The slopes determined by n points in the plane

0.42 MB·English

by  Jeremy L. Martin

THE SLOPES DETERMINED BY n POINTS IN THE PLANE
6
0
JEREMYL.MARTIN
0
2
n Abstract. Let m12, m13, ..., mn−1,n be the slopes of the n2 lines con-
Ja rneelcattiinognsnapmooinntgstihnegmeniejradlepfinoseistiaoncoinnfitghueraptliaonne.spTahceeicdaelaleldInthoe(cid:0)fsa(cid:1)lollpaelgveabrrieatiyc
4 ofthecompletegraph. WeprovethatInisreducedandCohen-Macaulay,give
anexplicitGro¨bnerbasisforit,andcomputeitsHilbertseriescombinatorially.
2
WeproceedchieflybystudyingtheassociatedStanley-Reisnersimplicialcom-

plex, which has an intricate recursive structure. In addition, we are able to
]
G answermanyquestionsaboutthegeometryoftheslopevarietybytranslating
them intopurelycombinatorialproblemsconcerningenumerationoftrees.
A
.
h
t
a
1. Introduction
m
[ lineLse.tTthheereclboseugrieveonf tnhe≥lo2cduisstoinfcatllpsoloinptes vinectthoersp(lamn1e2,,c.o.n.,nmecnt−ed1,nin) apraiisrisngbyfron2m
3 (cid:0) (cid:1)
some such configuration is an irreducible algebraic variety of dimension 2n 3,
v −
called the affine slope variety of the complete graph (as we soon explain). This
6
0 slope variety turns out to have an unexpectedly rich combinatorial and geometric
1 structure. Our techniques for investigating its properties draw on combinatorics
2 (graph theory and recursive enumeration of trees), commutative algebra (Gr¨obner
0
basesandStanley-Reisnertheory),andalgebraicgeometry. Beforestatingthemain
3
theorem,wesetitincontextbygivinganoverviewofthetheoryofgraph varieties,
0
/ considered by the author in [10].
h
Let P2 be the projective plane over an algebraically closed field k, and let G
t
a be a graph with vertices V and edges E. A picture P of G consists of a point
m
P(v) for each vertex, and a line P(e) for each edge, subject to the conditions that
: P(v) P(e) whenever v is an endpoint of e. Thus the data of n points and n
v ∈ 2
lines described earlier is a picture of the complete graph K on n vertices.
i n
X (cid:0) (cid:1)
ThesetofallpicturesofGiscalledthepicturespace (G). Apictureisgenericif
X
r thepointsP(v)arealldifferent;theclosureofthelocusofgenericpicturesiscalled
a
the picture variety (G). This is an irreducible component of (G) of dimension
V X
2V . Passing to an affine open subset ˜(G) (G), and projecting onto an affine
| | V ⊂V
space A|E| whose coordinates correspondto the slopes of lines P(e), we obtain the
k
affine slope variety ˜(G), of dimension 2V 3.
S | |−
A rigidity circuit is a graph which admits a decomposition into two spanning
trees, and contains no proper subgraph with that property. The most important
rigiditycircuits arethe wheels: a wheel consistsofa cycle with anattachedcentral
vertex. ForeachrigiditycircuitC,thereisacorrespondingtreepolynomial τ(C), a
2000 Mathematics Subject Classification. 05C10,13P10,14N20.
Key words and phrases. graph, graph variety, slope, Stanley-Reisner ring, shellability, tree,
Gro¨bnerbasis.
SupportedinpartbyanNSFPostdoctoral Fellowship.
1
2 JEREMYL.MARTIN
sum of signed squarefree monomials corresponding to spanning trees appearing in
such decompositions; this polynomial is homogeneous and irreducible. The affine
slope variety ˜(G) is cut out set-theoretically in A|E| by the polynomials τ(C),
S
where C ranges over all rigidity circuit subgraphs of G. These facts were proven
in [10]. We can now state the main theorem of this paper.
Theorem 1.1. Let R = k[m ,...,m ], and let I be the ideal generated by
n 12 n−1,n n
the tree polynomials of all rigidity circuits in the complete graph K . Then:
n
(i) The affine slope variety ˜(K ) is defined scheme-theoretically by I . That
n n
S
is, In is a prime ideal, and S˜(Kn)∼=SpecRn/In.
(ii) The tree polynomials of the wheel subgraphs of K generate I , and form a
n n
Gro¨bner basis with respect to a certain graded lexicographic order.
