The n-Category Café

I’m afraid I’m still too distracted to push this Klein 2-geometry project forwards very quickly. But let me give it a nudge.

David wrote:

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.

That sounds right to me. We’re going by the seat of our pants here, but I think it all works out quite simply.

In fact my main worry now is that it’s too simple to be very interesting! We’ll need to ask some tougher questions about these Grassmannian to see if they have hidden depths. For example, let’s study the “incidence relations” between the $(n,0)$ sub-2-spaces and $(m,0)$ sub-2-spaces of $k^{p,q}$. Then we’ll finally be doing categorified incidence geometry.

Of course we’re using enough jargon and symbols now that nobody except us can possibly understand what we’re talking about. So, my claim that it’s “too simple” may strike them as unconvincing.

But maybe that’s how math advances: when you’re completely stuck and a problem seems incomprehensible, lots of people can understand what you’re whining about. But when you finally crack the problem and it seems completely lucid, nobody can understand you anymore!

To reach understanding it seems one must suffer confusion, even when there’s somebody standing there telling you anything you want to know. The neurons must be reconfigured.

I’ve been running hard to try and catch up with this discussion. Gosh, there’s a lot of material! But some of it is a record of confusion and clarification rather than finished new stuff, and it might confuse me to spend too long on it. Therefore, in an effort to avoid absorbing the whole lot, I want to try and take a shortcut, by taking some big intuitive leaps to get quickly to the point where I really need to start to actually think.

I think the trick with this sort of learning is to start with what one already understands and expand from there. With this in mind, and with your permission, I want to take a brief diversion, and momentarily go back to the last thing I think I thoroughly understand, namely the automorphism 2-group of the groupoid $S_3//A_3$ (the pair-of-triangles). And I want to think about spaces and their figures in the way I’m most familiar with, namely as collections of points, and generalise from there. Since emphasising points as concrete fundamental entities is kind of contrary to the spirit of Klein, I’ll call this exercise “semi-Klein geometry”. Craving your lordships’ indulgence … hopefullly this won’t take too long … ulp! …. here goes …

So I see the pair-of-triangles as a space of two points, each of which is internally a triangle. Symmetries let us permute the points arbitrarily and map their internal structures isomorphically or (in the weak case) by any equivalence. 2-symmetries relate symmetries which permute the points the same way but differ by internal automorphisms or (in the weak case) internal auto-equivalences.

So generalising this as much as possible without actually thinking, we can add more points, we can cut down on the number of ways we’re allowed to permute them, we can give them more complicated internal structures (i.e. bigger/smaller/different internal automorphism/auto-equivalence groups/2-groups), and we can give different points within the space different internal structures from each other.

To express this more categorificationally:

With semi-Klein 1-geometry, we have a set of points and a group of symmetries on them. With semi-Klein 2-geometry, we get a category of points or (in the weak case) a 2-category of points, and a 2-group of symmetries and 2-symmetries on them. A symmetry of the space can map a given point to another given point only if the two points are isomorphic to each other or (in the weak case) equivalent; and the mapping must be an isomorphism or (in the weak case) an equivalence. I.e. we not only care which points go where but also how they get there (or at least what they look like when they arrive). We also have 2-symmetries which relate certain symmetries to each other, namely those symmetries that agree on which points get mapped to which, but differ by internal automorphisms/equivalences of the points.

I guess there are various uses we could put this concept of 2-symmetry to, but one way, perhaps, is to think of the 2-symmetries as gauge transformations. Then symmetries are only defined as fully distinct entities “up to gauge transformations”.

Does any of this make any sense so far? I’m kind of worried I don’t really understand how the strict/weak distinction affects things, or not as well as I should. Plus no doubt lots of other things.

But assuming we can still keep going down this path, we can think about cutting down the symmetries and 2-symmetries to a sub-2-group of the full 2-group of the space. In a spirit of wild reckless abandon, i.e. continuing to avoid actual thinking, I’ll try to make this as intuitively “injective” as possible by simply throwing away morphisms and 2-morphisms and hoping we can tidy it all up later.

