What IS a superfield, really?

March 7, 2006

Posted by Urs Schreiber

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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.

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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) $\Phi^i(t,\theta) = A^i(t) + B^i(t)\theta \,,$

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

(2)

C^* := \mathrm{Hom}_\mathbf{Cat}(C,\mathbf{Set})

of functors from $C$ to $\mathbf{Set}$ as the dual of $C$ . (Just like $\mathrm{Hom}_{\mathbf{Vect}}(V,\mathbb{C})$ 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.

(3)

\array{ C &\to& \mathrm{Hom}_\mathrm{Cat}(C,\mathbf{Set}) \\ c_1 \\ \phi\downarrow\;\; &\mapsto& \mathrm{Hom}_C(\phi,--) \\ c_2 } \,.

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 $C = SM$ 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

(4)

(M,\mathcal{O}_M) \overset{(f,\phi)}{\to} (N,\mathcal{O}_N)

of supermanifolds are maps

(5)

M \overset{f}{\to} N

of the underlying topological spaces together with morphisms

(6)

\mathcal{O}_N \overset{\phi}{\to} f_*\mathcal{O}_M

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 $\mathbb{R}^{n|m}$ be the supermanifold whose underlying space is $\mathbb{R}^n$ and whose ring of global sections is $C^\infty(\mathbb{R}^n)\otimes \Lambda^\bullet \mathbb{R}^m$ . 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 $\mathbb{R}^{n|m}$ 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 $\mathbb{R}^{n|m}$ (or rather of its dual). This set is simply

(7)

\left\lbrace (t_i,\dots,t_n,\theta_i,\dots,\theta_m) | t_i \in \Gamma(\mathcal{O}_S)^{\mathrm{even}} \,, \theta_i \in \Gamma(\mathcal{O}_S)^{\mathrm{odd}} \right\rbrace \,.

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

(Actually, I think this is contrafunctorial 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

(8)

V = V_0 \oplus V_1

we obtain a functor $F_V$ from supermanifolds to sets by setting

(9)

F_V : S \mapsto (V\otimes \Gamma(\mathcal{O}_S))^\mathrm{even} = V_0\otimes \Gamma(\mathcal{O}_S)^\mathrm{even} \oplus V_1\otimes \Gamma(\mathcal{O}_S)^\mathrm{odd} \,.

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

(10)

\Phi(\{x_i,\eta_i\}) = \Phi_0(t_i) + \Phi_1(t_i)\eta_1 + \cdots

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

(11)

\mathrm{Hom}_{SM}( M,N ) \,.

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 $SM^*$ ) which acs as

(12)

S \mapsto \mathrm{Hom}_{SM}(M\times S,N) \,.

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.

Posted at March 7, 2006 9:01 PM UTC

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44 Comments & 4 Trackbacks

Re: What IS a superfield, really?

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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”?

Posted by: Aaron Bergman on March 8, 2006 2:36 AM | Permalink | Reply to this

Re: What IS a superfield, really?

Hi Aaron,

You’ve got the right idea, but there are some details missing. The definitions usually go as follows:

Let S be a (separated) scheme. The category of schemes over S has a Grothendieck topology. (Usually people use the etale topology or fpxx, since you can’t say much in the Zariski topology.) A “space over S” is a sheaf of sets on S. Obviously any scheme over S is a space over S. But not all sheaves of sets on S are schemes. One good class of examples is the quotient spaces: Suppose we have a scheme X over S , and an algebraic equivalence relation R on X. (This means that R is a subscheme of X \times X.) There’s no reason for the quotient space X/R to be a scheme, but it can happen that the X/R still defines a space over S. This is basically what we mean by algebraic space. It’s a sheaf which arises as a quotient of a scheme by an equivalence relation which is also a scheme.

Some Comments:
1) If you don’t like working with relative schemes, take S=pt or maybe Spec(Z).
2) Because these objects are sheaves, we can can work locally, and just check most properties on affine objects. Goes by the fancy name “descent theory”. Conversely, you can take a sheaf of sets defined on affine schemes, and build up a sheaf of sets on all schemes. But I don’t think this procedure produces schemes and nothing else. (Maybe a real expert will correct me, or maybe I’ll remember after I’ve slept on it.)
3) It’s not quite right to say we have a sheaf of sets on the category of rings. Affine schemes is the opposite category of rings, so a sheaf of sets on schemes gives rise to a covariant functor from rings to sets. Presheaves and sheaves are contravariant.
4) I’d be shocked if we were using _all_ sheaves on the category of Z_2-graded schemes. There are too many monsters in that category. But it’s probably safe to consider algebraic super-spaces. (Or algebraic superstacks, if you want to get all fancy.) Is this what was meant, Urs?

Posted by: A.J. on March 8, 2006 8:10 AM | Permalink | Reply to this

Re: What IS a superfield, really?

Do I have to turn in my “Licensed Physicist” card now? :)

Posted by: A.J. on March 8, 2006 8:15 AM | Permalink | Reply to this

Re: What IS a superfield, really?