(iii) ˜(K ) has dimension 2n 3 and degree
n
S −
(2n 4)!
− =(2n 5)(2n 7) (3)(1), (1)
2n−2(n 2)! − − ···
−
the number of perfect matchings on [1,2n 4] = 1,2,...,2n 4 . Fur-
− { − }
thermore, the Hilbert series of R /I is
n n
n−2
h(n,k)tk
kX=0 ,
(1 t)2n−3
−
where h(n,k) counts the number of perfect matchings on [1,2n 4] with
−
exactly k long pairs, that is, pairs not of the form i,i+1 .
{ }
(iv) The ring R /I and the affine slope variety ˜(K ) are Cohen-Macaulay.
n n n
S
We begin in Section 2 by describing the basic objects—graph varieties and tree
polynomials—in somewhat more detail. We do this both to make this paper more
self-contained,andbecauseseveraldetailsoftheconstructionswillbeofimportance
lateron. Thereaderisreferredto[10]and[9]formoreleisurelytreatmentsofthese
subjects.
Inthe firstmainpartofthe paper,we constructamonomialidealJ , generated
n
by the initial terms of tree polynomials τ(W) of wheels W K with respect
n
⊂
to a certain graded lexicographic term ordering. As mentioned previously, the
monomials of τ(W) correspond to coupled spanning trees of W, that is, whose
complements are also spanning trees. In order to identify the leading term of a
wheel polynomial, we need several technical facts about the valences of vertices
in coupled trees; these facts comprise Section 3. In Section 4, we introduce the
term ordering and, in Theorem 4.3, give a necessary and sufficient combinatorial
conditiononmonomials(somewhatakinto patternavoidancein permutations)for
membership in J .
n
ThesecondpartofthepaperconsistsofSections5–9. HerewestudytheStanley-
Reisner simplicial complex ∆(n) whose faces correspond to squarefree monomials
that do not belong to J . This simplicial complex has a surprising amount of
n
combinatorial structure. First, ∆(n) is pure, and all its facets may be built up re-
cursivelyfromfacetsofsmallerStanley-Reisnercomplexes. Second,thisrecurrence
may be translated into a bijection between facets and a combinatorially more nat-
ural set, the binary total partitions, which are enumerated by the double factorial
THE SLOPES DETERMINED BY n POINTS IN THE PLANE 3
numbers(1). These numbers thereforegivethe degree(ormultiplicity) ofthe ideal
J . Third, the description of facets leads to a proof that ∆(n) is shellable and
n
hence Cohen-Macaulay. This part of the paper (Section 8) is very technical; the
argumentsarecombinatoriallyelementarybutdorequirecarefulbookkeeping. The
shellingargumentleadsinturnto arecursivecomputationofthe h-vectorof∆(n);
the coefficients of the h-vectorenumerate perfect matchings by the number of long
pairs (a combinatorialproblem first considered by Krewerasand Poupard [8]).
At this point, we do not yet know that these results on ∆(n) correspond to
properties of the affine slope variety. The missing piece is to show that J is not
n
too small—precisely, that it is the initial ideal of an ideal defining ˜(K ) scheme-
n
S
theoretically. Asitturnsout,itisenoughtoshowthatthedoublefactorialnumbers
give a lower bound for the degree of ˜(K ). We prove this in Sections 10 and 11.
n
S
Our approach is to consider a nested family of algebraic subsets of ˜(K ) called
n
S
flattened slope varieties, whose degree can be bounded from below by a recursive
formula (Theorem 10.4). We show that this recurrence is equivalent to one enu-
meratingcombinatorialobjectscalleddecreasingplanartrees,which,likematchings
and binary total partitions, are enumerated the double factorials. Using the fact
that J is Cohen-Macaulay and has the appropriate codimension and degree, we
n
conclude that J is the (complete) initial ideal of I under an appropriate term
n n
ordering;equivalently,the wheelpolynomialsformaGr¨obnerbasis. Theassertions
of the main theorem follow more or less immediately.
Togetherwith[10],theseresultsconstitutetheauthor’sdoctoraldissertation[9].
The author thanks his thesis advisor, Mark Haiman, for his ongoing support.
2. Preliminaries: Graphs and Tree Polynomials
We assume that the reader is familiar with the elements of graph theory, for
which a good generalreference is [15]. We first fix some notation and terminology.
