A vector space $V$ of dimension $n$ has an associated determinant line $Det(V)$.

An element of $Det(V)$ is represented as a (formal limear combination) of expresstions of the form $v_1 \wedge v_2 \wedge \ldots \wedge v_n$, subject to the usual multilinearity and antisymmetry relations.

I'm wondering what is analog of the above fact/construction in the world of super vector spaces.

Let $V$ be a supervector space of dimension $n|m$. Then there is a line $Ber(V)$ called the Berezinian of $V$, that behaves like a super-determinant.

Here's a naive description of the Berezinian: for $V=V_0\oplus V_1$, it is given by $$Ber(V)=Det(V_0)\otimes Det(V_1)^*.$$ That's clearly not a good description of $Ber(V)$, as it relies on the decomposition of $V$ into even and odd parts, which is not a $GL(V)$-invariant thing to do.

I want to make sure that I don't get non-invariant answers. To ensure that, I'll do things in families (and thus make the question more complicated $-$ sorry for that):

Let $\Lambda=\Lambda(\theta_1,\ldots,\theta_n)$ be a Grassmann algebra (=exterior algebra) on $n$ variables, and let $V$ be a $Spec(\Lambda)$-parametrized family of super vector spaces, i.e., a super vector bundle $V\to Spec(\Lambda)$. How can I describe concretely a section of the associated line bundle $Ber(V)\to Spec(\Lambda)$?

For those who don't like the above language, I can translate into algebra. Let $\Lambda=\Lambda(\theta_1,\ldots,\theta_n)$, and let $V$ be a free $\Lambda$-module on $n$ even generators and $m$ odd generators. How can I describe concretely an even element of the rank one $\Lambda$-module $Ber(V)$?

notmean the field of rational functions over $\mathbb R$ in the $\theta_i$ here. As a notation for an $n$-dimensional Grassmann algebra, it is unknown to me. (Grassmann=exterior? There are many things that can be called Grassmann algebras...) – darij grinberg Sep 15 '11 at 21:00