\chapter{Grassmann variables and Berezin integration}\label{s:grassmann}

This section introduces the calculus of anticommuting variables that the rest of the paper uses. The supersymmetric method writes an average of a ratio of determinants of $H$ as an integral over commuting and anticommuting variables, in which the average over $H$ can be computed exactly. In \zcref{s:rep}, this gives the exact representation of the source ratio $T_N^{(k)}$ as an integral over supermatrices, and in \zcref{s:sigma} the same calculus describes the limiting sigma model. We write for a reader who has not used supersymmetry before.

The material is standard. Textbook accounts are given by Berezin \cite{berezin1987introduction}, Efetov \cite{efetov1997supersymmetry}, Wegner \cite{wegner2016supermathematics}, and, for mathematicians, Varadarajan \cite{varadarajan2004supersymmetry}. Introductions aimed at probabilists are given in \cite{spencer2012susy} and \cite[Chapter~11]{bauerschmidt2019introduction}, and Zirnbauer \cite{zirnbauer2006supersymmetry} reviews the method for Hermitian band matrices, with emphasis on the correctness of the mathematical reasoning. We prove the facts that later sections use, so that the paper can be read without these references. The one exception is the general change of variables formula, \zcref{t:grass-change}, for which we refer to the literature.

We begin in \zcref{s:grass-motivation} by explaining why anticommuting variables appear. \zcref{s:grass-algebra} defines Grassmann algebras and functions of even elements, and \zcref{s:grass-berezin} introduces the Berezin integral and computes Gaussian integrals. \zcref{s:grass-supermatrix} treats supermatrices, the supertrace, and the superdeterminant. With these tools, \zcref{s:grass-example} carries out the whole method for a matrix from the Gaussian unitary ensemble, which shows what the calculus is for. \zcref{s:grass-change} then discusses changes of variables and the boundary terms they can produce, and in \zcref{s:grass-susy}, we prove the localization theorem in flat space; the same mechanism normalizes the sigma model in \zcref{s:sigma}. The coefficient norms, with which \zcref{s:sigma} and Part IV estimate superfunctions, are defined in \zcref{s:sigma-norms}, where they are first used.

\section{Motivation}\label{s:grass-motivation}

Let $A$ be an $N \times N$ complex matrix whose Hermitian part $\Re A = (A + A^*)/2$ is positive definite. We will show in \zcref{p:grass-gauss} that
\begin{equation}\label{e:grass-motivation-boson}
\int_{\C^N} e^{-\zeta^* A \zeta} \dd \zeta = \frac{1}{\det A}, \qquad \int_{\C^N} \zeta_i \bar \zeta_j \, e^{-\zeta^* A \zeta} \dd \zeta = \frac{(A^{-1})_{ij}}{\det A},
\end{equation}
where $\dd \zeta$ is the Lebesgue measure on $\C^N$ divided by $\pi^N$. Let $H$ be an $N \times N$ Hermitian matrix, let $z = E + \iu \eta$ with $E \in \R$ and $\eta > 0$, and recall that $G(z) = (H - z)^{-1}$. We apply \eqref{e:grass-motivation-boson} with $A = \iu (H - z)$, for which $\Re A = \eta \Id$. Since $A^{-1} = -\iu G(z)$, we obtain
\begin{equation}\label{e:grass-motivation-ratio}
G_{ij}(z) = \iu \, \frac{\int \zeta_i \bar \zeta_j \, e^{-\iu \zeta^* (H - z) \zeta} \dd \zeta}{\int e^{-\iu \zeta^* (H - z) \zeta} \dd \zeta} .
\end{equation}
In this sense a resolvent entry is a covariance of a Gaussian weight on $\C^N$, although a complex one. For random $H$, the difficulty lies in the denominator of \eqref{e:grass-motivation-ratio}, which depends on $H$. Since the expectation of a ratio is not the ratio of the expectations, the average of $G_{ij}(z)$ cannot be computed by averaging the numerator and the denominator separately.

The denominator can be removed by a second Gaussian integral, over anticommuting variables $\chi_1, \bar \chi_1, \dots, \chi_N, \bar \chi_N$, which produces the determinant instead of its inverse. Indeed, we will show that $\int e^{-\chi^* A \chi} \dd \chi = \det A$ for every complex matrix $A$. Multiplying the two integrals, we will find in \zcref{l:grass-resolvent} that
\begin{equation}\label{e:grass-motivation-super}
\begin{aligned}
\frac{\det(H - w)}{\det(H - z)} &= \int e^{-\iu \zeta^* (H - z) \zeta - \iu \chi^* (H - w) \chi} \dd \zeta \dd \chi, \\
G_{ij}(z) &= \iu \int \zeta_i \bar \zeta_j \, e^{-\iu \zeta^* (H - z) \zeta - \iu \chi^* (H - z) \chi} \dd \zeta \dd \chi
\end{aligned}
\end{equation}
for all $w \in \C$. In these integrals, the normalizations of the two Gaussian integrals have canceled, and $H$ enters linearly in the exponent. When $H$ is Gaussian, its average can be taken inside the integral, where it gives an explicit exponential, and the average of a ratio of determinants becomes an integral of an explicit function. Derivatives in $w$, and products of such ratios at $E + \iu \eta$ and $E - \iu \eta$, generate products of resolvent entries.

This is the supersymmetric method. It was introduced by Efetov \cite{efetov1982supersymmetry,efetov1983supersymmetry} in the theory of disordered metals; superspace had been brought into statistical physics by Parisi and Sourlas \cite{parisi1979random}, in the study of random magnetic fields. Verbaarschot, Weidenm\"uller, and Zirnbauer \cite{verbaarschot1985grassmann} applied the method to random matrix models of compound-nucleus scattering. The name refers to a symmetry between the commuting and the anticommuting variables of integrals such as \eqref{e:grass-motivation-super}, which we discuss in \zcref{s:grass-susy}.

\section{Grassmann algebras}\label{s:grass-algebra}

We now define the algebras in which anticommuting variables take values. Throughout this section, $r \ge 0$ is an integer, and we write $[r] = \{1, \dots, r\}$.

\begin{definition}\label{d:grass-algebra}
We define the \emph{Grassmann algebra} $\Grass{r}$ to be the complex associative algebra with unit generated by $\xi_1, \dots, \xi_r$ subject to the relations $\xi_a \xi_b = - \xi_b \xi_a$ for all $a, b \in [r]$. For $I = \{i_1 < \dots < i_s\} \subset [r]$, we set $\xi^I = \xi_{i_1} \cdots \xi_{i_s}$, with $\xi^\emptyset = 1$, and we call $\xi^I$ the \emph{exterior monomial} of degree $|I|$.
\end{definition}

The monomials $\xi^I$ form a basis of $\Grass{r}$ (which is the exterior algebra of $\C^r$), and every $f \in \Grass{r}$ has a unique expansion
\begin{equation}\label{e:grass-expansion}
f = \sum_{I \subset [r]} f_I \, \xi^I, \qquad f_I \in \C .
\end{equation}
We call the numbers $f_I$ the \emph{coefficients} of $f$, the number $f_\emptyset$ the \emph{body} of $f$, and $f - f_\emptyset$ the \emph{soul} of $f$. We call $f$ \emph{even} if $f_I = 0$ whenever $|I|$ is odd, and \emph{odd} if $f_I = 0$ whenever $|I|$ is even. We call $f$ \emph{homogeneous} if it is even or odd, and we assign the parity $0$ to even and $1$ to odd elements.

The relations imply $\xi_a^2 = 0$. One checks on monomials that $f g = (-1)^{\alpha \beta} g f$ for homogeneous $f$ and $g$ of parities $\alpha$ and $\beta$. In particular, even elements commute with all elements of $\Grass{r}$, while odd elements anticommute with each other and have square zero. We call $f$ \emph{nilpotent} if $f_\emptyset = 0$. For nilpotent $f$, every monomial of $f^s$ has degree at least $s$, and at least $2s$ if $f$ is even. We conclude that
\begin{equation}\label{e:grass-nilpotent}
f^{r + 1} = 0, \qquad \text{and} \qquad f^{\lfloor r/2 \rfloor + 1} = 0 \quad \text{if $f$ is even.}
\end{equation}
It follows that $f$ is invertible if and only if $f_\emptyset \ne 0$, and then $f^{-1} = f_\emptyset^{-1} \sum_{s = 0}^{r} (-n/f_\emptyset)^s$ with $n = f - f_\emptyset$. The anticommuting variables of the physics literature are the generators $\xi_a$, and the commuting variables are complex numbers or, more generally, even elements.

We next allow the coefficients to depend on commuting variables. Let $m \ge 0$ and let $U \subset \R^m$ be open.

\begin{definition}\label{d:grass-superfunction}
We define the space of \emph{superfunctions} on the \emph{superdomain} $U^{m|r}$ to be $C^\infty(U^{m|r}) = C^\infty(U) \otimes \Grass{r}$. Its elements are the sums $f = \sum_I f_I \xi^I$ with complex-valued coefficients $f_I \in C^\infty(U)$, multiplied as in $\Grass{r}$, with pointwise products of the coefficients. We call $x = (x_1, \dots, x_m) \in U$ the \emph{even coordinates} and $\xi_1, \dots, \xi_r$ the \emph{odd coordinates} of $U^{m|r}$.
\end{definition}

The terms coefficient, body, soul, even, odd, and nilpotent are used for superfunctions as in \zcref{d:grass-algebra}; the body $f_\emptyset$ of a superfunction is a function on $U$, and $f(x) \in \Grass{r}$ denotes the value of $f$ at $x \in U$. The spaces $C^k(U^{m|r})$ are defined in the same way, and so are holomorphic superfunctions on $U^{m|r}$ for open $U \subset \C^m$, whose coefficients are holomorphic.

Several formulas below contain odd elements that are not coordinates, such as the odd entries of a supermatrix or odd sources. We regard such elements as further generators of a larger Grassmann algebra $\Grass{r + r'}$ containing $\Grass{r}$, and we call these further generators \emph{auxiliary}. Even coordinate derivatives and the integration over odd coordinates defined in \zcref{s:grass-berezin} act coefficientwise in the auxiliary generators. Here we always write a superfunction as $f = \sum_I f_I \xi^I$ with the coefficients $f_I$, which may contain auxiliary generators, to the left of the monomials $\xi^I$. For instance, if $\theta$ is an auxiliary odd generator and $r = 1$, then the coefficient of $\xi_1$ in $\theta \xi_1$ is $\theta$, and in $\xi_1 \theta = - \theta \xi_1$ it is $-\theta$. The odd coordinate derivatives defined below extend by graded linearity: for an auxiliary coefficient $a$ of parity $\alpha$, we have $\partial_{\xi_b}(a g) = (-1)^\alpha a \, \partial_{\xi_b} g$. In particular, $\partial_{\xi_1}(\theta \xi_1) = -\theta$.

For $i \in [m]$, the derivative $\partial_{x_i}$ acts on the coefficients. For $a \in [r]$, every superfunction can be written uniquely as $f = f' + \xi_a f''$, where $f'$ and $f''$ do not contain $\xi_a$, and we set $\partial_{\xi_a} f = f''$. One checks on monomials that $\partial_{\xi_a} \partial_{\xi_b} = - \partial_{\xi_b} \partial_{\xi_a}$ and that
\[
\partial_{\xi_a}(f g) = (\partial_{\xi_a} f) g + (-1)^{\alpha} f \, \partial_{\xi_a} g
\]
for homogeneous $f$ of parity $\alpha$. We call an operator $\mathcal V = \sum_i c_i \partial_{x_i} + \sum_a d_a \partial_{\xi_a}$, where $c_i$ and $d_a$ are superfunctions, a \emph{vector field} on $U^{m|r}$. We call $\mathcal V$ even if the $c_i$ are even and the $d_a$ are odd, and odd if the $c_i$ are odd and the $d_a$ are even. A homogeneous vector field $\mathcal V$ of parity $\beta$ satisfies $\mathcal V(f g) = (\mathcal V f) g + (-1)^{\alpha \beta} f \, \mathcal V g$ for homogeneous $f$ of parity $\alpha$.

