A generalization of the Graham-Pollak tree theorem to even-order Steiner distance

Joshua Cooper; Gabrielle Tauscheck

doi:10.1515/spma-2024-0018

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A generalization of the Graham-Pollak tree theorem to even-order Steiner distance

Joshua Cooper and Gabrielle Tauscheck

Published/Copyright: August 1, 2024

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From the journal Special Matrices Volume 12 Issue 1

Abstract

Graham and Pollak showed in 1971 that the determinant of a tree’s distance matrix depends only on its number of vertices, and, in particular, it is always nonzero. The Steiner distance of a collection of k vertices in a graph is the fewest number of edges in any connected subgraph containing those vertices; for k = 2 , this reduces to the ordinary definition of graphical distance. Here we show that the hyperdeterminant of the k th order Steiner distance hypermatrix is always nonzero if k is even, extending their result beyond k = 2 . Previously, the authors showed that the k -Steiner distance hyperdeterminant is always zero for k odd, so together this provides a generalization to all k . We conjecture that not just the vanishing, but the value itself, of the k -Steiner distance hyperdeterminant of an n -vertex tree depends only on k and n .

Keywords: hyperdeterminant; Steiner distance; tree graphs

MSC 2010: 05C12 (Primary) 05C50; 15A69 (Secondary)

1 Introduction

In an influential 1971 paper, Graham and Pollak [3] computed the determinant of the distance matrix of a n -vertex tree, i.e., the n × n matrix D whose ( i , j ) entry is the shortest-path distance between i and j . Strikingly, the formula only depends on n , and nothing more about the tree:

(1) det ( D ) = ( 1 − n ) ( − 2 ) n − 2 .

In particular, this quantity is nonzero for any n ≥ 1 .

Steiner distance generalizes the classical notion of distance in a graph G from pairs of vertices to any subset S ⊆ V ( G ) , as the minimum number of edges in any connected subgraph of G containing S (or possibly ∞ if there is no such subgraph). Therefore, partly inspired by recent developments in hypermatrices, Mao (q.v. [4]) questions how to generalize (1) to hyperdeterminants of order- k Steiner distances in trees. The present authors showed previously [1] that for k odd (and n ≥ 3 ), the Steiner distance hyperdeterminant is 0. In this study, we show that the Steiner distance is always nonzero for k even (and n ≥ 2 ), implying a weak generalization of the Graham-Pollak theorem to Steiner distance: for n ≥ 3 , whether the hyperdeterminant vanishes depends only on k . In fact, we conjecture a much stronger statement: the value of the hyperdeterminant (not just whether it vanishes) only depends on n and k in general. This has been checked computationally for all trees with ( k , n ) ∈ { ( 4 , 4 ) , ( 4 , 5 ) , ( 6 , 4 ) } and holds trivially for n = 2 and n = 3 . The following table shows the common values of the hyperdeterminant, factored into primes.

( k , n ) det ( S ) ( 4 , 2 ) − 2 2 ⋅ 7 ( 4 , 3 ) 2 12 ⋅ 7 ⋅ 2 3 4 ( 4 , 4 ) − 2 38 ⋅ 3 27 ⋅ 5 6 ⋅ 7 ⋅ 1 3 12 ( 4 , 5 ) 2 203 ⋅ 5 32 ⋅ 7 ⋅ 1 1 32 ⋅ 2 3 24 ⋅ 3 7 8 ( 6 , 2 ) − 1 1 2 ⋅ 31 ( 6 , 3 ) 2 14 ⋅ 3 16 ⋅ 1 1 4 ⋅ 31 ⋅ 1923 1 4 ( 6 , 4 ) − 2 82 ⋅ 3 17 ⋅ 1 1 8 ⋅ 31 ⋅ 4 1 12 ⋅ 7 1 6 ⋅ 8 9 6 ⋅ 15 1 24 ⋅ 25 7 24 ⋅ 151 1 12 ( 8 , 2 ) − 2 6 ⋅ 2 9 2 ⋅ 127 ( 8 , 3 ) 2 56 ⋅ 1 3 16 ⋅ 2 9 4 ⋅ 11 3 8 ⋅ 127 ⋅ 100 9 8 ⋅ 214 3 4 .

First, some notations and facts about hyperdeterminants are needed. The following theorem is a useful generalization of the fact that the matrix determinant is the unique irreducible polynomial in the entries of M whose vanishing coincides with the existence of nontrivial solutions to the linear system corresponding to M .