ThesymbolNdenotesthepositiveintegers. Weabbreviatetheset m,m+1,...,n
{ }
by [m,n].
A graph G is a pair (V,E), where V = V(G) is a finite set of vertices and
E = E(G) is a set of edges, or unordered pairs of distinct vertices e = v,w .
{ }
(Thus we do not allow loops or multiple edges.) For ease of use, we frequently
abbreviate v,w by vw. The vertices v,w are the endpoints of e. A subgraph of
{ }
G is a graph G′ = (V′,E′) with V′ V and E′ E. We use the symbols + and
⊂ ⊂
to denote addition and deletion of edges.
−
The valence of v with respect to an edge set E, written val (v), is the number
E
of edges in E incident to v. (This is more usually called the degree, but we wish to
reserve that term for a different usage.) A vertex of valence 1 is called a leaf. The
support of an edge set E is V(E)= v val (v)>0 . The complete graph K is
E V
{ | }
thegraphwithvertexsetV andeverytwoverticesadjacent;thus E(K ) = |V| .
| V | 2
We abbreviate K by K . For convenience, we frequently ignore the technical
[1,n] n (cid:0) (cid:1)
distinction between an edge set E and the graph (V(E),E).
Letv ,...,v bedistinctvertices. Theedgeset v v ,v v ,...,v v iscalled
1 k 1 2 2 3 k−1 k
{ }
a path from v to v , and if k 3 then the edge set v v ,v v ,...,v v ,v v
1 k 1 2 2 3 k−1 k k 1
≥ { }
is called a cycle or k-cycle. It is frequently convenient to describe a path or cycle
by listing its vertices in order.
A graph G is connected if every pair of vertices belongs to some common path;
it is a tree if it is connected and contains no cycle. Equivalently, a tree may be
4 JEREMYL.MARTIN
defined as a connected graph with E(G) = V(G) 1, or as a graph in which
| | | |−
every pair of vertices belongs to exactly one common path. A spanning tree of G
is a tree T with V(T)=V(G) and E(T) E(G). The connected components of a
⊂
graph are its maximal connected subgraphs.
A graph G = (V,E) is a rigidity pseudocircuit if E = T T′, where T,T′ are
⊔
spanningtreesandthesymbol denotesadisjointunion. Arigiditypseudocircuitis
⊔
arigidity circuit ifitcontainsnootherrigiditypseudocircuitasapropersubgraph.
(For the reasons behind this terminology, see [7].) A spanning tree T E is called
⊂
coupled ifitscomplementisalsoaspanningtree;wedenotethesetofcoupledtrees
of G by Cpl(G).
Let v ,v ,...,v be distinct vertices. The k-wheel W = W(v ; v ,...,v ) is
0 1 k 0 1 k
definedasthegraphwithedges v v ,v v ,...,v v v v ,...,v v ,v v . It
0 1 1 2 0 k 1 2 k−1 k k 1
{ }∪{ }
is easy to check that every wheel is a rigidity circuit [7, Exercise 4.13]; the figure
below shows a 2-tree decomposition of W(v ; v ,...,v ).
0 1 6
v
(cid:8)tH1 (cid:8)t Ht
(cid:8) H (cid:8) H
(cid:8) H (cid:8) H
v6 t(cid:8)HH (cid:8)(cid:8)Htv2 tH(cid:8)H (cid:8)(cid:8)t t Ht
H (cid:8) H (cid:8)
(cid:8)(cid:8)v0(cid:8)Ht(cid:8)HHH = (cid:8)(cid:8)(cid:8)H(cid:8)tHHH ⊔ t
v5 t(cid:8)HH (cid:8)H(cid:8)tv3 t(cid:8) Ht Ht H (cid:8)(cid:8)t
H (cid:8) H (cid:8)
Ht(cid:8) t H(cid:8)t
v
4
Thevertexv iscalledthecenter ofW,andtheverticesv ,...,v areitsspokes.
0 1 k
An edge joining two spokes is a chord of the wheel; an edge joining a spoke to the
center is a radius. We denote the sets of chords and radii by Ch(W) and Rd(W)
respectively. It is notationally convenient to set v = v , so that Ch(W) =
k+1 1
v v i [1,k] .
i i+1
{ | ∈ }
We now describe certain polynomials associated to rigidity circuits. Let n 2
≥
be anintegerandk analgebraicallyclosedfield. We will workoverthe polynomial
ring in n variables
2
R =k[m , ..., m ].