Do I have to turn in my “Licensed Physicist” card now? :)

Please send it to me. I have lost mine…

Posted by: urs on March 8, 2006 1:50 PM | Permalink | Reply to this

Re: What IS a superfield, really?

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I’d be shocked if we were using all sheaves on the category of $\mathbb{Z}_2$ -graded schemes.

(Here by a sheaf on a category you essentially mean a functor with that category as domain - I hope??)

Is this what was meant, Urs?

What I know about this is taken from Stolz/Teichner and E. Markert’s thesis. There it says that we regard all functors from the category of supermanifolds to sets as generalized supermanifolds. But I don’t think that they actually use the most general such functors. Maybe you are right that there is an implicit restriuction somewhere.

And is the category of supermanifolds (= locally ringed spaces with rings locally looking like $C^\infty(U)\otimes \Lambda V$ ) precisely the same thing as the category of $\mathbb{Z}_2$ -graded schemes? Aren’t there objects in one but not in the other category?

P.S. My reply below to Aaron’s post was posted before I saw your reply.

Posted by: urs on March 8, 2006 1:39 PM | Permalink | Reply to this

Re: What IS a superfield, really?

(Here by a sheaf on a category you essentially mean a functor with that category as domain - I hope??)

There are some gluing conditions, too.

Posted by: Aaron Bergman on March 8, 2006 3:25 PM | Permalink | Reply to this

Re: What IS a superfield, really?

Nope, supermanifolds and superschemes aren’t the same thing. But closed super-Riemann surfaces and (nice adjectives) super-curves probably match up nicely. I think this is probably the case Stolz and Teichner are interested in. OTOH, you probably don’t gain anything by switching to schemes.

Instead, just transport all of the definitions I just gave to the category of topological/smooth supermanifolds. The only problem is that the notation gets ugly: “algebraic super-manifold” is a terrible shorthand for “algebraic space in the category of supermanifolds”

Posted by: A.J. on March 8, 2006 3:35 PM | Permalink | Reply to this

Re: What IS a superfield, really?

But closed super-Riemann surfaces and (nice adjectives) super-curves probably match up nicely. I think this is probably the case Stolz and Teichner are interested in.

Probably. Remarkably, actually implementing this is still indicated as work in progress. Given that the (relatively “trivial”) situation for real 1-dimensional supermanifolds already fills an entire thesis this is maybe not too surprising.

But it makes me worry. In the end one would expect there to be a nice conception of superfunctor on super Riemann surfaces, not something intractibly involved.

There are other steps in their construction which look to me like they should eventually disappear. The fact that every superbordism in their formalism is to be decorated by a fermionic Fock space looks strange to me. I don’t see what that would correspond to physically. (Of course I know what fermionic Fock spaces are good for, but they should not appear as decoration of “time intervals”, as far as I am concerned.)

Related to this is the fact how they declare that for every closed Riemann surface we have to tensor that fermionic Fock space furthermore with the Pfaffian. This condition comes out of thin air. It seems to be justified solely by the fact that this way we enforce the functors on the decorated superbordisms to actually produce modular forms.

But it appears to be artificial to essentially impose that requirement by hand. And, in fact, if it were not for this Pfaffian it seems we could completely drop the Fock space decoration in Stolz/Teichner’s definition of superbordisms and loose nothing. (Am I wrong?) The condition that our functor shall be Clifford-linear can be imposed without decorating anything, it seems to me.

Posted by: urs on March 8, 2006 8:48 PM | Permalink | Reply to this

Re: What IS a superfield, really?

I made a major typo: A “space over S” is a sheaf of sets on the category of schemes over S. In particular, a space X over S assigns to every scheme Y over S the set X(Y), and does so in a coherent way.

BTW, Grothendieck’s relative point of view is just that it’s more natural to study, instead of a scheme Y, a scheme Y together with a morphism to a base scheme S. The idea comes from vector bundles: Y is a family of schemes over S, associating to s in S, the fiber Y_s. In particular, when S = Spec(R), you get back the usual schemes over R. This is often used in conjunction with the “spaces are sheaves” philosophy, but it’s not the same thing. The latter is usually called the “functor of points” perspective.

Posted by: A.J. on March 8, 2006 7:18 PM | Permalink | Reply to this

Re: What IS a superfield, really?

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more natural to study, instead of a scheme $Y$ , a scheme $Y$ together with a morphism to a base scheme $S$

This seems to be a useful trick in order to give the Hom-sets in a category of certain spaces desired additional structure. At least that’s what this is used for by Stolz/Teichner/Markert.

There the problem is that one is dealing with categories of superbordisms and wants the Hom-spaces to be supermanifolds themselves. But the naïve set of superbordisms $\mathrm{Hom}(B_1,B_2)$ between given boundaries $B_1$ and $B_2$ is not a supermanifold.

So what they do is to define not just a fixed Hom-set $\mathrm{Hom}(B_1,B_2)$ , but a map that assigns to each supermanifold $S$ the set of entire bundles of superbordisms over $S$ . This gives a (contravariant) functor from supermanifolds to sets and hence a “generalized superspace”.

