Recurrence conjecture for Eulerian magnitude homology of path trees

Complete proof, with an integral decomposition, torsion-freeness, and explicit initial ranks.

Martin–Sazdanović, arXiv:2503.11976, Conjecture 8.2, predicts

$$\operatorname{rank}EMH_{k,\ell}(P_n) =(n-k+1)\operatorname{rank}EMH_{k,\ell}(P_k),$$

where $P_n$ has vertex set $\{0,\ldots,n\}$. We prove the stronger integral statement

$$\boxed{EMH_{k,\ell}(P_n;\mathbb Z) \cong \bigoplus_{a=0}^{n-k}EMH_{k,\ell}(P_k;\mathbb Z)}$$

for every $n\ge0$, $0\le k\le n$, and $\ell\ge0$, and prove all these groups are free abelian.

An Eulerian magnitude $j$-chain is an injective word $w=(x_0,\ldots,x_j)$ of length

$$L(w)=\sum_{i=1}^{j}|x_i-x_{i-1}|.$$

Its differential deletes an interior letter $x_i$ exactly when

$$|x_{i-1}-x_i|+|x_i-x_{i+1}|=|x_{i-1}-x_{i+1}|,$$

equivalently, when $x_i$ lies strictly between its neighbors.

Define the turning frame $F(w)$ to consist of the endpoints and all strict local extrema, in their original order. Every allowable deletion preserves both $F(w)$ and its interval hull

$$[\min F(w),\max F(w)].$$

Thus the entire Eulerian chain complex splits as a direct sum over turning frames.

Fix a frame

$$F=(f_0,\ldots,f_r),\qquad a=\min F,\qquad b=\max F,\qquad d=b-a.$$

For each nonframe vertex

$$v\in V_F=\{a,\ldots,b\}\setminus\{f_0,\ldots,f_r\},$$

define the eligible-run set

$$D_v(F)=\left\{i\in\{1,\ldots,r\}: \min(f_{i-1},f_i)<v<\max(f_{i-1},f_i)\right\}, \qquad m_v=|D_v(F)|.$$

Every $m_v\ge1$, since the successive frame segments cover the entire interval hull.

A word with frame $F$ independently either omits each $v\in V_F$, or inserts it into exactly one of its $m_v$ eligible runs. Once the choices are made, monotonicity uniquely orders the letters in each run. Hence the frame subcomplex is the shifted augmented simplicial chain complex of

$$K_F=\underset{v\in V_F}{\ast}\,D_v(F),$$

the join of discrete sets of cardinalities $m_v$.

For completeness, order inserted vertices first by run, then by the monotone order within each run. If $S$ is the inserted-vertex set, orient its simplex with the additional factor

$$\varepsilon(S)=(-1)^{\sum_{v\in S}i(v)}.$$

A vertex in run $i(v)$ occurring after $t$ earlier inserted vertices has magnitude-boundary sign $(-1)^{i(v)+t}$, while its simplicial-boundary sign is $(-1)^t$. Removing it changes $\varepsilon$ by $(-1)^{i(v)}$, so this orientation intertwines the differentials integrally.

Each discrete set $D_v$ has augmented complex

$$0\longrightarrow\mathbb Z^{m_v} \xrightarrow{(1,\ldots,1)}\mathbb Z \longrightarrow0.$$

The integral Künneth theorem for joins therefore gives

$$\widetilde H_s(K_F;\mathbb Z) \cong \begin{cases} \mathbb Z^{\prod_{v\in V_F}(m_v-1)},&s=|V_F|-1,\\ 0,&s\ne |V_F|-1. \end{cases}$$

Since $|V_F|=d-r$, the corresponding frame homology is

$$\boxed{ H_j(C_F;\mathbb Z)\cong \begin{cases} \mathbb Z^{\prod_{v\in V_F}(m_v-1)},&j=d,\\ 0,&j\ne d. \end{cases}}$$

If some $m_v=1$, the frame complex is acyclic. If $V_F=\varnothing$, the empty product is one with the usual augmented degree-$(-1)$ convention.

Thus a frame contributes to homological degree $k$ if and only if its interval hull has span $k$. There are exactly

$$[0,k],[1,k+1],\ldots,[n-k,n]$$

such hulls in $P_n$, and translation identifies each corresponding frame subcomplex with the full-hull subcomplex of $P_k$. Summing these $n-k+1$ identical contributions proves the claimed integral decomposition and the source’s rank recurrence simultaneously for every $(n,k,\ell)$.

The argument also gives an explicit initial-rank formula:

$$\boxed{ \operatorname{rank}EMH_{k,\ell}(P_k) = \sum_{\substack{ F\text{ injective turning frame on }\{0,\ldots,k\}\\ \min F=0,\ \max F=k,\ L(F)=\ell}} \ \prod_{v\in\{0,\ldots,k\}\setminus F} \bigl(m_v(F)-1\bigr).}$$

For $k=0$, the unique singleton frame contributes $\mathbb Z$ at length zero, and $EMH_{0,0}(P_n)=\mathbb Z^{n+1}$, exactly as required.