Dissimilarity Vectors of Trees and Their Tropical Linear Spaces
A thesis presented to the faculty of
San Francisco State University
In partial fulfillment of
The Requirements for
The Degree
Master of Arts
In
Mathematics
by
Benjamin Iriarte
San Francisco, California
December 2010
Copyright by
Benjamin Iriarte
2010
CERTIFICATION OF APPROVAL
I certify that I have read Dissimilarity Vectors of Trees and Their Tropical Linear Spaces by Benjamin Iriarte and that in my opinion this work
meets the criteria for approving a thesis submitted in partial fulfillment
of the requirements for the degree: Master of Arts in Mathematics at
San Francisco State University.
Federico Ardila
Professor of Mathematics
Serkan Hoşten
Professor of Mathematics
Matthias Beck
Professor of Mathematics
Dissimilarity Vectors of Trees and Their Tropical Linear Spaces
Benjamin Iriarte
San Francisco State University
2010
We present an introduction to the study of weighted trees, from a combinatorial
perspective. In particular, we explore the connections of the subject with tropical
algebraic geometry and with tropical linear spaces.
Finally, we prove that the set of dissimilarity vectors of trees is contained in
the tropical Grassmannian, and then describe the complex of bounded faces of the
tropical linear space of a dissimilarity vector.
I certify that the Abstract is a correct representation of the content of this thesis.
Chair, Thesis Committee
Date
ACKNOWLEDGMENTS
This work has been supported by an NSF Grant through my advisor,
Federico Ardila. I need to give special thanks to Federico for the meetings
we have held during this year. Thanks to them, a lot of things took right
course, including this project, and I felt supported at all moments. The
capacity and willingness to listen to my sometimes unprecise ideas has
been invaluable, and has been the best form of advising I could ever get.
Thanks Federico, you have also been a teacher for life.
I also need to specially thank my family, for being there in the good
times, and the bad times as well. Most of this gratefulness goes to my
mother, an example of strength, drive, love and capacity to work.
I am very grateful to both of them for the chance to spend this year at
SFSU: I have learned a great deal of lessons.
Special thanks go also to Bernd Sturmfels. A series of conversations in
the middle of December were the thrust to get the final chapters of this
project started. He has been a great support this year. I am also grateful
with Serkan Hosten for the things that I’ve learned from him this year.
Thanks to Nathalia for her constant support. She was always there.
Thanks to Sarah Billey and Isabella Novik for allowing me to present
some of these results in the Combinatorics Seminar at UW, and also
v
to Luis Serrano and Sergey Fomin. I must also thank Filip Cools for
fruitful discussions.
vi
TABLE OF CONTENTS
1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1
1.1
Motivation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1
1.2
Basic Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2
2 Background . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5
2.1
The Tropical Grassmannians . . . . . . . . . . . . . . . . . . . . . . .
5
2.2
Tropical Linear Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . 11
2.3
Results About Trees . . . . . . . . . . . . . . . . . . . . . . . . . . . 20
3 Dissimilarity Vectors are Contained in the Tropical Grassmannian . . . . . 28
3.1
Column Reductions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31
3.2
Main Result . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32
4 The Tropical Linear Space of a Dissimilarity Vector . . . . . . . . . . . . . 39
4.1
Vertices of Td and Their Matroids . . . . . . . . . . . . . . . . . . . . 40
4.2
Higher Dimensional Faces of Td and Their Matroids . . . . . . . . . . 56
4.3
A Combinatorial Description of the Geometric Object Td . . . . . . . 72
4.4
4.3.1
Faces from (m, i)-good pairs. . . . . . . . . . . . . . . . . . . . 73
4.3.2
Faces from full subtrees of T . . . . . . . . . . . . . . . . . . . 79
Face Enumeration in the Simplest Cases . . . . . . . . . . . . . . . . 89
5 Further Directions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92
vii
Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 94
viii
1
Chapter 1
Introduction
1.1 Motivation
This writing is, more than anything else, an excursion into the world of weighted
trees. Recently, this has become a combinatorial branch of its own, know as T theory. We find it very motivating that weighted trees are amazingly simple objects to understand, yet, a large number of natural questions about them remain
unanswered.
The T -theory comes motivated by its applications in biology and computer science. However, it is not hard to notice the intrinsic combinatorial wealth and beauty
of the subject. In particular, it turns out that weighted trees provide a handful of
examples to tropical geometry, illustrating several central concepts and providing
further intuition, conjectures and problems.
Among the most common applications, we find the reconstruction of phylogenetic
trees in evolutionary biology, from a set of known data on the taxa. Usually, these
data are the results of analysis of DNA sequences alignments of several species.
Along the lines of the writing, we provide a personal development of the subject,
2
sometimes providing proofs of the main results. The proofs themselves will sometimes be original, sometimes not. Clearly, we will only cover the smallest part of a
more developed subject, narrowed to our interests.
The final chapters will be devoted to present our own results and contributions
to the subject. In particular, we will provide a proof of an amazing relation between
weighted trees and a central object of tropical algebraic geometry, the tropical Grassmannian. Then, we will describe the tropical linear spaces that arise from weighted
trees.
We expect that our results provide further intuition and knowledge about weighted
trees, and to contribute to future endeavors in the area.
1.2 Basic Definitions
To begin, we recall the significant definitions associated with this work.
E
For every finite set E, define m
as the collection of all subsets of E of size m.
We adopt the somewhat standard convention under which [n] = {1, 2, . . . , n}.
Under this convention, notice that [n]
is the collection of subsets of [n] with size
m
m.
Recall that a tree can be defined as a connected graph that has no cycles. Every
vertex of degree 1 in a tree is called a leaf. Every vertex of a tree that is not a leaf
is called an internal vertex.
Now, assume that we have a tree containing an internal vertex v satisfying the
3
following condition: for every pair of different leaves u and w of the tree, the minimal
path from u to v intersects the minimal path from w to v at v and only at v, that
is, there is no common edge shared by both paths. In that case, we say that our
tree is a v-path-star tree.
Suppose that we have a tree T with n ≥ 1 leaves, labeled by the set [n]. Suppose
also that T does not have internal vertices of degree 2. Moreover, assume that for
each edge e of T , there exists a unique real number ω(e) associated to e such that:
• ω(e) > 0 whenever e is an internal edge of T , that is, whenever e is adjacent
to two internal vertices.
• ω(e) ∈ R whenever e is not an internal edge of T , that is, whenever e is
adjacent to a leaf. No restrictions apply to this case.
We call ω(e) the weight of e and we say that our tree T is a weighted tree. We
will let E(T ) be the set of (weighted) edges of T , V(T ) be the set of vertices of T ,
and L(T ) be the set of (labeled) leaves of T .
To briefly convey all these conditions, we sometimes say that T is an n-tree, or
simply that T is a tree. In fact, every time we refer to a tree T in this writing, we
will always refer to a weighted tree in the sense described above. In particular, note
that this convention involves the use of the letter T .
4
For every tree T , let the total weight ω(T ) of T be defined by:
X
ω(T ) :=
ω(e).
e∈E(T )
If E(T ) = ∅, we adopt the convention ω(T ) = 0.
[n]
Now, for m ∈ [n], we define a vector d ∈ R( m ) associated with the tree T , which
we call the m-dissimilarity vector of T . Notice that we have not indexed our vector
d with the integer m. Most of the time, the value of m will be clear from the context
and whenever it’s not, we will explicitly refer to d as the m-dissimilarity vector of
our tree.
For each m-set {i1 , i2 , . . . , im } ∈
[n]
m
, let di1 i2 ...im be the total weight of the
minimal subtree (w.r.t number of edges) of T that contains the leaves i1 , i2 , . . . , im .
Equivalently, let di1 i2 ...im be the total weight of the subtree of T spanned by these
leaves.
[n]
In particular, when d is the 1-dissimilarity vector, notice that d = 0 ∈ R( 1 ) .
This case is not very interesting. In case d is the n-dissimilarity vector, we have
d = ω(T ) ∈ R( n ) . Neither is this very interesting.
[n]
5
Chapter 2
Background
2.1 The Tropical Grassmannians
We first mention the idea of classical Grassmannians. Let F be a field of characteristic 0. The space of all m-dimensional subspaces of F n is called the Grassmannian
Gm,n,F . Given m linearly independent vectors v1 , v2 , . . . , vm of F n spanning an mdimensional subspace V , form the m × n matrix M whose rows are the vectors
v1 , v2 . . . , vm .
[n]
Let P ∈ F ( m ) be the vector of all maximal minors of M where, for all {i1 , i2 , . . . , im } ∈
[n]
, Pi1 i2 ...im is the minor coming from columns i1 , i2 , . . . , im of M . Call P the vecm
tor of Plücker coordinates of V . The following theorem and proof can be found
in [15].
Theorem 2.1. The m-dimensional linear subspaces of F n are uniquely determined
by their vector of Plücker coordinates, modulo multiplication by a constant non-zero
scalar.
Proof. Consider the m×n matrix M just defined. Any other m×n matrix with equal
6
row space is obtained as ∆M for some invertible matrix ∆. Each maximal minor of
∆M is then det(∆) times the corresponding maximal minor of M . Now, suppose
that the maximal minors of M are a constant multiple λ of the corresponding
maximal minors of N . We show that M and N have equal rowspace. For some
ρ = {ρ1 < ρ2 < · · · < ρm } ∈ [n]
, the maximal submatrix Mρ of M obtained from
m
columns ρ satisfies det(Mρ ) 6= 0. Thus, det(Mρ ) = λ det(Nρ ) 6= 0, where Nρ is the
corresponding maximal submatrix of N . We can find ρ because the rank of both
M and N is precisely m. Define M 0 = Mρ−1 M and N 0 = Nρ−1 N . Clearly, the row
span of N 0 coincides with the row span of N and the same is true for M 0 and M .
The columns ρ of M 0 and N 0 form the identity matrix. Consider some % ∈ [n]
. We
m
have
det(M%0 ) = det(Mρ−1 M% ) = det(Mρ−1 )det(M% )
=
1
det(Nρ−1 )λdet(N% ) = det(Nρ−1 )det(N% )
λ
= det(Nρ−1 N% ) = det(N%0 ).
Therefore, the maximal minors of M 0 and N 0 are equal. Take a pair 1 ≤ i ≤ m
and 1 ≤ j ≤ n. Consider the columns ρ1 , . . . , ρbi , . . . , ρm and j. The minor given by
these columns is (−1)l (M 0 )ij in M 0 and (−1)l (N 0 )ij in N 0 , for l ∈ {0, 1}. But these
two are equal and so (M 0 )ij = (N 0 )ij , that is, M 0 = N 0 .
Now, suppose that we have a vector x ∈ F n and we want to check if x ∈ V . To
7
the matrix M , append the row x after the m-th row to obtain an (m + 1) × n matrix
M x . If x is a point in V , then all the maximal minors of M x have to be 0, clearly.
On the other hand, if x 6∈ V , then M x is a rank-(m + 1) matrix, so it must
have m + 1 linearly independent columns as well, which then give rise to a non-zero
maximal minor.
Thus, x ∈ V if and only if all the maximal minors of M x are 0.
Now, notice that the maximal minors of M x can be expressed using the Plücker
coordinates of V , doing cofactor expansion of the determinants along the last row.
To do this, for every {i1 , i2 , . . . , im } ∈ [n]
, recall that Pi1 i2 ...im is the maximal minor
m
of M coming from columns i1 , i2 , . . . , im : Pi1 i2 ...im is a Plücker coordinate of V . Then,
we can see that the maximal minors of M x are:
m+1
X
(−1)r Pi1 i2 ...ibr ...im im+1 xir for all 1 ≤ i1 < i2 < · · · < im < im+1 ≤ n.
r=0
Therefore, x is a point in V if and only if all these sums are 0. We will keep
this characterization in mind when defining the completely analogous notion of
tropical linear space. The tropical linear spaces will arise after tropicalizing the
sums presented above.
We are now ready to provide some background on tropical Grassmannians.
To begin, consider the field K = C{{t}} of dual Puiseux series. Recall that this
8
is the algebraically closed field of formal expressions
ω=
k=p
X
ck tk/q
−∞
where p ∈ Z, cp 6= 0, q ∈ Z+ and ck ∈ C for all k ≤ p. It is the algebraic
closure of the field of dual Laurent series over C. The field comes equipped with a
standard valuation val : K 7→ Q ∪ {−∞} by which val(ω) = p/q. As a convention,
val(0) = −∞.
Let X = (Xij ) be an m × n matrix of indeterminates and let K[X] denote the
polynomial ring over K generated by these indeterminates. Fix a second polynomial
n
ring in m
indeterminates over the same field: K[Y ] = K[Yi1 i2 ...im : {i1 , i2 , . . . , im } ∈
[n]
].
m
Let φm,n : K[Y ] 7→ K[X] be the homomorphism of rings taking Yi1 ...im to the
maximal minor of X obtained from columns i1 , . . . , im .
The Plücker ideal or ideal of Plücker relations will be the homogeneous
prime ideal Im,n =ker(φm,n ), which contains all the algebraic relations or syzygies
among the m × m minors of any m × n matrix with entries in K.
In particular, letting F = K during the first lines of this section, we can see
that the Plücker coordinates Pi1 i2 ...im with {i1 i2 . . . im } ∈ [n]
of the m-dimensional
m
vector space V considered, lie in the variety V (Im,n ) of Im,n .
For m ≥ 2, the Plücker ideal has a Gröbner basis consisting of quadrics; a
comprehensive study of these ideals can be found in Chapter 14 of the book by
9
Miller and Sturmfels [15] and in Sturmfels [19]. In particular, the tropicalization of
some of these quadrics, known as the three-term Plücker relations, will give rise
to the notion of tropical Plücker relations that we present in the following section.
We note that for the purposes of studies in evolutionary biology and phylogenetics, the max convention should be adopted for tropical geometry. We follow that
convention here.
Now, the Plücker ideal is a polynomial ideal in K[Y ]. We can therefore define the
notion of its tropical variety in the usual way, as we now recall. Let R = R ∪ {−∞}.
([n])
Consider a finite set A ⊆ Z≥0m and let:
f=
X
cα
α∈A
Y
Yραρ ∈ K[Y ].
ρ∈([n]
m)
The tropicalization of f is given by:
trop(f ) = max{val(cα ) +
α∈A
X
ρ∈(
[n]
m
αρ yρ }.
)
The tropical hypersurface T ( trop(f )) of f is the set of points in R
([n]
m)
where
trop(f ) attains its maximum twice or, equivalently, where trop(f ) is not differentiable. It is the analogous tropical notion of the classical hypersurface of 0’s in K( m )
[n]
of f . By the fundamental theorem of tropical algebraic geometry [20, Theorem 2.1],
the classical and tropical hypersurfaces are related in the expected way: the later
10
is the (closure of) the pointwise valuation of the former. The actual content of the
theorem says that the same will still hold for polynomial ideals and their varieties.
In Theorem 2.2, we present a narrower version of the fundamental theorem, suitable
for our purposes.
With these tools, we now define the tropical analog of the Grassmannian Gm,n,K .
The tropical variety T (Im,n ) of the Plücker ideal Im,n is denoted by Gm,n and is
called the tropical Grassmannian. It is defined as the intersection:
T (Im,n ) =
\
T ( trop(f )) .
f ∈Im,n
We have the following characterization of Gm,n , which is a direct application of
the fundamental theorem of tropical algebraic geometry:
Theorem 2.2. The following subsets of R
([n]
m)
coincide:
• The tropical Grassmannian Gm,n .
[n]
• The closure of the set {(val(cρ ))ρ∈([n]) : (cρ )ρ∈([n]) ∈ V (Im,n ) ⊆ K( m ) }.
m
m
An important consequence of this theorem will be the following. Suppose that
a vector P ∈ K( m ) arises as the set of maximal minors of a full rank m × n matrix
[n]
with entries from K. That is, suppose that P is the vector of Plücker coordinates of
some m-dimensional vector space in Kn , as V . Then, its componentwise valuation
p = val(P ) = (val(Pρ ))ρ∈([n]) will lie in the tropical Grassmanian Gm,n . Naturally, p
m
11
will also satisfy the tropical Plücker relations that we will describe soon, and it will
do so as much as P satisfies the classical Plücker relations, and the former are the
tropicalization of a subset of the latter.
2.2 Tropical Linear Spaces
In this section we introduce one of the main objects of this writing.
Let p ∈ R( m ) . We say that p satisfies the tropical Plücker relations with the
[n]
max convention if the maximum of the set
{pSij + pSkl , pSik + pSjl , pSil + pjk }
[n]
is attained at least twice for all S ∈ m−2
and {i, j, k, l} ⊆ [n]\S
.
4
For all {i1 , i2 , . . . , im } ∈ [n]
, define ei1 i2 ...im = ei1 + ei2 + · · · + eim , where the
m
sum takes place in Rn with its canonical basis {e1 , e2 , . . . , en }.
Let Hm be the m-hypersimplex of Rn . Recall that Hm is the polytope obtained
as the convex hull of all vectors ei1 i2 ...im with {i1 , i2 , . . . , im } ∈ [n]
. Each of these
m
vectors is a vertex of Hm .
[n]
Suppose that p ∈ R( m ) . We define a regular subdivision Dp of Hm from p. Let
R ⊆ Rn+1 be the polytope obtained as the convex hull of all points (ei1 i2 ...im , pi1 i2 ...im )
with {i1 , i2 , . . . , im } ∈ [n]
. Consider the face of R consisting of all points r ∈ R that
m
maximize a linear functional (−x, 1) · r with x ∈ Rn . This face is the convex hull of
12
the vertices (ei1 i2 ...im , pi1 i2 ...im ) that it contains. Now, let Bx be the collection of sets
{i1 , i2 , . . . , im } ∈ [n]
for which (ei1 i2 ...im , pi1 i2 ...im ) is a point in our face. This is the
m
collection of sets {i1 , i2 , . . . , im } ∈ [n]
for which pi1 i2 ...im − (xi1 + xi2 + · · · + xim ) is
m
maximal. Finally, construct a face Px of Dp obtained from x as the convex hull of
all ei1 i2 ...im with {i1 , i2 , . . . , im } ∈ Bx .
Now, let P be a subpolytope of Hm . To recall, this means that the vertices of P
are of the form ei1 i2 ...im with {i1 , i2 , . . . , im } ∈ [n]
. Let BP be the collection of sets
m
{i1 , i2 , . . . , im } ∈ [n]
for which ei1 i2 ...im ∈ P. We say that P is matroidal or that P
m
is a matroid polytope if the set BP is the collection of bases of a rank-m matroid
over [n], see Oxley [16, Chapter 1] for a general introduction to the language of
matroids.
The following is a theorem of David Speyer [21], which motivates the introduction
of tropical Plücker relations.
Theorem 2.3. The following assertions are equivalent:
[n]
• the vector p ∈ R( m ) satisfies the tropical Plücker relations;
• the one skeleta of Dp and Hm are the same;
• every face of Dp is matroidal.
[n]
Now, suppose that p ∈ R( m ) satisfies the tropical Plücker relations. Define
13
L(p) ⊆ Rn as the set of all x for which the maximum of the set
{pi1 i2 ...ibr ...im im+1 + xir |r ∈ [m + 1]}
is attained at least twice, for all {i1 , i2 , . . . , im , im+1 } ∈
[n]
m+1
. The set L(p) will be
called the m-dimensional tropical linear space associated to p.
For any x ∈ Rn , consider the subpolytope Px of Hm introduced earlier. Recall
that Px is a face of Dp obtained as the convex hull of all points ei1 i2 ...im for which
pi1 i2 ...im − (xi1 + xi2 + · · · + xim ) is maximal. Also, notice that BPx = Bx . As p
satisfies the tropical Plücker relations, we conclude that Bx is the set of bases of
some rank-m matroid over [n], which we call Mx .
We now present a remarkable theorem of Speyer [21].
Theorem 2.4. x ∈ L(p) if and only if Mx is a loopless matroid.
Recall that a matroid over a finite set E is said to be loopless if every element
of E is contained in at least one basis.
Proof. To begin, notice that L(p) can be described as the set of all x for which the
maximum of the set
{pi1 i2 ...ibr ...im im+1 − xi1 + xi2 + · · · + xc
ir + · · · + xim + xim+1 |r ∈ [m + 1]}
is attained at least twice, for all {i1 , i2 , . . . , im , im+1 } ∈ [n]m+1 .
14
So let x ∈ L(p) and suppose, without loss of generality, that [m] ∈ Bx . Take
any element of the set [n]\[m]. We can assume without loss of generality that this
element is m + 1. By the definition of L(p), the maximum of the set
{p12...br...m(m+1) − x1 + x2 + · · · + xbr + · · · + xm + x(m+1) |r ∈ [m + 1]}
is attained at least twice. This maximum is attained by p12...m −(x1 + x2 + · · · + xm ).
But then, any other choice of subindices for which the maximum is attained