\subsection{Functions of even elements}

Expressions such as $e^{-\zeta^* A \zeta}$, where $A$ may have even entries, require functions of even elements. Since the soul of an even element is nilpotent, a Taylor expansion around the body terminates.

\begin{definition}\label{d:grass-even-function}
Let $a = (a_1, \dots, a_p)$ be even superfunctions on $U^{m|r}$, with bodies $a_\emptyset = (a_{1, \emptyset}, \dots, a_{p, \emptyset})$ and souls $n = a - a_\emptyset$. Let $V \subset \R^p$ be open, let $g \in C^\infty(V)$, and suppose that the bodies are real-valued with $a_\emptyset(U) \subset V$. We define
\begin{equation}\label{e:grass-even-function}
g(a) = \sum_{\alpha \in \N_0^p} \frac{(\partial^\alpha g)(a_\emptyset)}{\alpha!} \, n^\alpha, \qquad n^\alpha = n_1^{\alpha_1} \cdots n_p^{\alpha_p}.
\end{equation}
If instead $V \subset \C^p$ and $g$ is holomorphic on $V$, we define $g(a)$ by \eqref{e:grass-even-function} with complex derivatives, for complex-valued bodies with $a_\emptyset(U) \subset V$.
\end{definition}

By \eqref{e:grass-nilpotent}, the sum has only finitely many nonzero terms, those with $|\alpha| \le r/2$. If $g$ is a polynomial, then Taylor's formula shows that $g(a)$ is the polynomial evaluated in the algebra. For complex coordinates $\zeta \in \C^m$, we use \eqref{e:grass-even-function} with the real and imaginary parts as variables. In terms of the Wirtinger derivatives, a shift of $\zeta$ by even nilpotent elements $u$ and of $\bar \zeta$ by even nilpotent elements $v$, which may be unrelated, reads
\begin{equation}\label{e:grass-wirtinger-shift}
g(\zeta + u, \bar \zeta + v) = \sum_{\alpha, \beta} \frac{\partial_\zeta^\alpha \partial_{\bar \zeta}^\beta g(\zeta)}{\alpha! \, \beta!} \, u^\alpha v^\beta .
\end{equation}

\begin{lemma}\label{l:grass-calculus}
Let $a$, $V$, and $g$ be as in \zcref{d:grass-even-function}.
\begin{enumerate}[label=(\alph*)]
\item\label{i:grass-calculus-product} For all $\tilde g \in C^\infty(V)$, we have $(g \tilde g)(a) = g(a) \tilde g(a)$.
\item\label{i:grass-calculus-chain} For every homogeneous vector field $\mathcal V$ on $U^{m|r}$, we have $\mathcal V(g(a)) = \sum_{j = 1}^p (\partial_j g)(a) \, \mathcal V a_j$.
\item\label{i:grass-calculus-identities} Every identity between smooth, or holomorphic, functions on $V$ remains true when the variables are replaced by even elements $a$ with $a_\emptyset(U) \subset V$. In particular $e^{a_1 + a_2} = e^{a_1} e^{a_2}$, $\det(e^K) = e^{\Tr K}$ for every square matrix $K$ with even entries, and $g(a)^{-1} = (1/g)(a)$ if $g$ has no zeros on $V$.
\item\label{i:grass-calculus-composition} Let $V' \subset \R^{p'}$ be open, let $\tilde h \in C^\infty(V; \R^{p'})$ with $\tilde h(V) \subset V'$, and let $g' \in C^\infty(V')$. Then $(g' \circ \tilde h)(a) = g'(\tilde h(a))$, where $\tilde h(a)$ denotes the even elements $(\tilde h_1(a), \dots, \tilde h_{p'}(a))$.
\end{enumerate}
The same statements hold in the holomorphic case, and \ref{i:grass-calculus-composition} also holds for complex-valued $\tilde h$ and $g'$ holomorphic on an open set $V' \subset \C^{p'}$.
\end{lemma}

\begin{proof}
We begin with \ref{i:grass-calculus-product}. We insert the two expansions \eqref{e:grass-even-function} and collect the terms with a given power $n^\gamma$, which is possible because the souls are even and commute. By the Leibniz rule, we have
\[
g(a) \tilde g(a) = \sum_\gamma \sum_{\alpha + \beta = \gamma} \frac{(\partial^\alpha g)(a_\emptyset)(\partial^\beta \tilde g)(a_\emptyset)}{\alpha! \, \beta!} n^\gamma = \sum_\gamma \frac{\partial^\gamma(g \tilde g)(a_\emptyset)}{\gamma!} n^\gamma = (g \tilde g)(a) .
\]

We now prove \ref{i:grass-calculus-chain}. Both sides are left linear in $\mathcal V$ over superfunctions, and $(\partial_j g)(a)$ is even. It suffices to consider $\mathcal V = \partial_{x_i}$ and $\mathcal V = \partial_{\xi_b}$. Using the chain rule for the coefficients and the Leibniz rule for the commuting factors $n_j$, we obtain
\[
\partial_{x_i} g(a) = \sum_\alpha \sum_{j = 1}^p \frac{(\partial_j \partial^\alpha g)(a_\emptyset)}{\alpha!} \, (\partial_{x_i} a_{j, \emptyset}) \, n^\alpha + \sum_\alpha \frac{(\partial^\alpha g)(a_\emptyset)}{\alpha!} \sum_{j = 1}^p \alpha_j \, n^{\alpha - e_j} \, \partial_{x_i} n_j .
\]
After the substitution $\alpha \mapsto \alpha + e_j$ in the second sum, each sum becomes an expansion of $(\partial_j g)(a)$, and we find
\[
\partial_{x_i} g(a) = \sum_{j = 1}^p (\partial_j g)(a) \, \partial_{x_i} (a_{j, \emptyset} + n_j) = \sum_{j = 1}^p (\partial_j g)(a) \, \partial_{x_i} a_j .
\]
The case $\mathcal V = \partial_{\xi_b}$ is the same computation without the first sum, since $\partial_{\xi_b}$ annihilates the bodies and $\partial_{\xi_b}(n_j^{\alpha_j}) = \alpha_j n_j^{\alpha_j - 1} \partial_{\xi_b} n_j$ for even $n_j$.

Finally, if $g = \tilde g$ on $V$, then all derivatives agree, and \eqref{e:grass-even-function} implies that $g(a) = \tilde g(a)$. This proves \ref{i:grass-calculus-identities}, and the examples follow with \ref{i:grass-calculus-product}. For instance, $K \mapsto \det(e^K)$ and $K \mapsto e^{\Tr K}$ are entire functions of the entries of $K$ that agree on complex matrices, and the determinant of a matrix with even entries is the same polynomial in the entries.

It remains to prove \ref{i:grass-calculus-composition}. We let $T_{b_0} u$ denote the Taylor polynomial of order $\lfloor r/2 \rfloor$ at $b_0$ of a smooth function $u$ defined near $b_0$. The definition \eqref{e:grass-even-function} says that $g(a) = (T_{a_\emptyset} g)(a)$, where the polynomial is evaluated in the algebra, since $n^\alpha = 0$ for $|\alpha| > r/2$. By the chain rule for Taylor polynomials, $T_{a_\emptyset}(g' \circ \tilde h)$ is obtained from the composition $(T_{\tilde h(a_\emptyset)} g') \circ (T_{a_\emptyset} \tilde h)$ by discarding the terms of degree larger than $\lfloor r/2 \rfloor$ in the variables centered at $a_\emptyset$. The discarded terms vanish at $a$, and therefore
\[
(g' \circ \tilde h)(a) = (T_{\tilde h(a_\emptyset)} g') \big( (T_{a_\emptyset} \tilde h)(a) \big) = (T_{\tilde h(a_\emptyset)} g')(\tilde h(a)) = g'(\tilde h(a)) ,
\]
where the last step uses that $\tilde h(a)$ has the body $\tilde h(a_\emptyset)$ and a nilpotent soul.
\end{proof}

For real coefficients, the proof of \cite[Theorem~4.3.1]{varadarajan2004supersymmetry} defines $g(a)$, there called a pullback, by the expansion \eqref{e:grass-even-function} and shows that $g \mapsto g(a)$ is a homomorphism of algebras, as in \zcref{l:grass-calculus}\ref{i:grass-calculus-product}; see also \cite{berezin1987introduction}.

\section{Berezin integration}\label{s:grass-berezin}

We now define integration over odd coordinates. The definition is algebraic, and the integral of a superfunction will be the ordinary integral of the coefficient of its top monomial.

\begin{definition}\label{d:grass-berezin}
For $f \in \Grass{r}$, we define the \emph{Berezin integral} $\int f \dd \xi = f_{[r]}$ to be the coefficient of $\xi_1 \xi_2 \cdots \xi_r$ in $f$. We call a superfunction $f$ on $U^{m|r}$ \emph{integrable} if all its coefficients belong to $L^1(U)$, and for integrable $f$ we define
\[
\int_{U^{m|r}} f \dd x \dd \xi = \int_U f_{[r]}(x) \dd x .
\]
If $f$ takes values in a Grassmann algebra with auxiliary generators, both integrals are taken coefficientwise in the auxiliary generators.
\end{definition}

Exchanging two generators changes the sign of the Berezin integral, which is linear but not positive in any sense. Other references use other orders, which differ from ours by signs. Note that $\int f \dd \xi$ is the body of $\partial_{\xi_r} \cdots \partial_{\xi_1} f$; in this sense the Berezin integral behaves more like a derivative than like an integral. We will need the following elementary properties; in \ref{i:grass-berezin-fubini}, the odd variables $\theta_1, \dots, \theta_{r'}$ are integrated over as well.

\begin{lemma}\label{l:grass-berezin}
Let $a \in [r]$ and $i \in [m]$.
\begin{enumerate}[label=(\alph*)]
\item\label{i:grass-berezin-odd-ibp} For all $f \in \Grass{r}$, we have $\int \partial_{\xi_a} f \dd \xi = 0$. Consequently, for homogeneous $f$ of parity $\alpha$ and all $g \in \Grass{r}$,
\[
\int (\partial_{\xi_a} f) \, g \dd \xi = -(-1)^{\alpha} \int f \, \partial_{\xi_a} g \dd \xi .
\]
\item\label{i:grass-berezin-even-ibp} Let $f$ be a superfunction on $\R^{m|r}$ of class $C^1$ such that $f$ and $\partial_{x_i} f$ are integrable. Then $\int \partial_{x_i} f \dd x \dd \xi = 0$.
\item\label{i:grass-berezin-linear} Let $B \in \C^{r \times r}$. For $f \in \Grass{r}$, let $f(B \xi)$ denote the element obtained from \eqref{e:grass-expansion} by replacing each $\xi_b$ with $\sum_c B_{b c} \xi_c$. Then $\int f(B \xi) \dd \xi = \det B \int f \dd \xi$.
\item\label{i:grass-berezin-odd-shift} Let $\theta_1, \dots, \theta_r$ be odd elements in the auxiliary generators, and let $f(\xi + \theta)$ denote the element obtained by replacing each $\xi_b$ with $\xi_b + \theta_b$. Then $\int f(\xi + \theta) \dd \xi = \int f \dd \xi$.
\item\label{i:grass-berezin-even-shift} Let $f$ be a superfunction on $\R^{m|r}$ whose coefficients are smooth and have all their partial derivatives integrable, and let $u_1, \dots, u_m$ be even nilpotent elements in the auxiliary generators. Then $\int f(x + u) \dd x \dd \xi = \int f \dd x \dd \xi$, where $f(x + u) = \sum_I f_I(x + u) \xi^I$ is defined by \eqref{e:grass-even-function}. The same holds on $\C^m$ for the independent shifts \eqref{e:grass-wirtinger-shift} of $\zeta$ and $\bar \zeta$.
\item\label{i:grass-berezin-fubini} Let $f \in \Grass{r + r'}$, with generators $\xi_1, \dots, \xi_r, \theta_1, \dots, \theta_{r'}$. Then the iterated Berezin integrals in the two orders satisfy
\[
\int \Big( \int f \dd \xi \Big) \dd \theta = (-1)^{r r'} \int \Big( \int f \dd \theta \Big) \dd \xi .
\]
In this paper, $r$ and $r'$ are even whenever two Berezin integrals are exchanged, and the sign is one.
\end{enumerate}
\end{lemma}

\begin{proof}
The top coefficient of $\partial_{\xi_a} f$ vanishes, since $\partial_{\xi_a} f$ does not contain $\xi_a$. Together with the Leibniz rule for $\partial_{\xi_a}(f g)$, this implies \ref{i:grass-berezin-odd-ibp}. Statement \ref{i:grass-berezin-even-ibp} is the identity
\[
\int_{\R^m} \partial_{x_i} f_{[r]} \dd x = 0
\]
for an integrable $C^1$ function with integrable derivative, which follows from Fubini's theorem and the fundamental theorem of calculus.

We now prove \ref{i:grass-berezin-linear}. Only the term $f_{[r]} \prod_{b = 1}^r (\sum_c B_{b c} \xi_c)$ of $f(B \xi)$ contains the top monomial, and
\[
\prod_{b = 1}^r \Big( \sum_{c = 1}^r B_{b c} \xi_c \Big) = \sum_{\pi} \prod_{b = 1}^r B_{b \pi(b)} \, \xi_{\pi(1)} \cdots \xi_{\pi(r)} = \det B \; \xi_1 \cdots \xi_r ,
\]
where the sum runs over the permutations $\pi$ of $[r]$, and we used $\xi_{\pi(1)} \cdots \xi_{\pi(r)} = \sgn(\pi) \, \xi_1 \cdots \xi_r$. To prove \ref{i:grass-berezin-odd-shift}, observe that $\prod_{b \in I} (\xi_b + \theta_b)$ contains the monomial $\xi_1 \cdots \xi_r$ only for $I = [r]$, and then with coefficient one.