(cid:0) (cid:1) n 12 n−1,n
To each edge set E E(K ) we associate the squarefree monomial
n
⊂
m = m . (2)
E e
e∈E
Y
For every rigidity circuit C in K , there is an irreducible polynomial, the tree
n
polynomial of C, defined up to sign and having the form
τ(C)= ε(T)m , (3)
T
T∈XCpl(C)
whereε(T) 1, 1 . By [10,Theorem5.4],the tree polynomialisirreducible and
∈{ − }
homogeneous of degree V(C) 1= E(C)/2.
| |− | |
We summarizethe constructionofτ(C); formoredetailsandexamples,see[10].
For1 i<j n,regardtheedge i,j asanorientededge(i,j),andformallyset
≤ ≤ { }
(j,i) = (i,j). Fix a spanning tree T E(C) (not necessarily coupled), and let
− ⊂
T =E(C) T. Foreveryedgee=vw T,theedgesetT+econtainsauniquecycle
\ ∈
P (e),whichwemayregardasasetoforientededges (v,w), (w,w ), ..., (w ,v) .
T 1 r
{ }
THE SLOPES DETERMINED BY n POINTS IN THE PLANE 5
LetM be the (V 1) (V 1)squarematrix withrowsindexedby edgese T
| |− × | |− ∈
and columns indexed by the edges f T, and whose (e,f) entry is
∈
m m if f P (e),
e f T
− ∈
M = m m if f P (e), (4)
e,f  f e T
− − ∈
0 otherwise.
Thetreepolynomialisthendefinedasτ(C)=detM. Uptosign,thisconstruction
is independent of the choice of the tree T.
The affine slope variety ˜(n) = ˜(K ) is defined as follows. Let p ,...,p be
n 1 n
S S
n distinct points in the affine plane A2, with no two points lying on the same
k
vertical line. Let m k be the slope of the unique line joining p and p . Thus
ij i j
∈
(m ,...,m ) is a point in affine n -space A2 =SpecR , and ˜(n) is defined
12 n−1,n 2 k n S
as the closure of the locus of all such points arising from the data (p ,...,p ).
1 n
(cid:0) (cid:1)
By [10, Theorem 5.6], the ideal generated by the tree polynomials
I = τ(C) C K is a rigidity circuit (5)
n n
| ⊂
cutsouttheaffineslopev(cid:10)ariety ˜(n)insideSpecR . Infact,I(cid:11) isgeneratedbythe
n n
S
tree polynomials of wheel subgraphs of K . We omit the proof of this fact, since
n
weshalleventuallyprovethe followingstrongerresult: the wheelpolynomialsform
a Gr¨obner basis for I with respect to any of a large class of term orderings.
n
3. Vertex Valences in Coupled Trees
Thissectioncontainsseveraltechnicalfactsconcerningthevalencesofverticesin
couplesspanningtreesofawheel. Theseobservationsleadeventuallytoanexplicit
identification of the initial monomial of a wheel polynomial τ(W) with respect to
a certain term ordering.
Throughout this section, we fix a k-wheel W = W(v ; v ,...,v ) E(K ). If
0 1 k n
⊂
T E(W) is a coupled spanning tree of W, we set T :=E(W) T.
⊂ \
For each i [1,k] and each coupled tree T Cpl(W), either val (v ) = 1 or
T i
∈ ∈
val (v )=2. Inaddition,notallspokesv havethesamevalence,sinceT isneither
T i i
Ch(W) nor Rd(W). Thus val may be regarded as a nonconstant function from
T
[1,k] to [1,2].
Lemma 3.1. Let T Cpl(W) and i,j [1,k]. Then T contains at least one of
∈ ∈
the following four edges: v v , v v , v v , and v v .
0 i 0 j i i+1 j−1 j
Proof. Suppose not. Let i,j be a counterexample such that j i is as small as
−
possible. If necessary, we may reindex the spokes so that i j. If j = i, then
≤
val (v )=3,whichisimpossible. Ifj i=1,thenT containsthecyclev ,v ,v ,v ,
T i 0 i j 0
−
and if j i=2, then T contains the cycle v ,v ,v ,v ,v .
0 i i+1 j 0
−
Now suppose j i>2. Since T contains the path v ,v ,v ,v ,v , it cannot
i+1 i 0 j j−1
−
containthe path v ,v ,...,v . Letk andℓ be the leastandgreatestindices,
i+1 i+2 j−1
respectively, such that v v T, v v T, and i<k ℓ<j. Now v v T,
k k+1 ℓ−1 ℓ 0 k
6∈ 6∈ ≤ 6∈
otherwiseT containsthecyclev ,v ,v ,...,v ,v . Forasimilarreason,v v T.