Theorem 1.1

[2, Theorem 1.3] The hyperdeterminant det ( M ) of the order-k, dimension-n hypermatrix M = ( M i 1 , … , i k ) i 1 , … , i k = 1 n is a monic irreducible polynomial which evaluates to zero iff there is a nonzero simultaneous solution to ∇ f M = 0 → , where

f M ( x 1 , … , x n ) = ∑ i 1 , … , i k M i 1 , … , i k ∏ j = 1 k x i j .

Next, we define Steiner distance.

Definition 1.2

Given a graph G and a subset S of the vertices, the Steiner distance of S , written as d G ( S ) or d G ( v 1 , … , v k ) where S = { v 1 , … , v k } , is the number of edges in the smallest connected subgraph of G containing S = { v 1 , … , v k } . We often suppress the variable v when referring to the vertices within S . Since such a connected subgraph of G witnessing d G ( S ) is necessarily a tree, it is called a Steiner tree of S .

Then, we have a “Steiner polynomial” which reduces to the quadratic form defined by the distance matrix for k = 2 .

Definition 1.3

Given a graph G , the Steiner polynomial of G is the k -form

p G ( k ) ( x ) = ∑ v 1 , … , v k ∈ V ( G ) d G ( v 1 , … , v k ) x 1 ⋯ x k ,

where we often suppress the subscript and/or superscript on p G ( k ) if it is clear from context.

Equivalently, we could define the Steiner polynomial to be the k -form associated with the Steiner hypermatrix:

Definition 1.4

Given a graph G , the Steiner k-matrix (or just “Steiner hypermatrix” if k is understood) of G is the order- k , cubical hypermatrix S G of dimension n whose ( v 1 , … , v k ) entry is d G ( v 1 , … , v k ) .

Throughout the sequel, we write D r for the operator ∂ ∕ ∂ x r , and we always assume that T is a tree.

Definition 1.5

Given a graph G on n vertices, the Steiner k-ideal – or just “Steiner ideal” if k is clear – of G is the ideal in C [ x 1 , … , x n ] generated by the polynomials { D j p G } j = 1 n .

Thus, the Steiner ideal is the Jacobian ideal of the Steiner polynomial of G . Equivalently, we could define the Steiner k -ideal via hypermatrices. We use the notation S ( x k − 1 , * ) to represent S ( x , x , … , x ︷ k − 1 , * ) . Note that S ( x k − 1 , * ) = S ( x k − 2 , * , * ) x where S ( x k − 2 , * , * ) is an n × n matrix whose entries are homogeneous polynomials of degree k − 2 . Then, the Steiner k-ideal of a graph G on n vertices is the ideal in C [ x 1 , … , x n ] generated by the components of S ( x k − 1 , * ) .

Definition 1.6

A Steiner nullvector is a point where all the polynomials within the Steiner ideal vanish. The set of all Steiner nullvectors – a projective variety – is the Steiner nullvariety.

2 Main results

We show that the order- k Steiner distance hypermatrix of a tree T on n + 1 ≥ 2 vertices has a nonzero hyperdeterminant for even k . Therefore, for the remainder of this study, we assume T is a tree on at least 2 vertices and k is even. For concision, let S = S ( x k − 2 , * , * ) as defined in Section 1, and write S u for the row of S indexed by u and S u , v for the v entry of S u . We also employ multi-index notation: for i = ( i 1 , … , i r ) ∈ { 0 , … , n } r and x = [ x 0 , … , x n ] , write x i ≔ ∏ j = 1 r x i j and ν t ( i ) ≔ ∣ { j : i j = t } ∣ .

Proposition 2.1

If e = { u , u + 1 } is an edge of the tree T on n + 1 vertices { 0 , … , n } , then S u − S u + 1 is a vector of the form x = [ x 0 , x 1 , … , x n ] where x 0 = x 1 = … = x u = − ∑ i = 0 u x i k − 2 and x u + 1 = … = x n = ∑ i = u + 1 n x i k − 2 with 0 , … , u lying in one component of T ′ ≔ T − e and u + 1 , … , n lying in the other component.

Proof

Note that

(2) S u w = ∑ i ∈ V ( T ) k − 2 d T ( u , w , i ) x i .

It is sufficient to show that for any w and w ′ that lie within the same component of T ′ ,

d T ( u , w , i ) − d T ( u + 1 , w , i ) = d T ( u , w ′ , i ) − d T ( u + 1 , w ′ , i )

holds for all i 1 , … , i k − 2 ∈ V ( T ) . For every case, let d = d T ( u , w , i ) and d ′ = d T ( u , w ′ , i ) .

