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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 $Ber(V)$?

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    $\begingroup$ Did you mean tensor product instead of direct sum in the formula for Ber(V)? Also it seems like V_2 should be replaced by V_1. $\endgroup$
    – Dmitri Pavlov
    Commented Sep 15, 2011 at 16:03
  • $\begingroup$ Are you aware of the description of Ber(V) as Ext*_{Sym(V*)}(R,Sym(V*))? (Here R is the ring of scalars and Sym(V*) acts on R by augmentation.) If you combine this description with any explicit description of Ext, would this count as an explicit description of the Berezinian? $\endgroup$
    – Dmitri Pavlov
    Commented Sep 15, 2011 at 16:14
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    $\begingroup$ @Dmitri: I was unaware of this description, and I like it! You should leave it (maybe with more details or references, please) as an answer, because I would like to vote it up. $\endgroup$
    – Theo Johnson-Freyd
    Commented Sep 15, 2011 at 16:58
  • $\begingroup$ Thank you Dmitri for the small typos. Yes, I was aware (since today) of that description. But it's so much more complicated than the definition of Det(V)... I suspect that there is some simple answer involving Schur functors. Namely, that Ber(V) is just some appropriate Schur functor applied to V. But I don't know whether that's true or not, nor do I know which Schur functor to apply. $\endgroup$
    – André Henriques
    Commented Sep 15, 2011 at 18:14
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    $\begingroup$ It took me a while to realize that $\mathbb R\left(\theta_1,...,\theta_n\right)$ does not mean 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...) $\endgroup$
    – darij grinberg
    Commented Sep 15, 2011 at 21:00

2 Answers 2

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There is Koszul complex introduced by Yu. Manin in his book "Gauge field theory and complex geometry". The cohomology of this complex yields the Berezinian.

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  • $\begingroup$ Welcome to MO!! $\endgroup$
    – Alexander Chervov
    Commented Jan 27, 2013 at 11:50
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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.

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  • $\begingroup$ Well, a complete answer should probably include a description of how such elements transform under a change of basis. (In the purely even case such a description amounts to saying that Ber(L)=Λ^top(L).) $\endgroup$
    – Dmitri Pavlov
    Commented Nov 5, 2011 at 10:56
  • $\begingroup$ I known that I'm being pedantic... but I can't help it: "how such elements transform" is physics lingo. In math, I would talk about the "linear relation that two such elements satisfy". $\endgroup$
    – André Henriques
    Commented Nov 5, 2011 at 16:00
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    $\begingroup$ From my point of view, bases are a part of physics lingo :-), so if you talk about bases it makes sense to use other parts of the physics lingo too. More seriously, didn't you want to get an expression of Ber(L) in terms of some Schur functor? Your current answer simply amounts to saying that Ber is functorial with respect to isomorphisms of vector spaces and dim Ber(R^{m|n}) = 1. $\endgroup$
    – Dmitri Pavlov
    Commented Nov 5, 2011 at 19:34
  • $\begingroup$ You're completely right. What I really wanted was some connection between the Berezinian and Schur functors... Even worse, my answer provides zero information on the Berezinian actually is (well... it gives $\varepsilon>0$ information: in your comment, you describe very precisely how much information it gives). However, strictly speaking, that quote from Deligne-Morgan does answer my original question as it provides a notation for elements of $Ber(L)$... $\endgroup$
    – André Henriques
    Commented Nov 5, 2011 at 21:08
  • $\begingroup$ Hi Andre. Did you figure this out yet? I didn't realize you had asked this and have just asked a similar question: mathoverflow.net/questions/81191/… $\endgroup$
    – David Carchedi
    Commented Nov 17, 2011 at 19:48

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