0 i i+1 k 0 0 ℓ
6∈
But then k,ℓ is a counterexample to the lemma, and ℓ k < j i, which
contradicts the choice of i and j. | − | | − | (cid:3)
6 JEREMYL.MARTIN
Lemma 3.2. For each T Cpl(W), at least one of the following conditions is
∈
true: either
for all i [1,k], v v T if and only if v v T,
0 i i i+1
∈ ∈ ∈
or
for all i [1,k], v v T if and only if v v T.
0 i i−1 i
∈ ∈ ∈
Proof. Suppose that both conditions fail. That is, there exists i [1,k] such that
∈
either v v ,v v T or v v , v v T. Interchanging T and T if necessary,
0 i i i+1 0 i i i+1
∈ ∈
we may assume the former. Moreover, there exists j [1,k] such that either
∈
v v , v v T or v v , v v T. The former is ruled out by Lemma 3.1, so
0 j j−1 j 0 j j−1 j
∈ ∈
the latter must hold; in particular i=j.
6
Sincev v ,v v T,theedgev v mustbelongtoT. InparticularT contains
0 i i i+1 0 i+1
∈
the path v ,v ,v ,v . Thus T does not contain the path v ,v ,...,v .
j−1 j 0 i+1 i+1 i+2 j−1
Let h be the largest number in [i+1,j 1] such that v v T. Since the path
h−1 h
− ∈
v ,v ,...,v ,v iscontainedinT,theedgev v mustbelongtoT. Itfollowsthat
h h+1 j 0 0 h
T contains the edges v v , v v , v v , and v v , contradicting Lemma 3.1. (cid:3)
0 i i i+1 h−1 h 0 h
Proposition 3.3. Let d : [1,k] [1,2] be a nonconstant function. Then there
→
exist exactly two coupled trees of W for which val =d.
T
Proof. To simplify the notation, write m for m . Also, we put v = v .
i,j vivj k+1 1
Putting T =Rd(W) in (4), the matrix M becomes
m m m m 0 ... 0
0,1 1,2 1,2 0,2
− −
0 m m m m ... 0
0,2 2,3 2,3 0,3
 − − 
... ... ... ...
 
m0,k−mk,1 0 ... ... mk,1−m0,1
 
Taking the determinant, we obtain
k k
τ(W) = (m m ) + ( 1)k−1 (m m )
0,i i,i+1 i,i+1 0,i+1
− − −
i=1 i=1
Y Y
k k
= (m m ) (m m ). (6)
0,i i,i+1 0,i+1 i,i+1
− − −
i=1 i=1
Y Y
The monomial m appears in both products in (6), once with coefficient +1
Ch(W)
and once with 1. The same is true for the monomial m . One may easily
Rd(W)
−
verify that no other cancellation occurs. Accordingly, to enumerate the number of
coupled trees by the valences of spokes, we may substitute z z for m and
i i+1 i,i+1
z for m in (6) (where the z are indeterminates) and change all the ’s to +’s.
i 0,i i
−
This yields the expression
k k k k
(z +z z ) + (z +z z ) 2 z + z2 = 2 zd(i),
i i i+1 i+1 i i+1 − i i! i
i=1 i=1 i=1 i=1 d
Y Y Y Y X
where the sum is taken over all nonconstant functions d:[1,k] [1,2]. (cid:3)
→
Remark 3.4. Proposition 3.3 has the following consequence, which may also be
obtained by direct counting: every k-wheel has exactly 2k+1 4 coupled spanning
−
trees.
THE SLOPES DETERMINED BY n POINTS IN THE PLANE 7
Definition 3.5. Let d : [1,k] [1,2] be a nonconstant function. The type of a
→
chord v v with respect to d is the pair of numbers d(i),d(i+1). The type of a
i i+1
radius v v is the pair d(i 1),d(i+1). If T Cpl(W), we define the type of an
0 i
− ∈
edge with respect to T to be its type with respectto d=val . For brevity, we will
T
speak of type-11 chords, type-12 radii, etc.
Lemma 3.6. Let T Cpl(W), and define the types of chords and radii of W with
∈
respect to the function val . Then:
T
(i) Every type-22 chord belongs to T.