must contain m + 1, and this set of subindices will be an element of Bx . Therefore,
Mx is loopless.
Now, suppose that Mx is loopless and we want to prove x ∈ L(p). Without loss
of generality, it is enough to show that the maximum of the set
{p12...br...m(m+1) − x1 + x2 + · · · + xbr + · · · + xm + x(m+1) |r ∈ [m + 1]}
is attained at least twice.
Let e[m+1] = e1 + e2 + · · · + em + em+1 . For all reals t ≥ 0, let Mt = Mx−te[m+1]
and let Bt = Bx−te[m+1] .
Notice that for large enough values t > 0, we have Bt ⊆ [m + 1]. Let t∗
be the infimum of these values. Then Mt = Mt∗ for all t ≥ t∗ . As t ranges
from 0 to ∞, the matroid Mt changes at a finite number of times t, which we call
0 = t0 < t1 < t2 < · · · < tq = t∗ , and remains constant throughout (t` , t`+1 ) for all
15
0 ≤ ` < q.
Let t ∈ (t` , t`+1 ). Then, notice that the set of bases of Mt coincides with the
subset of bases of Mt` that meet [m + 1] at a maximal number of points.
Now, suppose that every element of [m + 1] is contained in a basis of Mt` .
Let i be the maximal number of elements of [m + 1] contained in a basis of Mt` .
Assume that [i] ⊆ I for some I ∈ Bt` . If j ∈ ([m + 1]\[i]), find J ∈ Bt` for which
j ∈ J. Then, by the maximality of i and considering the bases I and J, we can see
that ((I − k) ∪ j) ∈ Bt` for some k ∈ [i]. Therefore, for all j ∈ [m + 1], there exists
J ∈ Bt` such that j ∈ J and |J ∩ [m + 1]| = i. That is, every element of [m + 1] lies
in basis of Mt` which meets [m + 1] at a maximal number of points.
Therefore, every element of [m + 1] is contained in a basis of Mt .
On the other hand, the set of bases of Mt coincides with the subset of bases of
Mt`+1 that meet [m + 1] at a minimal number of points. If every element of [m + 1]
is contained in a basis of Mt , then every element of [m + 1] is contained in a basis
of Mt`+1 .
As Mx = Mt0 is loopless, so in particular every element of [m + 1] is contained in
a basis, inductively we obtain that every element of [m + 1] is contained in a basis
of Mtq = Mt∗ .
But notice that x ∈ L(p) if and only if |Bt∗ | ≥ 2, which is what we just proved.
A different approach to tropical linear spaces is fully presented now. Let p ∈
16
[n]
R( m ) satisfy the tropical Plücker relations with the max convention. Consider the
nonempty unbounded n-dimensional polyhedron Pp := {x ∈ Rn |xi1 +xi2 +· · ·+xim ≥
pi1 i2 ...im for all {i1 , i2 , . . . , im } ∈ [n]
}.
m
The polyhedron Pd coincides with the notion of the envelope of Hm with
respect to p, as presented in [11].
Let the reduced tropical linear space of p be the set Pp0 := {x ∈ Pp | If y ≤
x componentwise for some y ∈ Pp , then we have y = x}.
Take x ∈ ∂(Pp ) and define Bx to be the collection of all sets {i1 , i2 , . . . , im } ∈
[n]
m
for which
xi1 + xi2 + · · · + xim = pi1 i2 ...im .
The set Bx is the set of bases of a matroid Mx with rank m, over the set [n].
To understand why this is true, consider the polytope R of Rn+1 previously
defined.
For each r ∈ R, notice that (−x, 1) · r ≤ 0. The set of all such points r for
which (−x, 1) · r = 0 is a face of R. Of course, this face is then the convex hull
of all vertices (ei1 i2 ...im , pi1 i2 ...im ) of R with {i1 , i2 , . . . , im } ⊆ Bx , and its projection
Px back in the m-hypersimplex Hm is the convex hull of the vertices ei1 i2 ...im with
{i1 , i2 , . . . , im } ⊆ Bx . Given that p obeys the tropical Plücker relations with the
max convention, we know that Px is matroidal, so indeed Bx is the set of bases of
a matroid Mx .
Now, notice that x ∈ Pp0 if and only if Mx is a loopless matroid.
17
Consider the set Lm,n of all loopless matroids of rank m on the set [n]. Following
the spirit of our previous notation, denote the set of bases of a matroid M ∈ Lm,n
by B(M ).
For each M ∈ Lm,n , define the set Pp (M ) := {x ∈ Pp |xi1 + xi2 + · · · + xim =
pi1 i2 ...im whenever {i1 , i2 , . . . , im } ∈ B(M )}.
Clearly Pp (M ) is a polyhedron, which can be empty for some choices of M .
Moreover, notice that Pp (M ) is a (closed) proper face of Pp or a face of the boundary
complex ∂(Pp ) of Pp . Notice also that we can write
Pp0 =
[
Pp (M ).
M ∈Lm,n
Therefore, the set Pp0 is a polyhedral subcomplex of ∂(Pp ). It has been proved
in [21] that Pp0 is pure (m − 1)-dimensional, or equivalently (as we shall see), that
L(p) is pure m-dimensional.
We have:
• For x ∈ Rn , x ∈ L(p) if and only if Mx is loopless.
• For x ∈ ∂(Pp ), x ∈ Pp0 if and only if Mx is loopless.
Now, notice that for any x ∈ Rn we may find a real t for which x + te[n] ∈ Pp .
In particular, for every x ∈ Rn , there exists a unique t for which x + te[n] ∈ ∂(Pp ).
On the other hand, notice that L(p) and the matroid Mx are invariant under
translation by e[n] .
18
Therefore, we conlude that L(p) and Pd0 are the same object if considered in the
tropical projective space TPn−1 = Rn /(1, . . . , 1)R, and Pp0 arises from L(p) after a
natural choice of representative for each class. Under this choice, the maximum in
the definition of L(p) is precisely 0. This allows us to write Pp0 = L(p)/(1, . . . , 1)R.
Let S ⊆ Rn . The subset S is said to be tropically convex if
(max{a + x1 , b + y1 }, max{a + x2 , b + y2 }, . . . , max{a + xn , b + yn }) ∈ S
for all x, y ∈ S and a, b ∈ R.
Now, notice that the set
{p12...br...m(m+1) + xr |r ∈ [m + 1]}
is tropically convex. By the definition, the intersection of two tropically convex sets
is tropically convex. We then see that the tropical linear space L(p) is tropically
convex. It is proved in [7] that tropically convex sets are contractible. Therefore,
we see that L(p) is a contractible subspace of Rn . It deformation retracts onto Pp0 ,
which is then also contractible in Rn . On the other hand, for x ∈ ∂(Pp ), notice
that if Mx contains a loop, then x belongs to the relative interior of an unbounded
face of ∂(Pp ). To see this, consider the loop in the defining equations of the face.
Therefore, the complex Pp0 can be obtained from the polyhedron Pp by removing the
relative interior of a number of unbounded faces, in particular, the interior of Pp .
19
Finally, we want to emphasize the fact that Pp0 may contain unbounded faces.
In any case, it is common practice to consider only the complex of bounded faces of
Pp0 . As noted, this coincides with the complex of bounded faces of the polyhedron
Pp , and it will be the main object of our study.
Let the tight span Td of p be the complex of bounded faces of Pd .
It has been proved by Herrmann and Joswig in [11, Proposition 2.3] that Tp
coincides with the bounded faces of the dual complex of the subdivision of Hm
induced by p.
The tight span Tp is also known to be a contractible space, see for example
Dress [8] or Hirai [12].
Several studies about the combinatorial structure of this object have been done,
in particular as it relates to an important problem in evolutionary biology: to determine trees from their metric structure. There have been important results and
conjectures on the face enumeration of Tp [10], among which we find most relevant
to us the ones from Speyer [21].
[n]
Conjecture 2.1 (The f -vector Conjecture). Let p ∈ R( m ) satisfy the tropical
Plücker relations. Then, the f -vector of Tp satisfies the following bounds:
n − 2 − 2i n − i − 2
fi ≤
for all i with 0 ≤ i ≤ m − 1.
d−1−i
i
In particular, series-parallel tropical linear spaces realize the f -vector of the
f -vector conjecture. An example of these are the tree-tropical linear spaces of
20
Speyer [21]. These spaces can be regarded as approximations to the tropical linear
spaces of dissimilarity vectors, which we study in Chapter 4. On the other hand,
tropical linear spaces which are realizable have f -vectors which are bounded by the
f -vector conjecture [22].
We present the main idea of this Section in a separate Proposition.
[n]
Proposition 2.5 (Speyer, Herrmann, Joswig). Let p ∈ R( m ) satisfy the tropical
Plücker relations with the max convention. Then Pp0 = L(p)/(1, . . . , 1)R and Tp
coincides with the subcomplex of bounded faces of the dual complex to the subdivision
of Hm induced by p.
2.3 Results About Trees
Consider a tree T with m-dissimilarity vector d.
In this writing, we prove that d satisfies the tropical Plücker relations. We will
later provide a proof of this fact by obtaining d as the componentwise valuation of a
vector of classical Plücker coordinates of an m-dimensional vector space of Kn . By
the fundamental theorem of tropical algebraic geometry, this will moreover imply
that d ∈ Gm,n . The present writing is then in reality an investigation of the objects
that we have associated to the tropical Plücker relations, as obtained from d.
To begin, we point out that a connection between phylogenetic trees and tropical
geometry had been noted some time ago. That these two subjects are mathemat-
21
ically related can be traced back to Pachter and Speyer [17], Speyer and Sturmfels [20], and Ardila and Klivans [2]. The precise nature of this connection has been
the matter of some recent papers by Bocci and Cools [3] and Cools [6]. Later, we
will describe a concise relation between m-dissimilarity vectors and tropical Grassmannians Gm,n .
In order to present the classical motivating example, we develop some concepts.
Consider the polynomial ring K[Yij : {i, j} ∈ [n]
]. In this ring, consider the
2
ideal of Plücker relations, or the Plücker ideal I2,n :
I2,n = hYij Ykl − Yik Yjl + Yil Yjk : 1 ≤ i < j < k < l ≤ ni.
The generators of I2,n presented are called the quadric relations, or simply
quadrics. As remarked earlier, they form a Gröbner basis of I2,n . Even more,
it is known that the quadrics form a tropical basis of I2,n , by which it is meant
that
G2,n =
\
T ( trop(Yij Ykl − Yik Yjl + Yil Yjk )) .
1≤i<j<k<l≤n
[n]
We can restate this by saying that G2,n is the set of points y ∈ R( 2 ) for which
the maximum of the set {yij + ykl , yik + yjl , yil + yjk } is attained at least twice, for
all {i, j, k, l} ∈ [n]
.
4
[n]
Notice that, as promised, a vector of R( 2 ) lies in the tropical Grassmannian G2,n
if and only if it satisfies the tropical Plücker relations.
22
The main motivating example for studying these tropical Grassmannians in connection with trees comes from the following theorem, also known as the four point
condition theorem. It is a theorem of Buneman [5], widely applicable in evolutionary biology.
Theorem 2.6 (Pachter and Sturmfels [18]). The set of 2-dissimilarity vectors of
trees is equal to the tropical Grassmannian G2,n .
The theorem of Buneman is used as a tool for reconstructing shapes of trees from
their metrics. The extent of its applicability relies on the fact that we can project
generic points in R( 2 ) on the tropical Grassmannian G2,n . See Ardila [1].
[n]
One of the main open problems in computational biology is to device a way to
use the higher m-dissimilarity vectors to carry on the same purpose. It is commonly
known that such approach would allow more reliable statistical solutions of the tree
reconstruction problem.
From here sprouts the natural problem of characterizing m-dissimilarity vectors
of trees.
In the case of m = 2, there are other characterizations of dissimilarity vectors.
Among these, we present a result of Dress [9] which relates tropical linear spaces
and 2-dissimilarity vectors.
[n]
Theorem 2.7. A vector d ∈ R( 2 ) is a 2-dissimilarity vector if and only if Td is a
tree.
23
Describing the object Td for general m-dissimilarity vectors will be the main
goal of Chapter 4. See also Sturmfels and Yu for pictures of tight spans of generic
metrics [23].
Now, with the theorem of Buneman as motivation, the question had been posed
of whether or not the set of m-dissimilarity vectors of trees is contained in the
tropical Grassmannian Gm,n , for all values of m. This was initially asked by Lior
Pachter and David Speyer [17].
The major obstacle in trying to prove such result comes mainly from the fact
that no tropical basis is known for the Plücker ideals Im,n for m ≥ 3. Notice that the
knowledge of such bases would reduce the problem to a finite number of verifications
for each value of m. We have not found tropical bases for these ideals.
A different approach to understanding m-dissimilarity vectors for m ≥ 3 is to
find a relation between them and the 2-dissimilarity vectors, which are all well
understood.
Along this line of thought, a key result of Cools is the existence of a map that,
from the 2-dissimilarity vector of a tree T , computes its m-dissimilarity vector. We
provide a combinatorial proof of this fact.
Proposition 2.8. Consider a tree T and let m ≥ 3. Let Cm ⊆ Sm be the set of
cyclic permutations. For any σ ∈ Cm , define
dσ = d1σ(1) + dσ(1)σ2 (1) + dσ2 (1)σ3 (1) + dσ3 (1)σ4 (1) + · · · + dσm−1 (1)σm (1) .
24
Then
d12...m =
1
· min dσ .
2 σ∈Cm
Proof. Without loss of generality, assume that m = n. Consider an internal vertex
v of T . This vertex induces a partition on the set of leaves of T under which two
leaves i and j belong to the same class if and only if the minimal path from i to v
and the minimal path from j to v share a common edge. But then, each of the sums
dσ above can be seen to consider at least twice every edge adjacent to v. Thus, dσ
considers every edge of T at least twice.
Now, let σ ∈ Cn be such that dσ is minimal. Suppose that some edge e adjacent
to v is considered more than twice in dσ . Notice that e must then be internal, so
ω(e) > 0.
Let e be the class of leaves of T containing e in their minimal path to v. Then,
there exist a∗ , a1 , . . . , ai ∈ e and b∗ , b1 , . . . , bj ∈ ([n]\e) such that
σ = (· · · , a∗ , b1 , · · · , bj , a1 , · · · , ai , b∗ , · · · ).
Now, define
σ 0 = σ(a∗ , ai )(a∗ , bj )
= (· · · , a∗ , a1 , · · · , ai , b1 , · · · , bj , b∗ , · · · ) ∈ Cn .
25
But then dσ0 ≤ dσ − 2ω(e). This contradicts the minimality of dσ .
Therefore, every edge is considered exactly twice in dσ , which proves the result.
Remark. We expand on the argument presented. Abusing the notation of the proof,
for every leaf i of T , let div be the total weight of the minimal path from i to v.
Then:
da∗ b1 = da∗ v + db1 v
da1 bj = da1 v + dbj v
dai b∗ = dai v + db∗ v
dai b1 = dai v + db1 v .
From here we obtain:
da∗ b1 + da1 bj + dai b∗ − da∗ a1 + dai b1 + dbj b∗
= ((da∗ v + da1 v ) − da∗ a1 )
+ dbj v + db∗ v − dbj b∗
≥ 2ω(e) + 0 = 2ω(e).
From this result we could try to study m-dissimilarity vectors and the way in
which they induce a metric on T . An initial attempt would go along the lines of
26
trying to invert the presented map. One immediate application of the result is
presented now.
Corollary 2.9. Suppose that we have a tree T . Then, the following identity holds
for all {i, j, k, l, p} ∈ [n]
:
5
1
dij = (2dijk + 2dijl + 2dijp
3
− dipl − dikl − dikp
− djpl − djkl − djkp
+ 2dplk ).
Corollary 2.9 is a direct consequence of the identity
dijk =
1
(dij + dik + djk ) for all {i, j, k} ∈
2
[n]
3
.
In order to visualize the computation involved, it could be useful to think of the
following construction.
Let A = {i, j, k, l, p}. Consider the subdivision of a 3-dimensional tetrahedron
into four tetrahedra, induced by selecting a point i in the interior of the tetrahedron.
Then, label the remaining vertices j, k, l, p. Note that A3 is the set of 2-faces of
the subdivision and A2 is the set of edges. Let C` be the free Q-module with
A
basis `+1
, and define a Q-module homomorphism ∆ : C2 7→ C1 under which
27
∆[ijk] = 21 ([ij] + [ik] + [jk]), for all [ijk] ∈
A
3
.
Then, carry on the computation
∆ (2[ijk] + 2[ijl] + 2[ijp] − [ipl] − [ikl] − [ikp] − [jpl] − [jkl] − [jkp] + 2[plk])
to obtain 3[ij].
To finish, we note that for m ≥ 4 the situation is rather unclear.
28
Chapter 3
Dissimilarity Vectors are Contained in the
Tropical Grassmannian
The result presented in this chapter is based on two papers of Cools [6] and Bocci
and Cools [3], where the cases m = 3, m = 4 and m = 5 are handled.
We answer the question of Pachter and Speyer affirmatively for all m:
Theorem 3.1. Let T be a tree with m-dissimilarity vector d. Then, d ∈ Gm,n .
This result has more recently been proved by Manon [14], using very interesting
ideas from representation theory.
In order to prove Theorem 3.1, we need to introduce a new family of graphtheoretical-trees with special combinatorial properties.
We will refer to this new kind of tree using the letter U . In particular, we do
this to stress the difference with the conventions adopted in Chapter 1 for the letter
T.
To start, let U be a tree with n leaves labeled by the set [n], with n ≥ 1. Suppose
ω
also that U has a weight function E(U ) −
→ R≥0 . The function ω will extend to a
29
function from the subtrees of U to the nonnegative reals in the natural way. In the
latter case, we will speak of total weights of subtrees, following our convention. In
particular, notice that this time we are allowing internal edges of weight 0, and not
allowing external edges with negative weight. The tree U will be refered to as
being ultrametric if it further satisfies that:
• U is trivalent;
• U is rooted;
• U is `-equidistant. That is, the total weight of the minimal path from every
leaf to the root is a constant `;
• ω induces a metric on [n]. In particular, it separates points.
Note that the fourth condition directly implies ` > 0.
On the other hand, there is also an independent notion of an ultrametric.
A metric space S with distance function d : S×S 7→ R≥0 is called an ultrametric
space if the following inequality holds for all x, y, z ∈ S:
dxz ≤ max{dxy , dyz }.
It is a well known fact that finite ultrametric spaces are realized by ultrametric
trees, see for example [4, Lemma 11.1].
30
Note that the condition on ultrametric spaces S is equivalent to saying that for
all x, y, z ∈ S, the maximum of the set {dxy , dxz , dyz } is attained at least twice.
Now, suppose that we have a tree T with 2-dissimilarity vector d. From here,
[n]
define a vector d0 ∈ R( 2 ) by:
d0ij = 2t∗ + dij − din − djn for all different i, j ∈ [n].
Let t∗ > 0 be sufficiently large, so that d0ij > 0 for all different i, j. Observe that
d0in = 2t∗ for all i ∈ [n − 1]. Now, pick {i, j, k} ∈ [n−1]
and consider the set
3
{dij − din − djn , dik − din − dkn , djk − djn − dkn }.
We want to compare the members of this set. Comparing them pairwise, we
obtain three new sets to study:
{dij + dkn , dik + djn }
{dij + dkn , din + djk }
{dik + djn , din + djk }.
Note that these sets involve only three numbers. By the theorem of Buneman, one
of these three sets contains the same number twice, and that number is greater than
or equal to the third one. This translates back to concluding that the maximum of
the set
{dij − din − djn , dik − din − dkn , djk − djn − dkn }
31
is attained at least twice. Therefore, d0 |([n−1]) induces an ultrametric on the set
2
[n − 1]. Hence, d0 |([n−1]) is realized by an ultrametric tree.
2
Following this ingenious idea, Cools shows somewhat technically that in order
to prove Theorem 3.1, it suffices to prove the following variation:
Theorem 3.2. Let U be an ultrametric tree with m-dissimilarity vector d. Then,
d ∈ Gm,n .
This is the theorem that we will prove. First, we will need to introduce some
terminology.
3.1 Column Reductions
Let n ≥ 3. Suppose that we are given integers 1 ≤ a, b ≤ n with a 6= b and let
ca,b be the operator acting on Puiseux matrices for which, for any n × n matrix M ,
ca,b (M ) is the matrix obtained from M by subtracting column b to column a. We
know that ca,b preserves the determinant, i.e. det (ca,b (M )) = det(M ). For l ≥ 1, let
(cal ,bl ◦ · · · ◦ ca2 ,b2 ◦ ca1 ,b1 ) (M ) be the matrix obtained from M by first subtracting
column b1 to column a1 , then subtracting column b2 to column a2 , and so on up to
subtracting column bl to column al . Call this matrix a column reduction of M if
the following conditions are met:
• 1 ≤ a1 , . . . , al , b1 , . . . , bl ≤ n;
• the numbers a1 , a2 , . . . , al are pairwise different;
32
• whenever 1 ≤ k ≤ l, the number bk is different from a1 , . . . , ak .
For simplicity, we will accept M as a column reduction of itself.
3.2 Main Result
We are now ready to prove Theorem 3.1. We do this by proving Theorem 3.2 for
m ≥ 3.
Proposition 3.3. Suppose that 3 ≤ m ≤ n. Let U be an ultrametric tree with
m-dissimilarity vector d, all of whose edges have rational weight.
For each edge e of U , denote by h(e) the total weight of the minimal path from
the top node of e to any leaf below e.
Also, let a1 (e), . . . , an−2 (e) be n − 2 generic complex numbers associated to e.
For all i ∈ [n − 2] and j ∈ [n], let tij be the sum of the monomials ai (e)th(e) ,
where e runs over all edges in the minimal path from leaf j to r.
Notice that tij ∈ K.
33
Then, define a matrix:

1
 1

 t
t12
 11


 (t11 )2 (t12 )2
M =

 t21
t22

 .
..
 ..
.


t(n−2)1 t(n−2)2
Finally, for all ρ ∈
[n]
m

...
...
...
...
..
.
...
1


t1n 



(t1n )2 
.

t2n 

.. 
. 


t(n−2)n
, let mρ be the m × m upper minor of M coming from
the columns ρ.
Then, val(mρ ) = dρ .
Proof. First, consider the following matrix for some p1 , p2 , . . . , pn−2 ∈ K:

1
1


 t +p
t12 + p1
11
1



 (t11 + p1 )2
(t12 + p1 )2

N =
 t21 + p2
t22 + p2


..
..

.
.


t(n−2)1 + pn−2 t(n−2)2 + pn−2

...
1
...
t1n + p1
...
(t1n + p1 )2
...
..
.
t2n + p2
..
.
. . . t(n−2)n + pn−2







.







Then, notice that the situation is completely analogous to proving that val (det(N )) =
ω(U ). However, by first expanding the quadric terms of the second row, then sub-
34
tracting multiples of the first row to all rows to cancel the p-terms, and then using
the second row to cancel the appropriate remaining terms in the third row, we see
that det(N ) = det(M ).
Therefore, there is no loss of generality if we prove that val (det(M )) = ω(U ).
This is the plan.
As U is trivalent with n leaves, we know that U has n − 2 internal nodes of
degree three, one node (the root) of degree two and 2(n − 1) edges.
Let ≤U be the tree order of U with respect to r, that is, the order on the set
of nodes of U by which v ≤U w if and only if v lies in the path from r to w in U .
Let v1 , v2 , . . . , vn−1 be the n − 1 internal nodes of U , numbered in such a way that
if vi ≤U vj , then j ≤ i. We must have vn−1 = r.
Consider an injective function α : vi 7→ ai from the set of internal nodes to the
leaves of U , so that vi ≤U ai for all i with 1 ≤ i ≤ n − 1. Now, for each of these
values of i, let bi be the unique leaf such that bi 6= aj for all j with 1 ≤ j ≤ i, and
such that vi ≤U bi .
To show the existence of α, we construct it succesively starting with α(v1 ), then
α(v2 ) and then continuing up until we define α(vn−1 ). Suppose that we have already
defined α(v1 ), . . . , α(vi−1 ) for some i < n − 1. Consider the maximal subtree Ui of
35
U whose root is vi , i.e. Ui is the subtree below vi . If this tree has m leaves, then
it has m − 1 internal nodes, including vi itself. So far, we have not defined α for
nodes between r and vi , but we have defined it for all internal nodes of Ui different
from vi . Therefore, there are exactly m − 2 leaves of the tree Ui which have been
assigned to some of v1 , . . . , vi−1 under α, so there are 2 leaves which we can assign
to vi : α(vi ) can be either one of them. Incidentally, this also gives us the existence
and uniqueness of the respective bi .
Now, we want to establish the equality
Pn−1
i=1
h(vi ) = ω(U ) − `. This equality is
clearly true when U has 2 or 3 leaves, so that n = 2 or n = 3. Let now n ≥ 4 and
suppose that we have proved the result for all trees with i leaves, with i < n. Recall
n is being taken as the number of leaves in U , which is rooted `-equidistant with
root r = vn−1 . We know that the equality holds for each of the subtrees U1 , . . . ,
Un−2 below v1 , . . . , vn−2 , respectively. Let Un−2 be `n−2 -equidistant and let Un−3 be
`n−3 -equidistant. There are two cases to distinguish. If vn−2 <U vi for all i < n − 2,
P
then n−2
i=1 h(vi ) = (ω(U ) − ` − (` − `n−2 )) − `n−2 = ω(U ) − 2` by induction, so
Pn−1
i=1 h(vi ) = ω(U ) − `. Otherwise, suppose that vn−2 and vn−3 are incomparable
in <U . Then Un−2 and Un−3 are disjoint graphs and we have

