, Meguro-ku, Tokyo, 152-0033, Japan} 
\email{inoue@@math.titech.ac.jp}


\thanks{Published July 12, 2000.}
\thanks{Partially supported by Monbusyo Grant-in-aid No.10640201.}
\subjclass{35F10, 35L45, 35Q40, 70H99, 81S40} 
\keywords{Quantization, good parametrix, spin, superanalysis} 

\begin{abstract}
Associating classical mechanics to a system of PDE,
we give a procedure for Feynman type quantization
of a ``Schr\"odinger type equation with spin."
Mathematically, we construct a ``good parametrix"
for the Weyl equation with an external electro-magnetic field.
Main ingredients are a new interpretation of the matrix structure
using superanalysis and a reinterpretation of the
method of characteristics as a quantization procedure
of Feynman type.
\end{abstract}
\maketitle

\numberwithin{equation}{section}
\newtheorem{thm}{Theorem}[section]
\newtheorem{lem}[thm]{Lemma}
\newtheorem{prop}[thm]{Proposition}
\newtheorem{cor}[thm]{Corollary}
\newtheorem{defn}{Definition}[section]
\def\proj{\operatorname{proj}}
\def\dist{\operatorname{dist}}
\def\sdet{\operatorname{\rm sdet}}
\def\str{\operatorname{\rm str}}
\def\Mat{\operatorname{\rm Mat}}
\def\GL{\operatorname{\rm GL}}
\def\tr{\operatorname{\rm tr}}
\def\sgn{\operatorname{\rm sgn}}
\def\ccsl{{/\kern-0.5em{\mathcal C}}}
\def\clsl{{/\kern-0.6em{\mathcal L}}} %5
\def\cssl{{/\kern-0.6em{\mathcal S}}} 
\def\cdsl{{/\kern-0.7em{\mathcal D}}} 
\def\kbar{\mathchar'26\mkern-9mu k}
\def\pdq{\frac{\partial}{\partial {q}}}
\def\pdqj{\frac{\partial}{\partial q_j }}
\def\pdtheta{\frac{\partial}{\partial \theta}}
\def\pdx{\frac{\partial}{\partial {x}}}
\def\pdt{\frac{\partial}{\partial {t}}}
\def\trs{|\kern-0.1em|\kern-0.1em|}
\def\inidata{\underline{x},\underline{\xi},\underline{\theta},\underline{\pi}}
\def\mixdata{\underline{x},\underline{\xi},\underline{\theta},\underline{\pi}}
\def\mixxdata{{\underline{x}},{\underline{\theta}},{\underline{\xi}},{\underline{\pi}}}

\section{Feynman's problem for spin}

\subsection{Feynman's path integral representation and his problem}
Feynman proposed to represent solutions of the Schr\"odinger equation
\begin{equation}\gathered  
  i\hbar {\frac{\partial}{\partial t}}
 u(t,q)=H(q,\partial_q)u(t,q),\\
 u(0,q)={\underline{u}}(q), \quad  q=(q_1,\cdots,q_m)\in{\mathbb R}^m,
\endgathered \label{Sch}
\end{equation}
via the expression, called Feynman's path integral (representation),
\begin{equation}
F(t,q,q')
=\int_{C_{t,q,q'}}d_F\gamma \,e^{ i{\hbar}^{-1}\int_0^t 
L(\gamma(\tau),\dot\gamma(\tau))d\tau}.
\label{Fe}
\end{equation}
Here $H(q,\partial_q)$, 
the Hamiltonian operator with mass ${M}$, is given formally as
$$
H(q,\partial_q)={\frac{-\hbar^2}{2{M}}}\Delta +V(\cdot),\quad 
\Delta=\sum_{\ell=1}^m{\frac{\partial^2}{\partial q_\ell^2}},
$$
and $d_F\gamma$ denotes the notorious Feynman measure on the path space
$$
C_{t,q,q'}=\{\gamma(\cdot)\in AC([0,t]:{\mathbb R}^m)\,|\,
\gamma(0)=q',\gamma(t)=q\}.
$$
Here, $AC$ denotes absolute continuity.
For any path $\gamma\in C_{t,q,q'}$, 
the classical action $S_t(\gamma)$ is defined by
\begin{equation}
S_{t}(\gamma)=\int_0^t d\tau \,
L(\gamma(\tau),\dot\gamma(\tau))
\label{a}
\end{equation}
where the Lagrangian function
$$
L(\gamma,\dot\gamma)
={\frac M2}|{\dot\gamma}|^2-V(\gamma)\in
C^\infty(T{\mathbb R}^m:{\mathbb R}),
%\label{2.6}
$$
corresponds to the Hamiltonian function
$$
H(q,p)=\frac{|p|^2}{2M}+V(q)\in
C^\infty(T^*{\mathbb R}^m:{\mathbb R}).
$$

On the other hand, Feynman noted (\cite{FH65}, page 355) that
\begin{quotation}
...... path integrals suffer grievously from a serious defect.
They do not permit a discussion of spin operators or other such operators 
in a simple and lucid way. They find their greatest use in systems 
for which coordinates and their
conjugate momenta are adequate.
Nevertheless, spin is a simple and vital part of real 
quantum-mechanical systems. 
It is a serious limitation that the half-integral spin 
of the electron does not find a simple and ready representation. 
It can be handled if the amplitudes and quantities 
are considered as quaternions instead of ordinary complex numbers,
 but the lack of commutativity of such numbers is a serious complication.
\end{quotation}

\noindent
[{\bf{Problem for system of PDE}}]:
We regard Feynman's problem 
as calling for a {\bf new methodology of solving systems of PDE}.
A system of PDE has {\bf two non-commutativities}, 
\par
(i) one from $[\partial_q,q]=1$
(Heisenberg relation),
\par
(ii) the other from $[A,B]\neq 0$ (matrix noncommutativity).
\par
Non-commutativity from the Heisenberg relation is 
nicely controlled by using Fourier transformations 
(the theory of pseudodifferential operators).
Here, we want to give a {\bf new method of treating matrix non-commutativity}; 
after identifying matrix operations as differential operators
and using Fourier transformations, we may develop a theory of
pseudodifferential operators for supersmooth functions on
superspace ${\mathfrak R}^{m|n}$.

{\it Opinion. }
For a given system of PDE, 
if we may reduce that system to scalar PDEs by diagonalization, 
then we doubt 
whether it is truly necessary to
use a matrix representation.
Therefore, 
if we need to represent some equations using matrices,
we should try to treat a system of PDE as it is, {\bf without diagonalization}.
(Recall the Witten model, 
which is represented by two independent-looking equations,
has supersymmetry if treated as a system.)
\par{\it Remark. }
We may consider the method employed here as
an attempt to {\bf extend the ``method of characteristics" 
to PDE with matrix-valued coefficients}.
 
\subsection{Method of Fujiwara}
Unfortunately, the Feynman measure does not exist.
On the other hand, Fujiwara \cite{Fu79,Fu80} constructed the parametrix and
the fundamental solution of \eqref{Sch} using Feynman's arguments conversely, 
that is,
he made a part of the argument of Feynman mathematically rigorous. 

Let
$
\displaystyle{\sup_{q\in{\mathbb R}^m}|D^\alpha V(q)|\le C_\alpha},
$
for $|\alpha|\ge2$. Then
there exists a unique path $\gamma_0$ in $C_{t,q,q'}$
such that
\begin{equation*}
\inf_{\gamma\in C_{t,q,q'}} S_t(\gamma)=S_t(\gamma_0)=
{S}_L(t,q,q'),
\label{aa}
\end{equation*}
which gives a solution of the {\bf Hamilton-Jacobi equation}:
$$
\frac{\partial}{\partial t}S+H\bigg(q,\frac{\partial S}{\partial q}\bigg)=0.
$$
Introducing the {\bf van Vleck determinant}
\begin{equation*}
{D}_L(t,q,q')=
\det \Big(\partial_{q_i} \partial_{q'_j}
{S}_L(t,q,q')\Big),
\label{c}
\end{equation*}
which is a solution of the {\bf continuity equation}
$$
\frac{\partial}{\partial t}D_L
+\frac{\partial}{\partial q}
\bigg(D_LH_p\bigg(q,\frac{\partial S}{\partial q}\bigg)\bigg)=0,
$$
he defined
\begin{equation}
(F_t u)(q)=(2\pi i\hbar)^{-m/2}\int_{{\mathbb R}^m} dq'\,{D}_L^{1/2}(t,q,q')
e^{ i\hbar^{-1}{S}_L(t,q,q')} u(q').
\label{pf}
\end{equation}
\begin{thm}[Fujiwara \cite{Fu79}]
 Fix $0<T<\infty$ arbitrarily.
Putting ${\mathbb H}=L^2({\mathbb R}^m:{\mathbb C})$ and denoting
the set of bounded linear operators on ${\mathbb H}$ by ${\mathcal B}({\mathbb H})$, 
we have the following:
\par
(1) $F_t$ defines a bounded linear operator in ${\mathbb H}$:
%That is, there exists a constant $C$ depending on $T$ such that 
%for any $ u\in L^2({\mathbb R}^m:{\mathbb C})$ and $ t\in [-T,T]$,
$$
\| F_t u \|\le C \| u \|.
$$ 
\par
(2)
For any $ u\in L^2({\mathbb R}^m:{\mathbb C})$, $t,\,s,\,t+s\in [-T,T]$, 
$$
\begin{gathered}
\lim_{t\to 0} \|F_tu-u \|=0 ,\\
{i\hbar}\left.{\frac {\partial}{\partial t}}
 (F_tu)(q)\right|_{t=0}
=H(q,\partial_q)u(q),\\
\| F_{t+s}-F_tF_s \|
\le
C({t^2}+{s^2}) .
\end{gathered}
$$
\par
(3)
Moreover, there exists a limit 
$\lim_{k\to\infty}( F_{t/k})^k=E_t$
in ${\mathcal B}({\mathbb H})$,
i.e., in the operator norm of $L^2({\mathbb R}^m:{\mathbb C})$, 
which satisfies 
the initial value problem below:
$$\gathered {i\hbar}{\frac {\partial}{\partial t}}
(E_tu)(q)=H(q,\partial_q)(E_tu)(q), \\
(E_0u)(q)={\underline u}(q).
\endgathered 
$$
\end{thm}

{\it  Remarks.}
(i) In the above,  
$L^2$-boundedness of the operator $F_t$ is crucial and
is proved using Cotlar's lemma. 
This usage makes it difficult to prove the boundedness of analogous operators 
in curved space.
Therefore, it is an open problem to 
``quantize" the Lagrangian on a curved manifold.
The above procedure of Fujiwara was modified for the heat equation
on a curved manifold by Inoue-Maeda \cite{IM85}
to explain mathematically the origin of the term $(1/12)R$, where $R$ is the scalar
curvature of the configuration manifold.
\newline
(ii) Though in his papers \cite{Fu79,Fu80}, 
Fujiwara allowed the time-dependence of $V(t,q)$,
but we disregard the time-dependence in Theorem 1.1 
for descriptional simplicity.

\vspace{2mm}
We call \eqref{pf} a ``{\bf good parametrix}" 
since it has the following properties:
\newline
(1) The parametrix has explicit dependence on the classical quantities, i.e., 
Bohr correspondence is exemplified.
As a by-product, we may check easily whether we may derive the heat equation 
from the Schr\"odinger equation by replacing $\hbar$ in \eqref{pf} with $-i$.
\newline
(2) The infinitesimal generator of the parametrix 
gives a Hamiltonian operator $H(q,\partial _q)$
which corresponds to the quantization 
of the Lagrangian $L(\gamma,\dot\gamma)$, or the Weyl quantization for
the symbol $H(q,p)$.

\par
[{\bf{Problem 1}}] 
In \eqref{Fe} and \eqref{pf}, the Lagrangian is used. 
How can we connect the above procedure 
directly to the Hamiltonian without using the Lagrangian?
(A partial solution of this problem is now given in \cite{In99-1}.)

\par
[{\bf{Problem 2}}] How we may proceed when
\newline
(1) $V$ has singularities like Coulomb potentials?  or
\newline
(2) there exist many paths connecting points $q$ and $q'$ like
the dynamics on the circle?

%[new result.tex]
\section{Problem and Result}
\noindent
[{\bf Problem}]:
Find a {\bf good representation} of
$$
{\mathbb R}\times{\mathbb R}^3\ni(t,q)\to\psi(t,q)=
{
\begin{pmatrix}
\psi_1(t,q)\\
\psi_2(t,q)
\end{pmatrix}
}\in{\mathbb C}^2$$
 satisfying the Weyl equation with the time-dependent
external electro-magnetic field
\begin{equation} \label{W} 
\gathered
i\hbar {\pdt} {\psi}(t,q)={\mathbb H}(t)\psi(t,q),\\ 
\psi(\underline{t},q)={\underline{\psi}}(q),
\endgathered
\end{equation}
where
$$
{\mathbb H}(t)=c\sum_{j=1}^3\pmb\sigma_j\bigg(\frac{\hbar} i\frac{\partial}{\partial q_j}
-\frac{\varepsilon}{c} A_j(t,q)\bigg)-\varepsilon A_0(t,q),
$$
and the Pauli matrices
$\{\pmb\sigma_j\}$
are represented by 
$$
\pmb\sigma_1=\begin{pmatrix}
0&1\\
1&0
\end{pmatrix},\quad
\pmb\sigma_2=\begin{pmatrix}
0&-i\\
i&0
\end{pmatrix},\quad
\pmb\sigma_3=\begin{pmatrix}
1&0\\
0&-1
\end{pmatrix}.
$$
\par
(i) {\bf What is the classical mechanics corresponding to \eqref{W}?}
\par
(ii) {\bf How about its quantization?}

\vspace{2mm}
\noindent
[{\bf Reduction}]:
We introduce {\bf new independent variables}, called {\bf odd}
 or {\bf fermion variables}, which are used to
represent matrix structure (see the appendix on elements of
superanalysis).

