# Berezinian

In mathematics and theoretical physics, the Berezinian or superdeterminant is a generalization of the determinant to the case of supermatrices. The name is for Felix Berezin. The Berezinian plays a role analogous to the determinant when considering coordinate changes for integration on a supermanifold.

## Definition

The Berezinian is uniquely determined by two defining properties:
* $\operatorname {Ber} (XY)=\operatorname {Ber} (X)\operatorname {Ber} (Y)$
* $\operatorname {Ber} (e^{X})=e^{\operatorname {str(X)} }\,$
where str(X) denotes the supertrace of X. Unlike the classical determinant, the Berezinian is defined only for invertible supermatrices.
The simplest case to consider is the Berezinian of a supermatrix with entries in a field K. Such supermatrices represent linear transformations of a super vector space over K. A particular even supermatrix is a block matrix of the form
    $X={\begin{bmatrix}A&0\\0&D\end{bmatrix}}$
Such a matrix is invertible if and only if both A and D are invertible matrices over K. The Berezinian of X is given by
    $\operatorname {Ber} (X)=\det(A)\det(D)^{-1}$
For a motivation of the negative exponent see the substitution formula in the odd case.
More generally, consider matrices with entries in a supercommutative algebra R. An even supermatrix is then of the form
    $X={\begin{bmatrix}A&B\\C&D\end{bmatrix}}$
where A and D have even entries and B and C have odd entries. Such a matrix is invertible if and only if both A and D are invertible in the commutative ring R$_0$ (the even subalgebra of R). In this case the Berezinian is given by
    $\operatorname {Ber} (X)=\det(A-BD^{-1}C)\det(D)^{-1}$
or, equivalently, by
    $\operatorname {Ber} (X)=\det(A)\det(D-CA^{-1}B)^{-1}.$
These formulas are well-defined since we are only taking determinants of matrices whose entries are in the commutative ring R$_0$. The matrix
    $D-CA^{-1}B\,$
is known as the Schur complement of A relative to ${\begin{bmatrix}A&B\\C&D\end{bmatrix}}.$
An odd matrix X can only be invertible if the number of even dimensions equals the number of odd dimensions. In this case, invertibility of X is equivalent to the invertibility of JX, where
    $J={\begin{bmatrix}0&I\\-I&0\end{bmatrix}}.$
Then the Berezinian of X is defined as
    $\operatorname {Ber} (X)=\operatorname {Ber} (JX)=\det(C-DB^{-1}A)\det(-B)^{-1}.$

## Properties

* The Berezinian of $X$ is always a unit in the ring R$_0$.
* $\operatorname {Ber} (X^{-1})=\operatorname {Ber} (X)^{-1}$
* $\operatorname {Ber} (X^{st})=\operatorname {Ber} (X)$ where $X^{st}$ denotes the supertranspose of $X$.
* $\operatorname {Ber} (X\oplus Y)=\operatorname {Ber} (X)\mathrm {Ber} (Y)$

## Berezinian module

The determinant of an endomorphism of a free module M can be defined as the induced action on the 1-dimensional highest exterior power of M. In the supersymmetric case there is no highest exterior power, but there is a still a similar definition of the Berezinian as follows.
Suppose that M is a free module of dimension (p,q) over R. Let A be the (super)symmetric algebra S*(M*) of the dual M* of M. Then an automorphism of M acts on the ext module
    $Ext_{A}^{p}(R,A)$
(which has dimension (1,0) if q is even and dimension (0,1) if q is odd)) as multiplication by the Berezinian.