## March 2, 2006

### Superpoints

#### Posted by Urs Schreiber

I am currently staying at Schloss Mickeln, attending the workshop on Stolz&Teichner’s approach to elliptic cohomology which I mentioned recently.

There is little time to blog with lots of things to be said. Here I’ll just mention a single cool factoid.

In his introductory talk Prof. Laures mentioned the existence of unpublished work in progress which extends Stolz&Teichner’s approach to dimension 0. Not many details are available, but it seems like people are about to make sense of the first line in the following table:

(1)$\begin{array}{cccc}\mathrm{physical}\phantom{\rule{thickmathspace}{0ex}}\mathrm{object}& \mathrm{propagation}& \mathrm{cohomology}& \mathrm{chromatic}\phantom{\rule{thickmathspace}{0ex}}\mathrm{level}\\ {D}_{-1}-\mathrm{brane}& 0-\mathrm{transport}& \mathrm{deRham}& 0\\ \mathrm{point}\phantom{\rule{thickmathspace}{0ex}}\mathrm{particle}& 1-\mathrm{transport}& K-\mathrm{theory}& 1\\ \mathrm{string}& 2-\mathrm{transport}& \mathrm{elliptic}\phantom{\rule{thickmathspace}{0ex}}\mathrm{cohomology}& 2\end{array}$

The second line says something like that the space of all 1-dimensional supersymmetric euclidean field theories ($\simeq$ supersymmetric quantum mechanics) “at grade $n$” is homotopy equivalent to the $n$-th space in the $\Omega$-spectrum representing periodic K-theory.

E. Markert
Connective 1-dimensional Euclidean Field Theories
$\to$ .ps.

This has been the toy model setup for the content of the third line. There is evidence that elliptic cohomology is similarly given by supersymmetric 2-dimensional field theories, hence by “superstrings” of one sort or another.

Understanding the details of this is the whole point of workshops like the one I am reporting from.

From the perspective of $n$-transport (which is my terminology) it is natural to wonder if there is also going on something interesting for 0-transport. The claim now is that, yes, when correctly set up, 0-functors which send super-points to some some superspace obtained from vector bundles over target space yield, in a similar fashion, ordinary de Rham cohomology.

This sounds like it should be very trivial. But it is not. Part of the subtlety is the super-aspect. The above table crucially depends on this. If you pass from supersymmetric $n$-transport to more general $n$-transport the moduli spaces of these $n$-dimensional field theories become contractible and hence topologically trivial.

So what is a superpoint? Physically, it should be something like the worldvolume of a super $\left(-1\right)$-brane. The right mathematical language to think about this is to conceive supermanifolds as ${ℤ}_{2}$-graded ringed spaces.

This is well known to those who know it well. It turned out, however, to be some entertaining exercise to translate physicist’s naïve but efficient handling of supermanifold’s to the precise language, which takes a little getting used to.

I’ll mention some key aspects. Thanks go in particular to Guy Buss for explaining aspects of this stuff to me. One relevant textbook is

Supersymmetry for Mathematicians
Courant Lecture Notes in Mathematics vol. 11
American Mathematical Society (2004)

The entire setup here is motivated from the scheme concept in algebraic geometry. We describe a supermanifold in terms of its ring of “functions” over it.

A locally ringed space $\left(M,O\right)$ is a topological space $M$ together with a sheaf of rings $O$ over it, such that each stalk is a local ring.

This means roughly that the maximal ideals of these rings correspond to the points in $M$.

An ordinary manifold is the same as a locally ringed space (M,O) such that for each $p\in M$ there is a neighborhood $U\ni p$ such that $O{\mid }_{U}$ is isomorphic to a ring of smooth functions over $U$.

This easily generalized. A supermanifold is a locally ringed space (M,O) such that for each $p\in M$ there is a neighborhood $U\ni p$ such that $O{\mid }_{U}$ is isomorphic to a ring of smooth functions tensored with the exterior algebra of some vector space.

In particular, the supermanifolds of the form ${ℝ}^{p\mid q}$ are nothing but ringed spaces $\left(M,O\right)$ where $M={ℝ}^{p}$ is simply $p$-dimensional Euclidean space and where $O$ is the sheaf of smooth functions of ${ℝ}^{q}$ times elements of exterior powers of ${ℝ}^{q}$.

A superpoint is a ${ℝ}^{0,q}$.

There is an obvious notion of morphism between ringed spaces. Such a morphism is simply a continuous map between the two topological spaces together with a morphism of sheaves from the sheaf of rings on the target space to the pushforward of the sheaf of rings on the source space.

While all this may sound obvious, it turns out to be instructive to rederive the local coordinate formulas for operations on ${ℝ}^{p\mid q}$ which you find in physics texts. The crucial concept needed to do so is that of $S$-points of a supermanifold.

Let $S$ be some supermanifold, then $\mathrm{Hom}\left(S,{ℝ}^{p\mid q}\right)$ is called the space of $S$-points of ${ℝ}^{p\mid q}$ (the “$S$-shaped loci in ${ℝ}^{p\mid q}$”).