Then, we may reduce \eqref{W} to the super Weyl equation on ${{\mathfrak R}}^{3|2}$:
\begin{equation}\label{SW} \gathered
i\hbar \pdt u(t,x,\theta)=\hat{\mathcal H}(t)u(t,x,\theta),\\ 
 u(\underline{t},x,\theta)=\underline{u}(x,\theta),
\endgathered
\end{equation}
where
$$
\hat{\mathcal H}(t)
=c\sum_{j=1}^3\sigma_j\bigg(\theta,\frac{\partial}{\partial\theta}\bigg)
\bigg(\frac{\hbar}i\frac{\partial}{\partial x_j}
-\frac{\varepsilon}{c}A_j(t,x)\bigg) -\varepsilon A_0(t,x).  
$$

{\it Remark. }
Pauli claimed that ``{\bf  There exists no classical counter-part 
for a quantum spinning particle.}"
In spite of this, we present another representation 
which exhibits the ``{\bf underlying Classical Mechanics.}"

\begin{thm}[A good parametrix for the Weyl equation] Assume that\\
$\displaystyle{\sup|(1+|q|)^{|\alpha|-1}\partial_q^\alpha
A_j(t,q)|<\infty}$. 
Then,
\begin{equation*}\aligned
\psi(t,q)&=\flat\Big((2\pi{i}\hbar)^{-3/2}\hbar\iint_{{\mathfrak R}^{3|2}}
d{\underline{\xi}}d{\underline{\pi}} 
{\mathcal D}^{1/2}(t,\underline{t};{\mixxdata}) \\
&\qquad\qquad\qquad\qquad\qquad
\times e^{i\hbar^{-1}{{\mathcal S}}(t,\underline{t};{\mixxdata})}
{\mathcal F}(\#{\underline{\psi}})({\underline{\xi}},{\underline{\pi}})\Big)
\Big|_{\underline{x}_{\mathrm B}=q},
\endaligned\end{equation*}
gives a {\bf good parametrix} for \eqref{W}.
Here, ${\mathcal S}(t,\underline{t};{\mixxdata})$ and 
${\mathcal D}(t,\underline{t};{\mixxdata})$ are solutions 
of Hamilton-Jacobi and continuity equations, respectively.
${\mathcal F}$ is the Fourier transformation of functions on ${{\mathfrak R}}^{3|2}$.
\end{thm}

%%[charaOHP.tex]
\section{Main ingredients}
\subsection{How can we regard the method of characteristics as quantization?}

On the region $\Omega$ in ${\mathbb R}^{m+1}$,
we consider the following initial value problem:
\begin{equation}\label{mc}\gathered
\pdt u(t,q)+\sum_{j=1}^m a_j(t,q){\pdqj}u(t,q)=b(t,q)u(t,q)+f(t,q),\\
u({\underline{t}},q)={\underline{u}}(q).
\endgathered
\end{equation}
Corresponding characteristics are given by
$$ \gathered
\frac d{dt} q_j(t)=a_j(t,q(t)),\\
 q_j({\underline{t}})={\underline{q}}_j
\quad (j=1,\cdots,m).
\endgathered 
$$
We denote the solution by
$$
q(t)=q(t,{\underline{t}};{\underline{q}})=(q_1(t), \cdots, q_m(t))\in{\mathbb R}^m.
$$
The following theorem is well-known.
\begin{thm} %{Theorem 2.1}
Let $a_j\in C^1(\Omega:{\mathbb R})$ and $b, f\in C(\Omega:{\mathbb R})$.
For any point $({\underline{t}},{\underline{q}})\in\Omega$,
we assume that ${\underline{u}}$ is  $C^1$ in a neighborhood 
of ${\underline{q}}$.
\par
Then, in a neighborhood of $({\underline{t}},{\underline{q}})$, 
there exists a unique
solution $u(t, q)$ of \eqref{mc}. More precisely, putting
$$
U(t,{\underline{q}})
=e^{\int_{{\underline{t}}}^t d\tau\, B(\tau,q)}
\left\{\int_{{\underline{t}}}^t ds\,
e^{-\int_{{\underline{t}}}^s d\tau\, B(\tau,{\underline{q}})}F(s,{\underline{q}})+
{\underline{u}}({\underline{q}})\right\},
$$
that solution is represented by
$$
u(t,\bar{q})=U(t,y(t,{\underline{t}};{\bar{q}}))
$$
where
$B(t,{\underline{q}})=b(t,q(t,{\underline{t}};{\underline{q}}))$,  
$F(t,{\underline{q}})=f(t,q(t,{\underline{t}};{\underline{q}}))$
and
${\underline{q}}=y(t,{\underline{t}};{\bar{q}})$ 
is an inverse function defined through
${\bar{q}}=q(t,{\underline{t}};{\underline{q}})$.
\end{thm}

To understand the above theorem, we take the simplest example:
\begin{equation}\label{mc2}\gathered
i\hbar\pdt u(t,q)=a\frac{\hbar}i{\pdq} u(t,q)+bqu(t,q),\\
u(0,q)={\underline{u}}(q).
\endgathered 
\end{equation}
From the right-hand side of above, we derive the (Weyl) symbol 
as a Hamiltonian:
$$
H(q,p)=e^{-i\hbar^{-1}qp}\Big(a\frac{\hbar}i{\pdq}+bq\Big)e^{i\hbar^{-1}qp}=ap+bq.
$$
The classical mechanics associated to that Hamiltonian is given by
$$\gathered
\dot q(t)=H_p=a,\\
\dot p(t)=-H_q=-b
\endgathered 
$$ with $\begin{pmatrix} q(0)\\ p(0) \end{pmatrix}
=\begin{pmatrix} {\underline{q}}\\ {\underline{p}} \end{pmatrix}$
which is readily solved as
$$
q(s)={\underline{q}}+as,\quad p(s)={\underline{p}}-bs.
$$
Putting $U(t)=u(t,q(t))$ for a solution $u(t,q)$ of \eqref{mc2},
we have 
$i\hbar \frac{d}{dt} U(t)=bq(t)U(t)$, therefore
$$
U(t,{\underline{q}})=
{\underline{u}}({\underline{q}})
e^{-i\hbar^{-1}(b{\underline{q}}t+2^{-1}ab{t^2})}.
$$
As the inverse function of ${\overline{q}}=q(t,{\underline{q}})$ 
is given by 
${\underline{q}}=y(t,\bar q)=\bar q-at$, we get
$$
u(t,{\overline{q}})=U(t,{\underline{q}})|_{{\underline{q}}=y(t,\bar q)}
={\underline{u}}({\overline{q}}-at)
e^{-i\hbar^{-1}(b{\overline{q}}t-2^{-1}ab{t^2})}.
$$
\par{\it Remark.} In the above, we used only the path $q(s)$ moving in
the configuration space ${\mathbb R}^m$.

\vspace{2mm}
\noindent
{\bf Another point of view from the ``Hamiltonian path-integral method"}:
Put
$$
S_0(t,{\underline{q}},{\underline{p}})
=\int_0^t ds \,[\dot q(s)p(s)-H(q(s),p(s))]
=-b{\underline{q}}t-2^{-1}ab{t^2},
$$
and
$$
S(t,{\overline{q}},{\underline{p}})
={\underline{q}}\,{\underline{p}}+S_0(t,{\underline{q}},{\underline{p}})
|_{{\underline{q}}=y(t,{\overline{q}})}
={\overline{q}}{\underline{p}}-a{\underline{p}}t-b{\overline{q}}t+2^{-1}ab{t^2}.
$$
Then, $S(t,{\bar{q}},{\underline{p}})$ 
satisfies the Hamilton-Jacobi equation.
$$\gathered
\pdt S+H({\bar{q}},\partial_{\bar{q}}S)=0,\\
S(0,{\bar{q}},{\underline{p}})={\bar{q}}\,{\underline{p}}.
\endgathered 
$$
On the other hand, the van Vleck determinant, a scalar in this example, is
$$
D(t,{\overline{q}},{\underline{p}})={\frac{\partial^2S(t,{\overline{q}},{\underline{p}})}
{\partial{\overline{q}}\partial{\underline{p}}}}=1.
$$
This quantity satisfies the continuity equation:
$$ \gathered 
\pdt {D}+\partial_{\bar{q}}({D} H_p)=0,\\
D(0,\overline{q},\underline{p})=1
\endgathered
$$
where $H_p=\frac{\partial H}{\partial {p}}
\big({\bar{q}},\frac{\partial S}{\partial \bar{q}}\big)$.

As an interpretation of Feynman's idea, 
we understand the transition from classical to quantum mechanics
by studying the quantity
$$
u(t,{\overline{q}})=(2{\pi}i\hbar)^{-1/2}\int_{{\mathbb R}} d{\underline{p}}\,
D^{1/2}(t,{\overline{q}},{\underline{p}})
e^{i\hbar^{-1}S(t,{\overline{q}},{\underline{p}})}
\hat{\underline{u}}({\underline{p}}).
$$
That is, in our case at hand, we should study the quantity defined by
$$
\begin{aligned}
u(t,{\overline{q}})
&=(2{\pi}i\hbar)^{-1/2}\int_{{\mathbb R}} d{\underline{p}}\,
e^{i\hbar^{-1}S(t,{\overline{q}},{\underline{p}})}
\hat{\underline{u}}({\underline{p}})\\
&=(2{\pi}i\hbar)^{-1}\iint_{{\mathbb R}^2} d{\underline{p}}d{\underline{q}}\,
e^{i\hbar^{-1}(S(t,{\overline{q}},{\underline{p}})-{\underline{q}}\,{\underline{p}})}
{\underline{u}}({\underline{q}})\\
&={\underline{u}}({\overline{q}}-at)
e^{i\hbar^{-1}(-b{\overline{q}}t+2^{-1}ab{t^2})}.
\end{aligned}
$$

[{\bf{Problem}}] 
Can we extend the above argument to a system of PDEs?
For example, Dirac, Weyl or Pauli equations,
quantum mechanical equations with spin.

{\it Remark. }
Using the representation theory of the Heisenberg group,
the procedure of geometric quantization produces the operator 
$e^{-i\hbar^{-1}t(a{\pmb P}+b{\pmb Q}+c{\pmb I})}$
corresponding to the Hamiltonian $ap+bq+c$.

\subsection{A new look for the matrix structure}
The fact ``Clifford algebra is represented on Grassmann algebra"
is embodied as follows (here, we restrict ourselves to $2\times 2$ matrices):

We decompose a $2\times 2$ matrix $A$ as follows:
$$
\begin{aligned}
A&={\begin{pmatrix}
a&b\\
c&d
\end{pmatrix}}\\
&=\frac{a+d}2 {\mathbb I_2}
+\frac{a-d}2
\begin{pmatrix}
1&0\\
0&-1
\end{pmatrix}
+\frac{b+c}2
\begin{pmatrix}
0&1\\
1&0
\end{pmatrix}
+\frac{b-c}2
\begin{pmatrix}
0&1\\
-1&0
\end{pmatrix}\\
&=\frac{a+d}2 {\mathbb I_2}
+\frac{a-d}2{\pmb\sigma}_3
+\frac{b+c}2{\pmb\sigma}_1+\frac{b-c}2i{\pmb\sigma}_2.
\end{aligned}
$$
where the Pauli matrices $\{{\pmb\sigma}_j\}$ satisfies
the {\bf Clifford relation}: 
${\pmb\sigma}_i{\pmb\sigma}_j+{\pmb\sigma}_j{\pmb\sigma}_i=2\delta_{ij}$.

Using the {\bf identification}
$$
{\begin{pmatrix} u_0\\u_1\end{pmatrix}}
\;\begin{matrix}{\overset{\sharp}{\rightarrow}}\\[-8pt]
{\underset{\flat}{\leftarrow}}\end{matrix}\;
u_0+u_1\theta_1\theta_2=u(\theta)$$
we have, for example,
$$
\flat\Big(\theta_1\theta_2
-\frac{\partial^2}{\partial\theta_1\partial\theta_2}\Big)\sharp
{\begin{pmatrix} u_0\\u_1\end{pmatrix}}
=\flat (u_1+u_0\theta_1\theta_2)=
{\begin{pmatrix} u_1\\u_0\end{pmatrix}}
=\pmb{\sigma}_1{\begin{pmatrix} u_0\\u_1\end{pmatrix}}.
$$
Therefore, the action of $A$ on a vector ${}^t (u_0,u_1) $ is regarded 
as the action of a differential operator
$$
{\mathcal A}\Big(\theta,\frac{\partial}{\partial\theta}\Big)
=\frac{a+d}2 %{\mathbb I}
+\frac{a-d}2\Big(1-\theta_1\frac{\partial}{\partial\theta_1}
-\theta_2\frac{\partial}{\partial\theta_2}\Big)+c\theta_1\theta_2
-b\frac{\partial^2}{\partial\theta_1\partial\theta_2}
$$
on a function $u(\theta)=u_0+u_1\theta_1\theta_2$.
Moreover, we may associate the complete Weyl symbol ${\mathcal A}(\theta,\pi)$ as
$$
{\mathcal A}(\theta,\pi)=\frac{a+d}2
-i\frac{a-d}2\hbar^{-1}\big(\theta_1\pi_1+\theta_2\pi_2\big)
+c\theta_1\theta_2+b\hbar^{-2}\pi_1\pi_2.
$$
This is our interpretation of the fact above.
(Though we may define
${\mathcal A}(\theta,\partial_\theta)$ 
and the Fourier transformations w.r.t. odd variables
such that the symbol ${\mathcal A}(\theta,\pi)$ doesn't contain $\hbar$,
we take this form in this paper.)

{\it Remark.}
${\sigma}_j(\theta,\partial_\theta)$ are taken not only to be even operators
but also to annihilate the set
$\{v_1\theta_1+v_2\theta_2\}$.