So we can

a) Cut down on which points can go to which points, possibly foliating the space into subspaces of some kind (?)
b) Cut down on the internal symmetries of (some of) the points, thereby
   i) Reducing the number of ways to get from one point to another, and/or
   ii) Making the space of points of different kinds fall apart (or further apart) into more isomorphism/equivalence classes of different kinds of points, and of course
   iii) Adding extra structure to the points themselves

and also

c) by cutting down on the 2-symmetries, we can subdivide the internal symmetries into more kinds.
   This will also affect whether and how we’re allowed to map between points.

I’m pretty sure I’m being somewhat “evil” here, at least in spirit, but I’m still hoping for “redemption” later, in the tidying-up phase. So to press on…

Now, since I’m thinking of the whole space as made of points, I want to think of figures in a pointy way too.

Figures in this semi-Klein 1-geometry I’ve been talking about are not the same as subgroups of the space’s symmetry group, but are made of points, namely the points in the orbit of a given point, under the appropriate subgroup. So in semi-Klein 2-geometry, we want to have figures which are “2-orbits” of points under sub-2-groups. In this case, a figure will be not a set of points, but a category of points or (in the weak case) a 2-category of points. This is because we care not only which points are in the figure, but also how we are allowed to get from one to another (via symmetries in the sub-2-group) and also which internal symmetries are isomorphic under (the remaining) 2-symmetries in the sub-2-group.

Then an incidence relation between figures becomes, depending on how virtuous we want to be, either a functor, or a set of functors, or a category of “functors together with natural transformations (or natural isomorphisms?) between them”.

And this makes sense because, in the full Klein geometry approach, the figures are sub-2-groups, and incidence relations are (closely related to?) inclusion relations among sub-2-groups, which we already know are functors, sets of functors or categories of functors. We are now in a position to analyse what we mean by “2-orbits” given the various kinds of “injectivity” and “surjectivity” conditions we want to impose on the functors, and given the ways we want functors to be naturally isomorphic or not. However, this requires me to think, so I will stop here.

Does any of this make any sense at all? (Or is it too mind-numbingly obvious? That would also be bad.)

The 2-Euclidean spaces that David was thinking about right at the beginning, where we make the base space $\mathbb{R}^2$ and the internal symmetries O(2) (or, more generally, the base space $\mathbb{R}^n$ and the internal symmetries O(n)), seem quite restrictive when viewed through this lens. The reason is that the internal space of a point contains all the information about the global geometry that isn’t coded in the location of the point. That is, to specify a symmetry of the space, we simply need one point (any point) and a morphism from that point. Is this true? Maybe the 2-symmetries allow a bit more wiggle-room. I guess it’s not surprising that a space specified as the quotient of two low-dimensional groups should be quite restrictive, and then mechanical categorification in the manner of going to the weak quotient doesn’t loosen things up much. Actually, that’s really mere oidisation rather than categorification in its full power and glory, so maybe there’s more to be done with these spaces (or spaceoids!)

Hmm, well, quickly looking at the posts on 2-vector spaces in the light of what I’ve just said, I think one way I may have been “evil” is by talking about individual points in the space as though they were well-defined entities, when maybe they should only be defined up to isomorphism or something. I suppose wondering if this is so might indicate progress of some sort in my understanding.

OK. I think I’ve now caught up and am ready to begin thinking.

We return you to your usual service.
Apologies for the interruption.

Since no one’s reading this anyway, I might as well keep chuntering away.

Looking at chain maps from the edge graph E

        $p\rightarrow q$

to the ring of four edges

        $a\rightarrow b\rightarrow c\rightarrow d \rightarrow a$

a little thought about the nature of Grassmannians as homogeneous spaces makes it clear that the space of chain maps as far as vertex space is concerned is either $SO(4)$ or some quotient of $SO(4)$ by $Z_2$, hence the 6 dimensions.

Trying to find a still simpler example to get a better picture of what’s going on, we consider maps from E to the graph

        $a\array{\rightarrow\\\leftarrow} b$

which also has dimension (1, 1).

