The ordered configuration space

records the positions of $n$ distinguishable particles on a graph $G$. Although a graph has free abelian integral homology, the requirement that particles remain distinct introduces relations not present in the underlying one-dimensional space. The question is whether these relations can create torsion: can a nonbounding integral cycle have a nonzero integer multiple that bounds? Throughout, a finite graph is identified with its topological realization; parallel edges are allowed.

The distinction between ordered and unordered configurations is essential. Write $U_n(G)=F_n(G)/S_n$. For a finite connected graph and $n\ge 2$, Ko and Park describe $H_1(U_n(G),\mathbb{Z})$ in terms of graph invariants: its torsion consists of copies of $\mathbb{Z}/2$ indexed by nonplanar triconnected components [5, Theorem 3.16]. By contrast, their calculation for two ordered particles gives torsion-free integral homology in every degree [5, Theorem 3.25 and the paragraph following it]. Thus torsion in the unordered quotient does not by itself answer the ordered question.

Chettih and Lütgehetmann proved that $H_q(F_n(G),\mathbb{Z})$ is torsion-free for every $n$ and $q$ when $G$ is a tree with loops, meaning an iterated wedge of stars and circles. They also construct geometric generators for these groups, and conjecture torsion-freeness for all finite graphs [2, Theorem A and Conjecture 3]. An earlier assertion of the general result was withdrawn because the chosen bases did not give the required splitting of a Mayer–Vietoris spectral sequence [2, p. 2446]. This leaves an integral issue: free abelian groups on the pages of a spectral sequence do not preclude torsion in quotients by images of differentials.

In their 2025 preprint, Hainaut, Knudsen, and Wawrykow report that no torsion example was known for ordered graph configuration spaces and restate torsion-freeness as a folklore conjecture [3, Section 1.1 and Conjecture 1.7]. Their Theorem 1.2 gives leading Betti-number asymptotics independent of the characteristic of the coefficient field, under its graph and degree hypotheses. The resulting asymptotic restriction on possible torsion [3, Corollary 1.8] does not exclude its occurrence at a fixed particle number. Computations of rational homology representations for several graph families [8] likewise leave integral torsion undetermined.

In their September 2026 preprint, Mamun, Nalikka, and Ramos [6, Section 4.3.1] again describe torsion-freeness for ordered configurations as an open conjecture and single out the existence of odd torsion as a central question in graph-configuration homology. Their results on anchored configuration spaces of cographs, where collisions are permitted, bound torsion exponents for fixed homological degree, particle number, and anchor count [6, Remark 4.14 and Corollary 4.18], but do not exhibit torsion in the ordinary ordered space.

We disprove the ordered torsion-freeness conjecture and answer the odd-torsion existence question affirmatively in the ordered setting. Both conclusions follow from a single degree-two class of exact order six for four labelled particles. Define $G$ by starting with $K_4$ on vertices $x_1,x_2,y_1,y_2$, adding vertices $a,b$ and edges $a{x}_1,a{x}_2,b{y}_1,b{y}_2$, and joining $a$ to $b$ by three distinct parallel edges. This graph has six vertices and thirteen edges; see Figure 1.

Subdividing two of the parallel edges once gives a homeomorphic simple graph: replace the three $ab$ edges by $ab,au,ub,av,vb$, where $u,v$ are new vertices. Thus the theorem also holds for a simple graph with eight vertices and fifteen edges. The graph $G$ contains $K_{3,3}$ with bipartition $\{a,y_1,y_2\}$ and $\{b,x_1,x_2\}$, using one $ab$ edge, and hence lies outside the tree-with-loops class.

The proof is a finite, exact calculation in the cubical model of Abrams. Rather than compute a complete integral homology decomposition, we construct three objects: a cycle $z$, a chain bounding $6z$, and a modulo-six cocycle evaluating to one on $z$. The bounding chain proves that the order divides six; the cocycle proves the reverse divisibility. We use the established cubical model and elementary unit cancellation to obtain this explicit certificate. Encoding chains over $\mathbb{Z}[S_4]$ reduces the size of the calculation without taking the quotient by particle permutations. Section 2 describes the certificate and its topological interpretation. Appendix A explains the construction and gives the complete program, including a verifier that works in the unreduced complex.