{\it Remark.}
We may regard $\theta_j\sim dz_j$ and 
$\frac{\partial}{\partial{\theta_j}}\sim\frac{\partial}{\partial z_j}\rfloor$,
 i.e., for $j=1,2$,
$$
\begin{gathered}
\theta_j(u_0+u_1\theta_1\theta_2)=u_0\theta_j\sim
d{z_j}{\wedge}(u_0+u_1dz_1{\wedge}dz_2)=u_0 dz_j,\\
\frac{\partial}{\partial{\theta_1}}(u_0+u_1\theta_1\theta_2)
=u_1\theta_2\sim
\frac{\partial}{\partial z_1}\rfloor(u_0+u_1dz_1{\wedge}dz_2)=u_1dz_2,\\
\frac{\partial}{\partial{\theta_2}}(u_0+u_1\theta_1\theta_2)
=-u_1\theta_1\sim
\frac{\partial}{\partial z_2}\rfloor(u_0+u_1dz_1{\wedge}dz_2)=-u_1dz_1.\\
\end{gathered}
$$


\section{Outline of our procedure}
For the sake of simplicity, we outline the case when ${\mathbb H}(t)$
is independent of time~$t$.
\par
(I) We identify a ``spinor"
$\psi(t,q)={}^t(\psi_1(t,q),\psi_2(t,q))
:{\mathbb R}\times{\mathbb R}^3\to {\mathbb C}^2$
with an even supersmooth function 
$
u(t,x,\theta)=u_0(t,x)+u_1(t,x)\theta_1\theta_2:
{\mathbb R}\times{\mathfrak R}^{3|2}\to {\mathfrak C}_{{\rm ev}}.
$ 
Here, 
$u_0(t,x)$, $u_1(t,x)$ are the Grassmann continuation of $\psi_1(t,q)$,
$\psi_2(t,q)$, respectively.

(II) We represent the matrices $\{\pmb\sigma_j\}$, 
which act on $u(t,x,\theta)$ as follows:
$$ \gathered \sigma_1\Big(\theta,
\frac{\partial}{\partial\theta}\Big)=
\theta_1\theta_2
-\frac{\partial^2}{\partial\theta_1\partial\theta_2},\\
\sigma_2\Big(\theta,
\frac{\partial}{\partial\theta}\Big)=
i\Big(\theta_1\theta_2
+\frac{\partial^2}{\partial\theta_1\partial\theta_2}\Big),\\
\sigma_3\Big(\theta,
\frac{\partial}{\partial\theta}\Big)=
1-\theta_1\frac{\partial}{\partial\theta_1}
-\theta_2\frac{\partial}{\partial\theta_2}.
\endgathered
$$

(III) Therefore, we may correspond the differential operator given by
\begin{equation}
{\mathcal H}\Big(x,\frac{\hbar}i \pdx,\theta,{\pdtheta}\Big)
=c\sum_{j=1}^3\sigma_j\Big(\theta,\frac{\partial}{\partial\theta}\Big)
\Big(\frac{\hbar}i\frac{\partial}{\partial
x_j}-\frac{\varepsilon}{c}A_j(x)\Big)+\varepsilon A_0(x)
\label{*}
\end{equation}
which yields the superspace version of the Weyl equation
\begin{equation} \gathered
i\hbar \pdt u(t,x,\theta)
={\mathcal H}\Big(x,\frac{\hbar}i \pdx,\theta,{\pdtheta}\Big)u(t,x,\theta),\\
u(0,x,\theta)=u(x,\theta).
\endgathered
\label{**}
\end{equation}
Moreover, the ``complete Weyl symbol'' of \eqref{*} is given by
$$
\begin{aligned}
{\mathcal H}(x,\xi,\theta,\pi)
=&c(\eta_1+i\eta_2)\theta_1\theta_2
+c\hbar^{-2}(\eta_1-i\eta_2)\pi_1\pi_2\\
&\qquad
-ic\hbar^{-1}\eta_3(\theta_1\pi_1+\theta_2\pi_2)+\varepsilon A_0(x),
\end{aligned} 
$$
where $\eta_j=\xi_j-({\varepsilon}/{c})A_j(x)$.

(IV) We consider the classical mechanics corresponding to 
${\mathcal H}(x,\xi,\theta,\pi)$ given by
$$ \gathered 
{\frac{d}{dt}} x_j=\frac{\partial{\mathcal H}(x,\xi,\theta,\pi)}{\partial \xi_j},\;
{\frac{d}{dt}} \xi_k=-\frac{\partial{\mathcal H}(x,\xi,\theta,\pi)}{\partial x_k},\\
{\frac{d}{dt}}\theta_l=-\frac{\partial{\mathcal H}(x,\xi,\theta,\pi)}{\partial \pi_l},\;
{\frac{d}{dt}}\pi_m=-\frac{\partial{\mathcal H}(x,\xi,\theta,\pi)}{\partial \theta_m}.
\endgathered
$$

\begin{prop} %\proclaim{Proposition}%\label{prop:existence}
There exists a unique global solution
$(x(t),\xi(t),\theta(t),\pi(t))$  of
above ODE with initial data 
$(x(0),\xi(0),\theta(0),\pi(0))=({\inidata})
\in{{\mathfrak R}}^{6|4}={\mathcal T}^*{{\mathfrak R}}^{3|2}$. %=(y,\eta,\omega,\rho)
\end{prop}

\begin{prop} %\proclaim{Proposition}%\label{prop:inverse}
For any fixed $(t,{\underline{\xi}},{\underline{\pi}})$, 
the map defined by 
$$
({\underline{x}},{\underline{\theta}})\to
(\underline{x}=x(t,{\inidata}),\underline{\theta}=\theta(t,{\inidata}))
$$ 
gives a supersmooth diffeomorphism from ${{\mathfrak R}}^{3|2}\to {{\mathfrak R}}^{3|2}$.
Therefore,
there exists the inverse map given by
$$
({\underline{x}},{\underline{\theta}})\to
(\underline{x}=y(t,{\mixdata}), \underline{\theta}=\omega(t,{\mixdata})), $$
which satisfies
$$ \gathered
{\underline{x}}=x(t,y(t,{\mixdata}),\underline{\xi},\omega(t,{\mixdata}),\underline{\pi}),\\
{\underline{\theta}}=\theta(t,y(t,{\mixdata}),\underline{\xi},\omega(t,{\mixdata}),\underline{\pi}),\\
{\underline{x}}=y(t,x(t,{\inidata}),\underline{\xi},\theta(t,{\inidata}),\underline{\pi}),\\
{\underline{\theta}}=\omega(t,x(t,{\inidata}),\underline{\xi},\theta(t,{\inidata}),\underline{\pi}).
\endgathered
$$
\end{prop}

Now, we put
$$
{\mathcal S}_0(t,{\inidata})=\int_0^t ds\,\{\langle \dot x(s)|\xi(s) \rangle
+\langle \dot\theta(s)|\pi(s)\rangle
-{\mathcal H}(x(s),\xi(s),\theta(s),\pi(s))\},
$$
and
$$
{\mathcal S}(t,{\mixdata})=\langle {\underline{x}}|{\underline{\xi}}\rangle+
\langle {\underline{\theta}}|{\underline{\pi}}\rangle+{\mathcal S}_0(t,{\inidata})
\Big|_{\underline{x}=y(t,{\mixdata}),\;
\underline{\theta}=\omega(t,{\mixdata})}
%\begin{Sb}\underline{x}=y(t,{\mixdata})\\
%\underline{\theta}=\omega(t,{\mixdata})\end{Sb}
$$
\begin{prop} %\proclaim{Proposition} %\label{prop:HJ}
${\mathcal S}(t,{\mixdata})$ satisfies the Hamilton-Jacobi equation:
$$ \gathered
\pdt {\mathcal S}(t,{\mixdata})+{\mathcal H}
\Big(x,\frac{\partial {\mathcal S}}{\partial\bar{x}},
\bar{\theta},\frac{\partial {\mathcal S}}{\partial \bar{\theta}}\Big)=0,\\
{\mathcal S}(0,{\mixdata})=\langle \bar{x}|{\underline{\xi}}\rangle+
\langle\bar{\theta}|{\underline{\pi}}\rangle. %\hbar\hbar^{-1}
\endgathered
$$
\end{prop}

{\it Remark.} This process of constructing solution of H-J equation is 
essentially due to Jacobi.

Now, we put
$$
{\mathcal D}(t,{\mixdata})=\sdet %(-1)^{3+2}\,
\begin{pmatrix}
\frac{\partial^2{\mathcal S}}{\partial \bar{x}\,\partial {\underline{\xi}}}&
\frac{\partial^2{\mathcal S}}{\partial \bar{x}\,\partial {\underline{\pi}}}\\
\frac{\partial^2{\mathcal S}}{\partial \bar{\theta}\,\partial {\underline{\xi}}}&
\frac{\partial^2{\mathcal S}}{\partial \bar{\theta}\,\partial {\underline{\pi}}}
\end{pmatrix}.
$$
Then, we get
\begin{prop} %\proclaim{Proposition} %\label{prop:continuity}
${\mathcal D}(t,{\mixdata})$ satisfies the following continuity equation:
$$\gathered
\pdt {\mathcal D} +\frac{\partial}{\partial\bar{x}}
\Big({\mathcal D}
\frac{\partial{\mathcal H}}{\partial {\xi}}\Big)
+\frac{\partial}{\partial \bar{\theta}}
\Big({\mathcal D}
\frac{\partial{\mathcal H}}{\partial {\pi}}\Big)=0,\\
{\mathcal D}(0,{\mixdata})=1.
\endgathered
$$
In the above, the argument of ${\mathcal D}$ is $(t,{\mixdata})$, and
those of $\frac{\partial{\mathcal H}}{\partial \xi}$ and 
$\frac{\partial{\mathcal H}}{\partial \pi}$ are $(\bar{x},\frac{\partial
{\mathcal S}}{\partial \bar{x}}, \bar\theta, 
\frac{\partial {\mathcal S}}{\partial \bar{\theta}})$,
respectively.
\end{prop}

From here, we change the order of variables 
$$
({\mixdata})\to({\mixxdata}).
$$ 

We define an operator
$$
({\mathcal U}(t)u)(\bar{x},\bar\theta)=(2{\pi}i\hbar)^{-3/2}\hbar\iint
d{\underline{\xi}}d{\underline{\pi}}\,
{\mathcal D}^{1/2}(t,{\mixxdata})
e^{i\hbar^{-1}{{\mathcal S}}(t,{\mixxdata})}
{\mathcal F}u({\underline{\xi}},{\underline{\pi}}).
$$
The function $u(t,\bar{x},\bar\theta)=({\mathcal U}(t){\underline{u}})(\bar{x},\bar\theta)$
will be shown to be a desired good parametrix for \eqref{SW}, 
identical with \eqref{**}.

(V) On the other hand, using Fourier transformation, we have
readily that 
$$
{\mathcal H}\Big(x,{\frac{\hbar}i}{\pdx},\theta,{\pdtheta}\Big)
={\hat{\mathcal H}}
$$
where ${\hat{\mathcal H}}$ is a (Weyl type) pseudo-differential operator with
symbol ${\mathcal H}(x,\xi,\theta,\pi)$, that is, 
$$\begin{aligned}
&({\hat{\mathcal H}}u)(x,\theta) \\
&=(2{\pi}i\hbar)^{-3}\hbar^2 \iint d\xi d\pi dyd\omega \,
e^{i\hbar^{-1}
(\langle x-y|\xi\rangle +\langle \theta-\omega|\pi\rangle)}
{\mathcal H}\bigg(\frac{x+y}2,\xi,\frac{\theta+\omega}2,\pi\bigg)u(y,\omega).
\end{aligned}$$

\begin{thm} 
(1) For $t\in{\mathbb R}$,
${\mathcal U}(t)$ is a bounded operator in 
$\clsl_{S\!S}^2({{\mathfrak R}}^{3|2})$ such that for $|t|, |s|\ll 1$,
$$
\Vert {\mathcal U}(t){\mathcal U}(s)-{\mathcal U}(t+s)
\Vert_{{\mathbb B}
(\,\clsl_{S\!S}^2({{\mathfrak R}}^{3|2}),\,\clsl_{S\!S}^2({{\mathfrak R}}^{3|2}))}\leq
C(|t|^2+|s|^2).
$$
\newline
(2) ${\mathbb R}\ni t\to {\mathcal U}(t)\in {\mathbb
B}(\,\clsl_{S\!S}^2({{\mathfrak R}}^{3|2}),\,\clsl_{S\!S}^2({{\mathfrak R}}^{3|2}))$ 
is continuous.
\newline
(3) There exists the limit ${\mathcal E}(t)=\lim_{n\to\infty}{\mathcal U}(t/n)$
for any $|t|<\infty$ in the uniform operator topology. 
\newline
(4) Put 
$u(t,\underline{x},\underline{\theta})=({\mathcal E}(t){\underline{u}})(\underline{x},\underline{\theta})$,
for ${\underline{u}}\in \ccsl_{S\!S}({\mathfrak R}^{3|2})$ 
with ${\underline{u}}_0(q)$, ${\underline{u}}_1(q)$ being compactly supported.
Then
$$ \gathered
i\hbar\pdt u(t,\underline{x},\underline{\theta})={\hat{\mathcal H}} u(t,\underline{x},\underline{\theta}),\\
 u(0,\underline{x},\underline{\theta})={\underline{u}}(\underline{x},\underline{\theta}).
\endgathered 
$$
\end{thm}

(VI) We interpret the above theorem using
the identification maps
$$
L^2({\mathbb R}^3:{\mathbb C}^2)
\;\begin{matrix}{\overset{\sharp}{\rightarrow}}\\[-8pt]%9
{\underset{\flat}{\leftarrow}}\end{matrix}\;
\clsl_{S\!S}^2({\mathfrak R}^{3|2}).
$$
That is, remarking 
$\flat{\hat{\mathcal H}}\sharp={\mathbb H}$
and putting ${\mathbb E}(t)=\flat{\mathcal E}(t)\sharp$,
we have
\begin{thm} %\proclaim{Theorem}
(1) For $t\in{\mathbb R}$,
${\mathbb E}(t)$ is a well defined unitary operator in $L^2({\mathbb R}^3:{\mathbb
C}^2)$. 
\newline
(2) 
(i) ${\mathbb R}\ni t\to {\mathbb E}(t)\in {\mathbb
B}(L^2({\mathbb R}^3:{\mathbb C}^2),L^2({\mathbb R}^3:{\mathbb C}^2))$ is continuous. 
\newline
(ii) ${\mathbb E}(t){\mathbb E}(s)={\mathbb E}(t+s)$ for any $t,s\in{\mathbb R}$.
\newline
(iii)
Put 
$\psi(t,q)=\flat\big({\mathcal E}(t)
\sharp{\underline{\psi}}\big)\big|_
{\underline{x}_{\mathrm B}=q}$, 
for ${\underline{\psi}}\in C_0^\infty
({\mathbb R}^3:{\mathbb C}^2)$. Then 
$$ \gathered
i\hbar\pdt \psi(t,q)={\mathbb H}\psi(t,q),\\
\psi(0,q)={\underline{\psi}}(q).
\endgathered
$$
\end{thm}

\begin{cor}
${\mathbb H}$ is an essentially self-adjoint operator 
in $L^2({\mathbb R}^3:{\mathbb C}^2)$.
\end{cor}