Clearly, as far as maps to the space of 0-chains are concerned, each vertex of E can be mapped to either point, which gives a 2-dimensional space of maps, and modding out by automorphisms of the space over that vertex, we get a copy of the projective line. So the space of projective maps is the product of two copies of the projective line over $k$. E.g. for $k=\mathbb{R}$, the space is topologically a torus.

The maps on 1-chains clearly also form a 2-dimensional space, projectively a projective line, and the boundary map sends a point $x\rightarrow(x-1, 1-x)$, I think.

Compare this now with the slightly different graph

        $a\array{\rightarrow\\\rightarrow}b$

we have drastically reduced the space of projective 0-chains to a single point, while the space of 1-chains remains the same.

OK, now we can go back up to more complicated graphs. If there are $p$ places we can map our edge graph E to, then the maps on the 0-chains will projectively form the space $kP^{p-1}\times kP^{p-1}$ while the maps on the 1-chains will projectively form the space $kP^{p-1}$. The boundary map will map it into each factor of the 0-chain map space in more or less the obvious way, but with one being in some way the mirror image of the other. Oh, wait … I need to distinguish how many target elements each source element can go to …

OK, more generally, suppose we have a source graph with some vertices and some edges. Suppose that the source graph can be mapped into the target graph in various ways such that a vertex $v$ can be mapped into $\vert v\vert$ different vertices of the target graph, and that an edge $e$ can be mapped into $\vert e\vert$ different edges of the target graph. Then the projective space of maps of 0-chains will be $\prod_{v}(kP^{\vert v\vert-1})$ and the projective space of maps of 1-chains will be $\prod_{e}(kP^{\vert e\vert-1})$. The boundary map will send the latter into the former factor by factor according to the structure of the source graph.

No doubt there are many interesting homotopic things to be said about these maps, all well-known but not to me!

The above assumes there are no non-trivial automorphisms of the source graphs, which would give rise to extra automorphisms of the chains and would need to be modded out. Since these appear to be what give rise to more general Grassmannians than mere projective spaces, they are obviously also worth looking at.

So categorified incidence geometry has finally come up! I tried to think about this with little success a few months ago. The problem was how to define the dimension function - in the form of usual incidence geometry (in dimension $n$) this is a function from the set of figures to the ordinal $\mathbf{n}$ (Better thought of as a functor from the lattice of figures to the lattice $\mathbf{n}$). I toyed with Heyting algebras and got in a mess. :-(

It wasn’t until I saw Batanin’s talk at AustMS06 that a better idea came: use the `categorified natural numbers’, plane trees of height 2 (Section 2.1 in math.CT/0301221). The dimension should have target one of these. Then of course we get dimension labelled by two integers. So here’s a question - can a plane tree of height 2 (call it a 2-ordinal, say) be thought of as a 2-category, as an ordinal can be thought of as a category? Or the real question: can it be thought of as a 2-category in such a way that works with our concepts of categorified projective geometry?

Let’s go back a step - incidence geometry comes with a set $S$ and a function $dim:S \to \mathbf{n}$ (and some conditions, but they are properties, I think, so leave them for later). So if we have a category and a functor to a 2-ordinal, what can we say about the set up? A 2-ordinal is essentially a pair of ordinals and an order preserving map between them.

$$T = \mathbf{k}_2 \to \mathbf{k}_1$$

Think of it as the category $\mathbf{2}$ with some extra info (actually an ordinal internal to the category of ordinals). Then considering the object and morphism components of the functor to $T$ we get a pair of numbers which are candidates for the dimension.

The major problem with this idea is that it matters how the branches of the tree are organised - not a problem with 1-ordinals.

Tim: I personally think incidence is nicely thought of as spans in the suboject lattice of the whole space. Keep up the good work - I too tried to jump on the moving car, but I think I’m hanging off the spoiler instead ;)

I’ll post this before I reply to various people about things they’ve said or asked.