It remains to prove \ref{i:grass-berezin-even-shift}. By the definition of $f_I(x + u)$ in \zcref{d:grass-even-function}, we have
\[
f_I(x + u) = \sum_\alpha \frac{\partial^\alpha f_I(x)}{\alpha!} \, u^\alpha ,
\]
and the terms with $\alpha \ne 0$ integrate to zero by \ref{i:grass-berezin-even-ibp}. The shifts \eqref{e:grass-wirtinger-shift} are treated in the same way, since the Wirtinger derivatives are linear combinations of real derivatives.

Finally, with the coefficients written to the left, the left side of \ref{i:grass-berezin-fubini} is the coefficient of $\theta_1 \cdots \theta_{r'} \xi_1 \cdots \xi_r$ in $f$, and the integral $\int (\int f \dd \theta) \dd \xi$ is the coefficient of $\xi_1 \cdots \xi_r \theta_1 \cdots \theta_{r'}$. These two monomials differ by the sign $(-1)^{r r'}$.
\end{proof}

Statement \ref{i:grass-berezin-linear} is the first sign that odd variables transform inversely to even ones: for even coordinates, the substitution $x \mapsto B x$ multiplies the integral by $|\det B|^{-1}$. We will see in \zcref{s:grass-change} that the two rules combine into the superdeterminant.

\subsection{Gaussian integrals}

The superspaces of this paper have complex even coordinates and pairs of odd coordinates.

\begin{definition}\label{d:grass-flat}
Let $m, n \ge 0$. We write $\C^{m|2n}$ for the superdomain whose even coordinates $\zeta = (\zeta_1, \dots, \zeta_m) \in \C^m$ are identified with points of $\R^{2m}$, and whose odd coordinates are $\chi_1, \bar \chi_1, \dots, \chi_n, \bar \chi_n$, in this order. Here $\bar \zeta_j$ is the complex conjugate of $\zeta_j$, while $\bar \chi_j$ is a generator independent of $\chi_j$. We let $\dd \zeta = \prod_{j = 1}^m \pi^{-1} \dd \Re \zeta_j \dd \Im \zeta_j$, and for an integrable superfunction $F$ on $\C^{m|2n}$ we set
\[
\int F \dd \zeta \dd \chi = \int_{\C^m} F_{\rm top}(\zeta) \dd \zeta ,
\]
where $F_{\rm top}$ is the coefficient of $\chi_1 \bar \chi_1 \chi_2 \bar \chi_2 \cdots \chi_n \bar \chi_n$ in $F$. For $A \in \C^{m \times m}$ and $B \in \C^{n \times n}$, we write $\zeta^* A \zeta = \sum_{i, j} \bar \zeta_i A_{i j} \zeta_j$ and $\chi^* B \chi = \sum_{i, j} \bar \chi_i B_{i j} \chi_j$, and we use the same notation for matrices with even entries. We let $\partial_{\zeta_j} = (\partial_{\Re \zeta_j} - \iu \partial_{\Im \zeta_j})/2$ and $\partial_{\bar \zeta_j} = (\partial_{\Re \zeta_j} + \iu \partial_{\Im \zeta_j})/2$ denote the Wirtinger derivatives.
\end{definition}

The products $\chi_j \bar \chi_j$ are even, so the order of the pairs in the top monomial does not matter. (The factor $\pi^{-1}$ in $\dd \zeta$ is chosen so that $\int e^{-|\zeta|^2} \dd \zeta = 1$ on $\C^m$.) Up to the normalization of $\dd \zeta$, the integral $\int F \dd \zeta \dd \chi$ is the integral of \zcref{d:grass-berezin} on $\R^{2m|2n}$. We write $\Re A > 0$ for a complex matrix $A$ if $\Re A = (A + A^*)/2$ is positive definite, and for a matrix $A$ with even entries, we write $\Re A_\emptyset > 0$ if its body has this property. For a square matrix $B$, we let $\operatorname{adj} B$ denote its adjugate, so that $(\operatorname{adj} B)_{i j} = (-1)^{i + j} \det B_{[j, i]}$, where $B_{[j, i]}$ denotes $B$ with row $j$ and column $i$ deleted. Recall that $\operatorname{adj} B = (\det B) B^{-1}$ if $B$ is invertible.

\begin{proposition}\label{p:grass-gauss}
Let $m, n \ge 1$, and let $i$ and $j$ be indices.
\begin{enumerate}[label=(\alph*)]
\item\label{i:grass-gauss-boson} Let $A$ be an $m \times m$ matrix with even entries such that $\Re A_\emptyset > 0$. Then $\det A$ is invertible, and
\[
\int_{\C^m} e^{-\zeta^* A \zeta} \dd \zeta = \frac{1}{\det A}, \qquad \int_{\C^m} \zeta_i \bar \zeta_j \, e^{-\zeta^* A \zeta} \dd \zeta = \frac{(A^{-1})_{i j}}{\det A} .
\]
\item\label{i:grass-gauss-fermion} Let $B$ be an $n \times n$ matrix with even entries. Then
\[
\int e^{-\chi^* B \chi} \dd \chi = \det B, \qquad \int \chi_i \bar \chi_j \, e^{-\chi^* B \chi} \dd \chi = (\operatorname{adj} B)_{i j} .
\]
\item\label{i:grass-gauss-source} Let $B$ be as in \ref{i:grass-gauss-fermion} and invertible, and let $\theta = (\theta_1, \dots, \theta_n)^T$ and $\bar \theta = (\bar \theta_1, \dots, \bar \theta_n)^T$ consist of odd elements in the auxiliary generators. Then
\[
\int e^{-\chi^* B \chi + \chi^* \theta + \bar \theta^T \chi} \dd \chi = \det B \; e^{\bar \theta^T B^{-1} \theta} .
\]
\item\label{i:grass-gauss-mixed} Let $A$ be as in \ref{i:grass-gauss-boson} and $B$ as in \ref{i:grass-gauss-fermion}. Then, on $\C^{m|2n}$,
\[
\int e^{-\zeta^* A \zeta - \chi^* B \chi} \dd \zeta \dd \chi = \frac{\det B}{\det A} .
\]
If $m = n$ and $B = A$, then $\int \zeta_i \bar \zeta_j \, e^{-\zeta^* A \zeta - \chi^* A \chi} \dd \zeta \dd \chi = \int \chi_i \bar \chi_j \, e^{-\zeta^* A \zeta - \chi^* A \chi} \dd \zeta \dd \chi = (A^{-1})_{i j}$.
\end{enumerate}
\end{proposition}

\begin{proof}
We begin with \ref{i:grass-gauss-boson} for complex $A$. The only step beyond linear algebra is the passage from Hermitian to non-Hermitian $A$, which we make by analytic continuation. We first let $A$ be Hermitian and positive definite. We write $A = O D O^*$ with $O$ unitary and $D = \diag(d_1, \dots, d_m)$. The substitution $\zeta = O \zeta'$ preserves $\dd \zeta$, and
\[
\int_{\C^m} e^{-\zeta^* A \zeta} \dd \zeta = \prod_{j = 1}^m \frac{1}{\pi} \int_{\C} e^{-d_j |\zeta'_j|^2} \dd \Re \zeta'_j \dd \Im \zeta'_j = \prod_{j = 1}^m \frac{1}{d_j} = \frac{1}{\det A} .
\]

Now let $A$ be a complex matrix with $\Re A > 0$. We write $A = A_1 + \iu A_2$, where $A_1 = \Re A$ and $A_2 = (A - A^*)/(2 \iu)$ are Hermitian. The set $\mathcal S_A$ of $t \in \C$ with $\Re(A_1 + t A_2) > 0$ is a vertical strip containing $0$ and $\iu$, because $\Re(A_1 + t A_2) = A_1 + (\Re t) A_2$. For $t \in \mathcal S_A$, the matrix $A_1 + t A_2$ is invertible, since $\Re(v^* (A_1 + t A_2) v) > 0$ for all $v \ne 0$. Moreover, for every compact set $\mathcal K \subset \mathcal S_A$ there exists $c > 0$ such that, for all $t \in \mathcal K$,
\[
\big| e^{-\zeta^* (A_1 + t A_2) \zeta} \big| = e^{-\zeta^* (A_1 + (\Re t) A_2) \zeta} \le e^{-c |\zeta|^2} .
\]
Consequently the functions
\[
t \mapsto \int_{\C^m} e^{-\zeta^* (A_1 + t A_2) \zeta} \dd \zeta, \qquad t \mapsto \det(A_1 + t A_2)^{-1}
\]
are holomorphic on $\mathcal S_A$. They agree for real $t \in \mathcal S_A$, where $A_1 + t A_2$ is Hermitian and positive definite, and the identity theorem implies that they agree at $t = \iu$.

The second formula of \ref{i:grass-gauss-boson} follows by differentiating the first in the entry $A_{j i}$, which is allowed by dominated convergence. Indeed, we have
\[
\partial_{A_{j i}} e^{-\zeta^* A \zeta} = - \bar \zeta_j \zeta_i \, e^{-\zeta^* A \zeta}, \qquad \partial_{A_{j i}} \det(A)^{-1} = -\det(A)^{-1} (A^{-1})_{i j} .
\]
Each side of the two formulas is a holomorphic function of $A$ on the open set $\{A : \Re A > 0\}$, and their derivatives of all orders in the entries of $A$ may be computed under the integral. For $A$ with even entries, the integrand $e^{-\zeta^* A \zeta}$ is, by \zcref{d:grass-even-function}, the Taylor expansion of $e^{-\zeta^* A' \zeta}$ in $A'$ around $A_\emptyset$, evaluated at the soul $A - A_\emptyset$. Integrating term by term and using \zcref{l:grass-calculus}\ref{i:grass-calculus-identities}, we obtain \ref{i:grass-gauss-boson} for even $A$, where $\det A$ is invertible because its body $\det A_\emptyset$ is nonzero.

We now prove \ref{i:grass-gauss-fermion}. Each term of $\chi^* B \chi$ has degree two, and only the term $(-\chi^* B \chi)^n / n!$ of $e^{-\chi^* B \chi}$ contributes to the top monomial. Using $- \bar \chi_i \chi_j = \chi_j \bar \chi_i$, we have
\[
\frac{(-\chi^* B \chi)^n}{n!} = \frac{1}{n!} \sum_{i_1, \dots, i_n} \sum_{j_1, \dots, j_n} \prod_{k = 1}^n B_{i_k j_k} \, \chi_{j_k} \bar \chi_{i_k} .
\]
A term is nonzero only if $(j_1, \dots, j_n)$ and $(i_1, \dots, i_n)$ are permutations of $[n]$. The factors $\chi_{j_k} \bar \chi_{i_k}$ are even, and ordering them by $j_k$ accounts for the factor $n!$. A term then becomes $\prod_{l} B_{\pi(l) l} \prod_{l} \chi_l \bar \chi_{\pi(l)}$ for a permutation $\pi$ of $[n]$. Exchanging two of the $\bar \chi$ in the last product changes its sign, and we find
\[
\prod_{l = 1}^n \chi_l \bar \chi_{\pi(l)} = \sgn(\pi) \prod_{l = 1}^n \chi_l \bar \chi_l, \qquad \big( e^{-\chi^* B \chi} \big)_{\rm top} = \sum_{\pi} \sgn(\pi) \prod_{l = 1}^n B_{\pi(l) l} = \det B .
\]

The second formula follows by differentiating in $B_{j i}$, since
\[
\partial_{B_{j i}} e^{-\chi^* B \chi} = - \bar \chi_j \chi_i \, e^{-\chi^* B \chi} = \chi_i \bar \chi_j \, e^{-\chi^* B \chi}, \qquad \partial_{B_{j i}} \det B = (\operatorname{adj} B)_{i j} ,
\]
where the second identity is Jacobi's formula. Each side of these formulas is a polynomial in the entries of $B$, and the formulas remain valid for even entries.

To prove \ref{i:grass-gauss-source}, we complete the square. Expanding the product and keeping the order of the odd factors, one checks that
\[
-\chi^* B \chi + \chi^* \theta + \bar \theta^T \chi = - (\chi^* - \bar \theta^T B^{-1}) \, B \, (\chi - B^{-1} \theta) + \bar \theta^T B^{-1} \theta .
\]
All terms are even, and by \zcref{l:grass-calculus}\ref{i:grass-calculus-identities} the exponential factors accordingly. The first factor is obtained from $e^{-\chi^* B \chi}$ by shifting the generators $\chi_j$ and $\bar \chi_j$ by odd elements. By \zcref{l:grass-berezin}\ref{i:grass-berezin-odd-shift}, its Berezin integral is that of $e^{-\chi^* B \chi}$, and the claim follows from \ref{i:grass-gauss-fermion}.