If you are a physicist, you are familiar with adressing a series of symbols of the form

(2)$\left(t,\theta \right)↦\left(t+t\prime +\theta \prime \theta ,\theta +\theta \prime \right)$

as a expressing a supertranslation in ${ℝ}^{1\mid 1}$.

In the language of locally ringed spaces, what this really means is something like the following.

Let $S$ be some generic higher-dimensional supermanifold. What is an $S$-point in ${ℝ}^{1\mid 1}$? In order to compute that, we map all of the underlying topological space of $S$ into $ℝ$ and push all the even and odd sectins of $S$ forward. By the above definition of morphisms of supermanifolds, we next need to pick a homomorphism of the sheaf of rings of ${ℝ}^{1\mid 1}$ with that of the pushed forward sheaf. Let’s just concentrate on some arbitrary global even section $s$ of ${ℝ}^{1\mid 1}$ and some global odd section $\eta$. Under the ring homomorphism they are mapped to, say $t$ and $\theta$

(3)$\left(s,\eta \right)↦\left(t,\theta \right)\phantom{\rule{thinmathspace}{0ex}}.$

This are the “variables” $t$ and $\mathrm{theta}$ in the above mentioned formula. We may do the same for some other “S”-point. This will give some other ring homomorphism

(4)$\left(s,\eta \right)↦\left(t\prime ,\theta \prime \right)\phantom{\rule{thinmathspace}{0ex}}.$

Adding the two $S$-points is what gives rise to the homomorphism described by the right hand side of the above formula. Note that in particular the term $\theta \prime \theta$ is in general nonvanishing because this is in general the product of two different odd sections of $S$. They both arise as images of the single odd section $\eta$ of ${ℝ}^{1\mid 1}$. And of cource $\eta \eta =0$.

The tangent space of a supermanifold $\left(M,O\right)$ is simply the supermanifold which is realized as the sheaf of derivations of the rings in $O$. One such derivation is the familar superderivative

(5)$D=\theta \frac{\partial }{\partial s}+\frac{\partial }{\partial \eta }\phantom{\rule{thinmathspace}{0ex}}.$

By suitably restricting and pulling this back we obtain a derivation

(6)$D=\theta \frac{\partial }{\partial t}+\frac{\partial }{\partial \theta }$

on $S$. Superexponentiating this yields the flow

(7)$\mathrm{exp}\left(-t\prime {D}^{2}\right)\left(1+\theta \prime D\right)$

which generates the above supertranslation.

The point here is that one has to carefully do all one’s supercaclulations properly using the language of ringed spaces. Doing so should tell you what precisely 0-transport of super-(-1)-branes with superpoint worldvolumes gives rise to.

Posted at March 2, 2006 4:53 PM UTC

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### Re: Superpoints

In particular, the supermanifolds of the form R^p|q are nothing but ringed spaces (M,O) where M=R^p is simply p-dimensional Euclidean space and where O is the sheaf of smooth functions of R^q times elements of exterior powers of R^q.

Hm. Is the ordinary manifold R^(p+q) the ringed space with M=R^p and O the smooth functions of R^q times elements of symmetric powers of R^q?

### Re: Superpoints

Is the ordinary manifold ${ℝ}^{\left(p+q\right)}$ the ringed space with $M={ℝ}^{p}$ and $O$ the smooth functions of ${ℝ}^{q}$ times elements of symmetric powers of ${ℝ}^{q}$?

Unless I misunderstand what you mean this is not so.

The algebra of (smooth, say) functions on ${ℝ}^{n}$ is ${C}^{\infty }\left({ℝ}^{n}\right)$. Same if $n=p+q$, obviously. And ${C}^{\infty }\left({ℝ}^{p+q}\right)$ is not isomorphic to ${C}^{\infty }\left({ℝ}^{p}\right)\otimes {ℝ}^{{\otimes }_{S}q}$ in general. But I wonder if that’s what you had in mind?

Posted by: urs on March 4, 2006 2:24 PM | Permalink | Reply to this

### Re: Superpoints

I was thinking about polynomials (or formal power series) rather than smooth functions. Since the difference between even and odd coordinates is that the former commute and the latter anti-commute, you should get ordinary manifolds from supermanifolds by replacing exterior products by symmetric products.

### Re: Superpoints

The point I was trying to make is trivial. Let p = q = 1. A formal power series f(x,y) can be written as a sum over terms of the form f_n(x) y^n. If y is odd this sum consists of two terms only, since y^2 = 0, whereas infinitely many terms may be needed if y is even. If you now think of y = dx, this gives you exterior powers in the odd case and symmetric powers in the even case.

### Re: Superpoints

A formal power series $f\left(x,y\right)$ can be written as a sum over terms of the form ${f}_{n}\left(x\right){y}^{n}$.

Now I see what you mean. Yes, this works under the right conditions. But in order to really get a manifold you will need to take some completion or something in order to obtain all smooth functions.

Posted by: urs on March 7, 2006 12:14 PM | Permalink | Reply to this

### Re: Superpoints

But in order to really get a manifold you will need to take some completion or something in order to obtain all smooth functions.

Of course. But if we only are interested in algebraic aspects, it usually does not matter so much what class of functions we consider. For analytic aspects it does matter, of course.

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