# Realising Seifert-van Kampen in 2-complexes

Suppose you have a group $$G$$ given by a finite presentation $$\langle X; \mathbf{r}\rangle$$, and also suppose you know that this group decomposes as a free product with amalgamation $$A\ast_CB$$ (with $$A$$, $$B$$ and $$C$$ all infinite, and I am happy to assume $$C\cong\mathbb{Z}$$ for simplicity).

Is there any way we can link the presentation complex $$\mathcal{C_P}$$ of $$G$$ (single vertex, edges labelled from $$X$$, 2-cells attached using $$\mathbf{r}$$) with the Seifert-van Kampen-esque 2-complex $$\mathcal{C}_{SvK}$$ underlying $$A\ast_CB$$?

My initial thought was to look at the cover associated to the amalgamating subgroup $$C$$, and ask if I get a complex with a subcomplex which satisfies Seifert-van Kampen. However, this doesn't seem to work even if I take the cover of the presentation complex of $$A\ast_CB$$.

• Never heard of the Seifert-van Kampen-eqs 2-complex associated to an amalgamated product. Could you define it? Feb 23 '16 at 15:46
• I really meant "some" not "the", and I don't know if such a thing always/ever exists. But I meant a 2-complex with subcomplexes $A^{\prime}$ and $B^{\prime}$ associated to $A$ and $B$ which meet over some subcomplex associated to $C$. Feb 23 '16 at 16:08
• Regarding $\mathcal{C}_{SvK}$, might you be referring to a "graph of spaces" in the sense of Scott-Wall? 392c.wordpress.com/2009/03/08/… Roughly speaking, you take (say) presentation complexes $C_A$ and $C_B$ for chosen presentations of $A$ and $B$, and then attach an annulus with one boundary circle mapped to $A$ and the other mapped to $B$. Feb 24 '16 at 13:39
• @LeeMosher After Daniel's comment I got thinking, and I my ideas were roughly what you are suggesting. So, yes. Thanks for the link+name. Feb 25 '16 at 10:40

Hatcher in Algebraic Topology discusses "graphs of groups" at the end of section 1.B. One example is the realization of an Eilenberg-MacLane space $$K(A*_CB,1)$$ from a $$K(A,1)$$, a $$K(B,1)$$, and a $$K(C,1)$$. The construction uses the fact that any homomorphism $$\pi_1(X)\to A$$ for a CW complex $$X$$ is induced by a map $$X\to K(A,1)$$, unique up to homotopy. The two homomorphisms $$C\to A$$ and $$C\to B$$ correspond to a pair of maps $$K(C,1)\to K(A,1)$$ and $$K(C,1)\to K(B,1)$$ that can be used in a mapping-cylinder-like construction. In particular, you take a mapping cylinder for each and then join them along the $$K(C,1)$$.

The two-skeleton of a $$K(A*_CB,1)$$ can be thought of as a presentation complex for this amalgamated product, what you were calling $$\mathcal{C}_{SvK}$$. The cell structure of the Eilenberg-MacLane space can be arranged to have only a single $$0$$-cell, if you so choose.

An isomorphism $$G\to A*_CB$$ gives rise to a homotopy equivalence $$K(G,1)\to K(A*_CB,1)$$, which we can take to be cellular (the $$n$$-skeleton maps into the $$n$$-skeleton). If you restrict to the two-skeleta, you get a map $$f:\mathcal{C}_P\to\mathcal{C}_{SvK}$$ (in your terminology) that induces an isomorphism on $$\pi_1$$. It's worth mentioning that a $$K(G,1)$$ can be built by extending a presentation complex for $$G$$ using only $$(n\geq 3)$$-cells.

What follows is the beginnings of locating $$C$$ in $$\mathcal{C}_P$$.

Let's consider a complex for $$\mathcal{C}_{SvK}$$ from the mapping cylinder construction. Call the subcomplexes from each group $$\mathcal{C}_A$$, $$\mathcal{C}_B$$, and $$\mathcal{C}_C$$, where there are some remaining $$1$$-cells and $$2$$-cells in the complement from attaching the $$\mathcal{C}_C$$ to either complex.