(That’s pp. 22-23 here, in case anyone is interested.)

BTW, you keep referring to these functors as sheaves. Being functors, they are certainly presheaves, but in the literature that I am currently looking at I nowhere see a requirement stated that we actually want a sheaf condition for functors that represent generalized objects.

I see why that would be natural, but I wonder why it is not being mentioned.

Posted by: urs on March 8, 2006 8:27 PM | Permalink | Reply to this

Re: What IS a superfield, really?

Urs, I’m gonna be pretty lazy about switching between calling things “schemes” or “supermanifolds”. The machinery works in both cases. (We could agree to call these things “spaces”, but unfortunately we’re already using that term for something else. )

Anyways, requiring the sheaf condition is more or less standard. It means that, if X: Schemes -> Sets is a functor, and Y is a scheme, you can work out X(Y) by covering Y with some open sets Y_i, and then gluing together elements of X(Y_i). The sheaf condition is what allows you to treat these functors as if they were geometric objects; it means you can do your geometry locally.

My guess is that Markert didn’t want to get bogged down in explaining/checking that Hom(M,N)(Y) = Hom(M x Y, N) actually forms a sheaf. But I think this step is the crucial one. Turing the category of supermanifolds into the category of enriched supermanifolds by replacing the hom-set Hom(M,N) with the sheaf Hom(M,N)(-) is a tautological substitution. However, this trick is only interesting if Hom(M,N)(-) is actually a geometric object of some kind. Ideally, it would be a representable sheaf, i.e. Hom(M,N)(Y) = Hom(Y,R(M,N)), for some supermanifold R(M,N), so that we can just think about the supermanifold R(M,N). But at the least, this functor should be a sheaf on the category of supermanifolds, so that you can do local geometry.

Maybe we should pester Markert about this? I’m willing to believe that this thing is a sheaf, but I’d like to know if/why it’s not representable.

Posted by: A.J. on March 8, 2006 11:37 PM | Permalink | Reply to this

Re: What IS a superfield, really?

Hah! I just read a bit further into the thesis. The functors in question are indeed representable. (Apply Hohnhold’s Theorem to bordisms between empty sets.) Markert didn’t bother to mention sheaf conditions because something much stronger is true: Hom(M,N) is the functor of points of a supermanifold.

Posted by: A.J. on March 9, 2006 12:01 AM | Permalink | Reply to this

Re: What IS a superfield, really?

So, morally, is there any particular reason why I should care whether or not a sheaf functor is representable? My impression has always been that pretty much anything you can do with a space, you can do with its functor of points.

Posted by: Aaron Bergman on March 9, 2006 2:28 AM | Permalink | Reply to this

Re: What IS a superfield, really?

Posted by: Aaron Bergman on March 9, 2006 2:29 AM | Permalink | Reply to this

Re: What IS a superfield, really?

In theory, there’s not much difference. Most things that are true for schemes are true for algebraic spaces. But in practice, there’s a lot of difference: the proofs become much harder.

Posted by: A.J. on March 9, 2006 7:43 AM | Permalink | Reply to this

Re: What IS a superfield, really?

[…] reason why I should care whether or not a sheaf functor is representable?

In theory, there’s not much difference. Most things that are true for schemes are true for algebraic spaces.

Hm, does this mean that every non-representable sheaf functor on schemes is an algebraic space?

Posted by: urs on March 9, 2006 7:51 AM | Permalink | Reply to this

Re: What IS a superfield, really?

Hm, does this mean that every non-representable sheaf functor on schemes is an algebraic space?

No, I’m being sloppy again. (This tends to happen when I post after bedtime.) Algebraic spaces are those sheaf functors on the (etale) site of schemes which are obtained as quotients of schemes. Not all sheaves of sets arise this way (although identifying the ones that don’t usually strikes me as an exercise in pathology).

Posted by: A.J. on March 9, 2006 8:17 AM | Permalink | Reply to this

Re: What IS a superfield, really?

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The functors in question are indeed representable.

So let me ask a really dumb question: Is every representable functor (on a site) a sheaf?

Another related question: On p.12 Elke Markert defines what is supposed to be the superspace $\mathbb{R}^{n|m}$ in terms of a functor to sets, namely that sending the supermanifold $S$ to the set

(1)

\left\lbrace (t_1,\dots,t_m;\theta_1,\dots,\theta_n) | t_i \in \Gamma(S)^\mathrm{ev}, \theta_i \in \Gamma(S)^\mathrm{odd} \right\rbrace

of $(n+m)$ -tuples of $n$ even and $n$ odd global sections of $S$ .

It is suggested that this functor is indeed represented by $\mathbb{R}^{n|m}$ . But I have trouble believing this. A morphism of locally graded ringed spaces from $S$ to the ordinary version of $\mathbb{R}^{n|m}$ seems to me to be an $(n+m)$ -tuple of global sections of the above kind, but with the additional requirement that all the odd sections in that $(n+m)$ -tuple square to zero. Otherwise the tuple of sections will not define a homomorphism of rings from the ring of global sections of $S$ to the ring of global sections of $\mathbb{R}^{n|m}$ .