(ii) Every type-11 chord belongs to T.
(iii) Every type-22 radius belongs to T.
(iv) Every type-11 radius belongs to T.
Proof. Let v v be a chord. If (i) fails, then the edges v v , v v , v v ,
i i+1 i−1 i i+1 i+2 0 i
v v allbelong to T. If (ii) fails, then those edges allbelong to T. In either case,
0 i+1
Lemma 3.1 is contradicted.
Now let v v be a radius Statements (iii) and (iv) are equivalent (switch T and
0 i
T), so we prove only (iii). If val (v ) = 2, then T contains the chords v v and
T i i i+1
v v by parts (i) and (ii) of the lemma, so the radius v v belongs to T. On
i−1 i 0 i
the other hand, if val (v ) = 1 and v v T, then T contains the chords v v
T i 0 i i i+1
∈
and v v . Since val (v ) = val (v ) = 2 by hypothesis, the edges v v ,
i−1 i T i+1 T i−1 i−2 i−1
v v , v v , v v must all belong to T, which contradicts Lemma 3.1. (cid:3)
i+1 i+2 0 i−1 0 i+1
Lemma 3.7. Let T Cpl(W) and 1 i<j k. Suppose that either
∈ ≤ ≤
val (v )=1, val (v )= =val (v )=2, val (v )=1 (7a)
T i T i+1 T j T j+1
···
or
val (v )=2, val (v )= =val (v )=1, val (v )=2. (7b)
T i T i+1 T j T j+1
···
Then exactly one of the two chords v v , v v belongs to T.
i i+1 j j+1
Proof. Supposethat(7a)holds. Thenthechordsv v ,v v ,...,v v all
i+1 i+2 i+2 i+3 j−1 j
belongtoT. Ifbothv v andv v belongtoT,thenthepathv ,v ,...,v
i i+1 j j+1 i i+1 j+1
is a connected component of T, which is impossible. On the other hand, if neither
of those chords belong to T, then v v and v v both belong to T. But then T
0 i+1 0 j
contains the cycle v ,v ,v ,...,v ,v , which is impossible, If we assume (7b)
0 i+1 i+2 j 0
instead of (7a), the same argument goes through, switching 2 with 1 and T with
T. (cid:3)
Analternateformulationofthislemmaisasfollows. Letanonconstantfunction
d : [1,k] [1,2] be given, and let T be a coupled tree with val = d. Traverse
T
→
the chordsofW inorder,coloringthe type-12chords(ofwhichthereareapositive
even number) alternately red and blue. Then either the red chords all belong to T
and the blue chords all belong to T, or vice versa. Moreover,choosing the color of
a single type-12 chordsuffices to determine the rest. Having made such a choice, a
radius v v belongs to T exactly when d(i) T v v , v v =1.
0 i i−1 i i i+1
−| ∩{ }|
Alternatively, if the function d = val is given, then to determine T uniquely
T
it suffices to specify whether a single type-12 radius v v belongs to T or to T.
0 i
Without loss of generality v v is of type 11 or 22, and v v is of type 12. The
i−1 i i i+1
value of d(i) determines whether or not v v belongs to T, so the rest of T is
i i+1
determined uniquely as in the preceding paragraph. That is:
8 JEREMYL.MARTIN
Proposition3.8. Let T Cpl(W). Definethe typeof each edge in W with respect
∈
to val . Then T contains all type-22 chords, all type-11 radii, half the type-12
T
chords, in alternation, and a corresponding half of the type-12 radii.
We conclude our technical preliminaries with two results describing the condi-
tions under which a pair of complementary spanning trees may swap edges.
Lemma 3.9. Let T Cpl(W). Suppose that v v and v v belong toT, so that
i−1 i i i+1
∈
v v T. Assume without loss of generality that the path in T from v to v passes
0 i i 0
∈
through v . Then W admits the 2-tree decompositions
i+1
E =T v v +v v , E =T v v +v v
1 0 i i i+1 2 i i+1 0 i
− −
and
F =T v v +v v , F =T v v +v v .
