The n-Category Café

[see improved version below]

Let me try to summarize our work on this Klein 2-geometry business. I wouldn’t want to deny our new customers the delight of reading our May, June, July and August conversations on this subject. But, some of them may have jobs - so they may not have time to read all that stuff! And, starting a new group blog is a good moment to take stock of what we’ve done.

Klein 2-geometry is an attempt to categorify Klein’s Erlangen program by replacing groups with 2-groups, which are like groups but categories instead of sets.

The idea of Klein’s original Erlangen program is to study geometries that have some group of symmetries, $G$. If this group acts transitively on some kind of geometrical figure - e.g. points, lines, etc. - the space of figures of this kind is $G/H$, where $H \subseteq G$ is the stabilizer of such a figure: that is, the subgroup of symmetries that preserve it.

To categorify this we’d like to replace $G$ with a 2-group, and replace $H$ with a “sub-2-group”. We then need to define the suitable analogue of the quotient $G/H$, and see in what sense $G$ acts transitively on this.

How do we define a “sub-2-group”?

For groups, we can think of a subgroup of $G$ as an equivalence class of one-to-one homomorphisms

$$i: H \to G$$

So, let’s try the same for 2-groups! We know what a homomorphism of 2-groups is, but we need to know the right analogue of “one-to-one”. Since 2-groups are categories, and their homomorphisms are functors, one natural guess is to look at homomorphisms

$$i: H \to G$$

that are one-to-one both on objects and morphisms. Alas, this definition is evil: it is not respected by equivalence of 2-groups.

After pondering this for a while, we realized there are four basic concepts relating to “one-to-one-ness” and “ontoness” for homomorphisms of 2-groups. The trick is to think of a 2-group as a connected pointed space with only $\pi_1$ and $\pi_2$ nontrivial. If this sounds scary, just take your 2-group $G$ and make these definitions:

• $\pi_1(G)$ is the group of isomorphism class of objects of $G$.
• $\pi_2(G)$ is the group of automorphisms of the unit object $1 \in G$.

The result is the same. The point is, just as homotopy groups are preserved by homotopy equivalence, these concepts are preserved by equivalence of 2-groups!

Any homomorphism

$$f : H \to G$$

between 2-groups induces homomorphisms

$$\pi_1(f) : \pi_1(H) \to \pi_1(G)$$

and

$$\pi_1(f) : \pi_1(H) \to \pi_1(G)$$

This lets us write down a quartet of definitions that generalize the twin concepts of “one-to-one” and “onto” from groups to 2-groups:

• $f$ is essentially surjective iff $\pi_1(f)$ is onto.
• $f$ is essentially injective iff $\pi_1(f)$ is one-to-one.
• $f$ is full on automorphisms if $\pi_2(f)$ is onto.
• $f$ is faithful iff $\pi_2(f)$ is one-to-one.

Some secondary definitions are also in common use:

• $f$ is full iff $\pi_1(f)$ is one-to-one and $\pi_2(f)$ is onto (it’s full on automorphisms and essentially injective).
• $f$ is an equivalence iff both $\pi_1(f)$ and $\pi_2(f)$ are one-to-one and onto.

There seems to be some evidence that a sub-2-group should be an equivalence class of homomorphisms

$$i : H \to G$$

that are faithful ($\pi_2(i)$ one-to-one). Namely, if we let $G$ act on some category, and look at the appropriately defined stabilizer of an object, that’s the kind of sub-2-group the stabilizer is!

On the other hand, we also noticed some evidence saying we want $i$ to be faithful and essentially injective ($\pi_1(i)$ and $\pi_2(i)$ one-to-one). Namely, if we categorify the concept of vector space, we get a concept of “vector 2-space”, which is a special kind of 2-group, just as a vector space is a kind of group. And, if we look at “sub-2-spaces” of a vector 2-space, we see they’re only “smaller” if their inclusion homomorphism is both faithful and essentially injective!

The issue of vector 2-spaces and their sub-2-spaces is important when we categorify projective geometry - a famous special case of Klein’s approach to geometry. So, 2-groups show up in two very different ways in our work, and don’t seem to want the same concept of “sub-2-group”. This got us scared, but I now think it’s okay. We should expect some things become a bit more complex when we categorify them, and this may be one. The point is, we’re no longer living in a world where “one-to-one” and “onto” rule the roost! These two concepts have split into four, so life is bound to be a bit more interesting.

One thing I’d enjoy doing is marching ahead and working out the theory of projective 2-geometry. We need some concrete examples under our belt. Anyone interested?

David writes:

So for a nice concrete example, we might look at the projective 2-space of (1,0) sub-2-spaces of, say, (3,1). Does this gives us a 2d projective space of objects, each with a 1d space of automorphisms?

Great! We’re getting in gear now…

But, if you don’t mind me switching into reverse for a moment, let me just check for backwards-compatibility: always a good thing when you’re categorifying. What about the dimension (1,0) sub-2-spaces of $k^{(n,0)}$?

Nota Bene: Some new notation is cropping up here. We say a 2-vector space has dimension $(p,q)$ if its 0th Betti number is $p$ and its 1st Betti number is $q$. The simplest 2-vector space of this dimension looks as follows when viewed as a chain complex:

$$k^{p} \stackrel{0}{\leftarrow} k^{q}$$

where $k$ is our ground field, and we call this guy $k^{p,q}$. Every 2-vector space of this dimension is equivalent, though not isomorphic, to this one.