Isn’t that right? If it is right, is the functor to sets described above representable at all? If it is not right, can anyone tell me what I am missing?

Posted by: urs on March 9, 2006 7:25 AM | Permalink | Reply to this

Re: What IS a superfield, really?

Is every representable functor (on a site) a sheaf?

This depends on the site. If the open coverings in the site come from inclusions of open subsets, it’s pretty easy. If this is not the case, it’s harder. That the functor of points of a scheme forms a sheaf over the etale site of schemes is one of Grothendieck’s theorems.

I don’t believe that a representable functor on an arbitrary site is automatically a sheaf. However, I don’t think anyone studies sites where this reult doesn’t hold. If you ever hear an algebraic geometer going on about “descent”, this is basically what they’re doing: making sure the functor of points actually defines a sheaf on some site.

Isn’t that right? If it is right, is the functor to sets described above representable at all? If it is not right, can anyone tell me what I am missing?

It’s right. I think it’s implicit in her notation: a section is odd if your anti-commuting condition holds.

Posted by: A.J. on March 9, 2006 8:27 AM | Permalink | Reply to this

Re: What IS a superfield, really?

That the functor of points of a scheme forms a sheaf over the etale site of schemes is one of Grothendieck’s theorems.

Great, thanks for all your replies, very helpful.

(Concerning that odd-section business, I was just being dense, never mind.)

Posted by: urs on March 9, 2006 9:52 AM | Permalink | Reply to this

Re: What IS a superfield, really?

Great, thanks for all your replies, very helpful.

My pleasure.

I should add another caveat: there’s considerably more to descent theory than just asking whether the functor of points is a sheaf. Schemes and their morphisms have all sorts of properties (e.g. reduced, separated, constant dimension,…), and one usually wants to know if these properties can be checked locally. A lot of the work in descent theory is showing that these properties can be checked on open covers.

Posted by: A.J. on March 9, 2006 5:55 PM | Permalink | Reply to this

Re: What IS a superfield, really?

I realize that I have more questions on this statement:

That the functor of points of a scheme forms a sheaf over the etale site of schemes is one of Grothendieck’s theorems.

What is known about sheaves on the site of schemes that do not arise as the functor of points of some scheme, i.e. that are not representable? Are there any? If yes, are there any “non-pathological” ones in some suitable sense?

This is related to Aaron’s question: Should I desire all my sheaves to be representable?

For a more concrete question along these lines see this one.

Posted by: urs on March 9, 2006 6:30 PM | Permalink | Reply to this

Re: What IS a superfield, really?

$MathML-enabled post (click for more details).$

What is known about sheaves on the site of schemes that do not arise as the functor of points of some scheme, i.e. that are not representable? Are there any? If yes, are there any “non-pathological” ones in some suitable sense?

Yes, there are plenty of them. All schemes are algebraic spaces, but not all algebraic spaces are schemes. In fact, algebraic spaces were invented precisely to deal with the fact that many of the functors we want to study in algebraic geometry are not representable! (Stacks were invented for the same reason: It turns out that many of those functors we want to study aren’t even algebraic spaces. :)

Maybe we should dwell on algebraic spaces for a few minutes. It’s very helpful to understand the relation between algebraic spaces and schemes by analogy with the relation between schemes and affine schemes. Remember that we obtain a general scheme by taking affine schemes and pasting them togther with transition functions. More formally: We have an affine scheme $U = \sqcup_i U_i$ , which represents the disjoint union of our coordinate patches, and we have some gluing data which tells us how the patches overlap. This means we have morphisms $U_i \cap U_j \to U_i$ and $U_i \cap U_j \to U_j$ . Or more formally, we can think of $R= \sqcup_{i,j} U_i \cap U_j$ as a relation, a subscheme of $U \times U$ . Given these two pieces of data: an affine scheme $U$ and a relation $R$ on $U$ , we can build up a general scheme. All schemes are attained this way, but not all schemes are affine.

The story with algebraic spaces goes the same way: We want to enlarge the category of schemes by allowing reasonable quotients of schemes. Most quotients of schemes – even surprisingly nice quotients – aren’t actually schemes. So we just enlarge the category of objects we’re considering.

So, yes, there are plenty of non-representable sheaves: any algebraic space which isn’t a scheme.

Aside from group quotients, the best place to look for these critters is in the theory of moduli spaces. The first example might still be the best: Suppose you have a scheme Y, and you want another scheme H which parametrizes closed subschemes of Y. And suppose further that you want this object to behave well in families: maps from a scheme S to H should be the set of families of closed subschemes of Y which are parametrized by S. This means, you want to study the functor

(1)

S \mapsto \{set of closed subschemes Z of Y \times S\}

(Technically, you also want the projections $Z \to S$ to be reasonably continuous.) This functor isn’t usually representable – you need some nice conditions on $Y$ – but you can still deal with it by thinkig of it as an algebraic space.

