Sign conjecture for Steiner distance hyperdeterminants

Let $T$ be a tree on $n\geq1$ vertices, and let $M$ be its order-$k$ Steiner distance tensor, where $k\geq2$. We prove that whenever $\det(M)\ne0$,

$$\operatorname{sgn}\det(M)=(-1)^{n-1}.$$

This settles the sign conjecture of Cooper–Du, Conjecture 1. The key is to use edge-cut coordinates: all but one of the gradient equations then separate into one-variable equations. Their nonreal solutions occur in conjugate pairs, while the unique real solution contributes a positive factor.

We use the cut identity from Cooper–Du, Theorem 2.1 and the normalized resultant and Poisson formula recalled in Zheng, Section 2, Theorems 2, 4 and 5. The known vanishing for odd order when $n\geq3$, and the case $n=2$ for every order, are already recorded in Cooper–Du; the latter is their Proposition 4.1. Thus it suffices to prove the assertion for even $k\geq2$ and $n\geq2$. For $n=1$, every entry is zero.

Cut coordinates and normalization

Put $m=n-1$ and $d=k-1$. Its entries count the edges of the smallest subtree containing the indexed vertices, with repetitions allowed. Write its associated form as

$$F(x)=\sum_{i_1,\ldots,i_k=1}^{n} M_{i_1\ldots i_k}x_{i_1}\cdots x_{i_k}.$$

The symmetric tensor determinant used here is

$$\det(M)=\operatorname{Res} \left(\frac1k\frac{\partial F}{\partial x_1}, \ldots,\frac1k\frac{\partial F}{\partial x_n}\right),$$

with the normalization $\operatorname{Res}(x_1^d,\ldots,x_n^d)=1$.

Root the tree at vertex $n$, and order the other vertices so that descendants precede ancestors. For each nonroot vertex $i$, let $a_i$ be the sum of $x_v$ over its rooted subtree, and set $s=\sum_vx_v$. The linear map

$$x\longmapsto(a_1,\ldots,a_m,s)$$

has a lower triangular matrix with diagonal entries all equal to $1$. In particular, its determinant is $1$.

An edge belongs to the subtree spanned by a tuple precisely when that tuple meets both sides of its cut. Summing over tuples therefore gives the known cut identity

$$\begin{aligned} \Phi(a,s)&=m s^k\\ &\quad-\sum_{i=1}^{m}\bigl(a_i^k+(s-a_i)^k\bigr), \end{aligned}$$

where $\Phi$ is $F$ expressed in the new coordinates. Its normalized gradient, in the variable order $(a_1,\ldots,a_m,s)$, consists of

$$P_i=(s-a_i)^d-a_i^d\qquad(1\leq i\leq m)$$

and

$$P_0=m s^d-\sum_{i=1}^{m}(s-a_i)^d.$$

Both the variable change and the corresponding inverse-transpose change of gradient equations have determinant $1$. The composition rule for normalized homogeneous resultants consequently gives the exact equality

$$\det(M)=\operatorname{Res}(P_1,\ldots,P_m,P_0).$$

The order of the equations matches the order of the variables, including $P_0$ for the last variable $s$.

The product and its sign

Because $k$ is even, $d$ is odd. At $s=0$ we have $P_i=-2a_i^d$ for $1\leq i\leq m$, so these leading forms have no common nonzero zero. Their normalized resultant is

$$\begin{aligned} R_\infty &=\operatorname{Res}(-2a_1^d,\ldots,-2a_m^d)\\ &=(-2)^{m d^{m-1}}. \end{aligned}$$

After setting $s=1$, the first $m$ equations separate as $p(a_i)=0$, where

$$p(z)=(1-z)^d-z^d.$$

This polynomial has degree $d$ and $d$ distinct roots. Indeed, $z=0$ is not a root, and its roots are exactly

$$z=\frac{1}{1+\zeta},\qquad \zeta^d=1.$$

The denominators are nonzero because $d$ is odd, and distinct $\zeta$ give distinct roots. Moreover, the only real root is $z=1/2$: on the real line, the odd power map is injective, so $(1-z)^d=z^d$ forces $1-z=z$.

Let $\mathcal R$ be this root set. The affine common zeros of $P_1,\ldots,P_m$ are exactly $\mathcal R^m$, each with multiplicity one: their Jacobian is diagonal with nonzero entries $p'(a_i)$. Applying the Poisson formula in the last variable $s$ gives

$$\begin{aligned} \det(M) &=(-2)^{m d^m} \prod_{a\in\mathcal R^m}h(a),\\ h(a)&=m-\sum_{i=1}^{m}(1-a_i)^d. \end{aligned}$$

Assume $\det(M)\ne0$. Every factor in this product is then nonzero. Complex conjugation preserves $\mathcal R^m$ and pairs each nonreal tuple $a$ with a distinct tuple $\overline a$. Since $h$ has real coefficients, each such pair contributes

$$h(a)h(\overline a)=|h(a)|^2>0.$$

There is just one real tuple, namely $(1/2,\ldots,1/2)$, and its factor is

$$h(1/2,\ldots,1/2)=m(1-2^{-d})>0.$$

Thus the entire product is positive. Since $d^m$ is odd, the prefactor has sign $(-1)^m$. Therefore

$$\operatorname{sgn}\det(M)=(-1)^m=(-1)^{n-1},$$

as required. Together with the previously established odd-order and two-vertex cases, this proves Conjecture 1 for every order $k\geq2$ and every tree whenever its Steiner distance hyperdeterminant is nonzero.