The vector 2-space $k^{n,0}$ is just a fancy new way of talking about the vector space $k^n$: since it only has identity morphisms, it’s a category that might as well be a set!

A subspace of dimension $(p,0)$ sitting inside $k^{n,0}$ is just the same as a $p$-dimensional subspace of $k^n$. Moreover these subspaces have no interesting automorphisms (I bet).

So, the 2-Grassmannian of $(p,0)$-dimensional subspaces of $k^{n,0}$ is just the usual Grassmannian of $p$-dimensional subspaces of $k^n$, decked out in fancy new clothes!

Good. Okay, now let’s switch back to first gear and move forwards slowly.

What’s a $(1,0)$-dimensional subspace of $k^{3,1}$? I’m going to jump the gun a bit and guess it’s any sub-chain-complex of $k^{3,1}$ (viewed as a chain complex) whose dimension is $(1,0)$.

This is just the same as a 1-dimensional subspace of $k^3$. In other words, it’s a point in the projective plane $kP^2$.

These guys are the objects of the projective 2-space David wants us to look at. We also need to figure out the morphisms. For those, I should probably not have jumped the gun quite so fast: I should have described a $(1,0)$-dimensional sub-2-space of $k^{3,1}$ in terms of a chain map

$$i: k^{1,0} \to k^{3,1},$$

namely an “inclusion”. This lets me study chain homotopies between such inclusions. The chain homotopies between $i$ and itself should (probably) be the automorphisms of the given point in our projective 2-space.

Okay - if this is what we should do, it’s easy to see what we get: chain homotopies are simple when the differentials in our 2-term chain complexes vanish! For one thing, a chain homotopy can only go from from a chain map to itself. For another thing, a chain homotopy is just any map from 0-chains of the source to 1-chains in the target.

So, a chain homotopy from $i$ to itself boils down to any map

$$h: k^1 \to k^1$$

from 0-chains in $k^{1,0}$ to 1-chains in $k^{3,1}$.

So, we get a line’s worth of automorphisms of any point in our projective 2-space.

Verdict: when we form the projective 2-space associated to $k^{3,1}$, its objects form the projective plane $kP^2$, and each object has a line’s worth of automorphisms. There are no morphisms between different objects: we’re working with a “skeletal” version of our projective 2-space.

So, David is right - with much less work than it took me!

Perhaps some clever person can guess the precise description of the projective space associated to $k^{p,q}$.

Unfortunately for the connoisseur of vector bundles, this particular one is a trivial bundle.

There’s been a principal bundle and gauge theory feel to our discussions all along. How will we feel if all this Kleinian 2-geometry turns out to be some known chapter of differential geometry?

A while ago we saw that categorified 2d Euclidean space was $E(2) \leftarrow 1//P \leftarrow 1$ in the notation we used ($P$ being the stabilizer of a point). Presumably real projective 2-space is $PGL(3,R) \leftarrow 1//Q \leftarrow 1$, $Q$ the stabilizer of a point. Presumably things get more interesting when we replace the 1 by something more exciting. There ought to be such a replacement which corresponds to the projective space associated with $R^{3,q}$.

I’m reassigning this comment to the main flow of the river rather than as part of a tributary.

John wrote:

I think we need the chain map route. But, that method actually isn’t much work.

Oh yes, I see. It looks like chain homotopies again correspond to any map between 0-chains of the source and 1-chains of the target. So the 2-space of ($n$, 0) sub-2-spaces of $k^{p,q}$ has the grassmanian $G_{n,p}$ as objects and linear maps from $k^n$ to $k^q$ as automorphisms.

While we’re at it, isn’t then the set of $(a,b)$ sub 2-spaces of $k^{c,d}$ just $G_{a,c} \times G_{b,d}$ of objects each with linear maps $k^a$ to $k^d$ worth of automorphisms?

Here’s something I’d like to check before proceeding. John wrote:

I should have described a $(1,0)$-dimensional sub-2-space of $k^{3,1}$ in terms of a chain map $$i: k^{1,0} \to k^{3,1},$$ namely an “inclusion”.

Does this take into account that a scalar multiple of an inclusion of $k^p$ into $k^q$ corresponds to the same subspace of $k^q$? We seem to be implicitly factoring out such equivalence of maps when we look at 2-Grassmanians, but have we then been careful enough with those chain maps going from 0-chains of the source to 1-chains of the target? Are those which differ by a scalar factor different?

John also wrote:

The space of morphisms in our projective 2-space form a vector bundle over the space of objects, which is a projective space. Indeed, any vector bundle gives a 2-space in this way! Unfortunately for the connoisseur of vector bundles, this particular one is a trivial bundle.

If we try to find the symmetry 2-group of this bundle, we have to respect the projective-ness of the space of objects. What kind of map must a map of fibres be?

In The heritage of S.Lie and F.Klein: Geometry via transformation groups, E. Ortacgil asks:

What happened to Erlanger Programm? The main reason for this question is that the total space $P$ of the principal bundle $P \rightarrow M$ does not have any group-like structure and therefore does not act on the base manifold $M$. Thus $P$ emerges in this framework as a new entity whose relation to the geometry of $M$ may not be immediate and we must deal with $P$ now seperately. Consequently, it seems that the most essential feature of Klein’s realization of geometry is given up by this approach. Some mathematicians already expressed their dissatisfaction of this state of affairs in literature with varying tones…