X
vi ∈V(Un−2 )

h(vi ) = ω(U ) − 

X
vj ∈V(Un−3 )

h(vj ) + `n−3  − (` − `n−3 ) − (` − `n−2 )−`n−2
36
by induction. Reordering we get:
X
X
h(vi ) +
vi ∈V(Un−2 )
h(vj ) = ω(U ) − 2`,
vj ∈V(Un−3 )
so if we add h(vn−1 ) = ` to both sides, we get our result.
Now, consider the column reduction M ∗ = can−1 ,bn−1 ◦ · · · ◦ ca2 ,b2 ◦ ca1 ,b1 (M )
Q
∗
with
of M . We claim that the valuation of all the nonzero monomials ni=1 Mi,σ(i)
σ ∈ Sn in the sum
det(M ∗ ) =
X
σ∈Sn
is precisely
Pn−1
i=1
sgn(σ)
n
Y
!
∗
Mi,σ(i)
i=1
h(vi ) + ` = ω(U ). To see this, notice that for all i, 1 ≤ i ≤ n − 1,
we have:
∗
= 0;
• M1a
i
∗
is ` + h(vi );
• the valuation of M3a
i
∗
• the valuation of Mja
is h(vi ) if j 6= 1 and j 6= 3;
i
• the only nonzero term in the first row of M ∗ is the 1 in column bn−1 .
Because of our generic choice of coefficients, we can find some monomial term in
the sum det(M ∗ ) with valuation ω(U ) that does not get cancelled.
37
r = v9
5
(p)
5
(q)
9
v7
v8
2
(g)
3
(f )
v5
v1
1
1
(a) (b)
1
2
(c)
2
1
v6 (z)
3
2
(x)
1
(h)
v2
1
1
(d) (e)
4
2
(y)
v3
1
1
(r) (s)
5
6
4
(w)
v4
1
1
(u) (v)
7
8
9
10
Figure 3.1:
A rooted 10-tree.
The injective function α
:=
{(v1 , 1), (v2 , 4), (v3 , 6), (v
,
8),
(v
,
3),
(v
,
7),
(v
,
2),
(v
,
9),
(v
,
5)}
is
depicted,
4
5
6
7
8
9
P
as well as the equality 9i=1 h(vi ) = 35 − 9.
Example 3.1. Consider the 9-equidistant 10-tree of Figure 3.1 with total weight
35. The second row of the matrix M associated to this tree is the following vector
with generic complex coefficients:
[at1 + f t4 + pt9
,bt1 + f t4 + pt9
,ct2 + gt4 + pt9
,
dt1 + ht2 + gt4 + pt9
,et1 + ht2 + gt4 + pt9
,rt1 + xt3 + zt4 + qt9
,
st1 + xt3 + zt4 + qt9
,ut1 + yt3 + zt4 + qt9
,vt1 + yt3 + zt4 + qt9
,
wt4 + qt9 ]
38
Using the operator (c5,10 ◦ c9,10 ◦ c2,5 ◦ c7,9 ◦ c3,5 ◦ c8,9 ◦ c6,7 ◦ c4,5 ◦ c1,2 ) suggested by
the figure we obtain the column reduction M ∗ whose second row is the vector:
[(a − b)t1
,
(b − e)t1 − ht2 + (f − g)t4
,
− et + (c − h)t2
,
(d − e)t1
,
et1 + ht2 + (g − w)t4 + (p − q)t9
,
(r − s)t1
,
(s − v)t1 + (x − y)t3
,
(u − v)t1
,
vt1 + yt3 + (z − w)t4
,
wt4 + qt9 ]
It has valuation vector:
(1, 4, 2, 1, 9, 1, 3, 1, 4, 9) =
(h(v1 ), h(v7 ), h(v5 ), h(v2 ), h(v9 ), h(v3 ), h(v6 ), h(v4 ), h(v8 ))
,
where v1 , v7 , v5 , v2 , v9 , v3 , v6 , v4 , v8 are the preimages of 1, 2, 3, 4, 5, 6, 7, 8, 9 under α,
P
respectively in that order. Also notice that 9i=1 h(vi ) = 35 − 9.
We have shown that the m-dissimilarity vector of a tree T gives a point in the
tropical Grassmannian Gm,n , and therefore satisfies the tropical Plücker relations.
Thus, it gives rise to a tropical linear space. We now study this tropical linear space,
in its tight span version.
39
Chapter 4
The Tropical Linear Space of a
Dissimilarity Vector
The main object of this chapter is a tree T with m-dissimilarity vector d. To keep
our discussion interesting, from now on and for the remaining part of this writing,
d will never be a 1-dissimilarity vector or an n-dissimilarity vector. The purpose of
the chapter is to describe, from both a geometrical and combinatorial point of view,
the space Td .
During the whole chapter, we will sometimes use the symbol ⊆st to denote the
relation of being a subtree of. We will also use (st to denote the relation of being
a proper subtree of. We consciously use these symbols sporadically, combining
them with the use of the words “is a subtree of”. The motivation of the use of this
symbol will be emphasizing the fact that we are comparing trees.
Consider the tree in(T ) ⊆st T that consists of the internal edges and vertices of
T . If T does not contain internal edges or vertices, as a technical convenience we let
in(T ) = Λ, where Λ is the abstract tree consisting of an empty set of vertices and
edges, and which we consider to be a proper subtree of every other tree. We call
40
in(T ) the internal tree of T .
If S ⊆st T , let T /v S be the tree obtained from T by collapsing the subtree S to
a single vertex v.
4.1 Vertices of Td and Their Matroids
Our first and main result on the combinatorial structure of the tight span Td of d
is a description of its vertices.
We prove that the set of vertices of Td is in bijection with the set of subtrees of
in(T ) that satisfy certain conditions.
To state these conditions, let S ⊆st in(T ). Let S be the maximal subtree of T
such that in(S) = S. Notice that S is the subtree of T consisting of all edges and
vertices of T that are in S or adjacent to S. By definition S (st S. We call S the
extended tree of S. It will also be convenient to refer to the set L(S) of leaves of
S as the set of extended leaves of S.
Computing degrees in T , the number of extended leaves |L(S)| of S is
X
v∈V(S)
deg(v) − 2(|V(S)| − 1).
41
The number of vertices of S is
X
deg(v) − 2(|V(S)| − 1) + |V(S)|.
v∈V(S)
We will say that S is an m-good subtree of T if S ⊆st in(T ), |L(S)| ≤ m − 1
and |L(S)| ≥ m + 1.
Using these definitions, we are now ready to state the main result of this section.
Theorem 4.1. Let T be a tree with m-dissimilarity vector d. The set of vertices of
the tight span Td is in bijection with the set of m-good subtrees of T .
In the case where T is a trivalent tree, that is, when each internal edge of T
is of constant degree 3, we can enunciate our result in a more compact way. In this
case deg(v) = 3 for all v ∈ V(S), so the number of extended leaves of S becomes
X
deg(v) − 2(|V(S)| − 1) = 3|V(S)| − 2(|V(S)| − 1) = |V(S)| + 2.
v∈V(S)
Also, the number of vertices of S becomes 2|V(S)| + 2 = (|V(S)| + 2) + |V(S)|.
Therefore, the main result for this case can be stated as follows.
Theorem 4.2. Suppose that T is a trivalent tree with m-dissimilarity vector d. The
set of vertices of the tight span Td is in bijection with the set of trees S ⊆st in(T )
such that |L(S)| ≤ m − 1 ≤ |V(S)|.
To facilitate the exposition of the proof, we introduce some additional notation.
42
Let S ⊆st in(T ). For each leaf i ∈ [n] of T , there is a unique minimal path from i
to S made up solely of edges from E(T )\E(S). Let ω(i, S) be the sum of the weights
of all edges in this minimal path. Notice that ω(i, S) coincides with the intuitive
notion of distance from the leaf i to the subtree S.
From now on, assume that S is an m-good subtree of T as in Theorem 4.1.
For all i ∈ [n], define the entries of x ∈ Rn in the following way:
xi = ω(i, S) +
ω(S)
.
m
Proposition 4.3. The point x ∈ Rn just defined is a vertex of Td .
Proof. It has already been noted that our statement is equivalent to proving that
x is a vertex of the polyhedron Pd . This is what we do. The proof is divided into
several steps.
Notice that for each m-set {i1 , i2 , . . . , im } ⊆ [n] we have:
xi1 + xi2 + · · · + xim = ω(S) + ω(i1 , S) + ω(i2 , S) + · · · + ω(im , S).
Let Ti1 i2 ...im be the subtree of T spanned by the leaves i1 , i2 , . . . , im .
Step I: We have xi1 + xi2 + · · · + xim ≥ di1 i2 ...im with equality if and only if S ⊆st
Ti1 i2 ...im and Ti1 i2 ...im /v S is a v-path-star tree.
To avoid the excessive use of subscripts, let us assume without loss of generality
that i1 = 1, i2 = 2, . . . , im = m. Let Tm = T12...m .
43
First, suppose that Tm does not intersect S at an edge or a vertex. Notice
that ω(Tm ) = d12...m by definition. There exists a unique path P = (vS , vm )
with vS ∈ V(S) and vm ∈ V(Tm ) which consists purely of edges from the set
E(T )\ (E(S) ∪ E(Tm )). This is the minimal path connecting S and Tm . Notice
that this minimal path is made up of internal edges of T , which therefore have
positive weight. As a consequence, ω(P ) > 0. Also, notice that
ω(1, P ) + ω(2, P ) + · · · + ω(m, P ) = ω(1, vm ) + ω(2, vm ) + · · · + ω(m, vm )
≥ d12...m .
However, ω(i, S) = ω(i, P ) + ω(P ) for all i ∈ [m], so
ω(1, S) + ω(2, S) + · · · + ω(m, S) > d12...m .
This implies x1 + x2 + · · · + xm > d12...m .
Now, suppose that S and Tm do intersect at some tree S ∩ Tm . For ease of
notation, let:
Te = T /v (S ∩ Tm ),
Se = S/v (S ∩ Tm ) ⊆st Te and
Tem = Tm /v (S ∩ Tm ) ⊆st Te.
44
Notice that Se is precisely v if S ⊆st Tm .
We can compute some of the new weights:
e = ω(S) − ω(S ∩ Tm ),
ω(S)
ω(Tem ) = ω(Tm ) − ω(S ∩ Tm ) = d12...m − ω(S ∩ Tm ).
e and ω(i, v) as computed in Te.
Now, for all i ∈ [m], define the values of ω(i, S)
We can see that
ω(S) + ω(1, S) + ω(2, S) + · · · + ω(m, S)
e + ω(S ∩ Tm ) + ω(1, S)
e + ω(2, S)
e + · · · + ω(m, S)
e
= ω(S)
e + ω(S ∩ Tm ) + ω(1, v) + ω(2, v) + · · · + ω(m, v).
= ω(S)
But looking at our tree Te, notice that
ω(1, v) + ω(2, v) + · · · + ω(m, v) ≥ ω(Tem ) = d12...m − ω(S ∩ Tm ).
Substituting this in our previous equality, we get
e + ω(S ∩ Tm ) + ω(1, v) + ω(2, v) + · · · + ω(m, v) ≥ ω(S)
e + d12...m .
ω(S)
45
This gives us finally the following inequality:
e = ω(S ∩ Tm ) + ω(1, v) + ω(2, v) + · · · + ω(m, v)
x1 + x2 + · · · + xm − ω(S)
≥ d12...m .
e = 0, that is,
In particular, notice that we can only have equality when ω(S)
whenever Se is a vertex. In that case, we know that Se can only be v, which means
that S = S ∩ Tm . Therefore, a necessary condition for equality is that S ⊆st Tm .
At this point, we have already proved that x ∈ Pd . It remains to study the cases
where we have equality in the defining inequalities of Pd .
We know that S ⊆st Tm . But now, in order to attain the equality in x1 + x2 +
· · · + xm ≥ d12...m , we require that
ω(1, v) + ω(2, v) + · · · + ω(m, v) = ω(Tem ).
This is also a sufficient condition. In order for it to hold, we need that in Te,
the minimal path from i to v and the minimal path from j to v to intersect at v
and only at v, for all different i, j ∈ [m]. That is tantamount to saying that Tem is a
v-path-star tree.
Therefore, x1 + x2 + · · · + xm = d12...m if and only if S ⊆st Tm and Tem is a
46
v-path-star tree. This settles the first step of the proof.
Let us now describe these conditions completely in terms of S. We will first need
to introduce some additional concepts. This is when all the previously developed
notation will begin to come in handy.
Consider the subtree S. It has a number of leaves |L(S)| ≤ m − 1. Pick a leaf
l ∈ L(S). There are some leaves of T whose minimal path to S meets S precisely
at l. These are all the leaves i ∈ [n] such that ω(i, S) = ω(i, l). Recall the extended
tree S of S, which has at least m + 1 leaves, that is |L(S)| ≥ m + 1. This motivates
introducing the following sets.
For all l ∈ L(S), let Hl be the set of leaves of T whose minimal path to S meets
l. Analogously, for all l ∈ L(S) let Rl be the set of leaves of T whose minimal
path to S meets l. Notice that the collection of sets Rl define a partition of [n] as
l ranges over all leaves of S. Let R be the set of classes of this partition, so that
R = {Rl }l∈L(S) . Define H to be the analogous object for S, H = {Hl }l∈L(S) . It is
worth noting that H is also a collection of pairwise disjoint subsets of [n], but not
necessarily a partition of [n].
Also, note that each set in the collection H can be written as a disjoint union of at
least two sets from the collection R, R is a refinement of H. Furthermore, we can say
P
something about the size of these collections: |R| = v∈V(S) deg(v)−2(|V(S)|−1) ≥
47
m + 1 and |H| = |L(S)| ≤ m − 1.
Step II: We have xi1 +xi2 +· · ·+xim = di1 i2 ...im if and only if |{i1 , i2 , . . . , im }∩Hl | ≥ 1
for all l ∈ L(S) and |{i1 , i2 , . . . , im } ∩ Rl | ≤ 1 for all l ∈ L(S).
In order for S to be a subtree of Ti1 i2 ...im , a necessary and sufficient condition
is that |{i1 , i2 , . . . , im } ∩ Hl | ≥ 1 for all l ∈ L(S). On the other hand, in order for
Ti1 i2 ...im /v (S ∩ Ti1 i2 ...im ) to be a v-path-star tree, a necessary and sufficient condition
is that |{i1 , i2 , . . . , im } ∩ Rl | ≤ 1 for all l ∈ L(S). This settles the second step of our
proof.
For each leaf i ∈ [n] of T , let i be the leaf of S such that i ∈ Ri . Notice that the
leaf i also represents the class of i in the partition of [n] induced by S, so we follow
the usual notation for partitions.
Step III: For every pair of different leaves i, j ∈ [n] = L(T ), there exists an (m − 1)set {k1 , k2 , . . . , km−1 } ⊆ L(T )\{i, j} such that xk1 +xk2 +· · ·+xkm−1 +xi = dk1 k2 ...km−1 i
and xk1 + xk2 + · · · + xkm−1 + xj = dk1 k2 ...km−1 j .
Using a relabeling of the leaves of T when necessary, let us prove without loss of
generality that for a pair of different leaves i, j ∈ L(T )\[m − 1] we can construct a
set [m − 1] ⊆ L(T ) such that x1 + x2 + · · · + xm−1 + xi = d12...(m−1)i and x1 + x2 +
· · · + xm−1 + xj = d12...(m−1)j .
We know that i ∈ Ri and j ∈ Rj and it might be the case that i = j. We do
this in a case by case fashion.
48
If i, j ∈ Hl for some l ∈ L(S), choose one leaf of T from each of the sets in
the collection H\Hl to form a set {1, 2, . . . , |L(S)| − 1} of leaves of T . Clearly
i, j 6∈ (R1 ∪ R2 ∪ · · · ∪ R|L(S)|−1 ). As |R\{Ri , Rj }| ≥ m − 1 and |L(S)| − 1 < m − 1,
we complete our set of leaves by picking a total of (m − 1) − (|L(S)| − 1) new
leaves from the sets in the collection R\{R1 , R2 , . . . , R|L(S)|−1 , Ri , Rj }, no two of
which come from the same member of the collection. We thus form a set of leaves
{1, 2, . . . , m − 1} = [m − 1] of T which satisfies our conditions:
x1 + x2 + · · · + xm−1 + xi = d12...(m−1)i and
x1 + x2 + · · · + xm−1 + xj = d12...(m−1)j .
If there does not exist some Hl with l ∈ L(S) such that i, j ∈ Hl , then pick one
leaf from each member of the collection of sets H to form a set {1, 2, . . . , |L(S)|} of
leaves of T , but do this in such way that [|L(S)|]∩Ri = ∅ and [|L(S)|]∩Rj = ∅. This
is possible because we can write each member of H as a disjoint union of at least
two members of R. Notice also that we are now using the fact that |L(S)| ≤ m − 1.
Otherwise, the output of our last construction would be a set of more than m − 1
leaves of T . As |R\{Ri , Rj }| ≥ m − 1, if |L(S)| < m − 1, we can pick one leaf from
each of any m − 1 − |L(S)| different sets of the collection R\{R1 , R2 , . . . , R|L(S)| } to
49
complete our set [m − 1] of leaves of T . Again
x1 + x2 + · · · + xm−1 + xi = d12...(m−1)i and
x1 + x2 + · · · + xm−1 + xj = d12...(m−1)j .
This settles the third step of the proof.
Step IV: The defining equations of Pd at which equality is attained by x are a rank-n
system of equations which define a unique point in Rn . Therefore, x is a vertex of
Pd .
By the previous step, we know that the defining equations of Pd at which equality
is attained let us determine the value of xi − xj for all different i, j ∈ [n]. They also
let us determine xi1 + xi2 + · · · + xim for some set of leaves {i1 , i2 , . . . , im } ⊆ L(T ).
Thus, they are a rank-n system of equations.
The natural question to ask now is whether two different m-good subtrees S and
S 0 of T give rise to different vertices of Td . This is indeed the case.
Proposition 4.4. Let S and S 0 be m-good subtrees of T . Consider the vertices x
50
and x0 of Td given by the equalities, for all i ∈ [n]:
ω(S)
,
m
ω(S 0 )
x0i = ω(i, S 0 ) +
.
m
xi = ω(i, S) +
Then, if S and S 0 are different trees then x and x0 are different vertices of Td .
Proof. To begin, assume that S and S 0 are different trees and suppose that ω(S 0 ) ≤
ω(S). Let x, x0 ∈ Rn be given, for all i ∈ [n], by
ω(S)
,
m
ω(S 0 )
x0i = ω(i, S 0 ) +
.
m
xi = ω(i, S) +
Following the proof of Proposition 4.3, let H and R be the collections of sets of
leaves of T associated to S, and let H0 and R0 be the collections of sets of leaves of
T associated to S 0 . Take a leaf l of S that is not a vertex of S 0 and such that the
minimal path from every element of Hl ∈ H to l does not meet S 0 . Using the ideas
of the proof, construct a set of leaves [m] of T (relabeling when necessary) such that
|[m] ∩ Hl | ≥ 2 and such that
x1 + x2 + · · · + xm = d12...m .
This construction is possible because S has at most m − 1 leaves and at least
51
m + 1 extended leaves.
Now, take two different leaves i, j ∈ ([m] ∩ Hl ) and consider the set Ri0 ∈ R0 .
Notice that i and j are both in Ri0 , so |[m] ∩ Ri0 | > 1. Therefore,
x01 + x02 + · · · + x0m > d12...m .
Combining these results we see that x 6= x0 .
The final step of our study needs to be concerned with making sure that all these
vertices of Td that we have exposed are actually all the vertices of Td .
In order to prove this, we need to change our perspective on tight spans. In
particular, Td can be described as the set of bounded faces of the dual complex of
the subdivision of the m-hypersimplex Hm induced by d. Under this perspective,
vertices of Td correspond to (n − 1)-dimensional faces of this subdivision. These
are the maximal faces of the subdivision because Hm lies in the affine space of Rn
consisting of points x such that x1 + x2 + x3 + · · · + xn = m.
Let us begin by understanding the (n − 1)-dimensional face associated to the
known vertices of Td that we have described. Formally, let S be an m-good subtree
of T . Let H and R be the collections of sets of leaves of T associated to S, as in
the proof of Proposition 4.3. Define x ∈ Rn as usual by the following equalities, for
all i ∈ [n]:
xi = ω(i, S) +
ω(S)
.
m
52
Recall that Bx is the collection of m-sets {i1 , i2 , . . . , im } ⊆ [n] for which xi1 +
xi2 + · · · + xim = di1 i2 ...im . The face Px of the subdivision Dd that corresponds to x
is the convex hull of all points ei1 i2 ...im with {i1 , i2 , . . . , im } ∈ Bx . Now, to describe
Bx we can use our knowledge of H and R. Indeed, notice that {i1 , i2 , . . . , im } ∈ Bx
if and only if
|{i1 , i2 , . . . , im } ∩ Hl | ≥ 1 for all l ∈ L(S),
|{i1 , i2 , . . . , im } ∩ Rl | ≤ 1 for all l ∈ L(S).
We would like to describe Px with inequalities.
Proposition 4.5. Let
Px0 ={y ∈ Hm |
X
yj ≥ 1 for all l ∈ L(S) and
X
yj ≤ 1 for all l ∈ L(S)}.
j∈Rl
j∈Hl
Then, Px = Px0 .
Proof. We study the vertices of Px0 . Let y ∈ Px0 and suppose that 0 < y1 < 1.
If there exists an element of R1 \1, say 2, such that 0 < y2 < 1, then y +
(δ, −δ, 0, 0, . . . , 0) is also a point of Px0 for any δ with |δ| sufficiently small. Thus, y
cannot be a vertex of Px0 .
Otherwise, we have 0 <
P
j∈R1
= y1 < 1. Because y ∈ Hm , there must exist
one element of [n]\1, say 2, such that 0 < y2 < 1. It better be the case that
53
0<
P
j∈R2
= y2 < 1 because otherwise, per the case already handled we would be
done. If there exists some l ∈ L(S) such that 1, 2 ∈ Hl , or if neither one among 1 and
2 is contained in some Hl with l ∈ L(S), then y + (δ, −δ, 0, 0, . . . , 0) is also a point
of Px0 for any δ with |δ| sufficiently small. If only one of them, say 1, is contained in
P
some Hl for l ∈ L(S) and if j∈Hl yj > 1, then again y + (δ, −δ, 0, 0, . . . , 0) is also a
P
point of Px0 for any δ with |δ| sufficiently small. If however j∈Hl yj = 1 then notice
that we must have |Hl | ≥ 2, so we can assume that 2 ∈ Hl , which takes us back to
a case already studied.
Therefore, all vertices of Px0 are 0-1 vectors. But notice that a 0-1 vector belongs
to Px0 if and only if it is equal to ei1 i2 ...im for some {i1 , i2 , . . . , im } ∈ Bx . As Px ⊆ Px0
clearly, we conclude Px = Px0 .
We use Proposition 4.5 to prove the following result:
Proposition 4.6. Let y be a generic point in the interior of Hm . Then, there exists
an m-good subtree S of T such that the point x ∈ Rn defined by
xi = ω(i, S) +
ω(S)
m
for all i ∈ [n], satisfies y ∈ Px .
Remark. Our generic choice of y means precisely that y ∈ Hm does not satisfy the
equalities that we now describe. Let v be a vertex of T , or v ∈ V(T ) in our usual
notation. The vertex v defines a partition Πv on the set of leaves [n] of T . Under
54
this partition, two different leaves i, j ∈ [n] belong to the same class if and only if
the minimal paths from i to v and from j to v contain a common edge adjacent to
v. The number of classes in this partition is equal to the degree of v if v ∈ V (in(T ))
and equal to 2 if v ∈ L(T ).
For each a ∈ Πv , let σa =
P
i∈a
yi . By the statement that y is generic, we require
specifically that σa 6= m − 1 for all v ∈ V(T ) and for all classes a ∈ Πv .
Proof. We first introduce a subset Vy of the set of internal vertices of T :
Vy = {v ∈ V (in(T )) |m − σa > 1 for all a ∈ Πv }.
We prove that Vy 6= ∅. To begin, notice that if we take a vertex v ∈ V (in(T )) \Vy
and there exist two different classes a, b ∈ Πv such that σa > m − 1 and σb > m − 1,
then m ≥ σa +σb > 2m−2 or m < 2. This contradicts one of the earliest assumptions
made in this chapter. Thus, for each v ∈ V (in(T )) \Vy there exists a unique a ∈ Πv
such that σa > m − 1.
If Vy = ∅, among all pairs (v, σa ) with v ∈ V (in(T )) and a ∈ Πv such that
σa > m − 1, take the one with σa minimal. The class a corresponds to a unique
edge e of T adjacent to v. Let u be the other vertex adjacent to the edge e. If
u ∈ V (in(T )) so that deg(u) ≥ 3, then by the minimality of σa and because y is an
interior point of Hm (in particular, y > 0 componentwise), it can only be the case
that σb > m − 1 where b ∈ Πu is the class corresponding to the edge e. However, in
55
this later case we obtain m = σa + σb > 2m − 2 or m < 2, a contradiction. Finally,
if u 6∈ V (in(T )), then u has to be a leaf of T . But then 1 > yu = σa > m − 1 or
m < 2, the first inequality coming from the fact that y ∈ Hm .
Therefore, Vy 6= ∅.
It is not difficult to check that Vy defines a tree S ⊆st in(T ). More precisely, if
we have three different vertices u, v ∈ Vy and w ∈ V (in(T )) such that w lies in the
minimal path from u to v, then w ∈ Vy . For example, if σc > m − 1 for some c ∈ Πw ,
then that implies that either σa > m − 1 or σb > m − 1 holds for some a ∈ Πv or
some b ∈ Πu .
Now, S must have at most m − 1 leaves. Otherwise, for every leaf l of S, let
al ∈ Πl correspond to the edge of S adjacent to l. Then:
m≥
X
l∈L(S)
(m − σal ) >
X
1 = |L(S)| ≥ m.
l∈L(S)
Also, S must have at least m + 1 extended leaves. To see this, notice that the
set of leaves L(S) of S (the set of extended leaves of S) and the set of vertices
V(S) = Vy of S are disjoint. For each leaf l ∈ L(S), let al ∈ Πl correspond to
the edge e adjacent to S. It has to be the case that σal > m − 1 or 1 > m − σal .
Otherwise, if we let v be the vertex of Vy adjacent to e and a ∈ Πv be the class
corresponding to the edge e, then we would have σa > m − 1, contradicting the fact
that v ∈ Vy . But then, as the set of leaves L(S) of S induces a partition of [n], we
56
see that:
X
m=
l∈L(S)
(m − σal ) <
X
1 = |L(S)| ≤ m.
l∈L(S)
To sum up these arguments, if l ∈ L(S) and a ∈ Πl is the class corresponding
to the edge of S adjacent to l, then m − σa > 1. On the other hand, if l ∈ L(S)
and a ∈ Πl is the class corresponding to the edge adjacent to both S and l, then
m − σa < 1. Therefore, if we let x be defined as in the statement of the theorem,
then y ∈ Px per the description of Px in terms of inequalities.
4.2 Higher Dimensional Faces of Td and Their Matroids
Let us now understand the remaining faces of the tight span Td associated to T .
The idea is to study the internal faces of Px for a vertex x of Td .
Suppose that we have a subtree S ⊆st in(T ), where |L(S)| ≤ m − 1. Suppose
also that there exists some A ⊆ L(S) with 0 ≤ |A| = i ≤ m − 1, and such that the
number of extended leaves of S that are not adjacent to an element of A is at least
m + 1 − i. Computing degrees in T , this final assumption is equivalent to having:
X
deg(v) − 2(|V(S)| − 1) + 2i ≥ m + 1.
v∈V(S)\A
We call the pair (S, A) an (m, i)-good pair whenever each of these conditions
is satisfied.
57
Notice that taking i = 0, the left hand side of our inequality becomes precisely
|L(S)|, the number of extended leaves of S. Thus, we regard the conditions as a
generalization of the conditions imposed on S when we studied the vertices of Td .
In fact, if we let (S, A) be a (m, i)-good pair, then, as deg(v) ≥ 3 for all v ∈ A, we
get:
X
v∈V(S)
X
deg(v) − 2(|V(S)| − 1) ≥
deg(v) + 3i − 2(|V(S)| − 1) ≥ m + 1 + i.
v∈V(S)\A
Therefore, if (S, A) is an (m, i)-good pair, then (S, ∅) is also a (m, 0)-good pair,
which is equivalent to saying that S is an m-good subtree of T . Furthermore, the
former is a stronger condition than the later whenever i > 0.
Actually, this last phenomenon is part of a more general situation: If (S, A) is
an (m, i)-good pair and B ⊆ A, then (S, B) is an (m, j)-good pair where |B| = j.
Lemma 4.7. Suppose that (S, A) is an (m, i)-good pair and let x be the vertex of
Td associated to S. Recall the subpolytope Px of Hm arising from x. Consider the
matroid polytope consisting of all y ∈ Px that satisfy the following i equalities:
X
yj = 1 for all l ∈ A.
j∈Hl
Then, the face of Td dual to this matroid polytope is i-dimensional.
Proof. Computing the degrees in T , note that for each l ∈ L(S), the number of
58
extended leaves of S adjacent to l is deg(l) − 1.
Also, note that the total number of extended leaves of S which are not adjacent
to a leaf of S is
X
X
deg(v)−(2(|V(S)| − 1) − |L(S)|) =
v∈V(S)\L(S)
deg(v)−2(|V(S)|−1)+|L(S)|.
v∈V(S)\L(S)
With these motivations, let
ϕ
RL(S)\A × R −
→ R,