Finally, using $e^{-\zeta^* A \zeta - \chi^* B \chi} = e^{-\zeta^* A \zeta} e^{-\chi^* B \chi}$ and \ref{i:grass-gauss-fermion}, we find
\[
\big( e^{-\zeta^* A \zeta - \chi^* B \chi} \big)_{\rm top} = \det(B) \, e^{-\zeta^* A \zeta} .
\]
Integrating over $\zeta$ with \ref{i:grass-gauss-boson}, we obtain \ref{i:grass-gauss-mixed}. The two insertions follow in the same way from the second formulas of \ref{i:grass-gauss-boson} and \ref{i:grass-gauss-fermion}, since $(\operatorname{adj} A)_{i j} / \det A = (A^{-1})_{i j}$.
\end{proof}

We now apply \zcref{p:grass-gauss} to resolvents, with the convention $G(z) = (H - z)^{-1}$. The sign in front of $\iu$ in the exponent must be chosen so that the Gaussian integral over $\zeta$ converges, and this sign depends on the sign of $\Im z$.

\begin{lemma}\label{l:grass-resolvent}
Let $H$ be an $N \times N$ Hermitian matrix, let $E \in \R$ and $\eta > 0$, and set $z_\pm = E \pm \iu \eta$. Then, on $\C^{N|2N}$, for all $w \in \C$,
\begin{equation}\label{e:grass-ratio}
\frac{\det(H - w)}{\det(H - z_\pm)} = \int e^{\mp \iu [\zeta^* (H - z_\pm) \zeta + \chi^* (H - w) \chi]} \dd \zeta \dd \chi ,
\end{equation}
and for all $i, j \in [N]$,
\begin{equation}\label{e:grass-resolvent-entry}
\begin{aligned}
G_{i j}(z_\pm) &= \pm \iu \int \zeta_i \bar \zeta_j \, e^{\mp \iu [\zeta^* (H - z_\pm) \zeta + \chi^* (H - z_\pm) \chi]} \dd \zeta \dd \chi \\
&= \pm \iu \int \chi_i \bar \chi_j \, e^{\mp \iu [\zeta^* (H - z_\pm) \zeta + \chi^* (H - z_\pm) \chi]} \dd \zeta \dd \chi .
\end{aligned}
\end{equation}
\end{lemma}

\begin{proof}
We set
\[
A = \pm \iu (H - z_\pm) = \pm \iu (H - E) + \eta \Id .
\]
Since $\pm \iu (H - E)$ is anti-Hermitian, we have $\Re A = \eta \Id > 0$. Applying \zcref{p:grass-gauss}\ref{i:grass-gauss-mixed} with this $A$ and with $B = \pm \iu (H - w)$, we obtain
\[
\int e^{-\zeta^* A \zeta - \chi^* B \chi} \dd \zeta \dd \chi = \frac{(\pm \iu)^N \det(H - w)}{(\pm \iu)^N \det(H - z_\pm)} ,
\]
which is \eqref{e:grass-ratio}. For \eqref{e:grass-resolvent-entry}, we take $w = z_\pm$, so that $B = A$, and we use
\[
(A^{-1})_{i j} = (\pm \iu)^{-1} G_{i j}(z_\pm) = \mp \iu \, G_{i j}(z_\pm) .
\qedhere
\]
\end{proof}

Multiplying a copy of \eqref{e:grass-ratio} with $z_+$ by a copy with $z_-$, we write a product of two ratios of determinants as one integral over $\C^{2N|4N}$, with the signs $-\iu$ and $+\iu$ in the exponent. In \zcref{s:rep}, these signs are collected in the matrix $\Lambda$, and the dictionary between resolvent entries and Gaussian insertions is \zcref{l:rep-dictionary}. Derivatives of \eqref{e:grass-ratio} in $w$ also generate resolvents; for instance, Jacobi's formula gives
\[
\partial_w \frac{\det(H - w)}{\det(H - z)} \Big|_{w = z} = - \Tr G(z) .
\]

\section{Supermatrices}\label{s:grass-supermatrix}

Linear changes of all coordinates of a superspace, and the quadratic forms in Gaussian superintegrals, are described by supermatrices. Throughout this section, $\mathcal R$ denotes a Grassmann algebra, possibly with auxiliary generators, or an algebra of superfunctions. The even part of $\mathcal R$ is commutative, and determinants of matrices with even entries are defined by the usual formula.

\begin{definition}\label{d:grass-supermatrix}
Let $m, n \ge 0$. A \emph{supermatrix} of size $(m|n)$ over $\mathcal R$ is a block matrix
\[
M = \begin{pmatrix} A & B \\ C & D \end{pmatrix}
\]
with blocks $A \in \mathcal R^{m \times m}$, $B \in \mathcal R^{m \times n}$, $C \in \mathcal R^{n \times m}$, and $D \in \mathcal R^{n \times n}$. We call $M$ even if the entries of $A$ and $D$ are even and those of $B$ and $C$ are odd, and odd if the entries of $A$ and $D$ are odd and those of $B$ and $C$ are even. We call $A$, $B$, $C$, and $D$ the BB, BF, FB, and FF blocks of $M$. For homogeneous $M$ of parity $\alpha$, we define the \emph{supertrace} $\Str M = \Tr A - (-1)^\alpha \Tr D$. For even $M$ with $A$ and $D$ invertible, we define the \emph{superdeterminant}, or \emph{Berezinian},
\[
\Sdet M = \frac{\det(A - B D^{-1} C)}{\det D} .
\]
\end{definition}

(The letters B and F stand for bosonic, or commuting, and fermionic, or anticommuting.) An even supermatrix maps columns $(v, \theta)^T$, whose first $m$ entries $v$ are even and whose last $n$ entries $\theta$ are odd, to columns of the same kind. Products of supermatrices are matrix products, and the product of homogeneous supermatrices of parities $\alpha$ and $\beta$ is homogeneous of parity $\alpha + \beta \pmod 2$. An even supermatrix $M$ has the body $M_\emptyset = \diag(A_\emptyset, D_\emptyset)$. Suppose that $M_\emptyset$ is invertible, and suppose that $\mathcal R$ has $r$ generators. Then $M = M_\emptyset (\Id + K)$, where $K = M_\emptyset^{-1}(M - M_\emptyset)$ has nilpotent entries and therefore satisfies $K^{r + 1} = 0$, and
\[
M^{-1} = \sum_{s = 0}^{r} (-K)^s M_\emptyset^{-1} .
\]
Conversely, the body of the identity $M M^{-1} = \Id$ shows that $M_\emptyset$ is invertible if $M$ is. We conclude that an even supermatrix is invertible if and only if $A_\emptyset$ and $D_\emptyset$ are invertible; in this case $\Sdet M$ is defined.

The following identities are standard, and their proofs are elementary. The only point at which the anticommutation of the odd entries matters is \eqref{e:grass-xy} below, which replaces the identity $\det(\Id - F F') = \det(\Id - F' F)$ for complex matrices $F$ and $F'$ by $\det(\Id - F F') = \det(\Id - F' F)^{-1}$ for matrices $F$ and $F'$ with odd entries.

\begin{lemma}\label{l:grass-sdet}
Let $M$ and $M'$ be even supermatrices of size $(m|n)$ over $\mathcal R$, and let $A, B, C, D$ be the blocks of $M$.
\begin{enumerate}[label=(\alph*)]
\item\label{i:grass-sdet-block} If $M$ is invertible, then $\Sdet M = \det A \, \det(D - C A^{-1} B)^{-1}$.
\item\label{i:grass-sdet-mult} If $M$ and $M'$ are invertible, then $\Sdet(M M') = \Sdet M \, \Sdet M'$.
\item\label{i:grass-sdet-trace} For homogeneous supermatrices $M$ and $M'$ of parities $\alpha$ and $\beta$, we have $\Str(M M') = (-1)^{\alpha \beta} \Str(M' M)$. In particular $\Str(M M') = \Str(M' M)$ for even $M$ and $M'$.
\item\label{i:grass-sdet-exp} The series $e^M = \sum_{s \ge 0} M^s / s!$ converges, $e^M$ is even and invertible, and $\Sdet(e^M) = e^{\Str M}$.
\item\label{i:grass-sdet-triangular} Suppose that the coordinates are split into two groups, each of which may contain even and odd coordinates, and that $M$ is invertible and block upper triangular with respect to this split, with diagonal blocks $M_1$ and $M_2$, which are even supermatrices. Then $\Sdet M = \Sdet M_1 \, \Sdet M_2$.
\end{enumerate}
\end{lemma}

In \ref{i:grass-sdet-triangular}, the rows and columns of $M$ are reordered so that the coordinates of the first group come first, and within each group, even coordinates precede odd ones. Statements \ref{i:grass-sdet-block} and \ref{i:grass-sdet-mult} are \cite[Corollary~3.6.3 and Theorem~3.6.2]{varadarajan2004supersymmetry}, and \ref{i:grass-sdet-trace} is stated in \cite[Section~3.5]{varadarajan2004supersymmetry}. For block diagonal supermatrices with complex entries, \cite[Section~3.6]{varadarajan2004supersymmetry} chooses the definition of the superdeterminant so that \ref{i:grass-sdet-exp} holds. For \ref{i:grass-sdet-block}--\ref{i:grass-sdet-triangular}, see also \cite{berezin1987introduction}.

\begin{proof}
We begin with \ref{i:grass-sdet-block}. We set $F = A^{-1} B$ and $F' = D^{-1} C$, whose entries are odd. Since $A - B D^{-1} C = A(\Id - F F')$ and $D - C A^{-1} B = D(\Id - F' F)$, it suffices to show that
\begin{equation}\label{e:grass-xy}
\det(\Id - F F') \, \det(\Id - F' F) = 1 .
\end{equation}
For $s \ge 1$, we write $(F F')^s = F Z F'$ with $Z = (F' F)^{s - 1}$, whose entries are even. Using the fact that the odd entries of $F$ and $F'$ anticommute and commute with the even entries of $Z$, we find
\[
\Tr\big((F F')^s\big) = \sum_{a, b, c} F_{a b} Z_{b c} F'_{c a} = - \sum_{a, b, c} F'_{c a} F_{a b} Z_{b c} = - \Tr\big((F' F)^s\big) .
\]

The matrices $F F'$ and $F' F$ have nilpotent even entries. The identity
\[
\det(\Id - K) = \exp\Big( - \sum_{s \ge 1} \frac{\Tr(K^s)}{s} \Big)
\]
holds for complex matrices $K$ near zero, and by \zcref{l:grass-calculus}\ref{i:grass-calculus-identities} and \ref{i:grass-calculus-composition} it holds for matrices $K$ with nilpotent even entries, for which the sum is finite. Applying it to $K = F F'$ and $K = F' F$ proves \eqref{e:grass-xy}.

We now prove \ref{i:grass-sdet-mult}. We first record three special cases, in which $A', B', C', D'$ are the blocks of $M'$. Let $P$ and $P'$ be invertible matrices with even entries of sizes $m$ and $n$, and let $F$ and $F'$ be matrices with odd entries of sizes $m \times n$ and $n \times m$. Using the definition of $\Sdet$ in the first two cases and \ref{i:grass-sdet-block} in the third, we find that
\begin{equation}\label{e:grass-sdet-special}
\begin{aligned}
\Sdet \begin{pmatrix} A' + F C' & B' + F D' \\ C' & D' \end{pmatrix} &= \frac{\det(A' + F C' - (B' + F D') D'^{-1} C')}{\det D'} = \Sdet M', \\
\Sdet \begin{pmatrix} P A' & P B' \\ P' C' & P' D' \end{pmatrix} &= \frac{\det P \, \det(A' - B' D'^{-1} C')}{\det P' \, \det D'} = \frac{\det P}{\det P'} \, \Sdet M', \\
\Sdet \begin{pmatrix} A' & B' \\ F' A' + C' & F' B' + D' \end{pmatrix} &= \frac{\det A'}{\det(F' B' + D' - (F' A' + C') A'^{-1} B')} = \Sdet M' .
\end{aligned}
\end{equation}
The three supermatrices on the left are the products $U_F M'$, $\diag(P, P') M'$, and $L_{F'} M'$, where $U_F$ and $L_{F'}$ denote the block unitriangular supermatrices with off-diagonal blocks $F$ and $F'$. Taking $M' = \Id$, we see that
\[
\Sdet U_F = \Sdet L_{F'} = 1, \qquad \Sdet \diag(P, P') = \frac{\det P}{\det P'} .
\]
 Multiplying out, one checks that $M = U_F \diag(P, P') L_{F'}$ with
\[
F = B D^{-1}, \qquad P = A - B D^{-1} C, \qquad P' = D, \qquad F' = D^{-1} C .
\]
Applying the three cases of \eqref{e:grass-sdet-special} in turn, we find $\Sdet(M M') = (\det P / \det P') \Sdet M'$, and the case $M' = \Id$ shows that $\det P / \det P' = \Sdet M$.