Posted by: A.J. on March 9, 2006 7:15 PM | Permalink | Reply to this

Re: What IS a superfield, really?

Stacks were represented for the same reason

Urgh. Another typo. That should be “invented”, not “represented”.

I’m going to go get some coffee.

Posted by: A.J. on March 9, 2006 7:19 PM | Permalink | Reply to this

Re: What IS a superfield, really?

Hello everyone.

There is an obvious ( if correct :-) ) remark that is relevant to this whole discussion, namely, that in contrast to the algebraic setting, in the differential setting we cannot enlarge the category in this fashion unless we allow for objects that are _not smooth_ (in some appropriate sense), at least, as long as we don’t go to stacks. At least it appears so to me. An analogical statement should hold for superspaces as well, so, as long as we don’t want to consider orbifolds (that are naturally described as stacks) or such we don’t really need all of that categorical jazz. Probably superorbifolds can be described without it as well.
In the algebraic setting it is much more important to consider stuff like “etale algebraic spaces” since the Zariski topology which is the only topology in the ordinary sense we can use is a very bad one, for instance, it produces quite useless cohomology groups. Therefore it pays off to use fancy Grothendick topologies such as etale or flat. For differential manifolds, on the other hand, the usual classical topology is pretty good. In fact I don’t know any interesting (other) Grothendick topology for differential manyfolds. That is one that is not equivalent to the usual one (i.e. produces a different category of sheaves).

Posted by: Squark on March 24, 2006 6:19 PM | Permalink | Reply to this

Re: What IS a superfield, really?

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Hello Squark,

nice to hear from you!

as long as we don’t want to consider orbifolds (that are naturally described as stacks) or such we don’t really need all of that categorical jazz.

It would help me if we could discuss your remark in terms of the central example which motivates these constructions in the context that I mentioned.

We have a $\mathbb{Z}_2$ -graded Hilbert space $H$ and hence a graded vector space of operators on that Hilbert space. We want to equip that space $V$ of operators with the structure of a supermanifold in order to be able to write the propagagtor of supersymmetric quantum mechanics as a morphism of supergroups:

(1)

(t,\theta) \mapsto \mathrm{exp}(-tD^2)(1+\theta D) \,,

where $D$ is some odd operator.

Hence we conceive $V$ as a generalized supermanifold by declaring that its set of $S$ points is

(2)

(\Gamma(S)\otimes V)^{\mathrm{ev}} \,,

i.e. the set of products of odd sections of the ordinary supermanifold $S$ with odd operators in $V$ and products of even sections of $S$ with even operators in $V$ .

As I have already said somewhere else in this comment section, I would like to know if this generalized supermanifold is representable. I guess if it is, the problems you have in mind disappear automatically.

But I tend to expect that this is not representable. In this case, could you maybe illustrate in terms of this example what problem with differentiability you have in mind?

Posted by: urs on March 28, 2006 8:10 AM | Permalink | Reply to this

Re: What IS a superfield, really?

I think it _is_ representable, more precisely, it is Spec Symm V* where Symm V* is the supersymmetric algebra of V* (the Fock space corresponding to V* if you wish). If V is a Hilbert space V* is naturally isomorphic to V though I don’t know why is it important here.
I don’t have any problem in mind, it’s more like a coproblem! What I’m saying is that you don’t get anything new from all of this categorical machinary as long as you’re working in the differential context. In other words, you don’t have any “sufficiently good” non-representable functors.

Posted by: Squark on March 30, 2006 11:32 PM | Permalink | Reply to this

Re: What IS a superfield, really?

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I think it is representable, more precisely, it is $\mathrm{Spec}\mathrm{Symm} V^*$ where $\mathrm{Symm}V^*$ is the supersymmetric algebra of $V^*$ (the Fock space corresponding to $V^*$ if you wish).

That’s interesting. Could you indicate in more detail how one sees that this represents the functor in question? Is it supposed to be obvious (I might be lacking some basic tools here)?

What I’m saying is that you don’t get anything new from all of this categorical machinary as long as you’re working in the differential context. In other words, you don’t have any “sufficiently good” non-representable functors.

I see. Thanks for pointing this out.

I would still expect that it is easier to work with the represented functors, than with the representing objects, but maybe I am wrong. It would be helpful if I understood why $\mathrm{Spec}(\mathrm{Symm}V^*)$ represents

(1)

S \mapsto \{ (\Gamma(S)\otimes V)^\mathrm{ev}\} \,.

Posted by: urs on March 31, 2006 10:08 AM | Permalink | Reply to this

Re: What IS a superfield, really?