1 0 i−1 i−1 i 2 i−1 i 0 i−1
− −
Proof. Clearly E E = F F = W. The edge set E is a tree because v is a
1 2 1 2 1 i
⊔ ⊔
leafof T, andE is a tree because v and v are in different connected components
2 i 0
ofT v v . Meanwhile,F isatreebecausev andv areindifferentconnected
i i+1 1 i i−1
−
components of T v v , and F is a tree because v and v are in different
0 i−1 2 i−1 0
connected compon−ents of T v v . (cid:3)
i−1 i
−
Lemma3.10. LetT Cpl(W). Supposethatv v T andthatv v , v v T,
i−1 i 0 i i i+1
∈ ∈ ∈
so that v v T. Then:
0 i+1
∈
(i) If T contains at least one radius other than v v , then W admits the 2-tree
0 i
decomposition
E =T v v +v v , E =T v v +v v .
1 0 i i−1 i 2 i−1 i 0 i
− −
(ii) If T contains at least one radius other than v v , then W admits the 2-tree
0 i+1
decomposition
F =T v v +v v , F =T v v +v v .
1 i i+1 0 i+1 2 0 i+1 i i+1
− −
Proof. ClearlyW isthedisjointunionofE andE (resp.F andF ),soitsuffices
1 2 1 2
to show that these edge sets are in fact trees.
(i) E isatreebecausev isaleafofT. Ifthepathfromv tov inT doesnot
2 i i−1 i
gothroughv , then itmust be Ch(W) v v . But thenT containsatleastk 1
0 i−1 i
− −
chords and two radii, which is impossible. Therefore v and v lie in different
i−1 i
connected components of T v v , and E is a tree.
0 i 1
−
(ii) The path from v to v in T is just v ,v ,v , so v and v belong to
0 i+1 0 i i+1 0 i+1
different connected components of T v v . Hence F is a tree. If the path from
i i+1 1
−
v to v in T does not go through v , then it must be Ch(W) v v , which is
i i+1 0 i i+1
−
impossible. So v and v are in different connected components of T v v .
i i+1 0 i+1
Hence F is a tree. − (cid:3)
2
4. The Leading Tree of a Wheel
The main result of this section is Theorem 4.3, in which we describe explicitly
the ideal generated by the initial terms of wheel polynomials. Fix once and for all
the following lexicographic order > on the variables m :
ij
m >m >...>m >m >...
12 13 1n 23
The corresponding total order for edges of K is
n
12>13>...>1n>23>... (8)
THE SLOPES DETERMINED BY n POINTS IN THE PLANE 9
Wenextextend>toatermorderingonR ,graded lexicographic order,which
n
is defined as follows: maij > mbij if either
i,j ij i,j ij
Q a > Q b , or
ij ij
(9)
Xa = Xb and a > b ,
ij ij kℓ kℓ
where m is the greateXst variableX(in lexicographic order) such that a =b .
kℓ kℓ kℓ
6
Associating edge sets with square-free monomials as in (2), we may regard the
termorderingonR asdefininganextensionoftheorderingonedges(8)toatotal
n
order on subsets of E(K ). Then (9) becomes the following: for E,F E(K ),
n n
⊂
E >F ifeither E > F ,orelse E = F andmax(E#F) E,wherethe symbol
| | | | | | | | ∈
# denotes the symmetric difference operator.
Given a wheel W E(K ), we wish to identify the leading tree LT(W) of W,
n
⊂
thatis,thecoupledtreeofW correspondingtotheleadingmonomialofτ(W)(with
respect to graded lex order). We begin by computing the valence of each spoke of
LT(W),using the toolsdevelopedinthe previoussection. By Proposition3.8, this
will rule out all but two possibilities for the leading tree.
Proposition 4.1. Let W =W(v ;v ,...,v ), with V =V(W) [1,n]. Then:
0 1 k
⊂
(i) Suppose that v =min(V). Then val (v )=k 1.
0 LT(W) 0
−
(ii) Suppose that v =max(V). Then val (v )=1.
0 LT(W) 0
(iii) Suppose that v min(V),max(V) . Then for all i [1,k],
0
6∈{ } ∈
1 if v >v ,
i 0
val (v )=
LT(W) i
(2 if vi <v0.
In particular, if v is the rth largest member of V, where r [2,k], then
0
∈
val (v )=r 1.
LT(W) 0
−
Proof. (i,ii) Suppose v = min(V). We will show that if T is a coupled tree with
0
val (v )<k 1, then T cannot be the leading tree of W. Note that T contains at
T 0
−
leasttwo radii, sayv v andv v . At leastone ofthe chordsv v , v v belongs
0 i 0 j i−1 i i i+1
to T. If both do, then by Lemma 3.9, at least one of
T =T v v +v v , T =T v v +v v
1 i−1 i 0 i 2 i i+1 0 i
− −
is coupled. If v v T and v v T, then T is coupled by Lemma 3.10 (i).
i−1 i i i+1 1
∈ ∈
But T >T, so T =LT(W) as desired. The proof of (ii) is analogous.