Next, we prove \ref{i:grass-sdet-trace}, using that $\Tr(F F') = (-1)^{\gamma \gamma'} \Tr(F' F)$ for matrices $F$ and $F'$ whose entries are homogeneous of parities $\gamma$ and $\gamma'$. Applying this to the four products in the diagonal blocks of $M M'$, we find that $\Str(M M')$ and $(-1)^{\alpha \beta} \Str(M' M)$ consist of the same four terms with the same signs. For instance, if $M$ and $M'$ are even, then $\Tr(B C') = - \Tr(C' B)$ and $\Tr(C B') = - \Tr(B' C)$, and
\[
\begin{aligned}
\Str(M M') &= \Tr(A A') + \Tr(B C') - \Tr(C B') - \Tr(D D') \\
&= \Tr(A' A) - \Tr(C' B) + \Tr(B' C) - \Tr(D' D) = \Str(M' M) .
\end{aligned}
\]
The other parities are treated in the same way.

To prove \ref{i:grass-sdet-exp}, note that the supermatrices with entries in a Grassmann algebra $\mathcal R$ form a finite-dimensional algebra. The operator norm of left multiplication on this algebra is submultiplicative, and the series for $e^M$ converges absolutely in it. (For superfunctions we argue at each point.) The supermatrix $e^M$ is even, and $e^{-M}$ is its inverse. We set $f(t) = \Sdet(e^{t M})$ for $t \in \R$. By \ref{i:grass-sdet-mult}, the definition of $\Sdet$, and the expansion $e^{t M} = \Id + t M + O(t^2)$, we have
\[
f(t + t') = f(t) f(t'), \qquad f(t) = 1 + t \Str M + O(t^2) .
\]
Therefore $f'(t) = f(t) \Str M$, the derivative of $e^{-t \Str M} f(t)$ vanishes, and $f(t) = e^{t \Str M}$, where we used that the even part of $\mathcal R$ is commutative. Setting $t = 1$ proves \ref{i:grass-sdet-exp}.

It remains to prove \ref{i:grass-sdet-triangular}. A reordering of the coordinates that keeps even coordinates even is a conjugation by an invertible even supermatrix, which does not change $\Sdet$ by \ref{i:grass-sdet-mult}. In the order in which the coordinates of the first group come first, we write
\[
M = \diag(M_1, M_2) (\Id + K) ,
\]
where $K$ has nonzero entries only in the rows of the first group and the columns of the second group. Then
\[
K^2 = 0, \qquad \Id + K = e^K, \qquad \Str K = 0 ,
\]
and \ref{i:grass-sdet-exp} implies that $\Sdet(\Id + K) = 1$. In the standard order, the four blocks of $\diag(M_1, M_2)$ are block diagonal, so that $\Sdet \diag(M_1, M_2) = \Sdet M_1 \, \Sdet M_2$ by the definition of $\Sdet$. By \ref{i:grass-sdet-mult}, this completes the proof.
\end{proof}

We can now compute Gaussian integrals whose quadratic form mixes commuting and anticommuting variables. For an even supermatrix $M$ of size $(m|n)$ with blocks $A, B, C, D$, we define the even superfunction
\[
(\zeta, \chi)^* M (\zeta, \chi) = \zeta^* A \zeta + \zeta^* B \chi + \chi^* C \zeta + \chi^* D \chi
\]
on $\C^{m|2n}$, where $\zeta^* B \chi = \sum_{i, b} \bar \zeta_i B_{i b} \chi_b$ and $\chi^* C \zeta = \sum_{b, j} \bar \chi_b C_{b j} \zeta_j$.

\begin{proposition}\label{p:grass-supergauss}
Let $M$ be an even supermatrix of size $(m|n)$ with blocks $A, B, C, D$, whose entries lie in the auxiliary generators, and suppose that $\Re A_\emptyset > 0$. Then
\[
\int e^{-(\zeta, \chi)^* M (\zeta, \chi)} \dd \zeta \dd \chi = \frac{\det(D - C A^{-1} B)}{\det A} ,
\]
which equals $\Sdet(M)^{-1}$ if $M$ is invertible.
\end{proposition}

\begin{proof}
The matrix $A$ is invertible, since $\det A_\emptyset \ne 0$. Expanding the product, one checks that
\[
(\zeta, \chi)^* M (\zeta, \chi) = (\zeta^* + \chi^* C A^{-1}) \, A \, (\zeta + A^{-1} B \chi) + \chi^* (D - C A^{-1} B) \chi .
\]
The entries of the columns
\[
u = A^{-1} B \chi, \qquad v = (\chi^* C A^{-1})^T
\]
are even and nilpotent. By \zcref{l:grass-calculus}\ref{i:grass-calculus-identities}, the exponential of minus the first term is obtained from $e^{-\zeta^* A \zeta}$ by shifting $\zeta$ by $u$ and $\bar \zeta$ by $v$, as in \eqref{e:grass-wirtinger-shift}, and the second term does not involve $\zeta$.

We integrate over $\zeta$ first, coefficientwise in the odd coordinates and the auxiliary generators. The coefficients of $e^{-\zeta^* A \zeta}$ are polynomials in $\zeta$ and $\bar \zeta$ times $e^{-\zeta^* A_\emptyset \zeta}$, and all their derivatives are integrable. By \zcref{l:grass-berezin}\ref{i:grass-berezin-even-shift}, the shifts by $u$ and $v$ do not change the integral over $\zeta$, and \zcref{p:grass-gauss}\ref{i:grass-gauss-boson} implies that
\[
\int_{\C^m} e^{-(\zeta, \chi)^* M (\zeta, \chi)} \dd \zeta = \frac{1}{\det A} \, e^{-\chi^* (D - C A^{-1} B) \chi} .
\]
Integrating over $\chi$ with \zcref{p:grass-gauss}\ref{i:grass-gauss-fermion} completes the proof, and \zcref{l:grass-sdet}\ref{i:grass-sdet-block} identifies the result with $\Sdet(M)^{-1}$ when $M$ is invertible.
\end{proof}

Applied to a supermatrix of the form $M \otimes \Id_N$, acting on $N$ copies of the coordinates, \zcref{p:grass-supergauss} yields the $N$th power of the integral for $M$. This is how the size of a matrix becomes a parameter in \zcref{s:grass-example}.

\section{An example: the Gaussian unitary ensemble}\label{s:grass-example}

We now carry out the supersymmetric method in the simplest case, to show the reader the steps that \zcref{s:rep} carries out for band matrices. In this section, $H$ denotes a Hermitian matrix from the Gaussian unitary ensemble (GUE), with centered complex Gaussian entries such that $\E H_{i j} H_{k l} = \delta_{i l} \delta_{j k} N^{-1}$. This is the covariance \eqref{e:intro-covariance} of the band matrix of \zcref{d:model}, with the variance profile $J = (\Id - W^2 \Delta)^{-1}$ replaced by its limit $N^{-1} \1 \1^T$ as $W \to \infty$.

\begin{proposition}\label{p:grass-gue}
Let $N \ge 1$, let $z \in \C$ with $\Im z > 0$, and let $w \in \C$. Then
\begin{equation}\label{e:grass-gue}
\E \frac{\det(w - H)}{\det(z - H)} = \frac{N}{2 \pi} \int_{\R^2} e^{-N (x^2 + y^2)/2} \, \frac{(w - \iu y)^{N - 1}}{(z - x)^{N + 1}} \big( (z - x)(w - \iu y) - 1 \big) \dd x \dd y .
\end{equation}
\end{proposition}

The proof has four steps. We represent the ratio as a superintegral over $N$ copies of $\C^{1|2}$, compute the average over $H$, decouple the resulting quartic exponent by Gaussian integrals over the entries of a supermatrix of size $(1|1)$, and integrate out the $N$ copies. The last step produces the $N$th power of a superdeterminant, and the dimension of the final integral does not depend on $N$.

\begin{proof}
We begin with the representation of the ratio. For $i, j \in [N]$, we set
\[
\Pi_{i j} = \bar \zeta_i \zeta_j + \bar \chi_i \chi_j ,
\]
and we will show in \zcref{s:grass-susy} that these bilinears are supersymmetric. Applying \zcref{l:grass-resolvent} with the upper sign and $z_+ = z$, and using $\det(w - H)/\det(z - H) = \det(H - w)/\det(H - z)$, we obtain
\begin{equation}\label{e:grass-gue-step1}
\frac{\det(w - H)}{\det(z - H)} = \int e^{\iu z \zeta^* \zeta + \iu w \chi^* \chi - \iu \sum_{i, j} H_{i j} \Pi_{i j}} \dd \zeta \dd \chi
\end{equation}
on $\C^{N|2N}$. The coefficients of the integrand are polynomials in $\zeta$, $\bar \zeta$, and the entries of $H$, times a factor of modulus $e^{-(\Im z) |\zeta|^2}$, and we may take the expectation inside the integral.

We next compute the average. For a complex matrix $\pi = (\pi_{i j})$, the variable $\sum_{i, j} H_{i j} \pi_{i j}$ is a complex linear combination of jointly Gaussian variables, and
\[
\E \Big( \sum_{i, j} H_{i j} \pi_{i j} \Big)^2 = \frac1N \sum_{i, j} \pi_{i j} \pi_{j i}, \qquad \E e^{-\iu \sum_{i, j} H_{i j} \pi_{i j}} = e^{-\frac{1}{2N} \sum_{i, j} \pi_{i j} \pi_{j i}} .
\]
Both sides of the second identity are entire in $\pi$, and their derivatives may be computed under the expectation. By \zcref{l:grass-calculus}\ref{i:grass-calculus-identities}, the identity also holds for $\pi = \Pi$. We set
\[
\alpha = \zeta^* \chi = \sum_{i = 1}^N \bar \zeta_i \chi_i, \qquad \beta = \chi^* \zeta = \sum_{i = 1}^N \bar \chi_i \zeta_i ,
\]
which are odd. Reordering the odd factors, we find
\[
\sum_{i, j} \bar \zeta_i \zeta_j \bar \chi_j \chi_i = \sum_{i, j} \bar \chi_j \zeta_j \bar \zeta_i \chi_i = \beta \alpha = - \alpha \beta, \qquad \sum_{i, j} \bar \chi_i \chi_j \bar \chi_j \chi_i = - (\chi^* \chi)^2 .
\]
Expanding $\Pi_{i j} \Pi_{j i}$, in which each of the two cross terms gives $-\alpha \beta$ after summation, we find
\[
\sum_{i, j} \Pi_{i j} \Pi_{j i} = (\zeta^* \zeta)^2 - 2 \alpha \beta - (\chi^* \chi)^2 ,
\]
and therefore
\begin{equation}\label{e:grass-gue-step2}
\E \frac{\det(w - H)}{\det(z - H)} = \int e^{\iu z \zeta^* \zeta + \iu w \chi^* \chi} \, e^{-(\zeta^* \zeta)^2/(2N)} \, e^{(\chi^* \chi)^2/(2N)} \, e^{\alpha \beta / N} \dd \zeta \dd \chi .
\end{equation}

We now decouple the three quartic factors by Gaussian integrals; this is the Hubbard--Stratonovich transformation. For complex numbers $a$ and $b$, we have
\[
e^{-a^2/(2N)} = \sqrt{\frac{N}{2 \pi}} \int_\R e^{-N x^2/2 - \iu x a} \dd x, \qquad e^{b^2/(2N)} = \sqrt{\frac{N}{2 \pi}} \int_\R e^{-N y^2/2 + y b} \dd y ,
\]
and by \zcref{l:grass-calculus}\ref{i:grass-calculus-identities} these identities hold for the even elements $a = \zeta^* \zeta$ and $b = \chi^* \chi$. For the odd factor, we introduce two new odd coordinates $\vartheta$ and $\bar \vartheta$, with the Berezin integral $\int \cdot \dd \vartheta$ given by the coefficient of $\vartheta \bar \vartheta$. Applying \zcref{p:grass-gauss}\ref{i:grass-gauss-source} with $n = 1$, $B = N$, $\theta = \beta$, and $\bar \theta = \alpha$, we obtain
\[
e^{\alpha \beta / N} = \frac{1}{N} \int e^{-N \bar \vartheta \vartheta + \bar \vartheta \beta + \alpha \vartheta} \dd \vartheta .
\]