Number the core vertices by

The following ordered list fixes the orientations and indices of the thirteen core edges, starting with index zero:

For the core edge of index $e$ with endpoints $(u,v)$, use the oriented path

and number its three edges $3e,3e+1,3e+2$. The resulting graph $\mathrm{\Gamma }$ has 32 vertices and 39 edges.

Let $D_4(\mathrm{\Gamma })$ be the cubical subcomplex of ${\mathrm{\Gamma }}^4$ consisting of products of four graph cells with pairwise disjoint closures. The ordered stable-equivalence theorem of Prue and Scrimshaw [7, Theorem 3.2], refining Abrams’ discretization theorem [1], applies to a connected graph with at least four vertices when every path between distinct vertices of degree other than two has at least three edges and every homotopically nontrivial closed path has at least five edges. In $\mathrm{\Gamma }$ the shortest such vertex-to-vertex path has three edges, and the girth is six. Therefore the inclusion $D_4(\mathrm{\Gamma })\hookrightarrow F_4(\mathrm{\Gamma })$ is a homotopy equivalence. Subdivision preserves the underlying topological graph, so

Write $C_q=C_q(D_4(\mathrm{\Gamma }),\mathbb{Z})$ and ${\partial }_q:C_q\to C_{q-1}$ for the cellular boundary map.

We orient each cube by increasing indices of its moving graph edges, rather than by increasing particle labels. For a $q$-dimensional cube $c$, its boundary is

where $c_{j,+}$ and $c_{j,-}$ replace the $j$th moving edge by its terminal and initial vertices, respectively. Labels are retained on every face. This choice of orientations makes relabelling act without an additional orientation sign.

For the finite encoding, tokens $0,\mathrm{\dots },31$ denote vertices of $\mathrm{\Gamma }$, and token $32+e$ denotes open edge $e$. A canonical cube is an increasing four-tuple of tokens with pairwise disjoint closures. For a permutation $p=(p(0),\mathrm{\dots },p(3))$ of $\{0,1,2,3\}$, write $[p;c]$ for the labelled cube assigning token $c_i$ to particle $p(i)+1$. To form a face, replace an edge token by an endpoint and sort the resulting tokens. If the sorted order lists old coordinate indices as $q(0),\mathrm{\dots },q(3)$, the new particle assignment is $p\circ q$. The surviving moving edges retain their relative order, so the sign is exactly that in (3).

The construction in Appendix A produces an integral 2-chain $z$, an integral 3-chain $b$, and a homomorphism $\varphi :C_2\to \mathbb{Z}/6$. Its input is only the graph in (1); lexicographic cell and permutation orders and deterministic pivot rules specify every coefficient. The resulting cycle has the form

where the program constructs $\mathcal{L}$, consisting of 409 canonical 2-cubes with coefficients ${ϵ}_c\in \{1,-1\}$, and verifies this alternating form. Thus $z$ has 9,816 nonzero labelled coefficients, all $\pm 1$. The supports of $b$ and $\varphi$ have sizes 236,992 and 214,457, respectively. Coefficients absent from a sparse representation are zero.

The identities verified by the program are

The first two are integral identities; the last two are in $\mathbb{Z}/6$. The function check constructs the boundaries of $z$ and $b$ directly from (3). To verify the cocycle condition, it enumerates each set of three mutually disjoint graph edges, each stationary vertex outside their endpoints, and all 24 assignments of particle labels. These are precisely all the three-dimensional cubes of $D_4(\mathrm{\Gamma })$: each such cube has three moving particles and one stationary particle. There are

labelled cubes to check. In particular, the cocycle test is not restricted to cubes in the support of $b$. Every equality is tested using exact integer arithmetic or reduction modulo six.

The certificate identifies a failure of saturation:

In other words, the boundary subgroup contains a nonzero multiple of a cycle but not the cycle itself. The modulo-six cocycle detects precisely the obstruction that disappears after passing to rational coefficients. This is an integral phenomenon in the ordered space, not torsion introduced by an unordered quotient.