Discriminant magnitude homology conjecture for trees

Counterexample, with a complete classification of trees.

In Martin–Sazdanović, arXiv:2503.11976, Conjecture 8.1 asserts that $DMH_{k,\ell}(T)=0$ for $k\ne\ell$ and every tree $T$. In fact,

$$\boxed{DMH_{*,*}(T)\text{ is diagonal}\quad\Longleftrightarrow\quad |V(T)|\le3.}$$

The source’s own short exact sequence induces

$$\cdots\longrightarrow EMH_{k,\ell}(T)\longrightarrow MH_{k,\ell}(T) \longrightarrow DMH_{k,\ell}(T) \xrightarrow{\delta}EMH_{k-1,\ell}(T) \longrightarrow MH_{k-1,\ell}(T)\longrightarrow\cdots.$$

Ordinary magnitude homology of every tree is diagonal. Therefore, for every $k<\ell$, both ordinary groups surrounding the connecting map vanish, giving the canonical integral isomorphism

$$\boxed{DMH_{k,\ell}(T)\cong EMH_{k-1,\ell}(T).}$$

The source’s Lemma 7.6 incorrectly concludes that surjectivity of $EMH_{k,\ell}\to MH_{k,\ell}$ forces $DMH_{k,\ell}=0$; this overlooks the connecting morphism into $EMH_{k-1,\ell}$. Giusti–Menara, arXiv:2403.09248, had already warned that the long exact sequence need not split.

Now let $T$ have $N\ge4$ vertices, and let

$$L=\max_{\pi}\sum_{i=0}^{N-2}d_T(v_i,v_{i+1}),$$

where $\pi=(v_0,\ldots,v_{N-1})$ ranges over vertex orderings. In a uniformly random ordering, each unordered pair is consecutive with probability $2/N$. Since $T$ has $N-1$ edges, the expected number of consecutive nonedges equals

$$\frac2N\left[\binom N2-(N-1)\right] =\frac{(N-1)(N-2)}N>1.$$

Thus some ordering contains at least two consecutive nonedges, and consequently $L\ge N+1$.

An injective $(N-1)$-chain of length $L$ exists by definition. There is no injective $N$-chain, and there is no injective $(N-2)$-chain of length $L$, since appending its omitted vertex would contradict maximality. Hence both adjacent Eulerian differentials vanish and

$$EMH_{N-1,L}(T)=EMC_{N-1,L}(T)\ne0.$$

Because $N<L$, the connecting isomorphism yields

$$\boxed{DMH_{N,L}(T)\cong EMH_{N-1,L}(T)\ne0.}$$

Therefore every tree with at least four vertices contradicts the conjecture. Conversely, trees on at most two vertices are complete, and the unique three-vertex tree $S_2$ has only an Eulerian off-diagonal group whose shifted discriminant group lies on the diagonal. This proves the stated classification.

More explicitly, the source’s valid Theorem 7.11 for the star $S_n=K_{1,n}$, combined with the connecting isomorphism, gives two infinite families omitted by its Corollary 7.12:

$$DMH_{r+1,\,2r-1}(S_n) \cong\mathbb Z^{\,2n!/(n-r)!}, \qquad n\ge r\ge3,$$

and

$$DMH_{r+1,\,2r-2}(S_n) \cong\mathbb Z^{\,(r-2)n!/(n-r)!}, \qquad n\ge r\ge4.$$

The smallest counterexample is the four-vertex star:

$$\boxed{DMH_{4,5}(K_{1,3})=\mathbb Z^{12}.}$$

A direct computation from the defining discriminant chain complex at length five gives

$$\dim DMC_{3,5}=12,\qquad \dim DMC_{4,5}=72,\qquad \dim DMC_{5,5}=54,$$

with $\operatorname{rank}\partial_4=12$ and $\operatorname{rank}\partial_5=48$; hence

$$\operatorname{rank}DMH_{4,5}=72-12-48=12.$$

The connecting isomorphism identifies this group integrally with $EMH_{3,5}(K_{1,3})=\mathbb Z^{12}$. Thus the conjecture fails universally beyond three vertices, and the precise error is the omitted connecting homomorphism.