ϕ(r, c) =
i+
X
rl (deg(l) − 1) + c 
l∈L(S)\A
Define k = (kl )l∈L(S)\A by letting kl =

X
deg(v) − 2(|V(S)| − 1) + |L(S)| .
v∈V(S)\L(S)
1
deg(l)−1
for all l ∈ L(S)\A and 1 =
(1, 1, 1, . . . , 1) ∈ RL(S)\A . Notice that 0 < k < 1 componentwise. We have:
ϕ(k, 0) = i +
X
1 + 0 = i + |L(S)\A| = i + (|L(S)| − i) = |L(S)|
l∈L(S)\A
≤ m − 1 < m.
59
Moreover:
X
ϕ(1, 1) = i +
l∈L(S)\A
X
=i+
X
(deg(l) − 1) +
v∈V(S)\L(S)
deg(l) − (|L(S)| − i) +
l∈L(S)\A
=
X
l∈L(S)\A
=
X
deg(v) − 2(|V(S)| − 1) + |L(S)|
deg(l) +
X
deg(v) − 2(|V(S)| − 1) + |L(S)|
v∈V(S)\L(S)
X
deg(v) + 2i − 2(|V(S)| − 1)
v∈V(S)\L(S)
deg(v) + 2i − 2(|V(S)| − 1) ≥ m + 1 > m.
v∈V(S)\A
That is, ϕ(k, 0) < m < ϕ(1, 1).
Therefore, there exists a vector r∗ ∈ RL(S)\A and a real number c∗ such that
(k, 0) < (r∗ , c∗ ) < (1, 1) componentwise, and such that ϕ(r∗ , c∗ ) = m.
Using this pair (r∗ , c∗ ), define a vector y∗ ∈ Rn by:



1/|Hl | if j ∈ Hl and l ∈ A,



yj∗ =
r∗l /|Rj | if j ∈ Hl and l ∈ L(S)\A,




 c∗ /|R | if j ∈
6 Hl for all l ∈ L(S).
j
Then y∗ ∈ Px satisfies the following and only the following i equalities among
all defining inequalities of Px , including those inherited from the description of Hm :
X
j∈Hl
yj∗ = 1 for all l ∈ A.
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The consequence worth stressing here is that y∗ is then an interior point of Hm .
But furthermore, we conclude that the set of all y ∈ Px satisfying the aforementioned
i equalities is an (m − 1 − i)-dimensional internal face of Px , of which y∗ is an
interior point. That is, y∗ lies in the relative interior of the defined face. In turn,
this face corresponds to its dual, an i-dimensional face of Td .
Remark. We will refer to the situation of the previous proof by saying that the
(m, i)-good pair (S, A) gives rise to an i-dimensional face of Td .
Let’s now describe the matroid associated to the i-dimensional face F of Td
arising from an (m, i)-good pair (S, A). Let x be the vertex of Td associated to S
and Px the subpolytope of Hm associated to x. The set of bases of the matroid of
F comes from the 0-1 vectors y ∈ Px satisfying the equalities
X
yj = 1 for all l ∈ A.
j∈Hl
These 0-1 vectors, in turn, correspond to all m-sets {i1 , i2 , . . . , im } ∈ Bx for
which
|{i1 , i2 , . . . , im } ∩ Hl | = 1 for all l ∈ A.
We then see that the set of bases of our rank-m matroid over [n] associated to
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F is precisely the collection BF of all m-sets {i1 , i2 , . . . , im } ⊆ [n] satisfying:
|{i1 , i2 , . . . , im } ∩ Hl | = 1 for all l ∈ A,
|{i1 , i2 , . . . , im } ∩ Hl | ≥ 1 for all l ∈ L(S)\A,
|{i1 , i2 , . . . , im } ∩ Rl | ≤ 1 for all l ∈ L(S).
A point w ∈ ∂(Pd ) belongs to F if and only if BF ⊆ Bw . In particular, notice
that x ∈ F .
Lemma 4.8. Let F be the face of Td arising from an (m, i)-good pair (S, A) with
i > 0. Let x0 be the vertex of Td arising from an m-good subtree S 0 of T . Then, a
necessary condition for having x0 ∈ F is that S 0 ⊆st S.
Proof. First, note that S is not a vertex of T and m > 2.
To S, associate the collections H = {Hl }l∈L(S) and R = {Rl }l∈L(S) of subsets of
[n]. To S 0 , associate the collections H0 = {Hl0 }l∈L(S 0 ) and R0 = {Rl0 }l∈L(S 0 ) of subsets
of [n].
Let BF be the set of bases of the matroid of the face F . That is, let BF be the
collection of all m-sets {i1 , i2 , . . . , im } ⊆ [n] satisfying
|{i1 , i2 , . . . , im } ∩ Hl | = 1 for all l ∈ A,
|{i1 , i2 , . . . , im } ∩ Hl | ≥ 1 for all l ∈ L(S)\A,
|{i1 , i2 , . . . , im } ∩ Rl | ≤ 1 for all l ∈ L(S).
62
Also, recall that the set of bases Bx0 of the matroid of x0 is the collection of all
m-sets {i1 , i2 , . . . , im } ⊆ [n] satisfying
|{i1 , i2 , . . . , im } ∩ Hl0 | ≥ 1 for all l ∈ L(S 0 ),
|{i1 , i2 , . . . , im } ∩ Rl0 | ≤ 1 for all l ∈ L(S 0 ).
Suppose that S 0 is not a subtree of S and let u ∈ L(S 0 )\V(S) be such that the
minimal path from any element of Hu0 to u does not meet S. We want to prove that
x0 is not a vertex of F , or that BF 6⊆ Bx0 . Carefully consider our set Hu0 . For each
l ∈ L(S), take one element from Hl \Hu0 to form a subset [|L(S)|] = {1, 2, . . . , |L(S)|}
of leaves of T disjoint from Hu0 -after a relabeling of the leaves if necessary. This
is possible because the leaves of S are at least of degree 3 in T . As (S, A) is an
(m, i)-good pair, the number of elements in L(S) not adjacent to an element of A
is at least m + 1 − i. If we let u ∈ L(S) be the vertex at which the minimal path
from u to S meets S, we see that the number of elements in L(S)\u not adjacent to
an element of A is then at least m − i. Moreover, as the elements of [|L(S)|] come
from |L(S)| different members of R, this shows that the number of extended leaves
l ∈ L(S)\u not adjacent to an element of A and such that [|L(S)|] ∩ Rl = ∅, is at
least m − |L(S)| = m − i − (|L(S)| − i). Therefore, we can complete our set [|L(S)|]
of leaves of T up to a set [m] such that [m] ∩ Hu0 = ∅ and [m] ∈ BF . But then,
[m] 6∈ Bx0 .
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Proposition 4.9. There is a one-to-one function from the set of (m, i)-good pairs
(S, A), to the set of i-dimensional faces of Td .
Proof. Let x be the vertex of Td coming from S and let Px be the associated
subpolytope of Hm . Map the (m, i)-good pair (S, A) to the face F of Td dual to
the matroid polytope of all y ∈ Px satisfying
X
yj = 1 for all l ∈ A.
j∈Hl
Per Lemma 4.7, F is i-dimensional. We need only be concerned with proving
that our map is injective.
Let (S, A) and (S 0 , A0 ) be two good pairs with corresponding i-dimensional faces
F and F 0 . Suppose that S and S 0 have, respectively, associated vertices x and x0 of
Td . Using Lemma 4.8, we see that a necessary condition for having x0 ∈ F is that
S 0 ⊆st S. Also, a necessary condition for having x ∈ F 0 is that S ⊆st S 0 . Therefore,
as x ∈ F and x0 ∈ F 0 , if F = F 0 then we must have S = S 0 .
However, Lemma 4.7 shows that if (S, A) and (S, A0 ) are (m, i)-good pairs with
A=
6 A0 , then the corresponding faces of Td are different. In that case, the relative
interiors of the internal faces of Px given by A and A0 are nonempty and disjoint.
If T is a trivalent tree, we can restate these results without further mention of
64
extended leaves. In that case, we have
X
deg(v)−2(|V(S)|−1)+2i = 3(|V(S)|−i)−2(|V(S)|−1)+2i = (|V(S)| − i)+2.
v∈V(S)\A
Therefore, whether (S, A) is a good pair or not does not depend on the specific
choice of i-set A, but only on the value i and the tree S. We can check the following
result:
Proposition 4.10. Let T be a trivalent tree, 0 ≤ i ≤ m − 1 and S ⊆st in(T ).
Then, a pair (S, A) with A ⊆ L(S) and |A| = i is an (m, i)-good pair if and only
if |L(S)| ≤ m − 1 and |V(S)| ≥ m − 1 + i. All such (m, i)-good pairs give rise to
i-dimensional faces of Td . Furthermore, each i-dimensional face of Td corresponds
to at most one such pair.
The first central theorem of this section establishes that good pairs give rise to
all i-dimensional faces of Td whenever 0 ≤ i < m − 1.
Theorem 4.11. Let T be a tree with m-dissimilarity vector d. Suppose that Td
is the associated tight span and let 0 ≤ i < m − 1. Then, there is a one-to-one
correspondence between (m, i)-good pairs (S, A) and i-dimensional faces of Td .
Proof. We need only prove that for every i-dimensional face F of Td with 0 ≤ i <
m − 1, there is a good pair giving rise to it.
Let x be a vertex of F and let S be the m-good subtree of T giving rise to x.
65
Let F ◦ be the face of the subdivision of Hm dual to F . We know F ◦ is an
(n − 1 − i)-dimensional internal face of Px , with 0 ≤ i < m − 1. There exist sets
A ⊆ L(S) with |A| = i1 and B ⊆ L(S) with |B| = i2 such that i = i1 + i2 , and such
that F ◦ can be described as the set of all y ∈ Px satisfying the equalities
X
yj = 1 for all l ∈ A,
j∈Hl
X
yj = 1 for all l ∈ B.
j∈Rl
It must be the case that the number of extended leaves of S that are not adjacent
to an element of A is at least m + 1 − i1 . Otherwise, as F ◦ is a subset of Hm , this
number would have to be precisely m − i1 , and then all y ∈ F ◦ would be forced to
satisfy the following equalities from the description of Px :
X
yj = 1 for all extended leaves l ∈ L(S) not adjacent to an element of A,
j∈Rl
X
yj = 1 for all l ∈ A.
j∈Hl
However, this is a rank-m system of equations, so F ◦ would have to live in an
(n − m)-dimensional space of Rn . This is false because F ◦ is (n − 1 − i)-dimensional
with 0 ≤ i < m − 1.
Therefore, the number of extended leaves of S that are not adjacent to an element
of A is at least m + 1 − i1 and furthermore, (S, A) is an (m, i1 )-good pair.
66
Let S B be the subtree of in(T ) obtained as the span of the set of vertices V(S)∪B.
Notice that S B arises after appending to S the i2 edges adjacent to both S and
B. Elements of B now become leaves of S B . Our tree S B does have to be a subtree
of in(T ) because of the assumed equalities for B in the description of F ◦ and the fact
that F ◦ is internal. Also, notice that no element of B is adjacent to an element of
P
A in T . Otherwise, we would have j∈Rl yj = 0 for some l ∈ L(S)\B, which again
contradicts that F ◦ is internal. As there are at least m + 1 − i1 extended leaves of S
which are not adjacent to an element of A and these leaves include the i2 elements
of B, we conclude that there must be at least m + 1 − i = m + 1 − i1 − i2 extended
leaves of S B which are not adjacent to elements of A ∪ B.
The set A is also a set of leaves of S B , so A ∪ B ⊆ L(S B ). These sets satisfy
A ∩ B = ∅, showing that S B has at least i leaves. Let HB = {HlB }l∈L(S B ) and
RB = {RlB }l∈L(S B ) be the collections of subsets of [n] = L(T ) associated to S B . By
construction L(S B )\ (A ∪ B) ⊆ L(S), so all elements y ∈ F ◦ must satisfy the
following conditions:
X
yj = 1 for all l ∈ A ∪ B,
j∈HlB
X
yj ≥ 1 for all l ∈ L(S B )\ (A ∪ B) .
j∈HlB
As F ◦ lies in Hm , we see from here that |L(S B )| ≤ m. If |L(S B )| = m,
then among these expressions, equality must hold in each and everyone of the
67
last |L(S B )| − i inequalities, giving us a rank-m system of equations which all elements of F ◦ must satisfy. As we have seen, this yields a contradiction because
dim(F ◦ ) = n − 1 − i > n − m.
Thus |L(S B )| ≤ m − 1. Moreover, we have now all the necessary ingredients to
conclude that (S B , A ∪ B) is an (m, i)-good pair.
To finish, let xB be the vertex of Td associated to S B and consider:
PxB ={y ∈ Hm |
X
yj ≥ 1 for all l ∈ L(S B ) and
j∈HlB
X
yj ≤ 1 for all l ∈ L(S B )}.
j∈RlB
Then, F ◦ is precisely the set of all y ∈ PxB that satisfy the following i equalities,
each of which comes from the description of PxB :
X
yj = 1 for all l ∈ A ∪ B.
j∈HlB
Therefore, (S B , A ∪ B) is an (m, i)-good pair giving rise to F .
It remains to study the pathological case of the (m − 1)-dimensional faces of Td .
We have had glimpses of it previously. For example, we already know that some
(m − 1)-dimensional faces of Td do arise from (m − 1)-good pairs (S, A). In that
case, notice that this means that S has exactly m − 1 leaves and that the number of
extended leaves of S that are not adjacent to leaves of S is at least 2. In the case of
trivalent trees, this translates to saying that S has exactly m − 1 leaves and a least
68
2(m − 1) vertices.
To see why these are not necessarily all (m − 1)-dimensional faces of Td , we first
need to introduce some additional terminology.
Suppose that S ⊆st in(T ) is such that the collection H of sets of leaves of T
associated to S is a partition of [n]. We then say that S is a full subtree of T .
This is a natural tag for S, as can be checked by doing some small examples. One
example of full subtrees that we have found all the way up to this point are the
subtrees of in(T ) that arise as extended trees.
One favorable aspect of a full m-subtree S of T is that any proper subtree of
S that properly contains in(S) is an m-good subtree of T . However, neither S nor
in(S) are m-good subtrees of T . In particular, notice that full m-subtrees of T do
not arise as extended trees of m-good subtrees of T .
We want to prove the following result:
Theorem 4.12. Let T be a tree with m-dissimilarity vector d and let Td be the
associated tight span. Let F be the union of the following sets:
• the set of (m, m − 1)-good pairs (S, A);
• the set of full m-subtrees of T .
Then, there is a one-to-one correspondence between the elements of F and the (m −
1)-dimensional faces of Td .
In the case of trivalent trees, this result can be stated in a more convenient way.
69
Theorem 4.13. Let T be a trivalent tree with m-dissimilarity vector d and let Td
be the associated tight span. Let F be the union of the following sets:
• the set of subtrees S of in(T ) with m − 1 leaves and at least 2(m − 1) vertices,
• the set of subtrees S of in(T ) with m leaves and 2(m − 1) vertices.
Then, there is a one-to-one correspondence between the elements of F and the (m −
1)-dimensional faces of Td .
To understand why this result holds and how full m-subtrees of T come into
play, let’s study an (m − 1)-dimensional face F of Td and the way in which it arises
from T . Let F ◦ be the dual of F in the subdivision of Hm . We know that F ◦ is
(n − m)-dimensional. Also, F ◦ is an internal face of some matroid polytope Px ,
where x is a vertex of F . Let S be the m-good subtree of T giving rise to x.
There exist pairs (A, B) of sets of vertices of T such that
• A ⊆ L(S) with |A| = i1 ;
• B ⊆ L(S) with |B| = i2 ;
• m − 1 ≤ i1 + i2 ;
70
• F ◦ can be described as the set of all y ∈ Px satisfying the equalities:
X
yj = 1 for all l ∈ A,
j∈Hl
X
yj = 1 for all l ∈ B.
j∈Rl
Among all such pairs, choose one with i1 + i2 maximal.
No element of B is adjacent in T to an element of A because F ◦ is an internal
face of Px . Therefore, we must have i1 + i2 ≤ m. Extend S to a tree S B as in
the proof of the theorem, and notice that A ∪ B is a set of i1 + i2 leaves of S B .
Notice also that S B ⊆st in(T ) because F ◦ is internal. If we consider the collections
HB = {HlB }l∈L(S B ) and RB = {RlB }l∈L(S B ) of subsets of [n] associated to S B , then
all points y ∈ F ◦ satisfy
(F1)
X
yj = 1 for all l ∈ A ∪ B,
j∈HlB
(F2)
X
yj ≥ 1 for all l ∈ L(S B )\ (A ∪ B) ,
j∈HlB
(F3)
X
yj ≤ 1 for all extended leaves l ∈ L(S B ) not adjacent to a leaf of S B .
j∈RlB
The family (F1) of expressions contains i1 + i2 equalities and the family (F2)
contains |L(S B )| − (i1 + i2 ) inequalities. Both (F1) and (F2) consist of a total
of |L(S B )| expressions and we check that |L(S B )| ≤ m. If i1 + i2 = m − 1 <
71
m = |L(S B )|, we can see that the family (F2) consists of precisely one inequality,
which is then forced to be an equality. As L(S B )\ (A ∪ B) ⊆ L(S), the pair