Case 1: Let i 1 , … , i k − 2 lie in the same component of T ′ as u .

Assume w and w ′ also lie within the same component as u . Then,
d T ( u , w , i ) − d T ( u + 1 , w , i ) = d − ( d + 1 ) = − 1 ,

d T ( u , w ′ , i ) − d T ( u + 1 , w ′ , i ) = d ′ − ( d ′ + 1 ) = − 1 .
Assume instead that w and w ′ lie in the same component as u + 1 . Then,
d T ( u , w , i ) − d T ( u + 1 , w , i ) = d − d = 0 ,

d T ( u , w ′ , i ) − d T ( u + 1 , w ′ , i ) = d ′ − d ′ = 0 .

Case 2: Let i 1 , … , i k − 2 lie in the same component as u + 1 .

Assume w and w ′ also lie within the same component as u + 1 . Then,
d T ( u , w , i ) − d T ( u + 1 , w , i ) = d − ( d − 1 ) = 1 ,

d T ( u , w ′ , i ) − d T ( u + 1 , w ′ , i ) = d ′ − ( d ′ − 1 ) = 1 .
Assume instead that w and w ′ lie in the same component as u . Then,
d T ( u , w , i ) − d T ( u + 1 , w , i ) = d − d = 0 ,

d T ( u , w ′ , i ) − d T ( u + 1 , w ′ , i ) = d ′ − d ′ = 0 .

Case 3: Suppose some of the vertices among i 1 , … , i k − 2 lie in the same component as u and some lie in the same component as u + 1 . Then,

d T ( u , w , i ) − d T ( u + 1 , w , i ) = d − d = 0 ,

d T ( u , w ′ , i ) − d T ( u + 1 , w ′ , i ) = d ′ − d ′ = 0 .

Therefore, the vector S u − S u + 1 has equal entries among the coordinates indexed by vertices lying in the same component of T ′ . To compute these entries, we combine the different cases. Note that Case 3 contributes zero to the sum in (2); therefore, we can omit those entries in the calculations. First, consider x u = S u , u − S u + 1 , u . Since u lies in the same component as itself, we are working with Cases 1(a) and 2(b). Case 2(b) contributes a zero to the summation; therefore, we need to only consider the vertices in Case 1(a), giving us x u = − ∑ i ∈ { 0 , … , u } k − 2 x i = − ∑ i = 0 u x i k − 2 .

Now, consider x u + 1 = S u , u + 1 − S u + 1 , u + 1 . Since u + 1 lies in the same component as itself, we are working with Cases 1(b) and 2(a). Case 1(b) contributes a zero to the summation; therefore, we need to only consider the vertices in Case 2(a), giving us x u + 1 = ∑ i ∈ { u + 1 , … , n } k − 2 x i = ∑ i = u + 1 n x i k − 2 . □

Let h u ( a , b ) = [ h 0 , h 1 , … , h n ] ∈ C n + 1 be a vector with at most two distinct entries: h 0 = … = h u = a and h u + 1 = … = h n = b .

Proposition 2.2

Let x = [ x 0 , x 1 , … , x n ] be a Steiner nullvector of a tree T. All entries that correspond to leaf vertices v are the same value, equal to the summation of every entry of x except x v .

Proof

Since T is a tree on at least 2 vertices, it must have a leaf vertex; call it 0, and let 1 be its unique neighbor. Proposition 2.1 gives a and b so that

( D 0 p T − D 1 p T ) ( x ) = h 0 ( a , b ) T x = h 0 − x 0 k − 2 , ∑ i = 1 n x i k − 2 T x = − x 0 k − 1 + ∑ i = 1 n x i k − 1 = 0 ,

so we may conclude that x 0 k − 1 = ∑ i = 1 n x i k − 1 . Since k − 1 is odd,

x 0 = ∑ i = 1 n x i .

From here, we see that ∑ i = 0 n x i = x 0 + x 0 so that x 0 = 1 2 ∑ i = 0 n x i . The choice of vertex 0 was made arbitrarily among the leaves; therefore, every leaf corresponds to this same entry in the nullvector, namely, ∑ i = 1 n x i .□

Proposition 2.3

Suppose e = { u , u + 1 } is an edge of the tree T on n + 1 vertices. Let 0 , … , u lie in one component of T − e and u + 1 , … , n lie in the other component. Let x = [ x 0 , x 1 , … , x n ] be a Steiner nullvector. Then, ∑ i = 0 u x i = x v where v is any leaf vertex.

