Berezinian

Definition

The Berezinian is uniquely determined by two defining properties:

• ${\displaystyle \mathrm{Ber}(XY)=\mathrm{Ber}(X)\mathrm{Ber}(Y)}$
• ${\displaystyle \mathrm{Ber}(e^X)=e^{\mathrm{s}\mathrm{t}\mathrm{r}(\mathrm{X})}}$

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

${\displaystyle X=\left[\begin{array}{cc}A& 0\\ 0& D\end{array}\right]}$

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

${\displaystyle \mathrm{Ber}(X)=det(A)det(D{)}^{-1}}$

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

${\displaystyle X=\left[\begin{array}{cc}A& B\\ C& D\end{array}\right]}$

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 R₀ (the even subalgebra of R). In this case the Berezinian is given by

${\displaystyle \mathrm{Ber}(X)=det(A-B{D}^{-1}C)det(D{)}^{-1}}$

or, equivalently, by

${\displaystyle \mathrm{Ber}(X)=det(A)det(D-C{A}^{-1}B{)}^{-1}.}$

These formulas are well-defined since we are only taking determinants of matrices whose entries are in the commutative ring R₀. The matrix

${\displaystyle D-C{A}^{-1}B}$

is known as the Schur complement of A relative to ${\displaystyle \left[\begin{array}{cc}A& B\\ C& D\end{array}\right].}$

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

${\displaystyle J=\left[\begin{array}{cc}0& I\\ -I& 0\end{array}\right].}$

Then the Berezinian of X is defined as

${\displaystyle \mathrm{Ber}(X)=\mathrm{Ber}(JX)=det(C-D{B}^{-1}A)det(-B{)}^{-1}.}$