You indicated V is a Hilbert space so an obvious question is whether it is finite dimensional.
1) If it is finite dimensional, the statement is true both in the differential and the algebraic context, but is easier to see formally in the algebraic.
1a) In the algebraic context
Spec Symm V* is an affine superscheme. Consider Mor(X, Spec Symm V*) where X is another affine superscheme. It is the same as Mor(Symm V*, Gamma(X)). Now, Symm V* is freely generated by V* as a supercommutative algebra hence Mor(Symm V*, Gamma(X)) is the same as Hom(V*, Gamma(X)) in the sense of supervector spaces. The later is the same as (Inner Hom(V*, Gamma(X)))^ev and Inner Hom(V*, Gamma(X)) is the same as Gamma(X) (x) V as easy to see. So, the functors coincide on affine superschemes. Hence, they coincide on all superschemes, since both are sheaves and any superscheme can be covered by affine ones.
1b) In the differential context Spec Symm V* is the supermanifold whose “underlying manfiold” is just V*^ev (regarded as a manifold int the obvious way) and the structure sheaf is O(U) = C^infinity(U) (x) Lambda(V*^odd). It should be not difficult to see it gives the correct functor.
2) If it is infinite-dimensional, the corresponding object should be an infinite-dimensional supermanifold. People are usally afraid even of ordinary infinite-dimensional manifolds (at least as long as they’re working rigorously rather than doing some voodoo like QFT :-) ). So I guess this case falls outside of the “sufficiently good” category I referred to earlier. It is possible that the point-functor point of view is an effective approach to tackle infinite dimensional (super)manifolds, better than a straightforward geometrical definition (as far as I know the later doesn’t exist except for some feeble attempts - the name “Banach manifold” comes to mind, not that I know what it is), but I don’t know.

Posted by: Squark on March 31, 2006 2:56 PM | Permalink | Reply to this

Re: What IS a superfield, really?

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Hi Squark,

thanks for the reply! And sorry for my (extremely) slow response.

At the time you wrote this I maybe didn’t quite grasp it. But now that I have actually begun learning some elements of algebraic geometry systematically ( $\to$ ) I see that what you write is actually quite obvious and simple. Thanks for taking the time.

Posted by: urs on June 27, 2006 11:45 AM | Permalink | Reply to this

Re: What IS a superfield, really?

$MathML-enabled post (click for more details).$

Let me come back to the representability issue.

The functors in question are indeed representable.

Ok, that’s the case for the functors describing the spaces of morphisms in the domain category.

Is the same true for the functors which give the hom-spaces of the target category in Markert’s thesis (equivalently in Stolz/Teichner)?

I am speaking of the hom-spaces defined in definition 27, p. 30 of Elke Markert’s thesis.

This is the standard construction which I mentioned also above in the entry. Given any $\mathbb{Z}_2$ -graded vector space $V$ , we obtain a pre-sheaf

(1)

S \mapsto (\Gamma(S)\otimes V)^\mathrm{even}

which assigns to any suoermanifold $S$ the set of pairs of even sections of $S$ tensored with an even vector in $V$ and odd sections of $s$ tensored with an odd vector in $V$ .

The obvious question is hence:

Under what conditions is this presheaf a representable functor and/or a sheaf?

Posted by: urs on March 9, 2006 6:20 PM | Permalink | Reply to this

Re: What IS a superfield, really?

$MathML-enabled post (click for more details).$

The Wikepedia entry on algebraic space gives the impression (to me) as if an algeraic space is something like an orbifold structure on an affine scheme, since it is defined as the space $X$ obtained from a scheme $U$ together with a subscheme $R$ of $U\times U$ such that $R$ is an equivalence relation, by $X = |U|/|R|$ .

For $U$ an ordinary space that would describe a groupoid with the space as its space of objects and at most one nontrivial morphism between any two points. $X$ would be the decategorification of that groupoid, which is one way to describe orbifolds $X$ .

That’s at least what I extract from the Wiki definition. I’d be grateful for corrections. Maybe there are even different things being called “algebraic spaces”??

an algebraic space is a functor from commutative rings to sets

In any case, given any concept of space (ordinary topological space, locally ringed space, scheme, whatever), we can, I guess, always pass to generalizations of these spaces by looking at functors from them to sets. Apparently that’s called in general the relative point of view.

Posted by: urs on March 8, 2006 1:26 PM | Permalink | Reply to this

Re: What IS a superfield, really?

This reminds me of mathematicians who claim that Dirac didn’t understand what a delta-function really was until Laurent Schwartz invented distributions. This is nonsense, of course. Dirac both could compute with delta-functions and had a mental picture of them (an infinitely high, infinitely thin peak), and that is pretty much what understanding is all about. That mathematicians didn’t understand, or liked, what Dirac was doing is another matter.

Sometimes it seems to me that the goal of mathematicians is to confuse physicists. You did a good job at confusing me about what superfields really are.

Another question: What IS a Lie superalgebra, really? The answer is in section 1.0.1 in math-ph/0202025.

Posted by: Thomas Larsson on March 8, 2006 5:28 AM | Permalink | Reply to this

Re: What IS a superfield, really?

Here are some random ramblings (easily ignored) from a layman’s perspective…

I think what Urs and others like him are trying to do is to discover the language in which god wrote the universe. When you have the correct language at your disposal, meaningful statements can be made with little or no effort. The immense linguistic effort that is put into explaining the simplest of physical concepts these days is an indication that there is a long way to go before we even begin to understand what ANYTHING really is.