I’ve made two very stupid mistakes stemming from one source, and I am still kicking myself vigorously over this, because, out of everything I have worked on so far, this one is perfectly simple and clear, and I should have understood it right from the beginning. I just didn’t bother to do a simple calculation properly or think through the consequences of what I was saying. I will do it properly here with painful explicitness so that even idiots like me can understand it.

We need the boundary map to commute with chain maps. This means we need to actually understand what it does, not sort of vaguely guess.

Consider two edges and their vertices:

D:      $a\rightarrow b$

E:      $p\rightarrow q$

We have a chain map from $E\rightarrow D$.

Let’s suppose that the edge function on $p\rightarrow q$ in E has a value $f$. Let’s suppose the effect of the edge map to D is to multiply this by $x$. Then the edge function on $a\rightarrow b$ in D gets the value $x f$.

Suppose the vertex map sends $p\rightarrow a$ and $q\rightarrow b$. Then the boundary map on D will assign the value $x f$ to $b$ and $-x f$ to $a$.

Doing it the other way round, the boundary map on E will assign the value $f$ to $q$ and $-f$ to $p$. If the chain map on vertices multiplies the value on $q$ by $y$, then the image of the value $f$ on $q$ gets sent to $y f$ on $b$. If the boundary map is to commute with the chain map, this is only possible if $y=x.$ Similarly for $p$ and $a$. So there is just one multiplying factor for the edge and both vertices.

Alternatively, suppose instead that the vertex map sends $p\rightarrow b$ and $q\rightarrow a$. The boundary map on D will still assign the value $x f$ to $b$ and $-x f$ to $a$.

But on the other branch of the commutative square, the boundary map on E will assign the value $f$ to $q$ and $-f$ to $p$. If the chain map on vertices multiplies the value on $q$ by $y$, then the image of the value $f$ on $q$ gets sent to $y f$ on $a$. This is only possible if $y=-x$. Similarly for $p$ and $a$. So the multiplying factor for the edge is minus that for both vertices.

What about edges sharing a vertex? Suppose we have two graphs

F:      $a\rightarrow b\rightarrow c$

G:      $p\rightarrow q\rightarrow r$

Considering the chain map $G\overset{X}{\rightarrow}F$, sending $p\rightarrow a$, $q\rightarrow b$ and $r\rightarrow c$. Assume the edge $p\rightarrow q$ has the value $f$ and gets multiplied by $x$ and the edge $q\rightarrow r$ has the value $g$ and gets multiplied by $y$. Finally, suppose the vertex $q$ gets multiplied by $z$. Looking at the value of $\partial\circ X$ on $b$, we get $xf-yg$. Looking at the value of $X\circ\partial$ on $b$, we get $z(f-g)$. Obviously, setting these things equal requires $x=y=z$.

Instead of F, consider H:

$a\rightarrow b\leftarrow c$

Consider the chain map $G\overset{Y}{\rightarrow}H$, sending $p\rightarrow a$, $q\rightarrow b$ and $r\rightarrow c$. Assume once again that the edge $p\rightarrow q$ has the value $f$ and gets multiplied by $x$ and the edge $q\rightarrow r$ has the value $g$ and gets multiplied by $y$, also that the vertex $q$ gets multiplied by $z$. Looking at the value of $\partial\circ Y$ on $b$, we get $xf+yg$. Looking at the value of $Y\circ\partial$ on $b$, we get $z(f-g)$. Obviously, setting all these three things equal requires $x=-y=z$.

Conclusion: in a single component of the source graph, all vertices get multiplied by the same factor $z$, and each edge gets multiplied either by $z$, if its orientation matches that of the target, or $-z$, if its orientation is reversed. I can’t believe I got this so wrong before. Grr! <Kicks self again. Take that, dolt!> It’s not like I’ve never calculated a boundary or coboundary before. Just apparently never with my brain switched on. <Sigh!> Ah well, this too will pass.

I don’t think it does any actual harm to restrict the chain maps to those which preserve (or indeed reverse) orientation on all arrows, but this under-represents the contents of the 2-vector space. (It’ll violate one of the kinds of surjectivity, I guess; maybe a little later I’ll work out which one.) After all, the 2-vector space is characterised by its dimension, and the homotopy underlying this sure doesn’t care about the direction of the arrows.