%%[n-elem-1.tex]
\appendix
\section{Elements from superanalysis}
In this appendix, we gather notion from superanalysis without proof. 
As is well-known, we use very heavily Taylor expansion, integration by parts,
change of variables under integral sign and Fourier transformations
when we perform real analysis on ${\mathbb R}^m$.
After defining non-commutative numbers ${\mathfrak R}$ and ${\mathfrak C}$, called supernumbers, 
we develop such tools on superspace ${\mathfrak R}^{m|n}$.
We must emphasize the reason why Rogers \cite{Ro80}
%J. Hoyos et al.\cite{HQRdU}, et al, 
used the Banach-Grassmann algebra $B_\infty$, which is obtained 
from the sequence space $\ell^1$ by defining the product with Grassmann relations.
%V.S. Vladimirov and I.V. Volovich \cite{VV83,VV84} is used the finite dimensional
%Grassmann algebra. 
It is because not only she but also many others try to apply 
the general theory of differential calculus on Banach space, 
which is not valid on Fr\'echet space in general. 
On the other hand, 
de Witt \cite{dW} introduced his non-Hausdorff topology (called coarse topology) 
on his supermanifolds, 
which is scarcely used by mathematicians, because it seems difficult
to find a firm foundation of differential calculus. 
Our topology is just between coarse and Banach topology,
which guarantees to develop that calculus by the help of grading
inherited from Grassmann relations, i.e., we develop the theory of
differential calculus on the Frech\'et space with Grassmann grading.
\subsection{Supernumbers, superspaces and linear algebra}
\subsubsection{Supernumbers}
For symbols
$\{\sigma_j\}_{j=1}^{\infty}$ satisfying the Grassmann relation
$$
\sigma_j \sigma_k+ \sigma_k \sigma_j=0,
\quad j,k=1,2,\cdots,
$$
we put
$$
{{\mathfrak C}}=\{X=\sum_{I\in\mathcal I} X_I\sigma^I\;\big|\; X_I\in{\mathbb C}\}
$$
where
$$
\begin{gathered}
{\mathcal I}=\{I=(i_k)\in \{0,1\}^{\mathbb N}\;\big|\;
|I|=\sum_k i_k < \infty \},\\
\sigma^I=\sigma_1^{i_1}\sigma_2^{i_2}\cdots,\quad I=(i_1,i_2,\cdots ),\quad
\sigma^{\tilde 0}=1, \quad {\tilde 0}=(0,0,\cdots )\in\mathcal I.
\end{gathered}
$$

Besides trivially defined linear operations of sums and scalar
multiplications, we have a product operation in ${\mathfrak C}$: 
For 
$$
X=\sum_{J\in\mathcal J} X_J\sigma^J, \quad 
Y=\sum_{K\in\mathcal I} Y_K\sigma^K,
$$ 
we put
$$
XY=\sum_{I\in\mathcal I} (XY)_I\sigma^I \quad\mbox{with}\quad
(XY)_I=\sum_{I=J+K}(-1)^{\tau(I;J,K)}X_JY_K.
$$ 
Here, $\tau(I;J,K)$ is an integer defined by
$$
\sigma^J \sigma^K =(-1)^{\tau(I;J,K)}\sigma^I, \quad
I=J+K.
$$

\begin{prop}[\cite{IM91,In92}]
${\mathfrak C}$ forms an $\infty$-dimensional Fr\'echet-Grassmann algebra over ${\mathbb C}$,
that is, an associative, distributive and non-commutative ring with degree,
which is endowed with the Fr\'echet topology.
\end{prop}

{\it Remark. }
(1) We call this ${\mathfrak C}$ as super(complex)numbers.
Degree in ${\mathfrak C}$ is defined by introducing subspaces
$$
{{\mathfrak C}}^{[j]}=\{X=\sum_{I\in\mathcal I,|I|=j} X_I\sigma^I\} \quad\mbox{for}\quad j=0,1,\cdots
$$
which satisfy
$$
{{\mathfrak C}}=\oplus_{j=0}^\infty {{\mathfrak C}}^{[j]},\quad
{{\mathfrak C}}^{[j]}\cdot {{\mathfrak C}}^{[k]} \subset {{\mathfrak C}}^{[j+k]}.
$$
\newline
(2) Define 
$$
\proj_I(X)=X_I \quad\mbox{for}\quad 
X=\sum_{I\in\mathcal I}X_I\sigma^I\in{{\mathfrak C}}.
$$
The topology in ${\mathfrak C}$ is given by
$X\to 0$ in ${\mathfrak C}$ if and only if $\proj_I(X)\to 0$ in ${\mathbb C}$,
for any $I\in\mathcal I$.

This topology is equivalent to the one introduced by the metric
$\dist(X,Y)=\dist(X-Y)$ where $\dist(X)$ is defined by
$$
\dist(X)=\sum_{I\in\mathcal I}\frac 1{2^{r(I)}} \frac
{|\proj_I(X)|}{1+|\proj_I(X)|} 
\quad\mbox{with}\quad
r(I)=1+\frac 12 \sum_{k=1}^\infty 2^k i_k \quad\mbox{for}\quad I\in\mathcal I.
$$
\newline
(3) We introduce parity in ${\mathfrak C}$ by setting
$$
p(X)=\begin{cases}
0 & \text{if $X=\sum_{I\in\mathcal I,|I|={\mathrm{ev}}}X_I\sigma^I$},\\
1 & \text{if $X=\sum_{I\in\mathcal I,|I|={\mathrm{od}}}X_I\sigma^I$},\\
\text{undefined} & \text{otherwise}.
\end{cases}
$$
$X\in{\mathfrak C}$ is called homogeneous if it satisfies $p(X)=0$ or $=1$.
We put 
$$ \gathered
{{\mathfrak C}_{{\rm ev}}} = \oplus_{j=0}^\infty {{\mathfrak C}}^{[2j]}
=\{X\in{{\mathfrak C}}\,|\,p(X)=0\},\\
{{\mathfrak C}_{{\rm od}}} = \oplus_{j=0}^\infty {{\mathfrak C}}^{[2j+1]}
=\{X\in{{\mathfrak C}}\,|\,p(X)=1\},\\
{{\mathfrak C}} \cong {{\mathfrak C}_{{\rm ev}}}\oplus{{\mathfrak C}_{{\rm od}}}\cong{{\mathfrak C}_{{\rm ev}}}\times{{\mathfrak C}_{{\rm od}}}.
\endgathered
$$
Analogous to ${\mathfrak C}$, we define, super(real)numbers as
$$\gathered
{{\mathfrak R}}=\{X\in{{\mathfrak C}}\,|\, \pi_{\mathrm B}X\in{\mathbb R}\},\;
{{\mathfrak R}}^{[j]}={{\mathfrak R}}\cap {{\mathfrak C}}^{[j]},\\
{{\mathfrak R}_{{\rm ev}}}={{\mathfrak R}}\cap{{\mathfrak C}_{{\rm ev}}},\quad {{\mathfrak R}_{{\rm od}}}={{\mathfrak R}}\cap{{\mathfrak C}_{{\rm od}}}={{\mathfrak C}_{{\rm od}}},\\
{{\mathfrak R}}\cong {{\mathfrak R}_{{\rm ev}}}\oplus{{\mathfrak R}_{{\rm od}}}\cong{{\mathfrak R}_{{\rm ev}}}\times{{\mathfrak R}_{{\rm od}}}.
\endgathered
$$
We introduced the body (projection) map $\pi_{\mathrm B}$ by 
$$
\pi_{\mathrm B}X=\proj_{\tilde 0}(X)=X_{\tilde 0}=X^{[0]}=X_{\mathrm B}
\quad\mbox{for any}\quad X\in{\mathfrak C},
$$
and the soul part $X_{\mathrm S}$ of $X$ as
$$
X_{\mathrm S}=X-X_{\mathrm B}=\sum_{|I|\ge 1}X_I\sigma^I.
$$

\subsubsection{Superspaces} 
We define the (real) superspace ${\mathfrak R}^{m|n}$ by
$$
{\mathfrak R}^{m|n}={{\mathfrak R}}_{\mathrm{ev}}^m \times {{\mathfrak R}}_{\mathrm{od}}^n.
$$
The distance between $X,Y\in{\mathfrak R}^{m|n}$ is defined by, 
$$
\dist_{m|n}(X,Y)=\dist_{m|n}(X-Y)
$$ 
with 
$$
\dist_{m|n}(X)=
\sum_{j=1}^m
\left( \sum_{I\in\mathcal I}{\frac 1{2^{r(I)}}}
{\frac{|\proj_I(x_j)|}{1+|\proj_I(x_j)|}} \right)
+\sum_{k=1}^n
\left( \sum_{I\in\mathcal I}
{\frac 1{2^{r(I)}}}
{\frac{|\proj_I(\theta_k)|}{1+|\proj_I(\theta_k)|}}\right).
$$
We use the following notation:
$$
\begin{gathered}
X=(X_A)_{A=1}^{m+n}=(x,\theta)\in{\mathfrak R}^{m|n}  \quad\mbox{with}\quad\\
 x=(X_A)_{A=1}^{m}=(x_j)_{j=1}^m\in{\mathfrak R}^{m|0},\quad
\theta=(X_A)_{A=m+1}^{m+n}=(\theta_k)_{k=1}^n\in{\mathfrak R}^{0|n},\mbox{etc}.
\end{gathered}
$$
We generalize the body map $\pi_{\mathrm B}$ from ${\mathfrak R}^{m|n}$ 
or ${\mathfrak R}^{m|0}$ to ${\mathbb R}^m$ by putting,
$$
X=(x,\theta)\in{\mathfrak R}^{m|n}\longrightarrow
\pi_{\mathrm B}X=X_{\mathrm B}=(x_{\mathrm B},0)
\cong x_{\mathrm B}
=\pi_{\mathrm B}x
=(\pi_{\mathrm B}x_1,\cdots, \pi_{\mathrm B}x_m)\in{\mathbb R}^m.
$$
We call $x_j\in{\mathfrak R}_{{\rm ev}}$ and
$\theta_k\in{\mathfrak R}_{{\rm od}}$ as
even and odd (alias bosonic and ferminionic) variable, respectively.

\par{\it Remark. } 
Though the differential calculus on Fr\'echet
spaces has some difficulties in general (e.g. Yamamuro \cite{Yam}),  
such calculus on Fr\'echet-Grassmann algebra holds safely in our case,
because of the grading of Grassmann generators.
For example,
the implicit and inverse function theorems, 
and the chain rule for differentiation.
See, Inoue and Maeda \cite{IM91}, Inoue \cite{In92,In98-1,In99-2}.

%\midspace{0.1em}

\subsubsection{Elementary Linear Algebra}
\begin{defn}
A rectangular array $M$,  whose cells are indexed by pairs consisting of
a row number and a column number, is called a supermatrix
and denoted by $M\in\Mat((m|n) \times (r|s):{{\mathfrak C}})$,
if it satisfies the following:
\begin{enumerate}
\item
A $(m+n) \times (r+s)$ matrix $M$ 
is decomposed blockwisely as
$M=
\begin{bmatrix}
A&C\\ 
D&B 
\end{bmatrix}$
where $A$, $B$, $C$ and $D$ are $m\times r$, $n\times s$, $m\times s$ and 
$n\times r$ matrices with elements in ${{\mathfrak C}}$, respectively.
\item
One of the following conditions is satisfied: Either
\begin{itemize}
\item
$p(M)=0$, that is, $p(A_{jk})=0=p(B_{uv})$ and $p(C_{jv})=1=p(D_{uk})$ or
\item 
$p(M)=1$, that is, $p(A_{jk})=1=p(B_{uv})$ and $p(C_{jv})=0=p(D_{uk})$. 
\end{itemize}
\end{enumerate}
We call $M$ is even 
denoted by $\Mat_{\mathrm{ev}}((m|n) \times (r|s):{\mathfrak C})$
(resp. odd denoted by $\Mat_{\mathrm{od}}((m|n) \times (r|s):{\mathfrak C})$) 
if $p(M)=0$ (resp. $p(M)=1$). Therefore, we have
$$
\Mat((m|n) \times (r|s):{{\mathfrak C}})=\Mat_{\mathrm{ev}}((m|n) \times (r|s):{\mathfrak C})
\oplus\Mat_{\mathrm{od}}((m|n) \times (r|s):{\mathfrak C}).
$$
Moreover, we may decompose $M$ as $M=M_{\mathrm B}+M_{\mathrm S}$ where
$$
M_{\mathrm B}= \begin{cases} 
 \begin{bmatrix} A_{\mathrm B}& 0\\  0&B_{\mathrm B}  \end{bmatrix}
& \text{ when $p(M)=0$} ,\\[5mm] 
 \begin{bmatrix} 0&C_{\mathrm B}\\  D_{\mathrm B}&0 \end{bmatrix}
& \text{when $p(M)=1$.}
\end{cases}
$$
\end{defn}

\vskip 3mm

The summation of two matrices in $\Mat_{\mathrm{ev}}((m|n) \times (r|s):{\mathfrak C})$
or in $\Mat_{\mathrm{od}}((m|n) \times (r|s):{\mathfrak C})$ is defined as usual,
but the sum of $\Mat_{\mathrm{ev}}((m|n) \times (r|s):{\mathfrak C})$ 
and $\Mat_{\mathrm{od}}((m|n) \times (r|s):{\mathfrak C})$ is not defined 
in itself except at least one of them being zero matrix.

