Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

On the first hand one can define a superdomain $U^{p|q}$ as the super ringed space $(U^p,\mathcal{C}^{\infty p|q})$ where $U^p\subset\mathbb R^p$ is open and $\mathcal{C}^{\infty p|q}$ is the sheaf of supercommuting rings defined by [ $$\mathcal{C}^{\infty p|q} : V\mapsto C^\infty(V) [\theta_1,\dots,\theta_q], $$ where $V\subset U$ is open and the $\theta_j$ are anticommuting variables satisfying $\theta_i\theta_j=-\theta_j\theta_i$. Its dimension is defined to be $p|q$. $\mathbb R^{p|q}$ is a superdomain.

On the other hand consider the Grassman algebra $G(\mathbb R)$ over $\mathbb R$ with infinitely many generators $1, l_1,l_2,\dots$. Each element of $G(\mathbb R)$ is a finite linear combination of monomials $l_{i_1}\wedge\dots\wedge l_{i_n}$. Such a monomial is said to be even (resp. odd) if $n$ is even (resp. odd). Let $G_0$ and $G_1$ be the subspaces of $G(\mathbb R)$ spanned by the even and odd monomials respectively. One then set $$\mathbb R^{p|q}=\lbrace (x_1,\dots,x_p | \theta_1,\dots,\theta_q) ; x_i\in G_0,\theta_j\in G_1 \rbrace. $$

These are two different definitions of $\mathbb R^{p|q}$ : can anybody explain why they are equivalent or not ?

share|improve this question
1  
It seems to me that you want to compare a sheaf of (super-commutative) rings with a set... –  DamienC Jun 26 '12 at 11:52
add comment

2 Answers

The question is about two mathematically rigorous methods of defining supermanifolds that exist in the literature (I have seen at least one more, using symbols like $G^\infty$, but I am not particularly familiar with it). One is the "sheaf of supercommutative rings on an ordinary smooth manifold" method, which seems to the more popular of the two, and an exposition can be found in the IAS QFT lectures. The other, due to Bryce DeWitt, uses a countably infinite collection of odd variables, and can be found in his book "Supermanifolds".

Both methods seem to give the sorts of results you might expect from a reasonable theory of supermanifolds. In particular, I'm told one can translate Witten(-Getzler)'s supersymmetric proof of the Atiyah-Singer index theorem into both formalisms. However, I have not seen an explicit comparison between them - there seem to be obvious obstructions to constructing anything like an equivalence of categories.

I have heard of advantages and disadvantages to both methods, although I don't completely understand them. The sheaf-theoretic method seems to have problems expressing "odd functions of an even variable" (I may have transcribed this completely wrongly - this was from a conversation a while ago). DeWitt's method requires you to drag around infinitely many variables that you will never actually use.

Unfortunately, I don't think I can answer your question in a satisfactory way. The two spaces you describe seem to be inequivalent, but the manifolds modeled on them behave rather similarly.

share|improve this answer
    
the "infinitely many Grassmann variables" in de Witt's approach are a bad hack: really one wants to be looking instead at presheaves on the category of super-points, hence work with arbitrary finite Grassmann variables and ensure that every construction is natural in change of Grassmann variables. This is nicely discusses in Christoph Sachse's thesis, referenced here: ncatlab.org/nlab/show/supermanifold#ReferencesOverSuperpoints –  Urs Schreiber Jun 27 '12 at 6:45
    
Oh alright, this makes the question make a lot more sense. (And maybe my answer seems silly now, so I apologize for my ignorance.) So, wouldn't a natural question then be the following: Is the full subcategory of DeWitt supermanifolds spanned by the DeWitt versions of $\mathbb{R}^{p|q}$ equivalent to the full subcategory of "ordinary (sheaf theoretic)" supermanifolds spanned by the $\mathbb{R}^{p|q}$s? If not, I would safely say "they are not equivalent." –  David Carchedi Jun 28 '12 at 16:01
add comment

You can define a super domain $U^{p|q}$ as a locally super-ringed space the way you do, and that is for sure the correct definition.

NOTE: However, a supermanifold is in no way a set with extra structure. But what you can do, is talk about a supermanifold's global functions. In fact all (paracompact Hausdorff 2nd countable) supermanifolds are affine, so, one can form an equivalent category by looking at their super $\mathbb{R}$-algebra of smooth functions (i.e. global sections of their structure sheaf) but with functions being algebra homomorphisms "in the wrong direction."

In the case of $\mathbb{R}^{p|q}$, there is a slight ambiguity of what you mean, only because it could refer to the supermanifold or to the super vector space. Every super vector space "has the structure of a supermanifold" and this is how:

There is a contravariant functor from super vector spaces to super commutative $\mathbb{R}$-algebras which sends a super vector space $V_\cdot$ to $S\left(V_\cdot\right)$- the (super) symmetric algebra on $V_\cdot$. $S\left(V_\cdot\right)$ should be thought of as the algebra of polynomial functions on $V_\cdot.$ If the super dimension of $V_\cdot$ is $p|q,$ this is the free super commutative algebra with $p$ even generators and $q$ odd ones, i.e. $$\mathbb{R}[x_1,\cdots,x_p] \otimes \Lambda^q,$$ where $\Lambda^q$ is the Grassmannian algebra with $q$-generators. e. It shouldn't be called $\mathbb{R}^{p|q},$ but perhaps could be called $Poly(\mathbb{R}^{p|q}).$ I think is is what you mean by your "other definition" of $\mathbb{R}^{p|q},$ so I hope this clears things up a bit.

To get the "smooth" functions, a down-to-earth way is instead to consider the functor $$V_\cdot \mapsto C^\infty(V_0) \otimes_{S(V_0^*)} S(V^*_\cdot).$$

(Notice if $C^\infty$ is replaced by polynomials, it recovers the previous functor.)

Moreover, this gives rise to a sheaf on $V_0$ which assigns each open $U \subset V_0$ the super algebra $C^\infty(U) \otimes_{S(V_0^*)} S(V^*_\cdot).$ This gives $V_0$ together with this sheaf of algebras the structure of a supermanifold. If the super dimension of $V_\cdot$ is $p|q,$ this supermanifold is $\mathbb{R}^{p|q}$

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.