We insert these three identities into \eqref{e:grass-gue-step2}. The coefficients of the resulting integrand are bounded by a polynomial times $e^{-(\Im z) |\zeta|^2 - N(x^2 + y^2)/2}$, and we may integrate over $\zeta$ and $\chi$ first. Since
\[
\bar \vartheta \beta = - \sum_{i = 1}^N \bar \chi_i \bar \vartheta \zeta_i, \qquad \alpha \vartheta = - \sum_{i = 1}^N \bar \zeta_i \vartheta \chi_i ,
\]
we may write the exponent as
\[
- N \frac{x^2 + y^2}{2} - N \bar \vartheta \vartheta - \sum_{i = 1}^N (\zeta_i, \chi_i)^* M (\zeta_i, \chi_i), \qquad M = \begin{pmatrix} \iu (x - z) & \vartheta \\ \bar \vartheta & -(\iu w + y) \end{pmatrix} .
\]
The supermatrix $M$ is even, with a BB entry of real part $\Im z > 0$. \zcref[S]{p:grass-supergauss}, applied to $M \otimes \Id_N$ on $\C^{N|2N}$, yields
\[
\int e^{-\sum_i (\zeta_i, \chi_i)^* M (\zeta_i, \chi_i)} \dd \zeta \dd \chi = \Big( \frac{-(\iu w + y) - \bar \vartheta \vartheta / (\iu (x - z))}{\iu (x - z)} \Big)^N = \Big( \frac{w - \iu y}{z - x} + \frac{\bar \vartheta \vartheta}{(z - x)^2} \Big)^N .
\]

It remains to integrate over $\vartheta$, $x$, and $y$. Since $(\bar \vartheta \vartheta)^2 = 0$, we have
\[
e^{-N \bar \vartheta \vartheta} \Big( \frac{w - \iu y}{z - x} + \frac{\bar \vartheta \vartheta}{(z - x)^2} \Big)^N = \frac{(w - \iu y)^N}{(z - x)^N} + N \bar \vartheta \vartheta \, \frac{(w - \iu y)^{N - 1}}{(z - x)^{N + 1}} \big( 1 - (z - x)(w - \iu y) \big) .
\]
The first term on the right does not contain $\vartheta$ and has Berezin integral zero. Since the Berezin integral over $\vartheta$ is the coefficient of $\vartheta \bar \vartheta$, we have
\[
\int \bar \vartheta \vartheta \dd \vartheta = -1 .
\]
Collecting the factor $N^{-1} \cdot N/(2\pi)$ from the three Hubbard--Stratonovich identities, we obtain \eqref{e:grass-gue}.
\end{proof}

We checked \eqref{e:grass-gue} numerically for several values of $z$ and $w$, against integration over the eigenvalue density of the GUE for $N \le 3$ and against Monte Carlo simulation for $N = 4$, with the supplementary script \texttt{grassmann/gue\_check.py}. Exact formulas for general correlation functions of ratios of characteristic polynomials, valid for every unitarily invariant ensemble, were obtained by Fyodorov and Strahov \cite{fyodorov2003exact} with orthogonal polynomials; the point of the example is the method rather than the formula.

Three features of the proof reappear for band matrices. First, the supermatrix $M$ equals $\iu (R - \diag(z, w))$, where
\[
R = \begin{pmatrix} x & - \iu \vartheta \\ - \iu \bar \vartheta & \iu y \end{pmatrix}, \qquad \Str R^2 = x^2 + y^2 + 2 \bar \vartheta \vartheta .
\]
We have $\Sdet(c K) = \Sdet K$ for every number $c \ne 0$ and every invertible even supermatrix $K$ of size $(1|1)$, and the proof shows that
\[
\E \frac{\det(w - H)}{\det(z - H)} = \frac{1}{2\pi} \int e^{-N \Str(R^2)/2} \, \Sdet\big( \diag(z, w) - R \big)^{-N} \dd x \dd y \dd \vartheta ,
\]
where $\Sdet(K)^{-1}$ stands for $\det(D - C A^{-1} B)/\det A$, as in \zcref{p:grass-supergauss}, when the FF block of $K$ is not invertible. The measure $(2\pi)^{-1} e^{-N \Str(R^2)/2} \dd x \dd y \dd \vartheta$ has total integral one for every $N$, since the Gaussian factors of the even and odd entries cancel. In \zcref{s:rep}, the Hubbard--Stratonovich field becomes one supermatrix per site, and the fields at different sites are coupled through the inverse of the variance profile $J$. Second, the FF entry $\iu y$ of $R$ runs over the imaginary axis. The FF quartic factor $e^{(\chi^* \chi)^2/(2N)}$ has the sign opposite to the BB factor, and the factor $e^{-N \Str(R^2)/2}$ is integrable in the FF entry on the imaginary axis but not on the real axis. Third, the BB entry $x$ is real because only $\Im z > 0$ appears. When $E + \iu \eta$ and $E - \iu \eta$ appear together, the bosonic part of the quartic form is not definite, and the BB entries are integrated over the cycle of \zcref{s:rep-cycle} instead.

\section{Change of variables}\label{s:grass-change}

For even variables, the change of variables formula multiplies the integrand by the Jacobian determinant. Recall from \zcref{l:grass-berezin}\ref{i:grass-berezin-linear} that a linear change of odd variables multiplies it by the inverse determinant instead. Berezin's formula combines these rules for changes of coordinates that mix even and odd variables.

Let $U, V \subset \R^m$ be open, and let $y_1, \dots, y_m$ be even and $\xi'_1, \dots, \xi'_r$ odd superfunctions on $U^{m|r}$. For a superfunction $f = \sum_I f_I \xi^I$ on $V^{m|r}$, we write $f(y, \xi') = \sum_I f_I(y) \xi'^I$, where $f_I(y)$ is defined by \zcref{d:grass-even-function} and $\xi'^I$ is the product of the $\xi'_a$, $a \in I$, in increasing order. When the body map $y_\emptyset$ is a diffeomorphism from $U$ onto $V$, we regard $(y, \xi')$ as new coordinates, and we call the even supermatrix of size $(m|r)$
\begin{equation}\label{e:grass-jacobian}
\mathcal J = \begin{pmatrix} (\partial_{x_i} y_j)_{i, j} & (\partial_{x_i} \xi'_b)_{i, b} \\ (\partial_{\xi_a} y_j)_{a, j} & (\partial_{\xi_a} \xi'_b)_{a, b} \end{pmatrix}
\end{equation}
the \emph{Jacobian supermatrix} of the change of coordinates. The rows of $\mathcal J$ are indexed by the old coordinates and the columns by the new ones. The following lemma justifies the word ``coordinates''; it is the form of the inverse function theorem for superdomains that we need, and a local version for supermanifolds is \cite[Theorem~4.4.1]{varadarajan2004supersymmetry}.

\begin{lemma}\label{l:grass-coordinates}
Suppose that $y_\emptyset$ is a diffeomorphism from $U$ onto $V$ and that the body of the matrix $(\partial_{\xi_a} \xi'_b)_{a, b}$ is invertible at every point of $U$. Then $f \mapsto f(y, \xi')$ is an isomorphism of algebras from $C^\infty(V^{m|r})$ onto $C^\infty(U^{m|r})$, which maps even superfunctions to even ones and odd superfunctions to odd ones.
\end{lemma}

\begin{proof}
We write $\Phi(f) = f(y, \xi')$. By \zcref{l:grass-calculus}\ref{i:grass-calculus-product}, $\Phi$ is a homomorphism of algebras. It preserves parity, since the $y_j$ are even and the $\xi'_b$ are odd.

Let $\mathcal F_s$, for $s \ge 0$, denote the superfunctions, on $U^{m|r}$ or on $V^{m|r}$, whose monomials all have degree at least $s$. Each $\xi'_b$ lies in $\mathcal F_1$, with part of degree one $\sum_a (D_\emptyset)_{a b} \xi_a$, where $D_\emptyset = (\partial_{\xi_a} \xi'_b)_\emptyset$ is the body of the FF block of $\mathcal J$. The soul of each $y_j$ lies in $\mathcal F_2$. We therefore have, for $|I| = s$,
\[
\Phi(f_I \xi^I) = (f_I \circ y_\emptyset) \prod_{b \in I} \Big( \sum_a (D_\emptyset)_{a b} \xi_a \Big) \mod \mathcal F_{s + 1} ,
\]
where the product is taken in increasing order of $b$. Thus $\Phi$ maps $\mathcal F_s$ into $\mathcal F_s$, and on the monomials of degree exactly $s$, modulo $\mathcal F_{s + 1}$, it acts as the composition with $y_\emptyset$ followed by the linear change of generators with the invertible matrix $D_\emptyset(x)$. This induced map is bijective, with an inverse given by the composition with $y_\emptyset^{-1}$ and the inverse matrix, whose entries are smooth.

If $\Phi(f) = 0$ and $f \ne 0$, let $s$ be the smallest degree of a monomial of $f$. Then the part of $f$ of degree $s$ is mapped to zero modulo $\mathcal F_{s + 1}$, which contradicts the bijectivity of the induced map. Hence $\Phi$ is injective. To prove surjectivity, let $f' \in C^\infty(U^{m|r})$, and suppose that $f' \in \mathcal F_s$. By the bijectivity of the induced map, there exists $f_s$, a sum of monomials of degree $s$ with coefficients in $C^\infty(V)$, such that $f' - \Phi(f_s) \in \mathcal F_{s + 1}$. Repeating this with $f' - \Phi(f_s)$ in place of $f'$, we obtain $f = f_s + f_{s + 1} + \dots + f_r$ with $\Phi(f) = f'$, since $\mathcal F_{r + 1} = \{0\}$.
\end{proof}

By \zcref{l:grass-coordinates}, every superfunction on $U^{m|r}$ can be written uniquely as $f(y, \xi')$ with $f \in C^\infty(V^{m|r})$, and Berezin's formula compares the Berezin integrals in the two coordinate systems.

\begin{theorem}\label{t:grass-change}
In this setting, suppose that $y_\emptyset$ is an orientation-preserving diffeomorphism from $U$ onto $V$ and that the body of the matrix $(\partial_{\xi_a} \xi'_b)_{a, b}$ is invertible at every point of $U$. Then, for every superfunction $f$ on $V^{m|r}$ whose coefficients have compact support in $V$,
\[
\int_{V^{m|r}} f \dd y \dd \xi' = \int_{U^{m|r}} f(y, \xi') \, \Sdet \mathcal J \dd x \dd \xi .
\]
\end{theorem}

This is Berezin's change of variables formula; see \cite{berezin1987introduction} and, with another arrangement of the matrix of derivatives, \cite[Theorem~4.6.1]{varadarajan2004supersymmetry}. With the Berezinian of a change of coordinates written as in \cite[Definition~2.5]{alldridge2012berezin}, it is \cite[Theorem~2.7]{alldridge2012berezin}, which is credited there to Leites \cite{leites1980introduction}. Conventions for the arrangement of $\mathcal J$ and for the order of the odd coordinates differ between references, and the plain transpose of $\mathcal J$, with rows indexed by the new coordinates, gives a different, incorrect, formula. (The supplementary script \texttt{grassmann/berezin\_change\_check.py} checks the formula, in our arrangement, for a nonlinear change of coordinates on $\R^{1|2}$, and shows that the other arrangement fails.) Linear changes of the odd coordinates and translations of the even coordinates by nilpotent elements are the special cases proved in \zcref{l:grass-berezin}. For $y = x$ and $\xi'_b = \sum_a B_{b a} \xi_a$, we have $\Sdet \mathcal J = (\det B)^{-1}$, and for $y = x + u$ and $\xi' = \xi$, we have $\Sdet \mathcal J = 1$.

\subsection{Boundary terms}\label{s:grass-boundary}

For our purposes, boundary terms are the most delicate point of the calculus, and they determine the choice of coordinates in \zcref{s:sigma}. The hypothesis of compact support in \zcref{t:grass-change} cannot be dropped, even for changes of coordinates that have $\Sdet \mathcal J = 1$ and map $U$ onto itself. The following example is known as Rudakov's example (see \cite[Example~2.8]{alldridge2012berezin}). Let $U = (0, \infty)$, $r = 2$, and $g \in C_c^\infty(\R)$, and consider the coordinates $y = x + \xi_1 \xi_2$ and $\xi' = \xi$ on $U^{1|2}$. The Jacobian supermatrix has the blocks $1$, $0$, $(\xi_2, -\xi_1)^T$, and $\Id$, so that $\Sdet \mathcal J = 1$. The function $f = g$, a function of the even coordinate alone, has $\int_{U^{1|2}} f \dd y \dd \xi' = 0$. However, $f(y, \xi') = g(x) + g'(x) \xi_1 \xi_2$, and therefore
\[
\int_{U^{1|2}} f(y, \xi') \, \Sdet \mathcal J \dd x \dd \xi = \int_0^\infty g'(x) \dd x = - g(0) .
\]
The two sides differ by a term at the boundary point $x = 0$, unless $g(0) = 0$. The same computation on $U = \R$ gives zero on each side.