It is clear that if $M$ is the $(m+n) \times (r+s)$ matrix and $N$ is the 
$(r+s)\times (p+q)$ matrix, 
then we may define the product $MN$ and its parity $p(MN)$ as
$$
(MN)_{ij}=\sum_kM_{ik}N_{kj},\quad
p(MN)=p(M)+p(N)\mod 2.
$$

For notational simplicity, we put
$\Mat[m|n:{{\mathfrak C}}]=\Mat((m|n)\times(m|n) :{\mathfrak C})$.

\begin{defn}
Let $M=
\begin{bmatrix}
A&C\\
D&B
\end{bmatrix} \in \Mat[m|n:{\mathfrak C}]$. 
We define the supertrace of $M$ by
$$
\str M %=\sum_{i} (-1)^{(p(M)+1)p_{row}(i)}M_{ii}
= \tr A - (-1)^{p(M)} \tr B.
$$
\end{defn}
 
We get
\begin{prop}
(a) Let $M,N \in \Mat[m|n:{{\mathfrak C}}]$ such that $p(M)+p(N)\equiv 0 \bmod 2$. 
Then, we have
$$
\str (M+N)= \str M + \str N.
$$
(b) $M$ is a matrix of size $(m+n)\times (r+s)$ and
 $N$ is a matrix of size $(r+s)\times (m+n)$. Then,
$$
\str (MN)= (-1)^{p(M)p(N)} \str (NM).
$$
\end{prop}

If $M\in \Mat[m|n:{{\mathfrak C}}]$ is even, denoted by 
$M\in \Mat_{\mathrm{ev}}[m|n:{{\mathfrak C}}]$, then $M$ acts on ${\mathfrak R}^{m|n}$ linearly.
Denoting this by $T_M$, we call it super linear transformation on ${\mathfrak R}^{m|n}$
and $M$ is called the representative matrix of $T_M$.

\begin{prop}
Let $M\in \Mat_{\mathrm{ev}}[m|n:{{\mathfrak C}}]$ and assume 
%that the body $M_B$ of $M$ is non-singular, that is, 
$\det M_{\mathrm B}\neq 0$.
Then, for given $Y\in {\mathfrak R}^{m|n}$,
\begin{equation*}
T_M X=Y
%\label{LE}
\end{equation*}
has the unique solution $X\in{\mathfrak R}^{m|n}$, which is denoted by $X=M^{-1}Y$.
\end{prop}
\begin{defn}
$M\in \Mat_{\mathrm{ev}}[m|n:{{\mathfrak C}}]$ is called invertible or non-singular
if $M_{\mathrm B}$ is invertible, 
i.e., $\det A_{\mathrm B} \det B_{\mathrm B} \neq 0$,
and denoted by $M\in \GL_{\mathrm{ev}}[m|n:{{\mathfrak C}}]$.
% if $p(M)=0$ or
%$\det C_{\mathrm B} \det D_{\mathrm B} \neq 0$ if $p(M)=1$.
\end{defn}

\begin{defn}
 Let $B=(B_{jk})$ be $(\ell\times \ell)$-matrix with elements in ${{\mathfrak C}_{{\rm ev}}}$, 
denoted by, $B\in\Mat[\ell:{\mathfrak C}_{{\rm ev}}]$.
As ${{\mathfrak C}_{{\rm ev}}}$ is a commutative ring, we may define $\det B$ as usual:
$$
\det B=\sum_{\rho\in\wp_\ell} \sgn (\rho) B_{1\,\rho(1)}\cdots B_{\ell\,\rho(\ell)}.
$$
\end{defn}
Then, we have, as ordinary case,
\begin{equation}
\det(AB)=\det A\det B,\; \det(\exp A)=\exp( \tr A) \;\mbox{for}\; A,B\in\Mat[\ell:{\mathfrak C}_{{\rm ev}}].
\end{equation}

\begin{defn}
Let $M\in \Mat_{\mathrm{ev}}[m|n:{{\mathfrak C}}]$ be given. 
When $\det B_{\mathrm B}\neq 0$, we put
$$
\sdet M=(\det (A-CB^{-1}D))(\det B)^{-1}
$$
and call it superdeterminant or Berezinian of $M$.
\end{defn}

\begin{cor}
When $\det B_{\mathrm B}\neq 0$ and $\sdet M\neq 0$, 
then $\det A_{\mathrm B}\neq 0$.
\end{cor}

\begin{thm}
Let $M,N \in \Mat_{\mathrm{ev}}[m|n:{{\mathfrak C}}]$.
\par
(1)
If $M$ is invertible, then we have $\sdet M\neq 0$. 
Moreover, if $A$ is nonsingular, then 
\begin{equation*}
(\sdet M)^{-1}=(\det A)^{-1} (\det (B-DA^{-1}C)).
\end{equation*}

(2) Multiplicativity of $\sdet$ on $\GL_{\mathrm{ev}}[m|n:{{\mathfrak C}}]$:
\begin{equation*}
\sdet (MN)=\sdet M \sdet N.
\end{equation*}

(3)
$\str$ and $\sdet$ are matrix invariants on $\GL_{\mathrm{ev}}[m|n:{{\mathfrak C}}]$:
That is, for $M,\;N\in\GL_{\mathrm{ev}}[m|n:{{\mathfrak C}}]$,
$$\gathered
\str M=\str NMN^{-1} \\
% &\quad\text{if $N$ is invertible},(-1)^{p(M)+p(N)}
\sdet M= \sdet NMN^{-1}.
%&\quad\text{if $M\in\GL_{\mathrm{ev}}[m|n:{{\mathfrak C}}]$ and $N$ is invertible.}
\endgathered
$$ 

(4) Moreover, we have
\begin{equation*}
\exp(\str M)=\sdet(\exp M)\quad\text{for $M\in\GL_{\mathrm{ev}}[m|n:{{\mathfrak C}}]$}.
\end{equation*}
\end{thm}

%%[n-elem-2.tex]
\subsection{Elementary analysis I. Differential Calculus}
\subsubsection{Supersmooth functions}
\begin{defn}
For any
$f(q)\in C^\infty({\mathbb R}^m:{\mathbb C})$, 
we put, 
$$
{\tilde f}(x)=\sum_{|\alpha|=0}^\infty{\frac 1{\alpha!}}
\partial_q^\alpha f(x_{\mathrm B}) x_{\mathrm S}^\alpha
\in{\mathfrak C}_{\mathrm {ev}}
\quad\text{for $x=x_{\mathrm B}+x_{\mathrm S}$}\in
{\mathfrak R}^{m|0}
$$
which is called the Grassmann continuation (or extension) 
of $f(q)$, and denoted simply by $f(x)$.
\par
Moreover, if $g(q)=\sum_{I\in{\mathcal I}}g_I(q)\sigma^I\in 
C^\infty({\mathbb R}^m:{\mathfrak C})$, 
we define $g(x)=\sum_{I\in{\mathcal I}}g_I(x)\sigma^I$
where $g_I(x)$ is the Grassmann continuation of $g_I(q)$.
\end{defn}

\begin{defn}
We define functions $u\in\ccsl_{\mathrm{SS,ev}}({\mathfrak R}^{m|n})$,
or $u\in{\mathcal C}_{\mathrm{SS}}({\mathfrak R}^{m|n}:{\mathfrak C})$ by
$$
u(X)=u(x,\theta)=\sum_{|a|\le n}{\tilde u}_a(x)\theta^a
=\text{or simply}\;\sum_{|a|\le n}u_a(x)\theta^a,
$$
which are called supersmooth functions on ${{\mathfrak R}}^{m|n}$.
Here $u_a(x)$ is the Grassmann continuation 
of $u_a(q)\in C^\infty({\mathbb R}^m:{\mathbb C})$, or
$u_{a}(q)\in C^\infty({\mathbb R}^m:{\mathfrak C})$.
\end{defn}

%\midspace{0.1em}
 
{\it Example. } 
For $\xi=(\xi_1,\cdots,\xi_m)\in{{\mathfrak R}}^{m|0}={\mathfrak R}_{{\rm ev}}^m$, 
we define $|\xi|\in{\mathfrak R}_{{\rm ev}}$ as follows:
Putting 
$$
|\xi|=|\xi|_{\mathrm B}+|\xi|_{\mathrm S}\quad\mbox{with}\quad
|\xi|_{\mathrm S}=\sum_{|I|={\mathrm{even}}\ge2}|\xi|_{I}\sigma^I,
\quad |\xi|_{\mathrm B}\ge0,\;|\xi|_{I}\in{\mathbb R},
$$
we should have
$$
\begin{gathered}
|\xi|^2=\sum_{j=1}^m(\xi_{j,\mathrm B}+\xi_{j,\mathrm S})
(\xi_{j,\mathrm B}+\overline{\xi_{j,\mathrm S}})
=\sum_{j=1}^m\xi_{j,\mathrm B}^2+\sum_{j=1}^m\xi_{j,\mathrm B}
(\xi_{j,\mathrm S}+\overline{\xi_{j,\mathrm S}})
+\sum_{j=1}^m\xi_{j,\mathrm S}\overline{\xi_{j,\mathrm S}},\\
\xi_{j,\mathrm S}=\sum_{|I|={\mathrm{even}}\ge2} \xi_{j,I}\sigma^I,\;
\overline{\xi_{j,\mathrm S}}=\sum_{|I|={\mathrm{even}}\ge2}
\overline{\xi_{j,I}}\sigma^I
\end{gathered} $$
with $\overline{\xi_{j,I}}$ being the complex conjugate 
of $\xi_{j,I}$ in $\mathbb C$.
Therefore, $|\xi|_{\mathrm B}=\{\sum_{j=1}^m\xi_{j,\mathrm B}^2\}^{1/2}$
and
\begin{equation*} \aligned
2|\xi|_{K}|\xi|_{\mathrm B}
&+\sum_{I+J=K}|\xi|_{I}\overline{|\xi|_{J}}(-1)^{\tau(K;I,J)}\\
&\qquad
=\sum_{j=1}^m 2\xi_{j,\mathrm B}\,\Re\xi_{j,K}
+\sum_{I+J=K}\sum_{j=1}^m\xi_{j,I}\overline{\xi_{j,J}}(-1)^{\tau(K;I,J)}
\endaligned
\end{equation*}
which are solved by induction with respect to the length $|K|$.
For example,
if $|K|=2$, we have
$$
|\xi|_{K}=|\xi|_{\mathrm B}^{-1}
\sum_{j=1}^m\xi_{j,\mathrm B}\,\Re\xi_{j,K}.
$$
If $|K|=4$,
\begin{equation*}\aligned
2|\xi|_{K}&=|\xi|_{\mathrm B}^{-1}
\big(2\sum_{j=1}^m\xi_{j,\mathrm B}\,\Re\xi_{j,K}+
\sum_{I+J=K}\sum_{j=1}^m\xi_{j,I}\overline{\xi_{j,J}}(-1)^{\tau(K;I,J)}\\
&\qquad\qquad\qquad\qquad
-\sum_{I+J=K}\sum_{j=1}^m|\xi|_I|\xi|_J(-1)^{\tau(K;I,J)}\big),
\quad \mbox{etc}.
\endaligned
\end{equation*}

Now, we define $\sin|\xi|$ and $\cos|\xi|$ as 
$$
\sin|\xi|=\sum_{n=0}^\infty 
\frac 1{n!}\sin\big(|\xi|_{\mathrm B}+\frac{n\pi}2\big)|\xi|_{\mathrm S}^n,\quad
\cos|\xi|=\sum_{n=0}^\infty 
\frac 1{n!}\cos\big(|\xi|_{\mathrm B}+\frac{n\pi}2\big)|\xi|_{\mathrm S}^n.
$$
We may characterize these supersmooth functions as follows:


\begin{defn} A set $U=\pi_{\mathrm B}(U)\times{{\mathfrak R}}_{\mathrm{od}}^n$
is called a superdomain if $\pi_{\mathrm B}(U)$ is a domain in ${\mathbb R}^m$.
Let a function $f$ from a superdomain 
$U\subset{\mathfrak R}^{m|n}$ to ${\mathfrak C}$
be given.

