I wrote:

Maybe last weekend I had an idea for how to proceed here, maybe even a good idea. I am waiting for that idea to crystallize a little more,

Oops, looks like I am disregarding my own rule of conduct here. After all, the best way to help an idea crystallize is to make a fool of yourself and start talking about it to others. ;-)

So here is the question: “What is a Dirac operator, really? What is supergeometry, really?”

(Here “really” is defined like this: you know you *really* know what it is if and only if it can be blindly categorified.)

I had vaguely thought about this question every now and then in the past. But recently I was alerted again to the urgency of this question by the following observation:

*The superpoint is essentially nothing but the infinitesimal interval $T$ of synthetic differential geometry.*

This is in the sense that, if $\nearrow$
denotes either the superpoint or the infinitesimal interval, then,
for any space $X$, the morphism space
$T X := \mathrm{Hom}(\nearrow,X)$
behaves like the tangent bundle to $X$.

For $\nearrow = T$ the synthetic infinitesimal interval, this is true essentially by construction. For $\nearrow = \mathbb{R}^{0|1}$ the $N=1$ superpoint, this is a well known fact, appearing for instance as Lemma 3.2 on p. 29 of the thesis

Florin Dumitrescu, Superconnections and parallel transport.

The only slight lie about this statement is that
$\mathrm{Hom}(\mathbb{R}^{0|1},X)$ is really the *odd* tangent bundle. That is an important issue, which however I will no further get into here.

What I find exciting about this is that it suggests a way to naturally approach supergeometry in categorical terms. That’s because I think I know a nice way to approach synthetic differential geometry in categorical terms.

Here is how I conceive the “infinitesimal interval” this way: I think of it as the terminal 1-category, that with a single morphism. Accordingly, I shall argue that this is the $N=1$-superpoint for me.

More generally, I would then regard the $(N=n)$-superpoint as the terminal $n$-category: that with a single $n$-morphism.

I’ll try to convince you in a moment that this actually might make good sense. But before continuing, just notice that for $(N=0)$ this just says that the non-super point is just the 1-element set.

Which it is.

(If Toby is reading this, we could indulge in using this to favour the world with a discussion of $(N=-1)$ and $(N=-2)$ supersymmetry .)

The idea was this:

think of any space $X$ in terms of its strict $\infty$-groupoid of paths
$P(X)
\,.$
$n$-morphism are thin-homotopy classes of maps $[0,1]^n \to X$ cobounding two $(n-1)$-morphisms.

Then
$P_n(X) :=
\mathrm{Hom}_{n\mathrm{Cat}}
( \nearrow_n , P(X) )
\,,$
with slight abuse of notation, where I now write $\nearrow_n$ for my $(N=n)$-superpoint, as before.

What looks like notational overkill here is turned into a cool fact by using this cool fact: strict smooth $p$-functors
$f : P_p(X) \to \Sigma^p(U(1))$
are in bijection with differential $p$-forms.

But this tells us that in the world of categories, functors, transformations, modifications, etc (i.e. in the best of all worlds ;-) we are entitled to identitfy
$T^{\vee n} X :=
P_n(X)$
the $n$th exterior power of the tangent bundle with the $n$-path $n$-groupoid of $X$.

In fact, we can just as well use the $n$-groupoid $(U(1))^n$ coming from the $n$-crossed module $U(1) \to U(1) \to \cdots \to U(1)$ and find the entire exterior bundle up to degree $n$
$\oplus_{i=0}^n \Omega^n(X)
\simeq
\mathrm{Hom}_{n\mathrm{Cat}}( P_n(X),(U(1))^n)
\,.$

(Only thing I am still puzzled about is how to naturally see the wedge product operation on these functors. Must be something obvious, but I don’t see it yet).

Time is running out. I need to be more brief.

So what’s the superline then? Do functors on superlines reproduce superconnections? In particular, do we get supersymmetric quantum mechanics from functors on super-1-cobordisms?

Here is a brief sketch:

Like our superpoint is really a point with lots of paths emanating from it, with the magic of smooth functors on these paths ensuring that only the tangent vector of these paths at that point matters, so a supercurve is a superpoint in curve space, i.e. a supercurve with a tangent vector at every point.

As you have noticed, superification here is actually categorification. The super 0-point is actually a 1-category.