The same phenomenon occurs at the singular point of polar coordinates, and there it carries information. Let $g \in C_c^\infty([0, \infty))$, and consider $F = g(\bar \zeta \zeta + \bar \chi \chi)$ on $\C^{1|2}$. By \zcref{d:grass-even-function}, we have $F = g(|\zeta|^2) - g'(|\zeta|^2) \chi \bar \chi$, and therefore
\begin{equation}\label{e:grass-polar-example}
\int F \dd \zeta \dd \chi = - \frac{1}{\pi} \int_{\C} g'(|\zeta|^2) \dd \Re \zeta \dd \Im \zeta = - \int_0^\infty g'(t) \dd t = g(0) .
\end{equation}
However, on $(\C \setminus \{0\})^{1|2}$ we may use the new even coordinates
\[
\zeta' = \zeta \Big( 1 + \frac{\bar \chi \chi}{2 |\zeta|^2} \Big), \qquad \bar \zeta' = \bar \zeta \Big( 1 + \frac{\bar \chi \chi}{2 |\zeta|^2} \Big) ,
\]
which satisfy $\bar \zeta' \zeta' = \bar \zeta \zeta + \bar \chi \chi$, and keep the odd coordinates $\chi, \bar \chi$. In the coordinates $(\zeta', \chi, \bar \chi)$, the function $F = g(\bar \zeta' \zeta')$ is a function of the even coordinates alone, with Berezin integral zero. The odd coordinates are unchanged, so the BF block of the Jacobian supermatrix vanishes, its FF block is the identity, and $\Sdet \mathcal J$ is the determinant of its BB block. In real coordinates, $\zeta \mapsto \zeta'$ is a radial dilation by the factor $1 + c/(2|\zeta|^2)$ with $c = \bar \chi \chi$, whose Jacobian determinant $1 - c^2/(4|\zeta|^4)$ reduces to one, since $c^2 = 0$. Therefore $\Sdet \mathcal J = 1$. The change of coordinates is singular at $\zeta = 0$, where $F$ does not vanish, and the value $g(0)$ in \eqref{e:grass-polar-example} is a boundary term concentrated there. A computation of this kind in polar coordinates on a superdomain of dimension $(2|2)$ is \cite[Examples~4.7(ii)]{alldridge2012berezin}.

In the supersymmetric method, one often parametrizes supermatrices by eigenvalues and angles, and the boundary contributions of such changes of variables are known in physics as Efetov--Wegner terms (see \cite{kieburg2011efetov}). A general formula for them was given by Rothstein \cite{rothstein1987integration}, and a coordinate-free form by Alldridge, Hilgert, and Palzer \cite{alldridge2012berezin}. Zirnbauer \cite[Section~II.A]{zirnbauer1996riemannian} explains why they matter for the integrals of the supersymmetric method, whose integrands are not compactly supported. Berezin's formula is an identity for compactly supported integrands, and an integrand that does not vanish at the boundary of the body domain, or at a singular point of the coordinates, produces additional terms. In this paper, superintegrals over the symmetric superspace of \zcref{s:sigma} are written in global coordinates that are nonsingular everywhere (\zcref{s:sigma-superfunctions}). The body has no boundary, and the integrands have integrable coefficients, so that integration by parts produces no boundary terms. At rank one, \zcref{s:sigma-radial} also uses coordinates that are singular at a point, and there the contribution of the singular point is accounted for separately; the example \eqref{e:grass-polar-example} shows that localization can identify it.

\section{Supersymmetry and localization}\label{s:grass-susy}

Recall from \eqref{e:grass-ratio}, with $w = z_\pm$, that
\[
\int e^{\mp \iu [\zeta^* (H - z_\pm) \zeta + \chi^* (H - z_\pm) \chi]} \dd \zeta \dd \chi = 1
\]
for every Hermitian matrix $H$. This is forced by a symmetry that exchanges commuting and anticommuting variables. We now explain the mechanism, which also normalizes the sigma model in \zcref{s:sigma}. We work on the flat superspace $\C^{n|2n}$ of \zcref{d:grass-flat}.

\begin{definition}\label{d:grass-supercharge}
We define the odd vector fields
\[
\mathfrak q_+ = \sum_{j = 1}^n \big( \chi_j \partial_{\zeta_j} - \bar \zeta_j \partial_{\bar \chi_j} \big), \qquad \mathfrak q_- = \sum_{j = 1}^n \big( \bar \chi_j \partial_{\bar \zeta_j} + \zeta_j \partial_{\chi_j} \big)
\]
on $\C^{n|2n}$, and we define the \emph{supercharge} to be $\mathfrak q = \mathfrak q_+ + \mathfrak q_-$. We call a superfunction $F$ \emph{supersymmetric}, or \emph{$\mathfrak q$-closed}, if $\mathfrak q F = 0$, and \emph{$\mathfrak q_\pm$-closed} if $\mathfrak q_\pm F = 0$.
\end{definition}

We have $\mathfrak q \zeta_j = \chi_j$, $\mathfrak q \bar \zeta_j = \bar \chi_j$, $\mathfrak q \chi_j = \zeta_j$, and $\mathfrak q \bar \chi_j = - \bar \zeta_j$ on the coordinates; the field $\mathfrak q_+$ acts only on $\zeta_j$ and $\bar \chi_j$, and $\mathfrak q_-$ only on $\bar \zeta_j$ and $\chi_j$. Squares of odd vector fields are even vector fields, and computing them on the coordinates, one finds that $\mathfrak q_+^2 = \mathfrak q_-^2 = 0$ and
\[
\mathfrak q^2 = \mathfrak q_+ \mathfrak q_- + \mathfrak q_- \mathfrak q_+ = \sum_{j = 1}^n \big( \zeta_j \partial_{\zeta_j} + \chi_j \partial_{\chi_j} - \bar \zeta_j \partial_{\bar \zeta_j} - \bar \chi_j \partial_{\bar \chi_j} \big) .
\]
The operator $\iu \mathfrak q^2$ generates the phase rotations, which multiply $\zeta$ and $\chi$ by $e^{\iu t}$ and $\bar \zeta$ and $\bar \chi$ by $e^{-\iu t}$, for $t \in \R$. Every supersymmetric function $F$ is invariant under these rotations, since $\mathfrak q^2 F = \mathfrak q(\mathfrak q F) = 0$. Thus $\mathfrak q_+$ and $\mathfrak q_-$ have square zero, and so has $\mathfrak q$ on invariant functions. Sign conventions for these fields differ between references; ours is adapted to the quadratic forms $\zeta^* A \zeta + \chi^* A \chi$ of \zcref{l:grass-resolvent}.

\begin{lemma}\label{l:grass-supercharge}
Let $i, j \in [n]$, and set
\[
\Pi_{i j} = \bar \zeta_i \zeta_j + \bar \chi_i \chi_j, \qquad \sigma = \sum_{l = 1}^n \bar \zeta_l \chi_l, \qquad \sigma_+ = - \sum_{l = 1}^n \bar \chi_l \zeta_l
\]
on $\C^{n|2n}$, and let $\mathfrak q'$ be one of the fields $\mathfrak q$, $\mathfrak q_+$, and $\mathfrak q_-$.
\begin{enumerate}[label=(\alph*)]
\item\label{i:grass-supercharge-bilinear} We have $\mathfrak q' \Pi_{i j} = 0$, and $\mathfrak q' \, g(\Pi) = 0$ for every entire function $g$ of the $n^2$ entries of the matrix $\Pi = (\Pi_{i j})$.
\item\label{i:grass-supercharge-exact} We have $\mathfrak q \sigma = \mathfrak q_- \sigma = \mathfrak q_+ \sigma_+ = \zeta^* \zeta + \chi^* \chi$.
\item\label{i:grass-supercharge-divergence} For every superfunction $F$ on $\C^{n|2n}$ of class $C^1$ such that $F$, its first-order partial derivatives, and $\mathfrak q' F$ are integrable, we have $\int \mathfrak q' F \dd \zeta \dd \chi = 0$.
\end{enumerate}
\end{lemma}

\begin{proof}
Using that $\mathfrak q_\pm$ are odd derivations, we compute
\[
\begin{aligned}
\mathfrak q_+ (\bar \zeta_i \zeta_j) &= \bar \zeta_i \chi_j, & \mathfrak q_+(\bar \chi_i \chi_j) &= (\mathfrak q_+ \bar \chi_i) \chi_j - \bar \chi_i \, \mathfrak q_+ \chi_j = - \bar \zeta_i \chi_j , \\
\mathfrak q_- (\bar \zeta_i \zeta_j) &= \bar \chi_i \zeta_j, & \mathfrak q_-(\bar \chi_i \chi_j) &= (\mathfrak q_- \bar \chi_i) \chi_j - \bar \chi_i \, \mathfrak q_- \chi_j = - \bar \chi_i \zeta_j ,
\end{aligned}
\]
so that
\[
\mathfrak q_+ \Pi_{i j} = \mathfrak q_- \Pi_{i j} = \mathfrak q \Pi_{i j} = 0 .
\]
The second claim of \ref{i:grass-supercharge-bilinear} follows from \zcref{l:grass-calculus}\ref{i:grass-calculus-chain}. Next, we have
\[
\mathfrak q_+(\bar \zeta_l \chi_l) = 0, \qquad \mathfrak q_-(\bar \zeta_l \chi_l) = \bar \chi_l \chi_l + \bar \zeta_l \zeta_l, \qquad \mathfrak q_+(- \bar \chi_l \zeta_l) = \bar \zeta_l \zeta_l + \bar \chi_l \chi_l ,
\]
and summing over $l$ gives \ref{i:grass-supercharge-exact}.

To prove \ref{i:grass-supercharge-divergence}, note first that the assumed integrability of $\mathfrak q' F$ makes its Berezin integral well defined. The top coefficient of $\zeta_l \partial_{\chi_l} F$ vanishes, since this superfunction does not contain $\chi_l$; the same holds for $\bar \zeta_l \partial_{\bar \chi_l} F$. Up to sign, the top coefficient of $\chi_l \partial_{\zeta_l} F$ is the derivative in $\zeta_l$ of a coefficient of $F$, whose integral vanishes by \zcref{l:grass-berezin}\ref{i:grass-berezin-even-ibp}, and the same holds for $\bar \chi_l \partial_{\bar \zeta_l} F$. Each of the three fields is a combination of these four kinds of terms.
\end{proof}

Statement \ref{i:grass-supercharge-divergence} says that $\mathfrak q$, $\mathfrak q_+$, and $\mathfrak q_-$ have zero divergence for the flat Berezin integral, and statements \ref{i:grass-supercharge-bilinear} and \ref{i:grass-supercharge-exact} say that $\zeta^* \zeta + \chi^* \chi$ is closed and exact for each of them. The deformation step of the localization argument uses only these three properties, and we state it for a general odd vector field, which we also denote by $\mathfrak q$, and for a Berezin integral with a density, so that \zcref{s:sigma} can apply it on a curved superspace. For such a field, $\mathfrak q^2$ is in general a nonzero even vector field, and the lemma assumes only that $\mathfrak q (\mathfrak q \sigma) = 0$ for the exponent $\mathfrak q \sigma$, which would be automatic if $\mathfrak q^2 = 0$. For an even superfunction $\rho$ on $U^{m|r}$ and a superfunction $F$ such that $\rho F$ is integrable, we write $\int_\rho F = \int_{U^{m|r}} \rho F \dd x \dd \xi$.

\begin{lemma}\label{l:grass-deformation}
Let $\mathfrak q$ be an odd vector field on $U^{m|r}$, let $\rho$ be an even superfunction, and let $F$ and $\sigma$ be superfunctions of class $C^1$, with $\sigma$ odd. Suppose that $\mathfrak q F = 0$ and $\mathfrak q (\mathfrak q \sigma) = 0$. Let $\mathcal T \subset \R$ be an open interval, and suppose that the following conditions hold for all $t \in \mathcal T$.
\begin{enumerate}[label=(\roman*)]
\item\label{i:grass-deformation-int} The superfunctions $\rho F e^{-t \mathfrak q \sigma}$ and $\rho (\mathfrak q \sigma) F e^{-t \mathfrak q \sigma}$ are integrable, and the coefficients of the second are bounded, locally uniformly in $t$, by an integrable function.
\item\label{i:grass-deformation-div} We have $\int_\rho \mathfrak q(\sigma F e^{-t \mathfrak q \sigma}) = 0$.
\end{enumerate}
Then $t \mapsto \int_\rho F e^{-t \mathfrak q \sigma}$ is constant on $\mathcal T$.
\end{lemma}

\begin{proof}
The superfunction $\mathfrak q \sigma$ is even, and \zcref{l:grass-calculus}\ref{i:grass-calculus-chain} yields
\[
\mathfrak q e^{-t \mathfrak q \sigma} = -t \, e^{-t \mathfrak q \sigma} \, \mathfrak q (\mathfrak q \sigma) = 0 ,
\]
where we used the assumption $\mathfrak q (\mathfrak q \sigma) = 0$ in the last step.
Since $\mathfrak q$ is an odd derivation and $\sigma$ is odd, we have
\[
\mathfrak q(\sigma F e^{-t \mathfrak q \sigma}) = (\mathfrak q \sigma) F e^{-t \mathfrak q \sigma} - \sigma \, \mathfrak q(F e^{-t \mathfrak q \sigma}) = (\mathfrak q \sigma) F e^{-t \mathfrak q \sigma} ,
\]
where we used $\mathfrak q F = 0$. By \ref{i:grass-deformation-int}, we may differentiate under the integral, and we find
\[
\frac{\dd}{\dd t} \int_\rho F e^{-t \mathfrak q \sigma} = - \int_\rho (\mathfrak q \sigma) F e^{-t \mathfrak q \sigma} = - \int_\rho \mathfrak q(\sigma F e^{-t \mathfrak q \sigma}) = 0 ,
\]
where the second equality follows from the previous display and the third from \ref{i:grass-deformation-div}.
\end{proof}