We also need to think carefully about multiple components in the source graph. I had a moment of panic here, but I think it’s OK.

Suppose we are trying to map

      $p\rightarrow q$      $r\rightarrow s$

to

      $a\rightarrow b\rightarrow c\rightarrow d$

sending $p\rightarrow a$, $q\rightarrow b$, $r\rightarrow c$ and $s\rightarrow d$.

Let’s say the map is called $X$, that the edge $p\rightarrow q$ has the value $f$ and the edge $r\rightarrow s$ has the value $g,$ and that the $p\rightarrow q$ component gets multiplied by $x$ while the $r\rightarrow s$ component gets multiplied by $y$. Let’s suppose the value on $b\rightarrow c$ in the image is $h$.

Then, looking at the value of $\partial\circ Z$ on $a$, $b$, $c$ and $d$, we get

$\partial\circ Z(a)=-xf$
$\partial\circ Z(a)=xf-h$
$\partial\circ Z(a)=-yg+h$
$\partial\circ Z(a)=yg$

On the other hand,

$Z\circ\partial (a)=-xf$
$Z\circ\partial (b)=xf$
$Z\circ\partial (c)=-yg$
$Z\circ\partial (d)=yg$

So the only requirement is that $h=0$, which we expected anyway. Phew!

We also need to remember that two different chain maps may look identical on vertices (e.g. if more than one edge connects a given pair of vertices), and we should only count these once when looking at 0-chains.

Also, two chain maps which differ only in the orientation of an edge reduce to the same map on edges only, apart from the multiplicative factor, which disappears when we go to projective space.

There are two conclusions:

1) We need to include maps from source to target irrespective of the matching of orientation.
2) For each component $w$ of the source graph, if there are $\vert w\vert_0$ ways of mapping it to the target which are distinct on vertices, then it contributes a factor of $kP^{\vert w\vert_0-1}$ to the space of 0-chain maps; if there are $\vert w\vert_1$ ways of mapping it to the target which are distinct on edges, then it contributes a factor of $kP^{\vert w\vert_1-1}$ to the space of 1-chain maps.

The boundary map also maps factor by factor as before, but I need to think about this a bit.

OK, the mistake, though stupid, does not seem to have been quite as disastrous as I feared. Sometimes when I clear up a mistake, all the mathematically interesting behaviour evaporates.

I guess chain homotopies now work as before again, but I need to think this through properly!

Once again, if the source graph has non-trivial automorphisms, these will contribute higher Grassmannians.

Before I go on, I want to do one example of a boundary map on the space of chain maps, hopefully not completely trivial.

Consider the maps into the following graph:

$a\rightarrow b\rightarrow c$

from the following graph:

$p\rightarrow q$.

There are four maps. In this case we can distinguish them by where they send the vertices:

$(p,q)\rightarrow(a,b)$
$(p,q)\rightarrow(b,a)$
$(p,q)\rightarrow(b,c)$
$(p,q)\rightarrow(c,b)$

Let’s call the multiplying factors of these respectively $x_{a b}$, $x_{b a}$, $x_{b c}$ and $x_{c b}$.

Although the restriction of the maps to vertices is injective, the restriction to edges isn’t, since the $a b$ and $b a$ maps go to the same edge, as do the $b c$ and $c b$ maps.

The 0-chain map, represented in the basis $(p,q)\rightarrow(a,b,c)$ gives

$(h,k)\rightarrow(x_{a b} h+ x_{b a} k,x_{a b} k + x_{b a} h + x_{b c} h + x_{c b} k, x_{b c} k + x_{c b} h).$

The 1-chain map, represented in the basis $(p\rightarrow q)\rightarrow(a\rightarrow b,b\rightarrow c)$ gives

$(f)\rightarrow((x_{a b}- x_{b a})f,( x_{b c}- x_{c b})f).$

The boundary then gives

$\partial(f)\rightarrow(-f,f)$, which then gets sent to

$(-x_{a b} f+ x_{b a} f, x_{a b} f- x_{b a} f- x_{b c} f+ x_{c b} f, x_{b c} f- x_{c b} f)$

$=((-x_{a b}+ x_{b a}) f, (x_{a b}- x_{b a}) f+(- x_{b c}+ x_{c b}) f, (x_{b c} - x_{c b}) f)$

Comparing this to the expression for the 1-chain map, we see that in the image, we get

$\partial:(u,v)\rightarrow(-u,u-v,v)$ as expected.