Same for the supercurve now.

It’s a curve as usual
$\array{
\downarrow
\\
\downarrow
}$
but now we remember tangency classes of *surfaces* emanating from that (or rather: this is what our smooth 2-functors will see).

So, schematically, our supercuve looks like an *infinitesimally thin strip of surface*
$\array{
&\to&
\\
\downarrow &\Downarrow& \downarrow
\\
&\to&
\\
\downarrow &\Downarrow& \downarrow
\\
&\to&
\\
\downarrow &\Downarrow& \downarrow
\\
&\to&
\\
\downarrow &\Downarrow& \downarrow
}$
(rather, like a finite such surface, but our smooth 2-functors will only care about the tangents).

Suppose we think of this as the worldline of a superparticle. If it were an ordinary particle we’d want to look at smooth functors from the curve to $U(H)$, for $H$ some Hilbert space of states.

Now we need a 2-functor mimicking that. The obvious choice is to use instead the 2-group coming from the crossed module
$\mathrm{id} : U(H) \to U(H)
\,,$
i.e. the 2-group $\mathrm{INN}(U(H))$. Let’s assume that’s right.

Then what’s a smooth 2-functor applied to the above supercuve?

Now some magic happens, which is making me think that this is on the right track.

We know that such a smooth 2-transport comes from

- a 1-form $\tilde D$ on the strip (which represents our supercurve) with values in $u(H) = \mathrm{Lie}U(H)$

- a 2-form $\tilde H$ on the strip, with values in $u(H)$.

First thing to notice: it looks like these two differential forms take values in the same Lie algebra. But if we are sufficiently pedantic, then in fact they don’t! Rather, $\tilde D$ takes value in $u(H)$ regarded as being *odd* graded, while $\tilde H$ takes values in $u(H)$ regarded as being *even* graded.

The categorification of Lie algebras, and its equivalence to $L_\infty$-algebras naturally brings a grading into the game here. Without us having put this in by hand, the formalism knows that

$\tilde D$ is odd.

$\tilde H$ is even.

It’s god-given.

Second thing to notice:

the formalism tells us that these two differential forms are not independent.

Since we are thinking quantum mechanics here, and time-independent quantum mechanics for simplity, we assume this 1-form $\tilde D$ and 2-form $\tilde H$ are in fact constant along our curve. Then the compatibility condition (known sometimes as “fake flatness”) demands that
$\tilde H = \tilde D^2
\,.$

And that’s god-given. It’s not put in by hand.

Third thing to notice is this:

our 2-functor which we apply to the strip can be regarded as follows: it assigns a 1-form to each curve on the strip, and then assigns to the entire strip the path-ordered (in curve space!) exponential of that 1-form over the strip.

That 1-form turns out to be, in turn, the path-ordered exponential, along the curve, of
$H := \tilde H(t,\cdot)
\,,$
where $t$ is the horizontal tangent on our strip.

But since we are just looking at that one curve, and care about the strip emanating from it only in as far as this tangent vector is concerned, we finally find that
our 2-functor with values in $\mathrm{INN}(U(H))$ (here $H$ denoted a Hilbert space, recall) sends the supercurve $\gamma$ to
$e^{ |\gamma| H}$
where the Hamiltonian $H$ is an even operator that is required (that’s precisely what distinguishes our 2-functor on the strip with values in $\mathrm{INN}(U(H))$ from a mere 1-functor with values in $U(H)$!) to be that square of the odd operator $D$
$H = D^2
\,.$

And there we go: $U(H)$-transport along supercurves is in bijection with supersymmetric quantum mechanics.

Okay, so much for now.

## Re: Who’s on the Right Track?

I think there are really two issues here:

1) can we handle “nongeometric” spacetimes (i.e. spacetimes which are not manifolds, but something more general, like a noncommutative space)?

2) can we handle the quantum theory of all these nonperturbatively, i.e. without trying to expand the theory around any given point in the space of all these spacetimes?

It seems that when Connes and Marcolli write

they are referring to point 1), while when they say

they are referring to point 2).

There is a proposal in which for every 2-dimensional SCFT of central charge 15 we get a point in a space of possibly noncommutative/generalized geometries, and a prescription for how to make an expansion from there into the rest of the space of geometries.

But it is true that this proposal has been mainly studied at points where the given geometry is commutative.