Proof

Proposition 2.1 gives a and b so that

( D u p T − D u + 1 p T ) ( x ) = h u ( a , b ) T x = h u − ∑ i = 0 u x i k − 2 , ∑ i = u + 1 n x i k − 2 T x = − ∑ i = 0 u x i k − 1 + ∑ i = u + 1 n x i k − 1 = 0 ,

so we may conclude that ∑ i = 0 u x i k − 1 = ∑ i = u + 1 n x i k − 1 . Since k − 1 is odd,

(3) ∑ i = 0 u x i = ∑ i = u + 1 n x i .

Assume without loss of generality that 0 is a leaf vertex; then Proposition 2.2 gives that x 0 = ∑ i = 1 n x i . We can rewrite (3) to be x 0 + ∑ i = 1 u x i = x 0 − ∑ i = 1 u x i so that

∑ i = 1 u x i = 0 .

Therefore, ∑ i = 0 u x i = x 0 as desired.□

We are now in a position to describe all entries of a Steiner nullvector of a tree T .

Proposition 2.4

Suppose x = [ x 0 , … , x n ] is a Steiner nullvector of a tree T on n + 1 vertices. Then, for each t ∈ V ( T ) , x t = x v ( 2 − deg ( t ) ) , where v is any leaf vertex.

Proof

First fix a leaf vertex v ; note that Proposition 2.2 implies that x v does not depend on which leaf is chosen. Since deg ( v ) = 1 , we have x v = x v ( 2 − deg ( v ) ) .

Let t be any nonleaf vertex, with degree d > 1 . Since T is a tree on at least 2 vertices, t has at least one neighbor – call it u – and let e be the edge { t , u } . Let G be the set of all vertices that lie in the same component as t in T − e . Proposition 2.3 tells us that ∑ i ∈ G x i = x v . Partition G into sets { t } , G 0 , … , G d − 2 by taking G j to be vertices of the j th component induced by G \ { t } . (Note that there are only d − 1 subsets since the component of T − { t } containing vertex t + 1 is not included in G .) Once again, we can apply Proposition 2.3 to say that ∑ i ∈ G j x i = x v for 0 ≤ j ≤ d − 2 . Therefore,

x v = ∑ i ∈ G x i = x t + ∑ j = 0 d − 2 ∑ i ∈ G j x i = x t + ∑ j = 0 d − 2 x v = x t + x v ( deg ( t ) − 1 )

and x t = x v ( 2 − deg ( t ) ) as desired.□

We are now equipped to prove the main result.

Theorem 2.5

For k even, the Steiner k-matrix of a tree on at least 2 vertices has a nonzero hyperdeterminant.

Proof

Using Proposition 2.4, we prove the result by contradiction.

Consider any tree T on n + 1 vertices where n ≥ 1 . Assume without loss of generality that 0 is a leaf vertex of T whose neighbor is 1. Assume the hyperdeterminant is zero. Theorem 1.1 then implies that there exists a nontrivial common zero of the Steiner k -ideal; let x = [ x 0 , x 1 , … , x n ] be a Steiner nullvector of T . Since the generators of the Steiner k -ideal are homogeneous, and x 0 = 0 implies that x v = 0 for all v by Proposition 2.4, we may assume x 0 = 1 . Then,

D 1 p T ( x ) = ∑ i ∈ { 0 , … , n } k − 1 d T ( 1 , i ) x i = ∑ a = 0 k − 1 ∑ i ∈ { 0 , … , n } k − 1 ν 0 ( i ) = a d T ( 1 , i ) x i = ∑ a = 0 k − 1 k − 1 a x 0 a ∑ i ∈ { 1 , … , n } k − 1 − a d T ( 1 , i , 0 , 0 , … , 0 ︷ a ) x i = ∑ i ∈ { 1 , … , n } k − 1 d T ( 1 , i ) x i + ∑ a = 1 k − 1 k − 1 a ∑ i ∈ { 1 , … , n } k − 1 − a [ d T ( 1 , i ) + 1 ] x i

by partitioning the vectors i according to whether ν 0 ( i ) = 0 or a for some a > 0 . Continuing to rewrite the quantity D 1 p T ( x ) ,

