Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $T$ be a finite trivalent fatgraph - i.e. a graph with a cyclic order of the edges at each vertex. Then there are certain basic "moves" we can perform on $T$: an embedded edge can be collapsed and then uncollapsed in a different way (a "rotation", or "2-2 move"), or the circular order of the three edges incident at a vertex can be reversed (a "flip").

Define a set $\mathcal{T}$ whose elements are trivalent fatgraphs $T'$ homotopic to $T$ with a labeling of the edges from $1$ to $n$ and a labeling of the vertices from $1$ to $m$ (note $m = 2n/3$). A "move" is a pair $(T,c)$ where $T \in \mathcal{T}$, and $c$ is an element of the set $e1, e2, \cdots, en, v1, v2, \cdots, vm$. The move acts on the labeled fatgraph $T$, and turns it into a new labeled fatgraph $T'$ obtained from $T$ by performing a rotation on edge $ei$ if $c=ei$ or a flip on vertex $vi$ if $c=vi$. It is clear how a flip affects the labels (it doesn't). A rotation destroys one edge labeled by $ei$ and creates a new edge, so label this new edge $ei$.

Now define a marked fatgraph to be a labeled fatgraph (i.e. an element of $\mathcal{T}$) together with a homotopy class of homotopy equivalence to some fixed $K(\pi_1(T),1)$. The moves defined above generate a new groupoid on marked fatgraphs, by acting on the labeled fatgraph part. This groupoid - the groupoid acting on marked fatgraphs - I will denote by $V(T)$ (the notation $V(T)$ is suggested by the similarity to Thompson's group $V$).

This groupoid - or something like it - turns up in many different contexts, so as a preliminary question, it would be nice to know how it is referred to. (Or: does this construction even make sense?)

More substantially: what is known about the algebraic structure of $V(T)$? What can be said about the cohomology of its classifying space? What is the relation to the group $Out(F)$, where $F$ is the (free) fundamental group of $T$? (note that $Out(F)$ acts on marked fatgraphs in a way that commutes with $V(T)$, by acting by homotopy equivalences of the $K(\pi_1(T),1)$ and thereby changing the marking). Is there a good reference?


Since the answers I am getting are not really what I am after, I think I need to make the question more pointed. A marked fatgraph determines a certain amount of algebraic structure on a free group (i.e. the fundamental group of $T$), namely a pair $(l,e)$ where $l$ is a length function, and $e$ is a bounded 2-cocycle. The first part of the data comes from the "thin" underlying graph, and is just the translation length of each element on its axis. The second part of the data comes from the fattening, and is an explicit cocycle representing the Euler class of the thickened surface. The first kind of move affects $l$, the second kind affects $e$. Crucially, both $l$ and $e$ are integer valued (this is the point of discussing discrete combinatorial objects, namely fatgraphs, instead of eg. discrete faithful representations of $F$ into $PSL(2,R))$.

Many, many papers discuss length functions, and many, many papers discuss Euler classes, but I would like to have a (presumably homological) algebraic framework which treats the two components as a single object with, presumably, more structure. The question is: what is this structure? Is it something that is already well-studied? Is there a reference?

share|improve this question
    
i did not understand the second move. But first what is a fatgraph? –  Gil Kalai Nov 12 '09 at 11:04
    
A fatgraph is a graph together with a cyclic ordering on the edges incident to each vertex - i.e. the fatgraph admits a "local" planar embedding. Another way to think of it is as a graph embedded in a surface (with boundary) in such a way that the surface deformation retracts to the graph (the surface is obtained by "fattening" the "fat" graph). So the second move changes the cyclic ordering at a vertex. This move can change the topology of the (fattened) surface. –  Danny Calegari Nov 12 '09 at 14:55
add comment

3 Answers 3

Let FG be the graph of marked fat graphs. Let G be the graph of graphs obtained by forgetting the cyclic orders at the vertices. The forgetting map from FG to G is a quasi-isometry. This is because fibres are finite and the diameter of a fibre is finite. The bound on diameter holds because Out(Fn) acts co-finitely on G. Finally, the graph G is quasi-isometric to Out(Fn) as it is a coarsely dense subset of a spine for outerspace. Composing the quasi-isometries shows that FG is gi to Out(Fn).

But your question is more algebraic in nature. Perhaps Matthew Horak's thesis paper

http://front.math.ucdavis.edu/math.GT/0310328

is what you are looking for. There fat-graphs (called ribbon graphs) are used to realize outer space as a union of Teichmuller spaces.

share|improve this answer
    
Maybe I should explicitly say that the (coarse) geometry of the situation is reasonably clear; what I am interested in is the algebraic structure of the groupoid $V(T)$. Thanks for the reference to Horak's paper. –  Danny Calegari Nov 12 '09 at 16:57
add comment

It's referred to as the arc complex of a surface. Or if you want a longer name, it's the 1-skeleton of the Hatcher-Thurston arc complex of a punctured surface. The "fatgraph" is also called a ribbon graph and it has an underlying punctured surface. To recognize it as the arc complex, pass to the geometric dual of the ribbon graph to obtain an ideal triangulation (or ideal polygonal tiling) of the surface. Recall Hatcher's simplicial complex: It has a vertex for every isotopy class of arcs, and a family of arcs subtend a simplex when they can be made disjoint. You not only obtain all of the moves in this model, you also obtain ribbon graphs with lengths assigned to the edges, using barycentric coordinates on all of the simplices of the arc complex.


Well, it would have been called that if not for the flip move reversing the cyclic ordering at a vertex. I didn't notice that when I wrote the above. With this flip move, I am not sure why the graph is fat in the first place, though. If you take thin graphs instead, then at least the corresponding Teichmuller space does have a good name, "Outer Space". Karen Vogtmann et al have written much about it.

share|improve this answer
    
Greg, I know you know that I know about Outer space, the Hatcher-Thurston complex, etc. I think it is my fault for not making it clear what I am really interested in - apologies. I have had a stab at rewriting the question. Maybe there is still no good answer; c'est la vie. –  Danny Calegari Dec 1 '09 at 17:32
    
I simply didn't think about what you'd already know. Of course I should have! But yes, your real question reads rather differently. Also if you want people (other than me) to consider what you already know, you might give the hint of posting with your name. –  Greg Kuperberg Dec 1 '09 at 18:07
    
Hmm, I'll think about it. Thanks. –  Danny Calegari Dec 1 '09 at 18:35
add comment

This is a nice question and I find it very hard to make such things understandable so I'm impressed.

The relationship to Outer Space of course looks very strong, but you're just looking at the simplicies of maximal dimension (trivalent graphs), so to understand the relationship my first step would be to try to lift your moves to the whole of outer space.

Your edge moves get factored into edge expansion and edge contraction, these are well understood and studied, see any reference on outer space. As for the vertices this means having to alter fat vertices of higher degree and looking at the trivalent case I think that means that you have to allow moves between any two cyclic orderings. A first glance seems to indicate that there is some kind of compatibility between edge and vertex moves: essentially it does look to me that you can lift your concept of moves to the whole of "fattened outer space".

Now we've done this we can study your object in the formalism of Kontsevich's graph complexes (well the associated (semi-)simplicial structures). Fattened outer space becomes the associative graph complex but with some extra simplices (your vertex moves). We can characterise these as certain simplices coming from the kernel of the map from the associative graph complex to the commutative graph complex.

As such I don't think that your groupoid will mean much beyond outer space, but it may encode some information from the associative complex, although I'm guessing only something along the lines of enumeration of the kernel.

I've run out of time and need to leave, hopefully what I've said makes some sense and the details are fill-in-able. I'll address any comments tomorrow.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.