L(S B )\ (A ∪ B) ∪ A, B contradicts the maximality of i1 + i2 .
Therefore, i1 + i2 = |L(S B )| and the family (F2) of expressions must be void.
That is, L(S B ) = A ∪ B.
Assume that |L(S B )| = i1 + i2 = m − 1. The family (F3) must contain at
least one inequality. If it contains at least 2 inequalities, then (S B , A ∪ B) is an
(m, m − 1)-good pair giving rise to F . On the other hand, if it contains one and
only one inequality, then this inequality is forced to be an equality. That is, if the
family (F3) contains one expression, then there exists one and only one extended
P
leaf l of S B not adjacent to a leaf of S B , and for this l we have j∈RB yj = 1.
l
Moreover, l ∈ L(S)\B. Therefore, we see that the pair (A, B ∪ l) contradicts the
maximality of i1 + i2 .
Finally, assume that |L(S B )| = i1 + i2 = m. In that case, the family (F3) must
also be void because F ◦ is internal. We identify here the pathology we were looking
for. In this final case, notice that S B is a subtree of in(T ) which has m leaves and
such that all elements of L(S B ) are adjacent to elements of L(S B ). Notice that
this final condition is equivalent to saying that the collection HB associated to S B
defines a partition of [n] into m classes. The (n − m)-dimensional internal face of
72
the subdivision of Hm arising from S B is the set of all y ∈ Hm satisfying
X
yj = 1 for all l ∈ L(S B ).
j∈HlB
The matroid associated to this maximal face of Td is then the rank-m transversal
matroid whose bases are the transversals of the collection HB = {HlB }l∈L(S B ) .
Using our now fully developed methods, it can be checked that the following
holds. Let S be a full m-subtree of T and let H = {Hl }l∈L(S) be the collection of
subsets of [n] associated to S. Also, let S 0 be a tree such that in(S) (st S 0 (st S and
let x0 be the vertex of Td coming from S 0 . Then, the intersection of Px0 and the space
P
of all y ∈ Rn such that j∈Hl yj = 1 for all l ∈ L(S), is an (n − m)-dimensional
face of Px0 . As m > 1, for every such S we can find S 0 .
Therefore, every full m-subtree of T gives rise to an (m − 1)-dimensional face of
Td . Moreover, considering the matroids of the faces, no two different full m-subtrees
give rise to the same (m − 1)-dimensional face because the bases of the matroids
would be different.
4.3 A Combinatorial Description of the Geometric Object Td
We now use our results to study an i-dimensional face F of Td . Notice that F is a
face of Pd , so F maximizes some linear functional on Pd . Therefore, F is the convex
hull of its vertices and F is a polytope. Also, every vertex of Pd that lies in F must
73
be a vertex of F : restrict the domain of the linear functional on Pd maximized by
this vertex to F . On the other hand, every vertex of F is a vertex of Pd , as can be
seen from the description of both F and Pd in terms of inequalities. Thus, in order
to understand F , we can find all vertices of Td that are contained in F . This is our
initial plan.
We will omit any consideration of the empty face in the face lattice of a polytope,
so we speak of and consider vertices as “minimal” faces. The same rule will apply
to the construction of f -vectors, where we will omit the entry f−1 .
For a tree S ⊆st T and a set A ⊆ L(S), let SA be the subtree of S spanned by
the vertices in V(S)\A.
4.3.1 Faces from (m, i)-good pairs.
We begin studying a face F of Td coming from an (m, i)-good pair (S, A) with i > 0.
Lemma 4.14. Let F be a face of Td coming from an (m, i)-good pair (S, A) with
i > 0. If x0 is a vertex of F coming from an m-good subtree S 0 of T , then SA ⊆st
S 0 ⊆st S. Also, if a tree S 0 satisfies SA ⊆st S 0 ⊆st S, then S 0 is an m-good subtree
of T giving rise to a vertex x0 of F .
Proof. To S, associate the collections H = {Hl }l∈L(S) and R = {Rl }l∈L(S) of subsets
of [n]. To S 0 , associate the collections H0 = {Hl0 }l∈L(S 0 ) and R0 = {Rl0 }l∈L(S 0 ) of
subsets of [n].
Per Lemma 4.8, we know that if x0 is a vertex of F , then S 0 ⊆st S.
74
Let us now assume that S 0 ⊆st S and that x0 is a vertex of F .
We now see that S 0 must contain all leaves of S different from A. If there
exists some l ∈ L(S)\ (A ∪ V(S 0 )), then we must have |{i1 , i2 , . . . , im } ∩ Hl | = 1
for all m-sets {i1 , i2 , . . . , im } ∈ BF ∩ Bx0 . However, it is easy to construct a set
{i1 , i2 , . . . , im } ∈ BF for which |{i1 , i2 , . . . , im } ∩ Hl | = 2 because |L(S)| ≤ m − 1.
Therefore, BF 6⊆ Bx0 and x0 is not a vertex of F .
Let’s now see that S 0 must contain all the vertices of S adjacent to elements of
A. For any l ∈ A, let v be the vertex of S adjacent to l.
First, if S 0 = l, then |Rv0 ∩ {i1 , i2 , . . . , im }| ≥ 2 for all {i1 , i2 , . . . , im } ∈ BF
because l ∈ A and m > 2: S 0 cannot be equal to l. But then, S 0 must contain v.
Suppose on the contrary that v 6∈ V(S 0 ) and let u be the vertex at which the minimal
path from v to S 0 meets S 0 . The vertex v is of degree at least 3 in T . Choose an
edge (v, w) 6= (v, l) such that w does not lie in the minimal path from v to S 0 . Now:
• If w ∈ V(S), there exists a leaf l0 ∈ L(S)\l such that the minimal path from
l0 to (v, w) meets (v, w) precisely at w.
• If w 6∈ V(S), then w ∈ L(S).
In the first case, no element of BF lies in Bx0 because if {i1 , i2 , . . . , im } ∈ BF ,
then |{i1 , i2 , . . . , im } ∩ Ru0 | ≥ 2. In the second case, as |L(S)| ≤ m − 1, we can
construct an m-set {i1 , i2 , . . . , im } ∈ BF for which |{i1 , i2 , . . . , im } ∩ (Hl ∪ Rw ) | = 2,
so |{i1 , i2 , . . . , im } ∩ Ru0 | ≥ 2. Thus, in both cases, we conclude BF 6⊆ Bx0 .
75
This proves that if x0 is a vertex of F coming from an m-good subtree S 0 of T ,
then SA ⊆st S 0 ⊆st S.
For the other direction, let S 0 satisfy SA ⊆st S 0 ⊆st S. Note that the set of
extended leaves of S which are not adjacent to an element of A is contained in
L(SA ). Moreover, L(SA ) must be the union of this set and A. As the number of
extended leaves of S which are not adjacent to an element of A is at least m + 1 − i,
then |L(SA )| ≥ i + (m + 1 − i) = m + 1. If S 0 is a subtree of S containing SA , then
m + 1 ≤ |L(SA )| ≤ |L(S 0 )|,
|L(S 0 )| ≤ |L(S)| ≤ m − 1.
Therefore, S 0 is an m-good subtree of T that gives rise to a vertex x0 of Td .
Now, notice that L(S 0 ) ⊆ A ∪ L(S) . As a consequence, for {i1 , i2 , . . . , im } ∈
BF we have
|{i1 , i2 , . . . , im } ∩ Rl0 | ≤ 1 for all l ∈ L(S 0 ).
Furthermore, every leaf of S 0 must be a leaf of S or adjacent to an element of A.
In both cases, we check that if {i1 , i2 , . . . , im } ∈ BF , then
|{i1 , i2 , . . . , im } ∩ Hl0 | ≥ 1 for all l ∈ L(S 0 ).
Thus, for each such S 0 we have BF ⊆ Bx0 , or x0 ∈ F .
76
Remark. Notice that a tree S 0 satisfying SA ⊆st S 0 ⊆st S can be described in our
usual notation as (SA )B for some B ⊆ A, the subtree of S spanned by the set of
vertices V(SA ) ∪ B. Recall that (SA )B can also be seen as the tree arising from SA
after appending the edges adjacent to both SA and B. For example, (SA )A = S. But
then, note that every such tree (SA )B with B ⊆ A is a subtree of S which contains
SA , and there are a total of 2i = 2|A| of these trees. We conclude that the vertices
of F are given by subtrees (SA )B with B ⊆ A, and the number of such vertices is
2i = 2|A| .
Let us now study the faces of a face F coming from an (m, i)-good pair (S, A)
with i > 0.
The argument given at the beginning of this section generalizes to all faces F 0 of
F . If F 0 is a face of Pd contained in F , restricting the domain of the linear functonal
on Pd maximized by F 0 to F shows that F 0 is also a face of F . If F 0 is a face of
F , the description of both Pd and F in terms of inequalities shows that F 0 is also
a face of Pd . Therefore, to study the faces of F we need only study the faces of Pd
contained in F .
Lemma 4.15. Let F be a face of Td coming from an (m, i)-good pair (S, A) with
i > 0. For a triple C ⊆ B ⊆ A, the pair (SA )B , C is an (m, |C|)-good pair. The
set of triples C ⊆ B ⊆ A is in bijection with the set of faces of our i-dimensional
face F via the map taking the triple C ⊆ B ⊆ A to the face of Td arising from the
B
(m, |C|)-good pair (SA ) , C . Moreover, the set of k-dimensional subfaces of F is
77
in one-to-one correspondence with the triples C ⊆ B ⊆ A such that |C| = k via this
map.
Proof. Notice that if F 0 is a face of F , then F 0 must also come from an (m, j)-good
pair (S 0 , A0 ) with j ≤ i. For example, consider the dimension of F 0 to check this.
But the vertices of F 0 must be vertices of F as well, so:
• S 0 must be equal to (SA )B for some B ⊆ A;
B\C
0
• SA
for some C ⊆ B ⊆ A, and then A0 = B\ (B\C) =
0 must be equal to (SA )
C.
Therefore, the pair (S 0 , A0 ) must actually be equal to (SA )B , C .
To check that every pair (SA )B , C with C ⊆ B ⊆ A is indeed a good pair, it
is enough to check that (SA )B , B is a good pair. In order to do this, recall that
we want to verify that the number of extended leaves of (SA )B not adjacent to B
is at least m + 1 − |B|. This is equivalent to verifying |L(SA )\{B}| ≥ m + 1 − |B|.
As (S, A) = (SA )A , A is a good pair, by analogy we know that |L(SA )\{S}| ≥
m + 1 − |A| = m + 1 − i. But the containments B ⊆ A ⊆ L(SA ) show that
|L(SA )\{B}| = |L(SA )\{S}| + |A\B|
≥ (m + 1 − |A|) + (|A| − |B|)
= m + 1 − |B|.
78
Therefore, the pair (SA )B , C with C ⊆ B ⊆ A is a good pair. It gives rise to
a |C|-dimensional face of F .
Now, suppose that we have triples C ⊆ B ⊆ A and C 0 ⊆ B 0 ⊆ A. From them, we
0
obtain faces with associated pairs ((SA )B , C) and ((SA )B , C 0 ). These faces are equal
0
if and only if the pairs coincide, so that (SA )B = (SA )B and C = C 0 . Of course, then
B = B0 .
Using Lemma 4.15, let us now count the number of k-dimensional subfaces of
F . They are in one-to-one correspondence with the triples C ⊆ B ⊆ A such that
|C| = k. For the last triples, there are ki choices for C. Then, for a fixed C, there
are 2i−k choices for B. We conclude that the f -vector of F is given by
i−k
fk (F ) = 2
i
for all k with 0 ≤ k ≤ i.
k
The total number of faces of F is then 3i .
Now, for a triple C ⊆ B ⊆ A with associated k-dimensional face F 0 of F , we
describe all triples C 0 ⊆ B 0 ⊆ A giving rise to a face of F containing F 0 . Notice
that a necessary condition is that C ⊆ C 0 . Also, we require that B\C 0 = B 0 \C 0 . But
then, B 0 = (B 0 \C 0 ) ∪ C 0 = (B\C 0 ) ∪ C 0 = B ∪ C 0 . These conditions can be seen to be
sufficient conditions. Therefore, all faces of F containing F 0 come from triples
C 0 ⊆ (B ∪ C 0 ) ⊆ A with C ⊆ C 0 .
79
The number of such triples is then given by the number of sets C 0 ⊆ A containing
i−k
C, which is 2i−k . The number of such triples for which |C 0 | = p is then given by p−k
.
That is, the number of p-dimensional faces of F containing any given k-dimensional
i−k
face of F is p−k
. Letting k = 0 and p = i − 1, we see that the polytope F is simple.
To finish, let us expose the reason why we are obtaining these familiar results.
Theorem 4.16. Let T be a tree with m-dissimilarity vector d and let Td be the
associated tight span. If F is a face of Td coming from an (m, i)-good pair (S, A),
then F is an i-dimensional polytope and its face lattice is isomorphic to the face
lattice of the i-dimensional cube.
Proof. Consider the map
φ
{Triples of F } →
− {Faces of the 0-1 i-dimensional cube in RA },
φ
C⊆B⊆A→
− Convex hull of all 0-1 vectors with 0’s in A\B and 1’s in B\C.
The map φ induces an isomorphism between the lattice of faces of F and the
lattice of faces of the i-cube.
4.3.2 Faces from full subtrees of T .
We now want to develop a similar study for an (m − 1)-dimensional face F of Td
coming from a full m-subtree S of T . However, as any proper face F 0 of F is a
face of Pd with dimension less than m − 1, this will be considerably simplified if we
80
remember that then F 0 must come from a good pair, which is a case we have already
taken care of.
Let us first find the vertices of F . Notice that we no longer have restrictions on
the value of m, so m ≥ 2.
Lemma 4.17. Let F be an (m − 1)-dimensional face of Td coming from a full msubtree S of T . A tree S 0 satisfying in(S) (st S 0 (st S is an m-good subtree of T
and gives rise to vertex of F . All vertices of F are obtained in this way.
Proof. Let x0 be a vertex of F . Let S 0 be the m-good subtree of T giving rise to x0 .
If S 0 is not a subtree of S, let u ∈ L(S 0 )\V(S) be such that the minimal path from
any element of Hu0 to u does not meet S. Construct a transversal of the collection
H that does not contain any member of Hu0 . The construction is possible because
the element l ∈ L(S) at which the minimal path from u to S meets S is of degree
at least 3. This shows that BF 6⊆ Bx0 , so x0 6∈ F , yielding a contradiction.
If S 0 ⊆st S, then we must have also in(S) ⊆st S 0 . In case m = 2, we have
in(S) = Λ and the contaiment is trivial. If m ≥ 3 and in(S) is not a subtree of S 0 ,
there exists an element u ∈ L(S 0 ) such that for every transversal [m] of H we have
|Ru0 ∩ [m]| ≥ 2. This argument is completely analogous to the one given during the
related discussion for good pairs, only easier. Then, BF 6⊆ Bx0 and again we obtain
a contradiction.
Now, we know that S 0 cannot be S or in(S) because then (S 0 , ∅) would not be a
good pair.
81
Thus, if x0 is a vertex of F , the tree S 0 must satisfy that in(S) (st S 0 (st S.
In that case, as we noted earlier, (S 0 , ∅) is an (m, 0)-good pair and gives rise to
a vertex x0 of Td as required. Also, every transversal of H is now clearly a member
of the collection Bx0 , so we obtain BF ⊆ Bx0 or x0 ∈ F .
Now, every S 0 satisfying in(S) (st S 0 (st S can be described as (in(S))A with
∅ ( A ( L(S), and there are 2n − 2 of the latter. Thus, the number of vertices of
F is 2n − 2.
It is easy to check the following result:
Lemma 4.18. Let F be an (m − 1)-dimensional face of Td coming from a full