(i) It is called F-differentiable at $X=(X_A)_{A=1}^{m+n}$ 
in the direction $Y=(Y_A)_{A=1}^{m+n}$ if
there exist 
$$
\frac{d}{dt} f(X+tY)\bigg|_{t=0}=f_F'(X;Y).
$$
We denote $f_F'(X;Y_A)=f_{X_A}'(X;Y_A)$ for $A=1,\cdots,m+n$ 
and are called Fr\'echet derivatives.
\par(ii) $f$ is called G-differentiable at $X=(x,\theta)$ if
there exist $F_A(X)\in{\mathfrak C}$ and $R_A(X,Y)\in{\mathfrak C}$ such that
$$
f(X+Y)-f(X)=\sum_{A=1}^{m+n}Y_AF_A(X)+\sum_{A=1}^{m+n}Y_AR_A(X,Y),
$$
which satisfy
$$
d(R_A(X,Y),0)\to0\quad\mbox{when}\quad d_{m|n}(Y,0)\to0.
$$
\end{defn}

\begin{prop}
Let $f$ be a function from a superdomain $U\subset{\mathfrak R}^{m|n}$ to ${\mathfrak C}$.
Then, the following are equivalent:
\newline
(i) $f$ be supersmooth, denoted by ${{\mathcal C}_{{\rm SS}}} (U:{{\mathfrak C}})$,
\newline
(ii) $f$ is G-differentiable,
\newline
(iii) $f$ is F-differentiable and there exist functions $F_A(X)$ such that
$$
f_{X_A}'(X;Y_A)=Y_A F_A(X)  \quad\mbox{for any}\quad  A=1,\cdots,m+n,
$$
(iv) $f$ is F-differentiable and satisfies
$$
\begin{gathered}
Z_Af'_{X_A}(X;Y_A)-(-1)^{p(Y_A)p(Z_A)}Y_Af'_{X_A}(X;Z_A)=0\quad\mbox{for}\quad
p(X_A)=p(Y_A)=p(Z_A),\\
f'_{X_A}(X;Y_AZ_A)=Z_Af'_{X_A}(X;Y_A)\quad\mbox{for}\quad p(Z_A)=0 \quad\mbox{and}
\quad p(X_A)=p(Y_A).
\end{gathered}
$$
\end{prop}


To understand the meaning of supersmoothness, we consider the dependence with
respect to the `coordinate' more precisely.
\begin{prop}\label{3.5}
Let $ f=\sum_I f_I(X)\sigma^I\in{{\mathcal C}_{{\rm SS}}} (U:{{\mathfrak C}})$ 
where $U$ is a superdomain in ${\mathfrak R}^{m|n}$.
Let $X=(X_A)_{A=1}^{m+n}$ be represented by 
$X_{A}=\sum_{I} X_{A,I}\sigma^I$ where $A=1,\cdots, m+n$,
$ X_{A,I}\in {\mathbb C}$ for $|I|\ne 0$ and $ X_{A,0}\in {{\mathbb R}}$.
Then, $f(X)$, considered as a function 
of countably many variables $\{X_{A,I}\}$ with values in ${\mathfrak C}$,
satisfies the following (Cauchy-Riemann type) equations.
\begin{gather}
\frac{\partial}{\partial  X_{A,I}}f(X) =
\sigma^I {\frac{\partial}{\partial X_{A,0}}}f(X)
 \quad\mbox{for}\quad 1\le A \le m,\,\, |I|={\mbox{even}}, \label{II-3-12} \\
\sigma^K{\frac{\partial}{\partial X_{A,J}}}f(X)
+\sigma^J{\frac{\partial}{\partial X_{A,K}}}f(X)=0
\mbox{ for $m+1\le A \le m+n$  $|J|=|k|=$  odd. } \notag
\end{gather}

Here, we define
\begin{equation}
{\frac{\partial}{\partial  X_{A,I}}}f(X) =
{\frac d{dt}}f(X+tY_{(A,I)})\big|_{t=0},
\label{II-3-13}
\end{equation}
with
$Y_{(A,I)}=
(\overbrace{0,\cdots,0,1}^A,0,\cdots,0)\sigma^I\in{{\mathfrak R}^{m|n}}$ and
$Y_{(A,0)}=(\overbrace{0,\cdots,0,1}^A,0,\cdots,0)\in{{\mathfrak R}^{m|n}}$.
Conversely, let a function $f(X)=\sum_I f_I(X)\sigma^I$ be given such that
$f_I(X+tY)\in C^\infty([0,1]:{\mathbb C})$ for each fixed $X,Y\in U$ and
$f(X)$ satisfies the following equations:
\begin{equation}\gathered
\left.{\frac d{dt}}f(X+tY_{(A,I)})\right|_{t=0}=
\left.{\frac d{dt}}f(X+tY_{(A,0)})\right|_{t=0}\sigma^I
\quad\mbox{for}\quad 1\le A \le m,\,\, |J|={\mbox{even}}\\
\sigma^K \left.{\frac d{dt}}f(X+tY_{(A,J)})\right|_{t=0}+
\sigma^J \left.{\frac d{dt}}f(X+tY_{(A,K)})\right|_{t=0}=0\\
 \quad\mbox{for}\quad m+1\le A \le m+n,\,\, |J|={\mbox{odd}}=|K|.
\endgathered
\label{II-3-14}
\end{equation}
Then,  $ f \in {{\mathcal C}_{{\rm SS}}} (U:{{\mathfrak C}})$.
\end{prop}

\subsubsection{Derivations}
For a given supersmooth function $u(X)$ on ${{\mathfrak R}}^{m|n}$, 
we define its derivatives as follows:
For
$ j=1,2, \cdots ,m $ and $ k = 1,2, \cdots ,n$,
we put
$$\gathered
U_j(X)= \sum_{|a|\le n} \partial_{x_j} u_a(x) \theta^a ,\\
U_{k+m} (X)=\sum_{|a| \le n} (-1)^{l_k(a)} u_a(x) 
\theta_1^{a_1}\cdots\theta_{k}^{{a_k}-1}\cdots\theta_n^{a_n}
\endgathered
$$
where $l_k(a)= \sum_{j=1}^{k-1} a_j $ and
$\theta_{k}^{-1}=0$.
$U_{\kappa}(X)$ are called the partial derivatives of $u$ 
with respect to $ X_{\kappa} $ at
$ X = (x, \theta ) $ and are denoted by
$$\gathered
U_j (X) = {\displaystyle{\frac{\partial}{\partial x_j}}} u(x, \theta )
= \partial_{x_j} u(x, \theta ) \quad\mbox{for}\quad j=1,2, \cdots ,m , \\
U_{m+s}(X) = {\displaystyle{\frac{\partial}{\partial\theta_s}}}u(x,\theta )
= \partial_{\theta_s} u(x, \theta ) \quad\mbox{for}\quad s = 1,2, \cdots ,n 
\endgathered
$$
or simply by
$$
U_{\kappa}(X)={\partial}_{X_{\kappa}}u(X) \quad\mbox{for}\quad \kappa=1,\cdots,m+n.
$$
For 
$$
\begin{gathered}
{{\mathfrak a}}=(\alpha,a),\;
\alpha=(\alpha_1,\cdots,\alpha_m)\in{\mathbf 
 N}^m,\;
a=(a_1,\cdots,a_n)\in\{0,1\}^n,\\
|\alpha|=\sum_{j=1}^m \alpha_j, \;\;
|a|=\sum_{k=1}^n a_k, \;\;
|{\mathfrak a}|=|\alpha|+|a|,
\end{gathered}
$$
we put
$$
\begin{gathered}
\partial_X^{{\mathfrak a}}=\partial_x^\alpha \partial_\theta^a
\quad\mbox{with}\quad
\partial_x^\alpha =\partial_{x_1}^{\alpha_1} \cdots \partial_{x_n}^{\alpha_m},
\quad
\partial_\theta^a=\partial_{\theta_1}^{a_1}\cdots \partial_{\theta_n}^{a_n}.
\end{gathered}
$$

{\it Example. }
$
\partial_{\theta_2}\theta_1\theta_2\theta_3=-\theta_1\theta_3,\quad
\partial_{\theta_1}\partial_{\theta_3}\theta_1\theta_2\theta_3=\theta_2
\neq-\theta_2=\partial_{\theta_3}\partial_{\theta_1}\theta_1\theta_2\theta_3,
\quad \mbox{etc}.
$

\vspace{2mm}
{\bf Remarks for the need of $\infty$ number of Grassmann generators.}
\par
(i) Though  ${\mathfrak C}$ does not form a field because $X^2=0$ for any $X\in{\mathfrak C}_{{\rm od}}$,
but if  $X,Y\in{\mathfrak C}$ satisfy  $XY=0$ for any $Y\in{{\mathfrak C}_{{\rm od}}}$, then  $X=0$.
This property holds only when the number of generators is infinite.
By this, we may determine the derivative $\partial_X^{{\mathfrak a}}u(X)$ uniquely.
\par
(ii) In general, we need at least countable number of operations in doing analysis.
If the number of Grassmann generators is finite, 
then the effect of odd variables may vanish after finitely many operations.
\par
(iii) Rothstein \cite{Rot86} claims that the super Lie algebra 
{\it Der}$\,({\mathcal C}_{\mathrm{SS}}({\mathcal U})$) 
is the free module of 
${\mathcal C}_{\mathrm{SS}}({\mathcal U})$
only when the number of the Grassmann generators is infinite or null.
(Though Rothstein used the Banach-Grassmann algebra introduced by
Rogers \cite{Ro80}, but his argument is also applicable in our situation.)

\subsubsection{Taylor expansions and Implicit function theorem}
\begin{defn}\label{3.4}
 For a supersmooth function $f$, we define $df$ by
$$
df(X)=d_Xf(X)=\sum_{\kappa=1}^{m+n} 
dX_\kappa\frac{\partial f(X)}{\partial X_\kappa},
$$
or
$$
df(x,\theta)=
\sum_{j=1}^m dx_j\frac{\partial f(x,\theta)}{\partial x_j}
+\sum_{s=1}^n d\theta_s\frac{\partial f(x,\theta)}{\partial \theta_s}.
$$
\end{defn}

From Definition \ref{3.4}, we get readily
\begin{prop}\label{3.13}
Let $U$ be a superdomain in ${\mathfrak R}^{m|n}$.
For $f,g \in {{\mathcal C}_{{\rm SS}}} (U:{{\mathfrak C}})$,
the product $fg$ belongs to ${{\mathcal C}_{{\rm SS}}} (U:{{\mathfrak C}})$ and
the differentials $ {d_X} f(X)$ and ${d_X} g(X)$ 
may be regarded as continuous linear mappings from 
${\mathfrak R}^{m|n}$  
into ${{\mathfrak C}}^{m+n}$.
Moreover, they satisfy the following:
%\begin{enumerate}
\par (1) %\item
For any homogeneous elements $\lambda,\, \mu \in {{\mathfrak C}}$, we have
\begin{equation}
{d_X} ( \lambda f + \mu g)(X) =
(-1)^{p(\lambda)p(X)} \lambda {d_X} f(X) 
+ (-1)^{p(\mu)p(X)} \mu {d_X} g(X).
\label{3-21}
\end{equation}
\par (2) %\item
(Leibnitz formula)
\begin{equation}
\partial_{X_{\kappa}} [f(X)g(X)] 
= 
(\partial_{X_{\kappa}} f(X)) g(X)+ 
(-1)^{p( X_{\kappa} ) p(f(X))}
f(X) (\partial_{X_{\kappa}}g(X)).
\label{3-22}
\end{equation}
%\end{enumerate}
\end{prop}

\begin{prop}[Taylor's formula]
Let $X=(x,\theta),Y=(y,\omega)\in U\subset {\mathfrak R}^{m|n}$
satisfying $Y+t(X-Y)\in U$ for $0\le t\le 1$.
%Take $r_0>0$ sufficiently small such that
%$\{ x_{\mathrm B}\in{\mathbb R}^m;|x_{\mathrm B}-y_{\mathrm B}|<r_0\} 
%\subset \pi_{\mathrm B}(U)$. 
For $ f \in  {{\mathcal C}_{{\rm SS}}} (U:{{\mathfrak C}})$, Taylor's formula holds. 
That is, for any positive integer $p$, we have
\begin{equation}
f (x, \theta ) -
\sum_{|\alpha|+|a|\le p,\;|a|\le n} 
%\limits_{\scriptstyle|\alpha|+|a|\le p\atop\scriptstyle |a|\le n} 
{\frac{1}{\alpha ! }}
(x- y)^{\alpha} 
(\theta - \omega)^a 
\partial_x^{\alpha} \partial_\theta^a 
f(y,\omega ) 
= \tau_p (X,Y)
\label{3-14}
\end{equation}
where
\begin{eqnarray}
\tau_p (X,Y)&=& \sum_{|\alpha|+|a|=p+1,\; |a|\le n}
%\begin{Sb}|\alpha|+|a|=p+1,\\|a|\le n\end{Sb}
%\limits_{|\alpha|+|a|=p+1,\atop\scriptstyle |a|\le n}
(x-y)^{\alpha} (\theta - \omega)^a  \label{3-15} \\
&&\times \int_0^1 \! dt \, {\frac{1}{p!}} (1-t)^p \partial_x^{\alpha} 
\partial_\theta^a
f(y +t(x- y ),\omega+ t( \theta - \omega)). \nonumber
\end{eqnarray}
\end{prop}

\begin{defn}\label{3.14}
Let $U\subset{\mathfrak R}^{m|n}$ and $U'\subset {{\mathfrak R}}^{\!m'|n'}$
be superdomains 
and let $\varphi$ be a continuous mapping from $U$ to $U'$, denoted by
$$\varphi(X)=(\varphi_1(X),\cdots, \varphi_{m'}(X),
\varphi_{m'+1}(X),\cdots, \varphi_{m'+n'}(X))\in{{\mathfrak R}}^{\!m'|n'}.$$
$\varphi$ is called a supersmooth mapping from $U$ to $U'$
if each $\varphi_\kappa(X)\in {{\mathcal C}_{{\rm SS}}}(U:{{\mathfrak C}})$ for $\kappa=1,\cdots, m'+n'$ and
$\varphi (U)\subset U'$.
\end{defn}

\begin{prop}[Composition of supersmooth mappings]\label{3.15}
Let $U\subset{{\mathfrak R}}^{m|n}$ and $U'\subset {{\mathfrak R}}^{m'|n'}$
be superdomains and 
let $\Phi: U\to U'$ and $\Phi':U'\to{{\mathfrak R}}^{m''|n''}$
be supersmooth mappings.
{\par}
Then, the composition $\Psi=\Phi'\circ\Phi:U\to{{\mathfrak R}}^{m''|n''}$ 
gives a supersmooth mapping and
\begin{equation}
d_X\Psi(X)
=[d_Y\Phi'(Y)]\big|_{Y=\Phi(X)}[d_X\Phi(X)].
\label{3-23}
\end{equation}
\end{prop}

\begin{defn}\label{3.16}
Let $U\subset{\mathfrak R}^{m|n}$ and
$U'\subset{{\mathfrak R}}^{\!m'|n'}$ be superdomains and
let $\varphi:U\to U'$ be a supersmooth mapping represented by
$\varphi(X)=(\varphi_1(X),\cdots,\varphi_{m'+n'}(X))$ with
$\varphi_\kappa(X)\in {{\mathcal C}_{{\rm SS}}}(U:{{\mathfrak C}})$.
\newline
(1) $\varphi$ is called a supersmooth diffeomorphism if 
(i) $\varphi$ is a homeomorphism between
$U$ and $U'$ and
(ii) $\varphi$ and $\varphi^{-1}$ are supersmooth mappings. 
\newline
(2) For any $f\in {{\mathcal C}_{{\rm SS}}}(U':{{\mathfrak C}})$, 
$(\varphi^* f)(X)=(f\circ \varphi)(X)=f(\varphi(X))$,
called the {\it pull back} of $f$,
is well-defined and belongs to ${{\mathcal C}_{{\rm SS}}}(U:{{\mathfrak C}})$.
\end{defn}

\par
{\it Remarks}.
(1) It is easy to see that if $\varphi$ is a supersmooth diffeomorphism, then
$\varphi_{\mathrm B}=\pi_{\mathrm B} \circ \varphi$ is 
an (ordinary) $C^\infty$ diffeomorphism from $U_{\mathrm B}$ to $U'_{\mathrm B}$.
\newline
(2) If we introduce the topologies in
${{\mathcal C}_{{\rm SS}}}(U':{{\mathfrak C}})$ and ${{\mathcal C}_{{\rm SS}}}(U:{{\mathfrak C}})$ properly,
$\varphi^*$ gives a continuous linear mapping from 
${{\mathcal C}_{{\rm SS}}}(U':{{\mathfrak C}})$ to ${{\mathcal C}_{{\rm SS}}}(U:{{\mathfrak C}})$.
Moreover, if $\varphi:U\to U'$ is a supersmooth diffeomorphism,
then $\varphi^*$ defines an automorphism from
${{\mathcal C}_{{\rm SS}}}(U':{{\mathfrak C}})$ to ${{\mathcal C}_{{\rm SS}}}(U:{{\mathfrak C}})$.