Of course, all this is done projectively, so really we are interested in the projective space of 0-chain maps $kP^3$ and the projective space of 1-chain maps $kP^1.$ (I’m ignoring the graph automorphism $(a b\rightarrow b a)$ for the moment.)

The boundary map is still simple to describe, since we mod out symmetries of the source and therefore are left with all the interesting stuff happening in the image. If we go over to projective space by dividing throughout by $v$, and setting $w=\frac{u}{v}$, we get

$\partial(w)\rightarrow(-w,w-1,1)$ as the map of the projective line of projective 1-chain maps into the projective 3-space of projective 0-chain maps.

I guess one projective line looks much like another, so this isn’t a particularly exciting result. When we get to look at homotopies, things might get slightly more interesting.

First, though: Grassmannians. I think I’ll put them in the next post.

I have asked the following elsewhere already, but maybe I should ask again:

Given an abelian category $C$, form the category $K(C)$ of complexes in $C$. Objects are complexes, morphisms are chain maps.

Here we are talking about 2-term complexes in $C = \mathrm{Vect}$.

However, we realize that 2-term complexes actually live in a 2-category, with 2-morphisms being chain homotopies between chain maps.

In general, $n$-term chain complexes form an $n$-category.

But people who write $K(C)$ usually regard it just as a 1-category #. They do however invent another trick to weaken things.

A chain map of chain complexes is said to be a quasi isomorphism if it becomes an isomorphism after one passes to the cohomology of the chain complex.

It’s a popular move to pass from $K(C)$ to the category $D(K(C))$ (the derived category of complexes #) by declaring all quasi isomorphisms to be true isomorphisms.

I wonder: How is this related to regarding $K(C)$ as an $\omega$-category and considering weakyl invertible morphisms?

There is probably a simple answer to this. If so, we should get a universe of interesting examples for applications of $n$-vector spaces and all things related by substituting all occurences of quasi isomorphism in the $K(C)$-literature by the appropriate notion of weak invertibility.

Before thinking about how 2-Grassmannians arise from graph maps, let’s think about how 1-Grassmannians arise from set maps.

In fact, before I do that, I just want to think of Grassmannians as spaces of subspaces in the most boring set-theoretical way.

We want to think about the space of $m$-dimensional subspaces of an $n$-dimensional space, $m\le n.$ We can do this by picking a basis of the $n$-dimensional space, ordering it and picking e.g. the first $m$ members of it to represent the $m$-dimensional subspace. Pick two such bases to represent two $m$-dimensional subspaces. Given the first basis, we can get to any other by a member of $GL(n)$. However, the whole thing works just as well if we stick an arbitrary metric on the space and use orthonormal bases, in which case we can get from one to another with a member of $O(n).$ Which I find a little easier to imagine for some reason, so I will use that if I can.

However, we don’t care about internal basis changes of the subspace, so we need to quotient by $O(m).$ And we also don’t care about basis changes that don’t affect the subspace, so we also need to quotient by $O(n-m).$ So the Grassmannian is the homogeneous space $\frac{O(n)}{O(m)\times O(n-m)}$. In fact, I guess we can probably cancel factors of $Z_2$ and get $\frac{SO(n)}{O(m)\times SO(n-m)}.$ For instance, with $m=1$, we have $\frac{SO(n)}{O(1)\times SO(n-1)}.$ Since $\frac{SO(n)}{ SO(n-1)}$ is geometrically $S^{n-1}$ and $O(1)$ is $Z_2$, we get $kP^{n-1}$ as expected.

