# In what sense does the Berezinian generalize the determinant?

One way of defining the determinant of a endomorphism of a vector space $\varphi:V \to V$ is by using the action of $End(V)$ on the underlying $\mathbb{Z}$-graded vector space of the exterior algebra $\bigwedge^{\cdot}\left(V\right)$.This action is given by taking the canonical action of $\varphi$ on the underlying graded vector space of the tensor algebra and restricting it to the subspace $\bigwedge^{\cdot}\left(V\right)$. If $n=dim(V)$, then $\bigwedge^{n}\left(V\right)$ is $1$-dimensional, and any choice of basis identifies this space with the ground field $K$ and the action of $\varphi$ with the determinant of $\varphi$ expressed in that basis.

Now let $V_\cdot$ be a $\mathbb{Z}_2$-graded vector space and $\varphi:V_\cdot \to V_\cdot$ be a (parity preserving) endomorphism. Does the Berezinian of $\varphi$ have a description anything like the lines above? It cannot work "as is", since if $V_\cdot$ is purely odd, then $\bigwedge^{\cdot}\left(V\right)$ can be identified with the symmetric algebra on $\Pi V_\cdot$, which is has no top power, as $\Pi V_\cdot$ is purely even, but is there any relation at all?

-
This is explained in the wikipedia page for the Berezinian en.wikipedia.org/wiki/Berezinian a suitable Ext module is 1 dimensional and one can do the same trick. This only orks over a field -- I am not sure that the Berezinian can be defined over rings. – kassabov Nov 17 '11 at 19:14
Ha! Wiki of all places... but thanks! – David Carchedi Nov 17 '11 at 19:18
@Kassabov: Do you have a reference for this btw? I mean, one in which I can read the proof. – David Carchedi Nov 17 '11 at 19:23
André Henriques has recently asked a slightly more general question: mathoverflow.net/questions/75522/…, and in comments there I cite a formula for the Berezinian that generalizes the top exterior power. – Dmitri Pavlov Nov 17 '11 at 20:54
Restricting to parity preserving endomorphisms is not a very good thing to do: you'll only learn that the Berezinian of $\varphi$ is the quotient of $det(\varphi_0)$ by $det(\varphi_1)$. Here's a better alternative: allow yourself to do arbitrary base-changes (by tensoring with a supercommutative ring) and only then pick a parity preserving endomorphism. Your question remains interesting as stated though. – André Henriques Nov 18 '11 at 4:45