What you say about a mathematicians job being to confuse physicists sounds like something a poet could say about a linguist. In fact, I think the analogy is a useful one. The ultimate goal is a means to communicate something real (whatever “real” means). There are certain languages that simply lack the structure to communicate certain thoughts. I remember a friend of mine from Japan who said the greatest thing about learning english was the ability she gained to easily express anger. The structure of any language will have some built in limitations as to what it can communicate efficiently.

I like what Professor Baez once said about the pre-category-theoretic ways to express mathematical/physical equations linearly, i.e. a = b instead of

a
\ \
b.

I think it is a step in the right direction when the language used to express physical concepts starts to actually resemble the geometry of the physical concept.

*back to lurking*

Eric

Posted by: Eric on March 8, 2006 7:56 AM | Permalink | Reply to this

Re: What IS a superfield, really?

Hi Eric,

yes, I can subscribe to this.

When you have the correct language at your disposal,

Apart from more subjective aesthetic or other reasons for preferring one language over the other, I think the main point of identifying the right language and the right definitions is that it allows you to move on to more complicated or more general situations.

So for instance, in the case at hand, the purpose of reformulating supersymmetric quantum mechanics as a superfunctor on supercobordisms is to find the right language in which to categorify to obtain the description of superconformal 2D CFT.

Posted by: urs on March 8, 2006 1:49 PM | Permalink | Reply to this

Re: What IS a superfield, really?

$MathML-enabled post (click for more details).$

[…] delta-function […]

Yes, good example, I agree. For many purposes you are perfectly fine with “just” knowing Dirac’s (how should we call it, “heuristic”?) “definition” of the delta-“function”. Even better, for a lot of computations having Dirac’s definition in mind is probably more efficient than having the “correct” definition in mind.

But, as we know, there are situations when it is necessary to know what a Dirac delta “really” is.

To my mind, the optimal scenario is where you do know the “true” definition of an entity, in order not to confuse your your concepts, and you know in which situations you can safely use a heuristic way of thinking about that entity, in order to be able to efficiently compute.

You did a good job at confusing me about what superfields really are.

Sorry.

Another question: What IS a Lie superalgebra, really? The answer is in section 1.0.1 in math-ph/0202025.

But what is given in that paper is precisely the definition I talked about! The authors define a Lie superalgebra as a Lie algebra object internal to precisely that category of generalized supermanifolds which I mentioned.

For instance, the space of $C$ -points of a Lie superalgebra which they mention on the top of page 8 is nothing but the set associated to the superalgebra under that functor from supermanifolds to sets. I called these sets “ $S$ -points”, instead of “ $C$ -points”.

Posted by: urs on March 8, 2006 9:00 AM | Permalink | Reply to this

Re: What IS a superfield, really?

To me the point is, IS a superfield a algebra of functions over a commutative space or over a noncommutative one? Because if you trick fermionic variables to anticommute with grassman ones, then you have the superfields are a commutative algebra, are they?. If so, which should the obstruction to recover an underlying space then under Gelfand/Naimark duality be?

Posted by: Alejandro Rivero on March 9, 2006 8:36 PM | Permalink | Reply to this

Re: What IS a superfield, really?

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Because if you trick fermionic variables to anticommute with grassman ones, then you have the superfields are a commutative algebra, are they?

Do I understand correctly that you are referring to the fact that a superfield like

(1)

\Phi(t,\theta) = A(t)+ B(t)\theta,

is (supposed to be) even, meaning that $A$ is “bosonic” and B is “fermionic”?

If so, I can answer this question:

The point is, such a superfield is to be regarded as an “S-point” of a (generalized) supermanifold. Whatever that means, it implies that you should not think of these superfields as constituting ordinary function algebras.

Instead, underlying the concept of S-points is that of ordinary (as opposed to possibly generalized) supermanifolds. These look locally like $\mathbb{R}^{n|m}$ and hence are locally - on a patch $U\subset \mathbb{R}^n$ , say - determined by the ring

(2)

C^\infty(U)\otimes \Lambda \mathbb{R}^m \,,

which is just the ring of sections of the trivial exterior bundle over $U$ , i.e. essentially just the ring given by 0-, 1-, 2-, … $m$ -forms on $U$ .

This ring is graded commutative, but not commutative. And it has one maximal ideal for each maximal ideal of $C^\infty(U)$ .

Posted by: urs on March 9, 2006 10:38 PM | Permalink | Reply to this

Re: What IS a superfield, really?

Yeah, that was my point, that B is fermionic. If it is a classical fermion field, then two of such B anticommute and the superfield is even.

It is not if I should see the superfield in one way or another, the question is if I can built a commutative C* algebra from the multiplication of superfields. That is purely technical issue; if I can, then I can use Gelfand duality to recover a space. If I can not but I can build a noncommutative C* algebra, I still have the tools of Noncommutative Geometry to work with this space. If I recover just a manifold U, then at least I know my algebra is morita equivalent to the one of fuctions over U. If I can deform the process to recover a multiple sheeted manifold then I could give a hint of susy breaking.

