
> 
> 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.




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Let $V$ be a supervector space of dimension $n|m$. Then there is a line $Ber(V)$ called the [Berezinian](http://en.wikipedia.org/wiki/Berezinian#Berezinian_module) 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 $Ber(V)$?



How can I write down a point in the Berezinian of a super vector space?
From page 61 of Deligne and Morgan's article *Notes on supersymmetry (following Joseph Bernstein)*:



> 
> "A basis $\{e\_1,\ldots,e\_p,e\_{p+1},\ldots,e\_{p+q}\}$ of $L$ defines a one-element
>  basis $[e\_1,\ldots,e\_p,e\_{p+1},\ldots,e\_{p+q}]$ of $Ber(L)$."
> 
> 
> 


That's an answer to the question.