In words, a factor $e^{-t \mathfrak q \sigma}$ with a $\mathfrak q$-exact and $\mathfrak q$-closed exponent can be inserted into the integral of a $\mathfrak q$-closed function without changing it. As $t \to \infty$, this factor concentrates the integral near the zeros of the body of $\mathfrak q \sigma$, where it can be computed. Schwarz and Zaboronsky \cite[Theorem~1]{schwarz1997supersymmetry} prove a general localization theorem of this kind on compact supermanifolds, and they explain that the dimensional reduction of Parisi and Sourlas \cite{parisi1979random} can be obtained from such theorems. The superspaces of this paper are not compact, and the decay of the integrands replaces compactness. In flat space, the localization theorem reads as follows. For the field $\mathfrak q$, it is \cite[Theorem~11.4.5]{bauerschmidt2019introduction}, where it is formulated for differential forms, and the same argument on the hyperbolic superplane is \cite[Appendix~C, Lemma~16]{disertori2010quasi}. We include the short proof, since it also covers $\mathfrak q_\pm$ and \zcref{s:sigma} repeats its steps in a curved setting.

\begin{proposition}\label{p:grass-localization}
Let $\mathfrak q'$ be one of the fields $\mathfrak q$, $\mathfrak q_+$, and $\mathfrak q_-$. Let $F$ be a superfunction on $\C^{n|2n}$ of class $C^1$ such that $F$ and its first-order partial derivatives are integrable, and suppose that $\mathfrak q' F = 0$. Then
\[
\int F \dd \zeta \dd \chi = F_\emptyset(0) .
\]
\end{proposition}

\begin{proof}
Let $\sigma' = \sigma_+$ if $\mathfrak q' = \mathfrak q_+$, and $\sigma' = \sigma$ otherwise, with $\sigma$ and $\sigma_+$ as in \zcref{l:grass-supercharge}. By \zcref{l:grass-supercharge}\ref{i:grass-supercharge-exact}, we have
\[
\mathfrak q' \sigma' = \zeta^* \zeta + \chi^* \chi = \sum_{j = 1}^n \Pi_{j j} ,
\]
and \zcref{l:grass-supercharge}\ref{i:grass-supercharge-bilinear} therefore implies that $\mathfrak q'(\mathfrak q' \sigma') = 0$. We first apply \zcref{l:grass-deformation} with $\mathfrak q'$ and $\sigma'$ in place of $\mathfrak q$ and $\sigma$, $\rho = 1$, and $\mathcal T = (0, \infty)$. The $\chi_j \bar \chi_j$ are even and $(\chi_j \bar \chi_j)^2 = 0$, so that
\begin{equation}\label{e:grass-loc-gauss}
e^{-t \mathfrak q' \sigma'} = e^{-t |\zeta|^2} \prod_{j = 1}^n (1 + t \chi_j \bar \chi_j) .
\end{equation}

Since $|\zeta|^p e^{-t |\zeta|^2}$ is bounded for $p \le 3$, the coefficients of $F e^{-t \mathfrak q' \sigma'}$, of $(\mathfrak q' \sigma') F e^{-t \mathfrak q' \sigma'}$, and of $\sigma' F e^{-t \mathfrak q' \sigma'}$, and the first derivatives of the last, are bounded by a constant times sums of the moduli of the coefficients of $F$ and of their first derivatives, with a constant that is locally uniform in $t > 0$. This verifies \ref{i:grass-deformation-int}. Moreover, the identity
\[
\mathfrak q'\big(\sigma' F e^{-t \mathfrak q' \sigma'}\big) = (\mathfrak q' \sigma') F e^{-t \mathfrak q' \sigma'}
\]
from the proof of \zcref{l:grass-deformation} shows that this $\mathfrak q'$-derivative is integrable as well. Thus \zcref{l:grass-supercharge}\ref{i:grass-supercharge-divergence}, applied to $\sigma' F e^{-t \mathfrak q' \sigma'}$, verifies \ref{i:grass-deformation-div}. Therefore $\int F e^{-t \mathfrak q' \sigma'} \dd \zeta \dd \chi$ is constant in $t > 0$.

For $K \subset [n]$, let $F_{(K)}$ denote the coefficient of $\prod_{j \in K} \chi_j \bar \chi_j$ in $F$, so that $F_{(\emptyset)} = F_\emptyset$ and $F_{([n])} = F_{\rm top}$. By \eqref{e:grass-loc-gauss}, the top coefficient of $F e^{-t \mathfrak q' \sigma'}$ is
\begin{equation}\label{e:grass-loc-top}
(F e^{-t \mathfrak q' \sigma'})_{\rm top} = e^{-t |\zeta|^2} \sum_{K \subset [n]} t^{|K|} F_{([n] \setminus K)} .
\end{equation}
We now let $t \downarrow 0$. The right side of \eqref{e:grass-loc-top} converges pointwise to $F_{\rm top}$, and for $t \le 1$ it is bounded by $\sum_K |F_{(K)}|$, which is integrable. By dominated convergence, we have
\[
\lim_{t \downarrow 0} \int F e^{-t \mathfrak q' \sigma'} \dd \zeta \dd \chi = \int F \dd \zeta \dd \chi .
\]

Next, we let $t \to \infty$. For every integrable function $\tilde g$ on $\C^n$ that is continuous at zero, we have
\[
\lim_{t \to \infty} t^n \int_{\C^n} e^{-t |\zeta|^2} \tilde g(\zeta) \dd \zeta = \tilde g(0) ,
\]
where we used that $t^n e^{-t |\zeta|^2} \le t^n e^{-t}$ for $|\zeta| \ge 1$ and that
\[
t^n \int_{\C^n} e^{-t |\zeta|^2} \dd \zeta = 1 .
\]
We apply this to the terms of \eqref{e:grass-loc-top}. The term with $K = [n]$ converges to $F_\emptyset(0)$, while each term with $|K| < n$ carries a factor $t^{|K| - n}$ in front of a convergent quantity and tends to zero. Therefore
\[
\lim_{t \to \infty} \int F e^{-t \mathfrak q' \sigma'} \dd \zeta \dd \chi = F_\emptyset(0) .
\]
Since the integral is constant in $t$, this completes the proof.
\end{proof}

(For $n = 1$ and $F = g(\bar \zeta \zeta + \bar \chi \chi)$ with $g \in C_c^\infty([0, \infty))$, the function $F$ is supersymmetric by \zcref{l:grass-calculus}\ref{i:grass-calculus-chain} and \zcref{l:grass-supercharge}\ref{i:grass-supercharge-bilinear}, and \zcref{p:grass-localization} gives the value $g(0)$ found in \eqref{e:grass-polar-example}.)

The proof uses three properties: the divergence identity of \zcref{l:grass-supercharge}\ref{i:grass-supercharge-divergence}, the exactness of the localizing exponent, and the invariance of the Berezin integral under the rescaling $\zeta \mapsto t^{-1/2} \zeta$, $\chi \mapsto t^{-1/2} \chi$, $\bar \chi \mapsto t^{-1/2} \bar \chi$, whose Berezinian is one. The last property appears in the proof as the cancellation of $t^{|K|}$ against $t^{-n}$ in the term with $K = [n]$. The odd vector fields generated by the odd symmetries of the sigma model on the symmetric superspace of \zcref{s:sigma} have the same properties, and the proof of the localization lemma \zcref{l:sigma-localization} follows the same three steps.

In applications, the integrand is often a polynomially bounded function times a supersymmetric Gaussian. The following consequence of \zcref{p:grass-localization} treats this case; it is the flat-space prototype of the zero-source normalization \zcref{c:sigma-normalization}. For $A \in \C^{n \times n}$ with $\Re A > 0$, we let $\lambda_A > 0$ denote the smallest eigenvalue of $\Re A$.

\begin{corollary}\label{c:grass-normalization}
Let $A \in \C^{n \times n}$ with $\Re A > 0$, and let $\mathfrak q'$ be one of the fields $\mathfrak q$, $\mathfrak q_+$, and $\mathfrak q_-$. Let $f$ be a superfunction on $\C^{n|2n}$ of class $C^1$ such that $\mathfrak q' f = 0$, and suppose that there exist $C > 0$ and $\gamma < \lambda_A$ such that the coefficients of $f$ and their first-order partial derivatives are bounded by $C e^{\gamma |\zeta|^2}$. Then
\[
\int f \, e^{-\zeta^* A \zeta - \chi^* A \chi} \dd \zeta \dd \chi = f_\emptyset(0) .
\]
In particular, $\int P(\Pi) \, e^{-\zeta^* A \zeta - \chi^* A \chi} \dd \zeta \dd \chi = P(0)$ for every polynomial $P$ in the entries of $\Pi$, and $\int e^{-\zeta^* A \zeta - \chi^* A \chi} \dd \zeta \dd \chi = 1$.
\end{corollary}

\begin{proof}
Observe that
\[
\zeta^* A \zeta + \chi^* A \chi = \sum_{i, j = 1}^n A_{i j} \Pi_{i j} ,
\]
so $e^{-\zeta^* A \zeta - \chi^* A \chi}$ is an entire function of $\Pi$, which is $\mathfrak q'$-closed by \zcref{l:grass-supercharge}\ref{i:grass-supercharge-bilinear}. Since $\mathfrak q' f = 0$, the product $F = f e^{-\zeta^* A \zeta - \chi^* A \chi}$ satisfies $\mathfrak q' F = 0$. The coefficients of $e^{-\chi^* A \chi}$ are constants, and
\[
|e^{-\zeta^* A \zeta}| = e^{-\zeta^* (\Re A) \zeta} \le e^{-\lambda_A |\zeta|^2} .
\]
The coefficients of $F$ and their first derivatives are therefore bounded by a constant times $(1 + |\zeta|) e^{-(\lambda_A - \gamma) |\zeta|^2}$, which is integrable. Then \zcref{p:grass-localization} implies the first claim, since $F_\emptyset(0) = f_\emptyset(0)$. By \zcref{l:grass-supercharge}\ref{i:grass-supercharge-bilinear}, the polynomial $P(\Pi)$ is closed for each of the three fields. The coefficients of $P(\Pi)$ are polynomials in $\zeta$ and $\bar \zeta$, and $P(\Pi)_\emptyset(0) = P(0)$.
\end{proof}

For $f = 1$, the corollary also follows from \zcref{p:grass-gauss}\ref{i:grass-gauss-mixed}. For $f = \Pi_{j i}$, it says that the two insertions of \zcref{p:grass-gauss}\ref{i:grass-gauss-mixed} have opposite contributions, since $\bar \zeta_j \zeta_i = \zeta_i \bar \zeta_j$ and $\bar \chi_j \chi_i = - \chi_i \bar \chi_j$. By \zcref{l:grass-berezin}\ref{i:grass-berezin-linear}, a linear change of the odd coordinates multiplies the Berezin integral by a constant, so the corollary also identifies normalized Gaussian integrals in other coordinates. For instance, in the real coordinates $\zeta_j = x_j + \iu y_j$ and the odd coordinates $\alpha_j = \bar \chi_j$ and $\beta_j = \chi_j/2$, we have
\[
\zeta^* \zeta + \chi^* \chi = \sum_{j = 1}^n \big( x_j^2 + y_j^2 + 2 \alpha_j \beta_j \big), \qquad \mathfrak q_+ = \sum_{j = 1}^n \big( \beta_j (\partial_{x_j} - \iu \partial_{y_j}) - (x_j - \iu y_j) \partial_{\alpha_j} \big) .
\]
With $A = \frac12 \Id$, or with a diagonal $A$ with $\Re A > 0$, the corollary then says that the normalized Gaussian integral of a polynomially bounded $\mathfrak q_+$-closed function is the value of its body at the origin. This is the form in which localization is used for the normal oscillator in \zcref{s:tube}.