= ∑ a = 0 k − 1 k − 1 a ∑ i ∈ { 1 , … , n } k − 1 − a d T ( 1 , i ) x i + ∑ a = 1 k − 1 k − 1 a ∑ i ∈ { 1 , … , n } k − 1 − a x i = ∑ a = 0 k − 1 k − 1 a ∑ i ∈ { 1 , … , n } k − 1 − a d T ( 1 , i ) x i + 1 + ∑ i = 1 n x i k − 1 − ∑ i = 1 n x i k − 1 = ∑ a = 0 k − 1 k − 1 a ∑ b = 0 k − 1 − a ∑ i ∈ { 1 , … , n } k − 1 − a ν 1 ( i ) = b d T ( 1 , i ) x i + 2 k − 1 − 1 = ∑ a = 0 k − 1 k − 1 a ∑ b = 0 k − 1 − a k − 1 − a b x 1 b ∑ i ∈ { 2 , … , n } k − 1 − a − b d T ( 1 , i ) x i + 2 k − 1 − 1 = ∑ a + b = 0 k − 1 k − 1 a , b , k − 1 − a − b x 1 b ∑ i ∈ { 2 , … , n } k − 1 − a − b d T ( 1 , i ) x i + 2 k − 1 − 1 = ∑ j = 0 k − 1 k − 1 j ∑ b = 0 j j b x 1 b ∑ i ∈ { 2 , … , n } k − 1 − j d T ( 1 , i ) x i + 2 k − 1 − 1 = ∑ j = 0 k − 1 k − 1 j ( x 1 + 1 ) j ∑ i ∈ { 2 , … , n } k − 1 − j d T ( 1 , i ) x i + 2 k − 1 − 1 ,

where j has been introduced to replace a + b . If T is a tree on 2 vertices, then D 1 p T ( x ) = 2 k − 1 − 1 ≠ 0 . Therefore, the only Steiner nullvector of a tree on 2 vertices is the trivial one, and the hyperdeterminant must be nonzero. We continue by assuming that T is a tree on at least 3 vertices.

Upon deleting the leaf vertex 0, we obtain a new tree T ′ on n vertices. Let x ′ = [ x 1 ′ , x 2 ′ , … , x n ′ ] be a Steiner nullvector of T ′ . Note that deg T ( 1 ) = deg T ′ ( 1 ) + 1 and deg T ( 2 ) = deg T ′ ( 2 ) , … , deg T ( n ) = deg T ′ ( n ) . Using Proposition 2.4 and the definition of x , by setting x ′ v = x v for some leaf vertex common to both T and T ′ , we conclude that x ′ = [ x 1 + x 0 , x 2 , … , x n ] = [ x 1 + 1 , x 2 , … , x n ] . Then,

D 1 p T ′ ( x ′ ) = ∑ i ∈ { 1 , … , n } k − 1 d T ( 1 , i ) x ′ i = ∑ j = 0 k − 1 ∑ i ∈ { 1 , … , n } k − 1 ν 1 ( i ) = j d T ( 1 , i ) x ′ i = ∑ j = 0 k − 1 k − 1 j ( x ′ ) j ∑ i ∈ { 2 , … , n } k − 1 − j d T ( 1 , i ) x i = ∑ j = 0 k − 1 k − 1 j ( x 1 + 1 ) j ∑ i ∈ { 2 , … , n } k − 1 − j d T ( 1 , i ) x i .

Since we assumed x and x ′ were Steiner nullvectors, D 1 p T ( x ) = 0 = D 1 p T ′ ( x ′ ) . Therefore,

0 = D 1 p T ( x ) − D 1 p T ′ ( x ′ ) = ∑ j = 0 k − 1 k − 1 j ( x 1 + 1 ) j ∑ i ∈ { 2 , … , n } k − 1 − j d T ( 1 , i ) x i + 2 k − 1 − 1 − ∑ j = 0 k − 1 k − 1 j ( x 1 + 1 ) j ∑ i ∈ { 2 , … , n } k − 1 − j d T ( 1 , i ) x i = 2 k − 1 − 1 ≠ 0 .

Therefore, the only nullvector is the trivial one, and the hyperdeterminant must be nonzero.□

Acknowledgements

The authors are grateful for the reviewers’ careful reading, and to Zhibin Du for helpful suggestions.

Funding information: The authors state no funding involved.
Author contributions: All authors prepared and accepted responsibility for the entire content of this manuscript, consented to its submission to the journal, reviewed all the results, and approved the final version of the manuscript.
Conflict of interest: The authors state no conflict of interest.
Data availability statement: Data sharing is not applicable to this article as no datasets were generated or analyzed during the current study.

References

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Received: 2024-02-23

Accepted: 2024-06-14

Published Online: 2024-08-01

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https://doi.org/10.1515/spma-2024-0018

Keywords for this article

hyperdeterminant; Steiner distance; tree graphs

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