We can do it independently of our tastes about meaning. The doubt I have is if the algebra can get an adequate norm to be a C* one.

Posted by: Alejandro Rivero on March 10, 2006 4:51 PM | Permalink | Reply to this

Re: What IS a superfield, really?

$MathML-enabled post (click for more details).$

Let’s see. Do I understand correctly that you would like, in terms of the above example, to think of a product given by something like

(1)

(A(t) + B(t)\theta) (\tilde A(t) + \tilde B(t)\theta) = A(t)\tilde A(t) + (A(t)\tilde B(t) + \tilde A(t)B(t)) \theta

In terms of pairs $(A,B)$ of components this reads

(2)

(A,B)(\tilde A,\tilde B) = (A\tilde A, A\tilde B + \tilde A B) \,.

Right, I guess that should define a strictly commutative (instead of graded commutative) algebra structure on pairs $(A,B)$ of functions of one ordinary variable.

What’s the spectrum? It contains the spectrum of the $A$ component. Anything else?

What norm might you put on this algebra?

For vanishing $B$ it should reduce to the supremum norm (“uniform norm”) on the $A$ component, I guess.

Maybe we first fix a star involution. The most natural thing seems to be

(3)

(A,B)^* := (A^*,B^*) \,,

where the star on the right is complex conjugation.

It seems that if we define

(4)

||(A,B)|| := ||A|| \,,

where on the right we have the ordinary supremum norm

(5)

||A|| = \mathrm{sup}(|A(t)| : \forall t)

we do get a $C^*$ -algebra. No?

I don’t have an intuition right now where this should lead to. So maybe there is a more natural choice of norm and/or star structure.

Posted by: urs on March 10, 2006 7:06 PM | Permalink | Reply to this

Re: What IS a superfield, really?

This lack of intuition is the one I suffer when considering mathematical objects; it seems mathematicians get a subtle training to recognise the Objects in their pursuits. As physicist, I can only think of a math object in pragmatical terms: can I use it to formulate supersymmetry breaking, for instance? If instead of the spectrum of just the A component we were able to locate the spectrum of it times Z2 or so, we could break susy into things as the old Connes Lott models. It seemed so obvious to me that I was induced to think I was missing some fundamental weakness. Urs, let me to thank you warmly for reproduce my thinking, now I am a bit more sure of the soundness of this possibility.

(But the question of the thread was, I see, mathematical, asking “what is”, not “what use is for”. So I am afraid I have not been so helpful in exchange)

Posted by: Alejandro Rivero on March 12, 2006 6:23 PM | Permalink | Reply to this

Re: What IS a superfield, really?

$MathML-enabled post (click for more details).$

If instead of the spectrum of just the $A$ component we were able to locate the spectrum of it times $\mathbb{Z}_2$ or so, we could break susy into things as the old Connes Lott models.

That sounds interesting, but I cannot say that I see what you have in mind. Could you maybe explain in a little more detail how you think supersymmetry breaking would be related to what we have talked about?

I would think that in order for something like susy breaking to be meaningful, we’d first need some some action of some susy group on something. This seems to be more structure than we have talked about so far. So far we just have talked about supermanifolds and maps between them.

now I am a bit more sure of the soundness of this possibility

Good. But I am not sure which possibility you actually mean. Above I was just trying to guess in which way you wanted to think of superfields as forming a commutative algebra.

My point was that even if you have a commutative algebra structure on superfields it does not seem to be natural to try to think of this particular algebra as defining a supermanifold or any other space.

Maybe it is. But I don’t see it at the moment.

Posted by: urs on March 12, 2006 7:21 PM | Permalink | Reply to this

Re: What IS a superfield, really?

$MathML-enabled post (click for more details).$

Of course all that bounces in my head is the affinity between a grassman variable and an infinitesimal variable. If in a superfield $F(x,\theta)$ we forget about the grassman particularities, we get a function in an extra dimension. On the other hand, Connes-Lott models have an extra dimension but it is a discrete one so it seems a case sort of “intermediate” between grassman and standard variables. One can phantasize that in some limit where the distance in the Connes-Lott extra dimension goes to zero we should recover the structure of a superfield. And thus reciprocally, that Connes-Lott extra dimensional spaces are deformed, or broken, versions of superspaces.

In order to implement these phantsies, a first step is to see describre superspace in terms of the algebra of functions over superspace, which happens to be the algebra of superfields.

Posted by: Alejandro Rivero on March 13, 2006 12:40 PM | Permalink | Reply to this

Re: What IS a superfield, really?

[…] these phantsies […]

Ok, thanks. Now I am beginning to see what you have in mind. There is possibly an interesting idea here, though it remains a little vague at this point. I’ll keep it in mind.

Posted by: urs on March 14, 2006 9:58 AM | Permalink | Reply to this

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