%\vskip 2mm

\begin{prop}[Inverse function theorem]\label{3.17} 
Let $U$ be  a superdomain in ${\mathfrak R}^{m|n}$ and let
$G(X): U \subset {{\mathfrak R}^{m|n}}\rightarrow \, { {\mathfrak R}^{m|n}}$
be a supersmooth mapping. 
We assume the super matrix $[d_X G(X)]$
is invertible at $X={\tilde X}_{\mathrm B} \in \pi_{\mathrm B}(U)$.
{\par}
Then, there exists a superdomain $ U'$, a neighborhood of 
${\tilde Y} = G( {\tilde X} ) $ 
and a unique supersmooth mapping $F$
satisfying $F(G(X)) = X $ and we have
%${\tilde X}= F({\tilde Y})$, 
\begin{equation}
\left.{d_Y}F(Y)= ({d_X} G(X))^{-1} \,\right|_{X=F(Y)} 
\quad \text{in}\quad U^\prime .
\label{3-31}
\end{equation}
\end{prop}

Moreover, we have
\begin{prop}[Implicit function theorem]\label{3.18}
Let $\Phi (X,Y): U \times U' \to {{\mathfrak C}}^{m'|n'}$ 
be a supersmooth mapping and $( {\tilde X }, {\tilde Y}) \in U \times U' $,
where $U$ and $U'$ are superdomains of ${{\mathfrak R}^{m|n}}$ 
and ${{\mathfrak R}}^{\!m'|n'}$,
respectively.
Suppose $\Phi ( {\tilde X }, {\tilde Y} )=0$ and
$ {\partial}_Y \Phi
=[\partial_{y_j} \Phi , \partial_{\omega_r} \Phi ]$
is a continuous and invertible supermatrix 
at $({\tilde X}_{\mathrm B},{\tilde Y}_{\mathrm B})
\in \pi_{\mathrm B}(U) \times \pi_{\mathrm B}(U')$.
{\par}
Then, there exist a superdomain $V \subset U $ 
satisfying ${\tilde X}_{\mathrm B}\in \pi_{\mathrm B}(V)$ 
and a unique supersmooth mapping $Y=f(X)$ on V 
such that $ {\tilde Y} = f( {\tilde X} )$ and $ \Phi (X,f(X))=0 $ in V.
Moreover, we have
\begin{equation}
{\partial_X} f(X) 
= -\left.{[ \partial_Y \Phi(X,Y)]^{-1}}
[\partial_X \Phi(X,Y)] \,\right|_{Y=f(X)}.
\label{3-46}
\end{equation}
\end{prop}


%\midspace{0.1em}

\subsection{Elementary analysis II. Integral Calculus}

\subsubsection{Integration (even case)}
Now, we define the integration of a {supersmooth} function $u(x)$ 
on an even superdomain $U_{\mathrm{ev}} \subset {\mathfrak R}^{m|0}$, 
which is similar to the integral of holomorphic functions on a complex domain.
(See, Rogers \cite{Ro80,Ro86-2}.) 
\begin{defn}\label{4.1} 
Let $u(x)$ be a {supersmooth} function defined on a even super domain 
$U_{\mathrm{ev}}\subset{{\mathfrak R}}^{\!1|0}$.
Let $ \lambda $= $\lambda_{\mathrm B} + \lambda_{\mathrm S}$, 
$\mu $=$ \mu_{\mathrm B} + \mu_{\mathrm S} \in U_{\mathrm{ev}} $ 
and let a continuous and piecewise $C^1 $-curve 
$ c : [ \lambda_{\mathrm B} , \mu_{\mathrm B} ] \,\rightarrow \, U_{\mathrm{ev}} $
be given such that 
$c( \lambda_{\mathrm B} ) = \lambda $, 
$c( \mu_{\mathrm B} )  = \mu $.
We define
\begin{equation}
\int_c  dx \, u(x)
=
\int_{\lambda_{\mathrm B}}^{\mu_{\mathrm B}}\!\!dt\,u(c(t)){\dot c}(t)\in {{\mathfrak C}}
\label{4-1}
\end{equation}
and call it the {\rm integral of $u$} along the curve $c$.
\end{defn} 

Using the integration by parts, we get the following fundamental result (see
\cite{dW}). 
\begin{prop}\label{4.2}
Let $u(t) \in C^{\infty} ([\lambda_{\mathrm B},\mu_{\mathrm B}]:{{\mathfrak C}}) $ 
and let $u(x)$ be 
the Grassmann continuation of $u(t)$.
Suppose that there exists a function
$U(t) \in C^{\infty} ([\lambda_{\mathrm B},\mu_{\mathrm B}]:{{\mathfrak C}}) $ 
satisfying $ U' (t)=u(t) $ on
$ [ \lambda_{\mathrm B} , \mu_{\mathrm B} ] $.
{\par}
Then, for any continuous and piecewise
$C^1 $-curve $ c : [ \lambda_{\mathrm B} , \mu_{\mathrm B} ] 
\, \rightarrow \, U_{\mathrm{ev}}\subset{{\mathfrak R}}^{\!1|0}$
such that $c( \lambda_{\mathrm B})=\lambda$, $c(\mu_{\mathrm B})=\mu $,
we have
\begin{equation}
\int_c \!dx \, u(x)= U( \lambda)- U(\mu).
\label{4-2}
\end{equation}
\end{prop}

\begin{cor}\label{4.3}
Let $u(x)$ be a {supersmooth} function defined on a even superdomain 
$U_{\mathrm{ev}} \subset {{\mathfrak R}}^{\!1|0}$ into ${{\mathfrak C}}$.
Let $ c_1 , c_2 $ be continuous and piecewise $C^1 $-curves from 
$[\lambda_{\mathrm B},\mu_{\mathrm B}] \, \rightarrow \, U_{\mathrm{ev}}$
such that $\lambda =c_1 ( \lambda_{\mathrm B} ) =c_2 ( \lambda_{\mathrm B} ) $ 
and $ \mu = c_1 ( \mu_{\mathrm B} ) =c_2 ( \mu_{\mathrm B} ) $.
If $c_1 $ is homotopic to $c_2 $, then 
\begin{equation}
\int_{c_1}\!dx \, u(x)
= \int_{c_2}\!dx \, u(x).
\label{4-3}
\end{equation}
Thus, if 
$[\lambda_{\mathrm B},\mu_{\mathrm B}] \subset \pi_{\mathrm B}(U_{\mathrm{ev}})$, 
we have
\begin{equation}
\int_{\lambda}^{\mu}\! dx \, u(x)
=\int_{\lambda_{\mathrm B}}^{\mu_{\mathrm B}}\! dt \, u (t).
\label{4-4}
\end{equation}
\end{cor}
Because of \eqref{4-4}, we have
\begin{defn}\label{4.4}
(1) Let $I_{\mathrm{ev}}$ be a even superdomain in ${\mathfrak R}^{m|0}$
such that $\pi_{\mathrm B}(I_{\mathrm{ev}})$
$=\prod_{j=1}^m (a_j,b_j){}\subset {{\mathbb R}^m}$
with $-\infty<a_j<b_j<\infty$, which is called a even supercube.
For $u\in {{\mathcal C}_{{\rm SS}}}(I_{\mathrm{ev}}:{{\mathfrak C}})$, we define
\begin{equation}
\int_{I_{\mathrm{ev}}} dx\,u(x)
=\int_{a_1}^{b_1} dq_1\cdots \int_{a_m}^{b_m} dq_m \,u(q_1,\cdots,q_m)
=\int_{\pi_{\mathrm B}(I_{\mathrm{ev}})} \!\! dx_{\mathrm B}\,u(x_{\mathrm B}).
\label{4-5}
\end{equation}
(2) For any even superdomain $U_{\mathrm{ev}} \subset {\mathfrak R}^{m|0}$ such that
$\pi_{\mathrm B}(U_{\mathrm{ev}})$ is of definite area, we may put
\begin{equation}
\int_{U_{\mathrm{ev}}}\!\! dx\, u(x)
=\int_{\pi_{\mathrm B}(U_{\mathrm{ev}})}\!\!dx_{\mathrm B}\,u(x_{\mathrm B})
\label{4-6}
\end{equation}
for $u\in {{\mathcal C}_{{\rm SS}}}(U_{\mathrm{ev}}:{{\mathfrak C}})$.
\end{defn}

\subsubsection{Integration (odd case)}
It seems natural to put formally
$$
d\theta_j=\sum_{I\in{\mathcal I},|I|={\mathrm{od}}}d\theta_{j,I} \sigma^I
\quad\mbox{for}\quad \theta_j=\sum_{I\in{\mathcal I},|I|={\mathrm{od}}}\theta_{j,I} \sigma^I.
$$
Therefore, we have
$d\theta_j\wedge d\theta_k=d\theta_k\wedge d\theta_j$
for $j\neq k$.
This suggests the integration w.r.t. odd variables is quite different 
from the one w.r.t. ordinary variables. In fact, it is defined as follows:

Let $v$ be a polynomial of odd variables
$\theta$=
$(\theta_1 ,\cdots ,\theta_n ) \in$
${{\mathfrak R}}_{\mathrm{od}}^n$ such that
$$
v(\theta_1,\cdots ,\theta_n ) = \sum_{|b| \leq n} v_b \theta^b
\quad \text{with homogeneous $v_b\theta^b \in {{\mathfrak C}}$ for each $b$}.
$$
Denote by 
$ P_n ({{\mathfrak C}}) $ 
the set of all $v$ as above.
\vskip 2mm
\begin{defn}\label{4.5}
For $ v \in P_n ( {{\mathfrak C}}) $, we put 
$$ 
\int_{{\mathfrak R}^{0|n}}\!\!\!\! d \theta \, v( \theta ) 
= \int_{{{\mathfrak R}^{0|n}}} \!\!\!\! 
d \theta_n \cdots d \theta_1 \, v(\theta_1,\cdots ,\theta_n ) 
= (\partial_{\theta_n} \cdots 
\partial_{\theta_1} v)(0)
\label{4-7}
$$
and we call it the integral of $v$ on  $ {\mathfrak R}^{0|n}$.
\end{defn}
Especially for odd integration, we have the following 
curious looking but well-known relations
$$
\int_{{\mathfrak R}^{0|n}} \!\!\!\!d \theta_n \cdots  d \theta_1 
\,\theta_1 \cdots \theta_n =1 \quad\mbox{and}\quad
\int_{{\mathfrak R}^{0|n}} \!\!\!\!d \theta_n \cdots  d \theta_1 \, 1=0
\quad\text{(Berezin integral)}.
$$

Moreover, we have
\begin{prop}\label{4.6}
Given
$ v, \, w \in P_n ({{\mathfrak C}}) $, we have the following:
\begin{enumerate}
\item
($ {{\mathfrak C}}$-linearity ) 
For any homogeneous $ \lambda ,\, \mu \in {{\mathfrak C}}$,
$$
\int_{{\mathfrak R}^{0|n}} \!\!\!\! d \theta( \lambda v + \mu w)( \theta ) =
(-1)^{np(\lambda)}\lambda \int_{{\mathfrak R}^{0|n}} 
\!\!\!\! d\theta \,v( \theta ) +
(-1)^{np(\mu)}\mu \int_{{\mathfrak R}^{0|n}}\!\!\!\! d \theta \,w( \theta ) .
\label{4-9}
$$
\item
(Translational invariance) For any $\rho\in{\mathfrak R}^{0|n}$, we have
$$
\int_{{\mathfrak R}^{0|n}} \!\!\!\! d \theta \,v( \theta + \rho )=
\int_{{\mathfrak R}^{0|n}} \!\!\!\! d \theta \,v( \theta ).
\label{4-10}
$$
\item
(Integration by parts)
For $ v \in P_n ( {{\mathfrak C}})$ such that $p(v) = 1$ or 0, we have
$$
\int_{{\mathfrak R}^{0|n}}\!\!\!\!d\theta 
\,v( \theta ) \partial_{\theta_s} w( \theta ) =
-(-1)^{p(v)}
\int_{{\mathfrak R}^{0|n}}\!\!\!\!d\theta \,(\partial_{\theta_s} v( \theta )) w( \theta ).
\label{4-11}
$$
\item
(Linear change of variables)
Let $A=(A_{jk})$ with $A_{jk}\in {{\mathfrak R}_{{\rm ev}}}$ be invertible.
Then, 
$$
\int_{{\mathfrak R}^{0|n}} \!\!\!\!d \theta \,v( \theta )
= (\det A )^{-1} \int_{{\mathfrak R}^{0|n}} \!\!\!\!d\omega \,v(A\cdot\omega ).
\label{4-12}
$$
\item
(Iteration of integrals)
$$
\int_{{\mathfrak R}^{0|n}} \!\!\!\! d \theta \,v( \theta )
= \int_{{\mathfrak R}{}^{0|n-k}} \!\!\!\!d \theta_n  \cdots  d \theta_{k+1}
 \left( \int_{{\mathfrak R}{}^{0|k}} \!\!\!\!d \theta_{k}  \cdots d \theta_1
\,v(\theta_1, \cdots , \theta_k , \theta_{k+1} , \cdots , \theta_n ) \right).
\label{4-13}
$$
\item
(Odd change of variables)
Let $\theta=\theta(\omega)$ be an odd change of variables such that
$\theta(0)=0$ and
$\det\left.
\displaystyle{{\frac{\partial\theta(\omega)}{\partial \omega}}}
\right|_{\omega=0}\ne 0$.
Then, for any  $ v \in P_n ({{\mathfrak C}})$,
$$
\int_{{\mathfrak R}^{0|n}} \!\!\!\! d \theta \, v(\theta)=
\int_{{\mathfrak R}^{0|n}} \!\!\!\! d \omega \, v(\theta(\omega))
\,{\det}^{-1}{\frac{\partial \theta(\omega)}{\partial \omega}}.
\label{4-14}
$$
\item  
For $ v \in P_n ({{\mathfrak C}})$ and $ \omega \in {\mathfrak R}^{0|n} $,
$$
\int_{{\mathfrak R}^{0|n}} \!\!\!\! d \theta 
\,( \theta_1 - \omega_1 ) \cdots ( \theta_n - \omega_n ) v( \theta ) 
= v( \omega ) .
\label{4-15}
$$  
\end{enumerate}
\end{prop}

