# Is the configuration space of ordered triples of distinct points in the four-edge banana graph homotopy equivalent to a surface of genus 13?

If $X$ is a topological space, write $C_n(X)$ for the configuration space of distinct ordered tuples of points in $X$: $$C_n(X) = \{(x_1, \ldots, x_n) \in X^n \mbox{ so that i \neq j \implies x_i \neq x_j } \}.$$ Define the banana graph $\beta_k$ to be the suspension of the discrete space $\{1, \ldots, k\}$. Finally, write $\Sigma_g$ for the closed orientable surface of genus $g$.

A) If $X = \beta_4$ is the four-edge banana graph, is there a homotopy equivalence $C_3(X) \simeq \Sigma_{13}$?

As background, the thesis of Aaron Abrams (available on his website http://home.wlu.edu/~abramsa/publications/index.html) gives homotopy equivalences

\begin{align*} C_2(K_5) & \simeq \Sigma_6 \\ C_2(K_{3,3}) & \simeq \Sigma_4. \end{align*} Abrams also proves that a related combinatorial approximation of configuration space $C_n^{approx}(X)$ satisfies \begin{align*} C_3^{approx}(K_5) & \simeq \Sigma_{16} \\ C_4^{approx}(K_{3,3}) & \simeq \Sigma_{37}, \end{align*} where $K_5$ is a complete graph and $K_{3,3}$ is complete bipartite. (Thank you to user j.c. for pointing out the difference between $C_n^{approx}$ and $C_n$, which I had not understood.) So this sort of thing has happened before!

Also, using an explicit simplicial model of $C_3(\beta_4)$ that Sage tells me has 336 vertices and 840 facets, I am able to compute that $$H^*(C_3(\beta_4) \, ; \mathbb{Z}) \simeq H^*(\Sigma_{13} \, ; \mathbb{Z}),$$ and that the cup product in rational cohomology gives a non-degenerate pairing on $H^1$.

B) How might I check if a finite simplicial complex has the homotopy type of some $\Sigma_g$?

I say "might" because it's probably not computable in general. Finally, I'll ask what might be a tricky question:

C) For what graphs $G$ and $n \in \mathbb{N}$ does $C_n(G)$ have the homotopy type of a surface?

At Ryan Budney's suggestion, I have collapsed many free faces. (The algorithm I used comes from the paper https://arxiv.org/pdf/1303.6422.pdf by Benedetti and Lutz). The result is a complex with 120 vertices and 288 facets. It is a pseudomanifold! Sage computes a presentation for $\pi_1$ $$\langle a,b,c,d,e,f,g,h,i,j,k,l,m,n,o,p,q,r,s,t,u,v,w,x,y,z \rangle / \langle \omega \rangle$$ where $\omega$ is the impressive-looking word $$y^{-1}a^{-1}bf^{-1}en^{-1}ml^{-1}x^{-1}wo^{-1}vkg^{-1}hpu^{-1}tdh^{-1}ic^{-1}e^{-1}car^{-1}qlod^{-1}fq^{-1}i^{-1}s^{-1}b^{-1}xt^{-1}m^{-1}p^{-1}rsk^{-1}uznjv^{-1}w^{-1}yz^{-1}j^{-1}g$$ in which every variable appears exactly twice, once with an inverse. So $\pi_1$ is also correct!

• Does your simplicial complex have any free faces? If so, put it into a software package that does basic simple-homotopy equivalences / Whitehead moves and try some greedy simplifications. – Ryan Budney May 4 '17 at 19:54
• @RyanBudney Good idea! It would be helpful for other things as well to has a smaller model. Any suggestions about what package would be easy to use? – John Wiltshire-Gordon May 4 '17 at 19:56
• Off the top of my head I'm not certain. I imagine you'd want to have a package to manipulate arbitrary (mostly) 2-dimensional CW-complexes, which is pretty much equivalent to groupoid presentations. Your cell complexes, for these configuration spaces, do they have any 3-cells or have they had those simplified-away? – Ryan Budney May 4 '17 at 20:20
• If there's no 3-simplices, perhaps the least-energy thing you could do is collapse a maximal forest and feed the group presentation to some code that does small cancellation theory. The software Heegaard is very powerful in this regard. The group presentation code in Regina (mostly written by me) is quite capable but often Heegaard is more powerful. If you can find it, the software Magnus (available for some linux systems) is very versatile and powerful. There is probably other sortware out there, but that's where I would start. – Ryan Budney May 4 '17 at 21:48
• If you could show the higher homotopy groups to be zero, that would be sufficient. – W. Cadegan-Schlieper May 6 '17 at 2:41

Since $C_n(G)$ (for any finite $n$ and finite connected $G$ with at least one cycle or vertex of valence at least $3$) and $\Sigma_g$ (for $g\ge 1$) are both $K(\pi,1)$ spaces homeomorphic to CW complexes, showing they have isomorphic fundamental groups suffices to show that they are homotopy equivalent. The fact that configuration spaces of graphs are $K(\pi,1)$ is due to Ghrist.
Since you have calculated that $C_3(\beta_4)$ and $\Sigma_{13}$ have isomorphic fundamental groups, this answers your question A in the affirmative.
which appeared in July 2017. They use a combinatorial model due to Swiatkowski to show that $C_3(B_3)$ is a surface of genus 13 glued from 144 2-cells.