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

Question

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?

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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!

Answer 1

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.

This gives a criterion (probably not practical in most cases) to answer question C.

Since you have calculated that $C_3(\beta_4)$ and $\Sigma_{13}$ have isomorphic fundamental groups, this answers your question A in the affirmative.

Answer 2

It has come to my attention that this example has been computed independently (and more directly) by Safia Chettih and Daniel Lütgehetmann. The result is given as Proposition 4.3 in v2 of their preprint

https://arxiv.org/abs/1612.08290v2

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.