\subsubsection{Integration (mixed case)}
Finally, we define 
\begin{defn}\label{4.7}
Let $U=U_{\mathrm{ev}}\times{{\mathfrak R}}_{\mathrm{od}}^n \subset{\mathfrak R}^{m|n}$ 
be a superdomain
and let $u\in {{\mathcal C}_{{\rm SS}}}(U:{{\mathfrak C}})$,
that is, $u(x,\theta)=\sum u_a(x)\theta^a$ with 
$u_a(x)\in {{\mathcal C}_{{\rm SS}}}(U_{\mathrm{ev}}:{{{\mathfrak C}}})$.
Then, we define
$$
\begin{aligned}
\int_{{\mathfrak R}^{m|n}}dxd\theta \,u(x,\theta)&=
\int_{{\mathfrak R}^{m|0}}\!\!\!\!dx\left\{\int_{{\mathfrak R}^{0|n}} 
\!\!\!\!d\theta\,u(x,\theta)\right\}\\
&=\int_{{\mathbb R}^m}\!\!\!\!dX_{\mathrm B}
(\partial_{\theta_n} \cdots 
\partial_{\theta_1} u)(X_{\mathrm B})
\quad (\pi_{\mathrm B}({\mathfrak R}^{m|0})={\mathbb R}^m)\\
&=\int_{{\mathfrak R}^{0|n}}\!\!\!\!d\theta\left\{\int_{{\mathfrak R}^{m|0}}\!\!\!\!dx 
\,u(x,\theta)\right\}
=\int_{{\mathfrak R}^{m|n}}d\theta dx\,u(x,\theta).
\end{aligned}
$$
\end{defn}

\subsubsection{Change of variables under integral sign}
\begin{thm}\label{4.8}
Let
$$
x=x(y,\omega), \quad \theta=\theta(y,\omega)
\label{4-17}
$$
be a {supersmooth} diffeomorphism from 
${\mathfrak R}^{m|n}_Y$ to ${\mathfrak R}^{m|n}_X$.
Putting
$$
M= \begin{bmatrix}
A&C\\ D&B \end{bmatrix},\quad
A={\frac {\partial x}{\partial y}},\quad
B={\frac {\partial \theta}{\partial \omega}},\quad
C={\frac {\partial x}{\partial \omega}},\quad
D={\frac {\partial \theta}{\partial y}},
\label{4-18}
$$
we assume that either
$\det A|_{\omega=0}$ and $\det (B-DA^{-1}C)|_{\omega=0}$,
or
$\det B|_{\omega=0}$ and $\det (A-CB^{-1}D)|_{\omega=0}$,
are non-zero for all $y$.
Then, for any function 
$f\in{{\mathcal C}_{{\rm SS}}}({{\mathfrak R}{}^{m|n}_X}:{{\mathfrak C}})$ with compact support,
we have the change of variables formula
$$
\int_{{\mathfrak R}{}^{m|n}_X}\!\!dxd\theta \, f(x,\theta)
=\int_{{\mathfrak R}{}^{m|n}_Y}\!\!dyd\omega \, 
f(x(y,\omega),\theta(y,\omega))({\sdet} M)(y,\omega).
\label{4-19}
$$
\end{thm}

{\it Remark.}
In case when we consider the integral on a superdomain with boundary
and functions with support intersecting with that boundary,
there occurs some difficulty which is exemplified as follows:
Let $I={\widetilde{(0,1)}}\times{{\mathfrak R}}_{\mathrm{od}}^2\subset{{\mathfrak R}}^{1|2}$,
where
${\widetilde{(0,1)}}=
\{x\in{\mathfrak R}_{{\rm ev}}\,|\, \pi_{\mathrm B}(x)\in(0,1)\}\subset{{\mathfrak R}}_{\mathrm{ev}}$.
We have
$$
\int_Idy d\omega_2 d\omega_1\, y=0.
$$
On the other hand, 
using the change of coordinates for $0\neq\alpha\in{{\mathfrak R}_{{\rm ev}}}$ and for $j=1,2$
$$
y=x+\alpha\theta_1\theta_2,\; \omega_1=\theta_1,\;\omega_2=\theta_2,
$$
which maps $I$ to $I$ and 
yields $\sdet(\frac{\partial(y,\omega)}{\partial(x,\theta)})=1$, 
we have
$$
\int_Idx d\theta_2 d\theta_1\, (x+\alpha\theta_1\theta_2)
=\int_{{\widetilde{(0,1)}}}dx \alpha=\alpha.
$$

The resolution of this difficulty is due to Rothstein \cite{Rot87}, 
but we don't mention it here (see also 
Zirnbauer \cite{Zi96}, Martellini and Teofilatto \cite{MT88}, 
Inoue and Nomura \cite{IN99}).


\subsection{A few elements from real analysis}
\subsubsection{Scalar products and norms} 
Following \cite{In92}, we introduce
$$
\gathered
{\ccsl}_{\mathrm{SS,ev}}({\mathfrak R}^{m|n}) =
\{u(X)=\sum_{|a|={\mathrm{even}}\le n}u_a(x)\theta^a \,\big|\, 
u_a(q)\in C^\infty({\mathbb R}^m:{\mathbb C}) \quad\mbox{for any}\quad a\},\\
{\cdsl}_{\mathrm{SS,ev}}({\mathfrak R}^{m|n}) =
\{u(X)=\sum_{|a|={\mathrm{even}}\le n}u_a(x)\theta^a \,\big|\, 
u_a(q)\in {\mathcal D}({\mathbb R}^m:{\mathbb C}) \quad\mbox{for any}\quad a\},\\
{\cssl}_{\mathrm{SS,ev}}({\mathfrak R}^{m|n}) =
\{u(X)=\sum_{|a|={\mathrm{even}}\le n}u_a(x)\theta^a \,\big|\, 
u_a(q)\in {\mathcal S}({\mathbb R}^m:{\mathbb C}) \quad\mbox{for any}\quad a\},\;\mbox{etc.}
\endgathered
$$
Let another set of odd variables $\{{\bar{\theta}}_j\}_{j=1}^n$
satisfy $\theta_j\theta_k+\theta_k\theta_j
=\theta_j\bar{\theta}_k+\bar{\theta}_k\theta_j=
\bar{\theta}_j\bar{\theta}_k+\bar{\theta}_k\bar{\theta}_j=0$.
We define the conjugation ${\overline{u(x,\theta)}}=\sum_a{\overline{u_a(x)}}\,
{\overline{\theta^a}}$ where
${\overline{\theta^a}}=\bar{\theta}_n^{a_n}\cdots\bar{\theta}_1^{a_1}$ and 
${\overline{u_a(x)}}$ being the complex conjugate of $u_a(x)$. 
Then, we define
$$\gathered
(u,v)
=\int_{{\mathfrak R}^{m|2n}} dx d\theta\,d{\bar\theta}
\,e^{\langle \bar\theta|\theta\rangle}
{\overline {u(x,\theta)}}v(x,\theta)
=\sum_{|a|\le n}\int_{{\mathfrak R}^{m|0}} dx\, 
{\overline{u_a(x)}}v_a(x),\\
(\!( u,v)\!) _k
=\sum_{|{{\mathfrak a}} |\le k } (\partial_X^{{\mathfrak a}}u, \partial_X^{{\mathfrak a}}v)
=\sum_{|{\alpha} |+|a|\le k } 
(\partial_x^{\alpha}u_a,\partial_x^{\alpha}v_a),\\
(\!(\!( u,v)\!)\!)_k 
=\sum_{|{{\mathfrak a}} |+l\le k } 
( (1+|X_{\mathrm B} |^2 )^{l/2 }\partial_X^{{\mathfrak a}}u, 
(1+|X_{\mathrm B} |^2 )^{l/2 }\partial_X^{{\mathfrak a}}v)
\endgathered
$$
with
$$
\Vert u \Vert^2=(u,u),\quad 
\| u\|_k^2 =(\!( u,u)\!) _k,\quad
\trs u {\trs}_k^2=(\!(\!( u,u)\!)\!)_k.
$$
The space $\clsl_{\mathrm{SS,ev}}^2({\mathfrak R}^{m|n})$ 
is the completion of ${\cdsl}_{\mathrm{SS,ev}}({\mathfrak R}^{m|n})$ 
in the norm $\Vert \cdot \Vert$.

Generally, we may identify vectors $L^2({\mathbb R}^m:{\mathbb C}^{r})$ 
with supersmooth functions 
${\clsl}_{\mathrm{SS,ev}}^2({{\mathfrak R}}^{m|n})$ with suitably 
related $r$ and $n$.
Without specifying this relation, we consider only the case
$$
L^2({\mathbb R}^3:{\mathbb C}^2)
\;\begin{matrix}{\overset{\sharp}{\rightarrow}}\\[-8pt]%9
{\underset{\flat}{\leftarrow}}\end{matrix}\;
\clsl_{\mathrm{SS,ev}}^2({\mathfrak R}^{3|2}).
$$
(See, more precisely, \cite{In92,In98-1}.)
%\pagebreak
%\newline

\subsubsection{Fourier transformations} % (of second kind):
For $\hbar\in{\mathbb R}^{\times}$ and $\kbar\in{\mathbb C}^{\times}$,
$v(x), w(\xi)\in{\cssl}_{\mathrm{SS,ev}}({\mathfrak R}^{m|n})$ and
$v(\theta), w(\pi)\in P_n({\mathfrak C})$, we put
$$
\begin{gathered}
(F_e v)(\xi) =(2{\pi}i\hbar)^{-m/2}\int_{{\mathfrak R}^{m|0}} dx\,
e^{-i\hbar^{-1}\langle x|\xi\rangle}v(x),\\
({\bar F}_e w)(x) =(2{\pi}i\hbar)^{-m/2}\int_{{\mathfrak R}^{m|0}} d\xi\,
e^{i\hbar^{-1}\langle x|\xi\rangle}w(\xi),\\
(F_o v)(\pi) =\kbar^{n/2} \iota_n
\int_{{\mathfrak R}^{0|n}} d\theta\,
e^{-i\kbar^{-1}\langle\theta|\pi\rangle}v(\theta),\\
({\bar F}_o w)(\theta) =
\kbar^{n/2}\iota_n
\int_{{\mathfrak R}^{0|n}} d\pi\,
e^{i\kbar^{-1}\langle\theta|\pi\rangle}w(\pi)
\end{gathered}
$$ 
where 
$$
\langle{x}|\xi\rangle = \sum_{j=1}^m x_j \xi_j,
\quad
\langle\theta|\pi\rangle = \sum_{k=1}^n \theta_k \pi_k ,
\quad
\iota_n=e^{-\pi{i}n(n-2)/4}.
$$ 
We put
$$ 
\begin{gathered}
({\mathcal F}u)(\xi,\pi)
= c_{m,n}
\int_{{\mathfrak R}^{m|n}}\!\!dX\,e^{-i\hbar^{-1}\langle X|\Xi\rangle} u(X) 
=\sum_a[(F_eu_a)(\xi)][(F_o\theta^a)(\pi)],\\
({\bar {\mathcal F}}v)(x,\theta)
= c_{m,n}
\int_{{\mathfrak R}^{m|n}}\!\! d\Xi\,e^{i\hbar^{-1}\langle X|\Xi\rangle }v(\Xi)
=\sum_a[({\bar F}_ev_a)(x)][({\bar F}_o\pi^a)(\theta)]
\end{gathered}
$$
where 
$$
\langle X|\Xi\rangle=\langle x|\xi\rangle
+\hbar\kbar^{-1}\langle\theta|\pi\rangle\in{\mathfrak R}_{{\rm ev}},\quad 
c_{m,n}=(2{\pi}i\hbar )^{-m/2}\kbar^{n/2}\iota_n.
$$
Using these Fourier transformations, 
we may prove the Plancherel formula and 
define pseudo-differential operators, Fourier integral operators
analogously as the standard cases.
In \S 4 before, we take $\kbar=\hbar$.

We introduce useful constants $e(a,b)$ and $e(a)$ as follows:
$$
\theta^a\theta^b=(-1)^{e(a,b)}\theta^{a+b},\quad %\\
\int_{{\mathfrak R}^{0|n}}d\theta\,e^{-i\kbar^{-1}\langle\theta|\pi\rangle}\theta^a
=(-i\kbar^{-1})^{n-|a|}(-1)^{e(a)}\pi^{{\tilde 1}-a},
$$
with $e(a)\equiv n|a|-|a|-1+[(n-1-|a|)/2]+e(\bar 1-a,a)
\mod 2$ and $\bar 1=(1,\cdots,1)\in\{0,1\}^n$.


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