m-subtree S of T . The proper faces of F arise from the pairs (in(S))A , B with
∅ ( A ( L(S) and B ( A. All such pairs are good. In particular, the i-dimensional
proper faces of F correspond to the (m, i)-good pairs (in(S))A , B with ∅ ( A (
L(S), B ( A and |B| = i < m − 1.
The number of pairs in Lemma 4.18 can be computed easily. Pick first a set B,
for which there are mi choices. Then, for a fixed B, the number of choices for A is
2m−i − 2. Therefore, the f -vector of F for this case is given by
m−i
fi (F ) = 2
m
−2
for all i with 0 ≤ i ≤ m − 2,
i
fm−1 (F ) = 1.
The total number of faces of F is 3m − 2m+1 + 2.
82
Now, consider two good pairs:
A
• (in(S)) , B with ∅ ⊆ B ( A ( L(S) and 0 ≤ |B| = k < m − 1,
•
A0
(in(S)) , B
0
with ∅ ⊆ B0 ( A0 ( L(S) and k ≤ |B 0 | = i < m − 1.
Necessary and sufficient conditions for the second face to contain the first are
B ⊆ B 0 and A0 \B 0 = A\B 0 . These conditions are equivalent to B ⊆ B 0 and A0 =
A ∪ B 0 . In particular, when k = 0 and i = m − 2, the conditions break down
to finding, for a fixed A satisfying ∅ ( A ( L(S), all B 0 with |B 0 | = m − 2 and
|A\B 0 | = 1. For our polytope F , this time the nature of the answer does depend on
A.
If |A| = 1, then B 0 comes from picking any m − 2 elements of the set L(S)\A, so
there are m − 1 = m−1
possibilities. If |A| = m − 1, then B 0 comes from selecting
m−2
any |A| − 1 members of A and we again find m − 1 = m−1
possibilities. If however
m−2
1 < |A| < m − 1, then B 0 comes from picking |A| − 1 members of A and m − 1 − |A|
members of L(S)\A, and there are |A| (m − |A|) different ways of doing this. The
number of vertices on which the latter situation springs up is given by the number
of subsets of [m] with at least 2 elements and at most m − 2 elements, 2m − 2m − 2
if m ≥ 3 and 0 if m = 2.
Remark. In particular, if m = 2 or m = 3 then our polytope F is simple, but not
anymore if m ≥ 4.
Let us now describe the cover relations of the lattice of faces of F . We restrict
the lattice to faces of dimension at most m − 2. To simplify the notation, we will
83
consider L(S) = [m] and we will always let A be a nonempty proper subset of [m].
Vertices of F are now pairs (A, ∅). Faces of F of dimension at most m − 2 are now
pairs (A, B), where B will always be assumed to be a proper subset of A. The face
lattice is ranked. The rank of a pair (A, B) is |B|.
If B = ∅, (A, B) is minimal.
If |B| = m − 2, (A, B) is maximal.
If |B| < m − 2, we describe the cover of (A, B).
If |A| − |B| = 1, then |A| < m − 1. The cover of (A, B) are exactly all pairs
(A ∪ i, B ∪ i) with i ∈ ([m]\A). There are m − |A| such coverings. The new pair
satisfies |A ∪ i| − |B ∪ i| = 1.
If |A| − |B| ≥ 2, we distinguish two cases:
• If |A| < m − 1, the cover of (A, B) are exactly all pairs (A ∪ i, B ∪ i) with
i ∈ ([m]\B). There are m − |B| such coverings. The new pair satisfies |A ∪
i| − |B ∪ i| = |A| − |B| if i 6∈ A, and |A ∪ i| − |B ∪ i| = |A| − |B| − 1 if i ∈ A.
• If |A| = m − 1, the cover of (A, B) consists of all pairs (A ∪ i, B ∪ i) with
i ∈ (A\B). There are |A| − |B| such coverings. The new pair satisfies |A ∪
i| − |B ∪ i| = |A| − |B| − 1.
Remark. In the case m = 2, F is a line segment. In case F = 3, F is isomorphic
to an hexagon. In case m = 4, F can be seen to be isomorphic to the rhombic
dodecahedron.
84
Finally, for a pair (A, B) of rank k with k < m − 1, we count the number of
pairs (A0 , B 0 ) of rank i with k < i < m − 1 greater than or equal to (A, B) in our
lattice. The pair (A0 , B 0 ) must satisfy B ⊆ B 0 and A0 = A ∪ B 0 . Letting C = B 0 \B,
the problem becomes that of counting all sets C ⊆ ([m]\B) with |C| = i − k such
that [m] ) A ∪ C ) B ∪ C. Now, A ∪ C ) B ∪ C if and only if (A\B) \C 6= ∅.
If k = m − 2, (A, B) is maximal.
Assume that k < m − 2.
For this exercise, we distinguish four cases:
• If i < |A| and i − k < m − |A|, then C is any (i − k)-subset of [m]\B. The
number of choices is m−k
.
i−k
• If i ≥ |A| and i − k < m − |A|, then C is any (i − k)-subset of [m]\B which
does not contain A\B. There are m−k
− m−|A|
choices.
i−k
i−|A|
• If i < |A| and i − k ≥ m − |A|, then C is any (i − k)-subset of [m]\B which
|A|−k
does not contain [m]\A. There are m−k
−
choices.
i−k
(i−k)−(m−|A|)
• If i ≥ |A| and i − k ≥ m − |A|, then C must be any (i − k)-subset of [m]\B
|A|−k
not containing [m]\A or A\B. There are m−k
− m−|A|
− (i−k)−(m−|A|)
i−k
i−|A|
choices.
These results give a hint of the precise nature of the polytope F . Recall that the
(m − 1)-dimensional pyrope Pm−1 , or (m − 1)-pyrope, can be realized as the
85
polytope conv ([−1, 0]m−1 ∪ [1, 0]m−1 ), the convex hull of the 0-1 and 0-(−1) cubes
in Rm−1 . See Joswig and Kulas [13] for a detailed description.
Theorem 4.19. An (m − 1)-dimensional face F of Td coming from a full m-subtree
S of T has a face lattice isomorphic to the face lattice of the (m − 1)-pyrope Pm−1 .
Proof. Fix a choice of l ∈ L(S). For every pair of sets (A, B) satisfying ∅ ( A (
L(S) and B ( A, consider the finite set VA,B ⊆ RL(S)\l that we now construct. For
each C ⊆ B:
• If l ∈ L(S)\ (A\C), define v ∈ RL(S)\l as the vector of 1’s in A\C and 0’s
everywhere else. Then, let v ∈ VA,B .
• If l ∈ A\C, define v ∈ RL(S)\l as the vector of 0’s in A\C and −1’s everywhere
else. Then, let v ∈ VA,B .
Consider the map
φ
{ Faces of ∂(F ) } →
− { Faces of ∂(Pm−1 ) },
under which the face of ∂(F ) arising from the pair in(S)A , B with ∅ ( A ( L(S)
and B ( A gets mapped to the convex hull of the set VA,B .
Using the ideas discussed up to this point, it is routine to check that φ is a lattice
isomorphism between the face lattices of ∂(F ) and ∂(Pm−1 ), where of course we
are considering the (m − 1)-pyrope as living in RL(S)\l . The map φ induces a lattice
isomorphism between the face lattices of F and Pm−1 .
86
To close this section, we prove a short result about the behaviour of the intersection of two faces F and F 0 of Td , each of which comes from a good pair. As we
will see, these intersections behave nicely.
Theorem 4.20. Let F and F 0 be faces of Td . Suppose that F comes from an
(m, |A|)-good pair (S, A) and that F 0 comes from an (m, |A0 |)-good pair (S 0 , A0 ).
Then, F ∩ F 0 6= ∅ if and only if
0
SA ⊆st S 0 and SA
0 ⊆st S.
Moreover, in the case when our faces meet, F ∩F 0 comes from the (m, |A∩A0 |)-good
pair (S ∩ S 0 , A ∩ A0 ). In particular, F ∩ F 0 is of dimension |A ∩ A0 |.
Proof. Suppose that F ∩ F 0 6= ∅. Then, there exist C ⊆ A and C 0 ⊆ A0 such that
0
0 C
0
0
0
(SA )C = (SA
0 ) . Letting B = A\C and B = A \C , we can restate this as: there
exist B ⊆ A and B 0 ⊆ A0 such that SB = SB0 0 . From here we see that indeed
0
SA ⊆st S 0 and SA
0 ⊆st S.
0
0
Now, suppose that SA ⊆st S 0 and SA
0 ⊆st S and we want to prove F ∩ F 6= ∅.
Observe that
SA\V(S 0 ) ⊆st S 0 ,
0
SA
0 \V(S) ⊆st S.
87
Also, V(SA\V(S 0 ) ) ∩ (A0 \V(S)) = ∅, so from the first containment we obtain
SA\V(S 0 ) = SA\V(S 0 )
A0 \V(S)
0
⊆st SA
0 \V(S) .
0
Analogously, from the second containment we obtain SA
0 \V(S) ⊆st SA\V(S 0 ) . Thus,
0
SA\V(S 0 ) = SA
0 \V(S) , which proves the first assertion.
For the second one, observe that
(A ∩ A0 ) ∩ (A\V(S 0 )) = ∅,
(A ∩ A0 ) ∩ (A0 \V(S)) = ∅.
0
0
Therefore, from the equality SA\V(S 0 ) = SA
=
0 \V(S) , we see that the pairs SA\V(S 0 ) , A ∩ A
0
0
SA
give rise to a common face of F and F 0 .
0 \V(S) , A ∩ A
Now
0
SA\V(S 0 ) = SA\V(S 0 ) ∩ SA
⊆st (S ∩ S 0 ) .
0 \V(S)
On the other hand V(S∩S 0 ) ⊆ V(S)\ (A\V(S 0 )), which implies (S ∩ S 0 ) ⊆st SA\V(S 0 ) .
The consequence is
0
SA\V(S 0 ) = (S ∩ S 0 ) = SA
0 \V(S) .
This shows that the pair (S ∩ S 0 , A ∩ A0 ) does give rise to a common face of F
and F 0 of dimension |A ∩ A0 |.
To see that this is the maximal common face, let a ∈ (A\A0 ) and then let v be
88
the vertex of S adjacent to a. Then, as SA ⊆ S 0 , v ∈ V(S 0 ).
By the choice of a and considering the edge (v, a), either (v, a) ∈ E(S 0 ) or
0
(v, a) 6∈ E(SA
0 ), but not both. Exactly one of them must hold. Thus, no subface of
F of the form (SA )B , C with a ∈ C ⊆ B ⊆ A can also be a subface of F 0 . This
can be restated by saying that if (SA )B , C gives a common face of F ∩ F 0 for some
C ⊆ B ⊆ A, then C ⊆ (A ∩ A0 ). Thus, the dimension of F ∩ F 0 can be at most
|A ∩ A0 |.
Remark. Notice that we actually have the containment
0
0
(SA ∪ SA
0 ) ⊆st (S ∩ S ) ,
and from there we can check that
0
0
(SA ∪ SA
0 ) = (S ∩ S )A∩A0 .
One application of this result is to decide whether or not F is a face of F 0 : we
0
have that F ⊆ F 0 if and only if A ⊆ A0 , S ⊆st S 0 , and SA
0 ⊆st SA . These results
generalize our study of the i-dimensional faces of Td .
Notice that if F is contained in F 0 , S 0 is obtained from S by appending some or
none of the edges adjacent to both S and L(S). In particular, we can only pick edges
that are not adjacent to an element of A. Now, recall that [n] is the set of leaves of
T , and then notice that if |L(S)| is the number of leaves of S and |L(S)| < m − 1,
89
for all l ∈ L(SA )\[n] we have that S l , A ∪ l is an (m, |A ∪ l|)-good pair. More
generally, for all B ⊆ L(SA )\[n] with |B| ≤ m − 1 − |L(S)|, the pair S B , A ∪ B is
an (m, |A ∪ B|)-good pair. Using these observations, some enumeration of the faces
of Td containing F could be possible for the case of trivalent trees, but we leave this
for a subsequent writing.
4.4 Face Enumeration in the Simplest Cases
In this brief section, we enumerate the faces of Td for some restricted values of m.
Let m = 2. The 0-dimensional faces are in bijection with the set of subtrees of
in(T ) with at most 1 leaf and at least 1 vertex. Therefore, the 0-dimensional faces
of Td correspond to the internal vertices of T . The 1-dimensional faces correspond
to either subtrees of in(T ) with exactly 1 leaf and at least 2 vertices, or to subtrees
of in(T ) with 2 leaves and 2 vertices. The former do not exist, and the latter are
the internal edges of T . We can see that Td is a tree isomorphic to in(T ).
For m = 2, Td has n − 2 vertices and n − 3 edges.
Let’s assume that T is trivalent.
Let m = 3. The 0-dimensional faces of Td correspond to subtrees of in(T ) with
at most 2 leaves and at least 2 vertices. These are all nontrivial paths in in(T ) and
are in bijection with 2-subsets of internal vertices of T . Therefore, there are n−2
2
vertices of Td . The 1-dimensional faces of Td correspond to the pairs (S, A) where
S is an internal subtree of T with at most 2 leaves and at least 3 vertices, and A
90
is a leaf of S. All such trees S must be internal paths of T with at least 3 vertices,
and they have exactly 2 leaves. The set of these trees is in bijection with 2-subsets
of internal edges of T , of which there are n−3
. Thus, the number of edges of Td
2
is (n − 3)(n − 4) = 2 n−3
, the factor 2 coming from the choices for A. Now, the
2
2-dimensional faces of Td correspond to either internal subtrees of T with 2 leaves
and at least 4 vertices, or to internal subtrees of T with 3 leaves and 4 vertices. To
count the first, consider 2-sets of internal edges of T , the number of which there are
n−3
. From here, discount the number of internal paths of T with 3 vertices, the
2
3-paths of in(T ).
Consider exclusively the tree in(T ) and orient all its edges towards one leaf l of
in(T ). We count many 3-paths by picking any vertex v 6= l not adjacent to l, and
then considering the 3-path containing v and contained in the oriented path from v
to l. There are n − 4 such paths. Now, computing the degrees in in(T ), if M is the
number of vertices of degree 3 in in(T ), then the total number of 3-paths in in(T )
is n − 4 + M . Therefore, the number of internal subtrees of T with 2 leaves and
at least 4 vertices is n−3
− (n − 4) − M . Now, the number of internal subtrees of
2
T with 3 leaves and 4 vertices corresponds to the number of vertices of degree 3 in
in(T ), computing the degrees in in(T ). There are M of them. We conlude that the
number of 2-dimensional faces of Td is n−4
= n−3
− (n − 4) − M + M .
2
2
We see that Td attains the f -vector of the f -vector conjecture for the case of 2
and 3-dissimilarity vectors of trivalent trees. This will no longer hold for sufficiently
91
big trees T for m ≥ 4.
92
Chapter 5
Further Directions
There is a result of Pachter and Speyer [17] which claims that the metric of T can
only be reconstructed from its m-dissimilarity vector d if and only if n ≥ 2m − 1.
We would like to understand how this result can be seen in our work. In the case
of trivalent trees, we believe there is a simple explanation for this fact.
Now, we have found that the faces of the tight span Td carry with them a
wealth of information about the tree T . Specifically, we refer to the collections H
and R. These collections give rise to a family of matroids that can be described using
trees. Very nicely, for the pathological faces of maximal dimension, these matroids
degenerate to transversal matroids. It is of our interest to fully understand these
matroids.
For example, notice that the 2-circuits of the matroids carry information about
the collections R. We do not know how to identify the collections H.
This has some interesting applications. It could be possible to devise an efficient
way to compute the vertices of the polytope Pd . In that case, we would know a lot
of information about the collections R and H. Is this information enough to recover
completely the collections? If so, can these collections be used to recover the shape
93
of T ?
From a different perspective, it might also be possible to obtain a theoretical
characterization of m-dissimilarity vectors of trees, for example, by requiring that
the family of matroids associated to Dd satisfy certain properties.
In general, we have shown a combinatorial richness associated to Td . The question is, how can we use it?
94
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