1
6
(iii) Suppose that v is neither the minimum nor the maximum element of V.
0
Let T be a coupled tree of W such that val (v ) = 1 for some v < v . To show
T i i 0
that T = LT(W), we will construct a tree T′ Cpl(W) with T′ > T. There are
6 ∈
two cases to consider.
Case 1: v v T. Then T contains the chords v v and v v . Without loss
0 i i−1 i i i+1
∈
ofgenerality,we may assumethat the path fromv to v in T passes throughv .
i 0 i+1
Thenv v T. ByLemma3.9, the treeT′ =T v v +v v is coupled,and
0 i−1 0 i−1 i−1 i
∈ −
v v >v v , so T′ >T.
i−1 i 0 i−1
Case 2: v v T. Without loss of generality, v v T and v v T, so
0 i i i+1 i−1 i
∈ ∈ ∈
v v T. LetT′ =T v v +v v . Note that T′ >T. If T′ is coupled, then
0 i+1 0 i+1 i i+1
∈ −
wearedone. Otherwise,Lemma 3.10(i)implies thatv v is the uniqueradiusin
0 i+1
T, that is, T is the path v , v , v , ..., v , v . Since v = max(T), we may
0 i+1 i i−2 i−1 0
6
10 JEREMYL.MARTIN
choosej suchthatv >v ;notethatj =i. ThenthetreeT′ =T v v +v v
j 0 j j+1 0 j+1
is coupled, and T′ >T. 6 − (cid:3)
Proposition 4.2. Let W E(K ) be a k-wheel with vertices V = V(W) and
n
⊂
center v .
0
(i) Suppose v =min(V). Label the spokes so that W =W(v ;v ,...,v ) with
0 0 1 k
v =max v ,...,v and v >v . Then
1 1 k 2 k
{ }
LT(W)=Rd(W) v v +v v .
0 1 k k+1
−
(ii) Suppose v =max(V). Label the spokes so that W =W(v ;v ,...,v ) with
0 0 1 k
v v =min(Ch(W)) and v >v . Then
1 2 1 2
LT(W)=Ch(W) v v +v v .
1 2 0 2
−
Proof. (i) ByProposition4.1(i), LT(W)containsexactlyonechord. Sincev v >
0 i
v v for all i,j, the unique radius not in LT(W) must be min(Rd(W)) = v v .
j j+1 0 1
This implies the desired result because v v >v v .
k k+1 1 2
(ii) For each 1 [1,k], define
∈
T =Ch(W)+v v min(v v , v v ).
i 0 i i−1 i i i+1
−
By Proposition 4.1 (ii), LT(W) contains exactly one radius, so LT(W) = T for
i
somei. NotethatT =Ch(W)+v v v v fori=1,2;inparticularmax(T #T )=
i 0 i 1 2 1 2
−
v v T . On the other hand, for i>2, we have
0 2 2
∈
max(T #T ) = max(v v , v v , v v , min(v v ,v v ))
i 2 0 i 0 2 1 2 i−1 i i i+1
= min(v v , v v ) T .
i−1 i i i+1 2
∈
We conclude that LT(W)=T . (cid:3)
2
In the case that v min(V),max(V) , we have by Proposition 4.1 (iii)
0
6∈{ }
1 if v >v
i 0
val (i)=
LT(W)
(2 if vi <v0.
ByProposition3.3,thereareexactlytwocoupledtreesT,T′ Cpl(W)satisfying
∈
these conditions. Moreover
T T′ = chords of type 22 radii of type 11 ,
∩ { }∪{ }
T#T′ = chords of type 12 radii of type 12 .
{ }∪{ }
Define the critical edge of W to be the maximum element of T#T′. Thus LT(W)
is whichever of T,T′ contains the critical edge.
Theorem 4.3. Let T be a tree with V(T) [1,n]. Then the following are equiva-
⊂
lent:
(i) There exists a wheel W K such that T =LT(W).
n
⊂
(ii) T contains a path (v ,...,v ) satisfying the conditions
1 k
k 4,
≥
max(v ,...,v )=v ,
1 k 1 (10)
max(v ,...,v )=v ,
2 k k
v >v .
2 k−1