Berezinian

From Wikipedia, the free encyclopedia

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.

[edit] 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 even 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. An even supermatrix is then 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

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

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 R₀ (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 R₀.

[edit] Properties

The Berezinian of X is always a unit in the ring R₀.
$\operatorname{Ber}(X)^{-1} = \operatorname{Ber}(X^{-1})$
$\operatorname{Ber}(X^T) = \operatorname{Ber}(X)$ where $X T$ denotes the supertranspose of X.
$\operatorname{Ber}(X\oplus Y) = \operatorname{Ber}(X)\oplus\mathrm{Ber}(Y)$