### What IS a superfield, really?

#### Posted by Urs Schreiber

In the context of functorial transport over supermanifolds, I mentioned recently that sorting out the *true nature* of supermanifolds leads to concepts that may look quite sophisticated, compared to the familiar carefree handling of Grassmann coordinates. Actually, I think they are quite elementary but most of us are not used to thinking in terms of them. Since supermanifolds are really just a simple example for way more general generalized spaces that are bound to play a major role in (formal) high energy physics - like for instance noncommutative geometries - it seems worthwhile to spend a couple of minutes sorting out the answer to the question in the title of this entry.

The following is an exercise in *physics made difficult*. It reminds me of this piece of wisdom I once heard.

Before you understand, a tree is a tree.When you understand, a tree is no longer a tree.After you have understood, a tree is a tree.

What does the following paragraph mean

$\Phi $ is a 3-component superfield on a 1-dimensional superspace with coordinates $(t,\theta )$. Hence in components it reads

(1)$${\Phi}^{i}(t,\theta )={A}^{i}(t)+{B}^{i}(t)\theta \phantom{\rule{thinmathspace}{0ex}},$$where the ${A}^{i}$ are even and ${B}^{i}$ is odd.

?

Of course every (high energy) physicist knows what this means *operationally*. But what does it mean *conceptually*? What *is* $\Phi $? What is $\Phi $ in a language that does not use (super)coordinates, for instance?

The answer is a two step-process.

1) In the first step we realize spaces as ideals of rings, conceiving them as “algebraic spaces”.

2) In the second we pass to the *dual* of these algebraic spaces – but to the dual in the sense of the Yoneda lemma…

I had indicated how the first step works before in the recent entry on superpoints.

1) A supermanifold is a locally ringed space where the rings are ${\mathbb{Z}}_{2}$-graded.

But it turns out that this is a lie! The supermanifolds that we need in physics are actually more general than that. See section 2.1

E. Markert,

Connective 1-dimensional Euclidean Field Theories

for a “toy”(!) example for why this is so.

We really need to pass to the “cateorical dual” of the category of locally ringed spaces.

Let $C$ be any category. For the purpose of efficiently thinking about this business, I want to address the category

of functors from $C$ to $\mathrm{Set}$ as the *dual* of $C$. (Just like ${\mathrm{Hom}}_{\mathrm{Vect}}(V,\u2102)$ is the vector space dual to $V$.)

The **Yoneda lemma**, a fundamental result with an elementary proof, says that every category can be *embedded* into its dual category (such that it forms a full subcategory $C\subset {C}^{*}$). This embedding is mediated again by the $\mathrm{Hom}$-functor.

So $C$ sits inside of ${C}^{*}$. But, in general, ${C}^{*}$ is considerably larger than $C$. This is quite familiar from the behaviour of (infinite dimensional) dual vector spaces. And for similar reasons, actually.

So we know, for instance from quantum mechanics, that in many situations we start out being interested in a vector space $V$, but then realize that what we are really interested in is the (possibly larger) space ${V}^{*}$.

The same happens here. Let $C=\mathrm{SM}$ be the category of supermanifolds. As I mentioned in that previous entry, the objects $(M,{\mathcal{O}}_{M})$ in that category are topological spaces $M$ equipped with sheaves ${\mathcal{O}}_{M}$ of graded rings in a certain way. Morphisms

of supermanifolds are maps

of the underlying topological spaces together with morphisms

of sheaves. It may be worthwhile to notice that

1) Sheaves are pushed forward along $f$ (instead of pulled back) since *pre*-images of open sets are open sets,

2) that a morphism of sheaves is a natural transformation (since a sheaf is a presheaf is a functor from a category of open subsets to something, to rings in our case)

3) which, in particular, means that such a morphism assigns a ring homomorphism from a ring in ${\mathcal{O}}_{N}$ to a ring in ${\mathcal{O}}_{M}$ to each open subset

4) that this morphism really has to go the way it does (which looks “contra”) in order to reproduce ordinary maps between ordinary manifolds.

Here is the example to keep in mind. Let ${\mathbb{R}}^{n\mid m}$ be the supermanifold whose underlying space is ${\mathbb{R}}^{n}$ and whose ring of global sections is ${C}^{\mathrm{\infty}}({\mathbb{R}}^{n})\otimes {\Lambda}^{\u2022}{\mathbb{R}}^{m}$. Since it suffice to define a ring homomorphism on the generators of the source ring, it follows that a map from any supermanifold $S$ to ${\mathbb{R}}^{n\mid m}$ is the same as a collection of $n$ even global sections of $S$ and $m$ odd global sections of $S$.

Notice that this gives us a hands-on way to describe the *set* which is associated to any supermanifold $S$ by the supermanifold ${\mathbb{R}}^{n\mid m}$ (or rather of its dual). This set is simply

Crucially, this assignment of sets is functorial. For any ordinary of supermanifolds form $S$ and $S\prime $ we get an induced map on these sets.

(Actually, I think this is *contra*functorial due to the fact that the above ring homomorphism sort of point in the “opposite” direction, but this does not change any of the statements here and in the following, either way.)

The point now is that we can define *any* functor from supermanifolds to sets and regard that functor itself as (representing) a *generalized supermanifold*. This may sound abstract, but the following class of examples of such generalized supermanifolds actually reproduces nothing but the most basic intuition about how superspaces should work.

Namely, given *any* ${\mathbb{Z}}_{2}$-graded vector space

we obtain a functor ${F}_{V}$ from supermanifolds to sets by setting

You can (and should) think of this functor as nothing but the naïve physical idea of what it would mean to have a superfield $\Phi $ on $S$ with values in $V$. If $({x}_{i},{\eta}_{i})$ are local supercoordinates on $S$ you would write

where ${\Phi}_{0}$ is an even graded element of $V$, ${\Phi}_{1}$ an odd graded, and so on. This way of cooking up components of a superfield is precisely the way in which the above functor cooks up sets (of collections of components of superfields) from given supermanifolds.

So that’s essentially the point.

Another important example for the need of “generalized” supermanifolds in physics is the consideration of superstructures on the spaces of maps from one supermanifold to the other. When you map a superstring into (super or non-super) spacetime, you want the space of these maps (the moduli space, maybe) to be a supermanifold itself. But it is certainly not an ordinary supermanifold, but a generalized one, given by a functor from ordinary supermanifolds to sets.

So let $M$ and $N$ be supermanifolds and consider the space of maps

This is just a *set*, nothing more. Certainly not a supermanifold itself. But we can enlarge this a little and make it a supermanifold, again essentially by the same old dualization trick.

Namely, for every pair of supermanifolds as above, we obtain a functor from supermanifolds to sets (an object of ${\mathrm{SM}}^{*}$) which acs as

So this assigns to $S$ not quite the original set of morphisms from $M$ to $N$, but a set of morphism *parameterized* by $S$. It is easy to see that this assignment is indeed (contra-)functorial and hence defines a well defined generalized supermanifold, which is the *supermanifold of supermaps* from $M$ to $N$.

This is important.

## Re: What IS a superfield, really?

This sounds very much like the definition of an algebraic space (which I don’t actually know).

I think the rough idea is that an algebraic space is a functor from commutative rings to sets that’s also a sheaf (maybe someone can help me out here?). You don’t need a functor from schemes to sets because every scheme can be covered by affines.

Anyways, so I guess the definition here is that a superscheme (or whatever) is a functor from ${\mathbb{Z}}_{2}$ graded rings to sets that’s a “sheaf”?