OK, so far, so elementary, this is just me thinking aloud, sorry to be so slow and cumbersome.

Next, let’s consider maps from a set $M$ with $m$ elements into a set $N$ with $n$ elements, $m\le n$.

The space of functions to a field $k$ on $M$ is an m-dimensional vector space over $k$, isomorphic to $k^m$. Likewise, the space of functions on $N$ is isomorphic to $k^n$.

OK, consider the space of maps from $k^m\rightarrow k^n$.

Well, as far as set maps are concerned, we only care about which subset of $N$ we end up with. We can get from any $m$-element subset to any other by an automorphism (i.e. permuatation) of $N$, but we don’t care about the internal structure of the subset, so we mod out by permutations of that, and we don’t care about the internal structure of its complement, so we mod out by permuations of that.

So far, so good. Now as far as the set of maps $M\rightarrow N$ is concerned, we can get from any map to any other by applying an automorphism to both $M$ and an automorphism to $N$. The automorphisms on $M$ handle automorphisms of the image subset, but the automorphisms on $N-Im(M)$ need to be done separately.

OK, so I guess this carries over to the vector spaces of functions from $M, N\rightarrow k.$

So we need to try and translate this to 2-vector spaces and 2-Grassmannians. We conclude that we need to mod out not only by automorphisms of the source chain generated by automorphisms of the source graph, but also by ones generated by automorphisms of the complement of its image.

Or something like that. OK, that’s enough for now.

Before pressing on, I want to pause to take stock of what we’ve done so far.

As JB said, I’ve been dealing with a 2-functor from one 2-category to another. The target 2-category is the well-known one of 2-term chain complexes, chain maps between them, and chain homotopies of chain maps. On the other hand, I’ve been more than slightly vague about the source 2-category, so I should now be a bit more precise.

The objects of this category are directed graphs. Curiously, although the directions on the graph edges are used in forming the 2-functor to the 2-category of 2-term chain complexes, the directions don’t actually do anything very much in the graph category, since two graphs which differ only in the directions of their arrows are always isomorphic.

The morphisms are maps between graphs which map vertices to vertices, edges to edges, and agree about (i.e. commute with) the assignment of vertices as the boundaries of edges.

There are various possible choices one can make about which maps are legitimately morphisms of this category. I’ve been working with the choice that the maps are injective on both vertices and edges.

One could also allow them to merge vertices under certain circumstances. I think this will work OK if there is always an edge from each vertex to itself, and we require any two points joined by an edge in the source to still be joined by an edge in the target. We could also allow edges to merge if they have the same boundary vertices (either only in the target or in the source as well)

We could also allow an in-between set of morphisms which can merge vertices only if they are “isomorphic” in the sense of being linked to exactly the same other vertices (and, optionally, in exactly the same way, i.e. via isomorphic sets of edges).

But I’ll stick with the purely injective maps.

A directed graph generates a category in the obvious way, with vertices as objects and catenations of edges as morphisms, and in this case the morphisms between graphs generate functors between categories.

In addition, I want 2-morphisms. There are various ways to do this, but I’ll pick what seems to me to be one of the simplest. We start by sticking a graph structure on the set of morphisms. Two morphisms $f$ and $g$ are connected by an edge if, $\forall x:f_v(x)~g_v(x).$ where $f_v$ and $g_v$ are the vertex maps of $f$ and $g$ and the relation $~$ indicates that two vertices are connected by an edge. In fact, we have an edge between $f$ and $g$ for every combination of ways of putting an edge between two of their target points, so we get a potentially huge cartesian product. (I don’t think there’s an obvious way of assigning a direction to these edges.)

Then this graph of morphisms generates a category of morphism whose morphisms are the 2-morphisms of the original category. We could stick extra conditions on this but I think this would make things a bit boring.

Just as the original graphs generate categories, and their morphisms generate functors between those categories, so these 2-morphisms generate natural transformations between the functors.