Now, let's suppose that each $$2$$-complex's $$2$$-cells' attaching maps are "taut," in that there is no backtracking or pausing and it is basically piecewise linear (the boundary of the $$D^2$$ is subdivided into intervals for how it is glued into the $$1$$-skeleton, just like the usual identification diagrams). We can make the image of $$f$$ transverse to $$\mathcal{C}_C$$ in a sense. By a homotopy of $$f$$, we can make the $$0$$-skeleton avoid $$\mathcal{C}_C$$, and instead land in the $$0$$-skeleta of $$\mathcal{C}_A$$ and $$\mathcal{C}_B$$. Now consider the image of a $$1$$-cell of $$\mathcal{C}_P$$ through $$f$$, which we may assume is taut after a homotopy of $$f$$. Since each $$1$$-cell has an incident $$2$$-cell from the mapping cylinder construction, we can homotope $$f$$ so that the $$1$$-cell crosses through $$\mathcal{C}_C$$ at a $$0$$-cell, without the path spending any time in $$\mathcal{C}_C$$. Lastly, we can assume the images of $$2$$-cells are taut, and then if a $$2$$-cell spends any time on a $$2$$-cell of $$\mathcal{C}_C$$, the boundary of that $$2$$-cell has an image on $$\mathcal{C}_A$$ and $$\mathcal{C}_B$$, so the map can be modified without changing the fact that $$f_*$$ is an isomorphism on $$\pi_1$$ (this amounts to homotoping the $$\mathcal{C}_P\to K(A*_CB,1)$$ map, which still had $$3$$-cells). By another homotopy, we can make it so the $$2$$-cells contain entire $$1$$-cells of $$\mathcal{C}_C$$ if they intersect the interior of the $$1$$-cell at all.

Then, $$f^{-1}(\mathcal{C}_C)$$ consists of points of the $$1$$-skeleton (finitely many per $$1$$-cell) along with loops and arcs on the $$2$$-cells, with the arcs connecting these points. There is a modification of $$f$$ that can remove these loops: since they bound disks, the image of the disk can be made to avoid the $$\mathcal{C}_C$$, essentially by a homotopy of the original $$\mathcal{C}_P\to K(A*_CB,1)$$. If one could modify $$f$$ to make the preimage connected, then this would decompose $$\mathcal{C}_C$$ into two pieces, with the $$\pi_1$$-image of the common intersection lying in the image of $$C$$ in $$A*_CB$$.

Example. Let $$\Sigma$$ be a genus-$$g$$ closed surface, which is already a presentation complex of $$\pi_1(\Sigma)$$. If $$\pi_1(\Sigma)$$ were represented as an amalgamated product, the preceding would give that there is a map $$f:\Sigma\to \mathcal{C}_{SvK}$$ with $$f^{-1}(\mathcal{C}_C)$$ being a collection of disjoint curves, none of which bound disks. There are two kinds of curves: non-separating and separating, where the latter must be of the type that represent a connect-sum decomposition. By some theory of non-separating curves on surfaces, since any closed loop must intersect transversely the separating curves an even number of times (since $$\mathcal{C}_{SvK}-\mathcal{C}_C$$ is two pieces), they must come in homotopic pairs, and so by a modification of $$f$$ we can eliminate the pairs one at a time while still giving an isomorphism on $$\pi_1$$. So, using the fact that there are at most $$g-1$$ pairwise non-homotopic separating curves that are not nullhomotopic: the $$g=1$$ case implies that $$\mathbb{Z}^2$$ is not a non-trivial amalgamated product, and the $$g=2$$ case implies every non-trivial amalgamated product comes from a connect sum of two tori (i.e., $$\pi_1(\Sigma_2)=F_2*_{\mathbb{Z}}F_2$$ with $$1\mapsto aba^{-1}b^{-1}$$ is the only nontrivial splitting, up to isomorphism). I'm not sure what can be said about $$g>2$$, though! I couldn't figure out how to join separating curves together into one, if that is even possible.