I’m more familiar with the lower-level case where, instead of a set (or a, um, 0-tangle? i.e. set of oriented points) of edges between two points, we have just a truth value of edges, i.e. a symmetric relation (possibly also reflexive, if we want to abandon the requirement of injectivity on vertices), so I’ll mention a couple of things I know about this.

In this case, the automorphisms of a graph form a group which is also a graph in a compatible way, namely that multiplication on the left and the right are graph automorphisms. (There ought to be a way to define this group in terms of a map from a product of a symmetric relation with itself, to itself. But when I tried this, my definition of product gave rise to a more boring sort of relation on the group, which had to be transitive as well as symmetric and reflexive. This arises if we use the stiffer condition on morphism relations that $f~g$ iff $\forall x \forall y: x~y \Rightarrow f(x)~g(y)$.)

This group has some properties reminiscent of topological groups, e.g. the identity component is a normal subgroup and is generated by a ‘neighbourhood of the identity’, i.e. any set containing everything linked to the identity. I guess that it gives rise to a (strict) 2-group. There’s probably some way of weakening this, e.g. by considering maps that are only automorphisms ‘up to the relation $~$’, but I don’t know what the exact right definition is, if there is one.

[Grr, I thought I’d posted this but it hasn’t appeared. Apologies if it now appears twice.]

Of course, our graph is not our space. The points (or “0-points”) of our space are functions from the graph vertices to $k$, while the “1-points” are pairs consisting of a function from the vertices to $k$ and a function from the edges to $k$, which acts as a morphism from the 0-point.

So, I think it’s time to look at the automorphisms groups and automorphism 2-groups of these things.

Let’s consider first automorphisms which do not arise from internal graph automorphisms. If there is only one component, then the automorphism group is simply the multiplicative group of $k,$ for both vertices and edges, and the homotopy group is trivial. If there are $n$ multiple isomorphic components, with only one isomorphism between any given pair, the automorphism group is $GL_k(n)$ for both vertices and edges, and again the homotopy group is trivial. If the set of components is partitioned into $m$ isomorphism classes, $F_i$, each of size $\vert F_i\vert$, $1\lt i\lt m$, then the automorphism group will be $\prod_i GL_k(\vert F_i\vert).$ The only 2-automorphisms are, again, trivial.

Now we need to think about internal automorphisms of a 1-component graph.

If the automorphisms induce $\vert A_v\vert$ distinct vertex automorphisms, we get a factor of $GL_k(\vert A_v\vert)$ in the vertex automorphisms. Likewise, if the automorphisms induce $\vert A_e\vert$ distinct edge automorphisms, we get a factor of $GL_k(\vert A_e\vert)$ in the edge automorphisms

In this case, the graph homotopies induce chain map homotopies. Maybe we can simply form the automorphism group graph of the graph, and consider chain map automorphisms over this …

Maybe some examples would help.

I seem to be walking around these 2-Grassmannians looking at them from every angle without actually constructing them. I’d like to think I’m circling my prey before moving in for the kill, but possibly I’m just wandering around in circles because I’m lost. There seems to be a lot going on in this example, with generalisations stretching as far as the eye can see in multiple directions. I feel a long, careful think is about due.

So where has our projective 2-geometry arrived at? We were going to form the projective 2-space associated to a vector 2-space, find the 2-group of projective linear transformations, then look for interesting sub 2-groups. We seem to have that the projective 2-space associated to a $(p,q)$ 2-space has $kP^{p-1}$ worth of objects each with $k^n$ automorphisms, for some $n$ that I’m not sure we quite worked out correctly, but which relates to the chain homotopies.

This gave rise to a worry that we had only a trivial modification of ordinary geometry. It would be nice to know if that’s right. How would the axioms of ordinary projective geometry need to be modified?

A) Given two distinct points, there exists a unique line that both points lie on.

B) Given two distinct lines, there exists a unique point that lies on both lines.

C) There exist four points, no three of which lie on the same line.

D) There exist four lines, no three of which have the same point lying on them.

Would the 2-group $PGL(V)$ be uninteresting? Is the trouble that there’s no twisting going on in the bundles? What if we’d looked at (1,1) sub 2-spaces instead?