On the geometry of real or complex supersolvable line arrangements

0.17 MB·English

by  Benjamin Anzis

ON THE GEOMETRY OF REAL OR COMPLEX SUPERSOLVABLELINE ARRANGEMENTS
BENJAMINANZISANDS¸TEFANO.TOHAˇNEANU
ABSTRACT. Givenarank3realarrangementAofnlinesintheprojectiveplane,theDirac-Motzkinconjecture
5 (provedbyGreenandTaoin2013)statesthatfornsufficientlylarge,thenumberofsimpleintersectionpoints
1 of A is greater than or equal to n/2. With a much simpler proof we show that if A is supersolvable, then
0 the conjecture is true for any n (a small improvement of original conjecture). The Slope problem (proved
2 by Ungar in 1982) states that n non-collinear points in the real plane determine at least n−1 slopes; we

y show that this is equivalent to providing a lower bound on the multiplicity of a modular point in any (real)
a supersolvable arrangement. Inthesecond part wefindconnections between thenumber of simplepoints of
M a supersolvable line arrangement, over any field of characteristic 0, and the degree of the reduced Jacobian
schemeofthearrangement. OverthecomplexnumberseventhoughtheSylvester-Gallaitheoremfailstobe

9 true,weconjecturethatthesupersolvableversionoftheDirac-Motzkinconjectureistrue.

]
O
C
1. INTRODUCTION
.
h
t Let bealinearrangementinP2. Supposeℓ1,...,ℓn K[x,y,z]arethedefiningequationsofthelines
a A ∈
of ,andassumedim Span(ℓ ,...,ℓ ) = 3(i.e.that hasfullrank,equalto3).
m K 1 n
A A
An intersection point P of any two of the lines of is a singularity, denoted by P Sing( ). The
[ A ∈ A
numberoflinesof thatintersect atasingular pointP Sing( )iscalledthemultiplicity ofP,denoted
A ∈ A
2 m(P, ), or just m if the line arrangement is clear. The multiplicity of , denoted m( ), is m( ) =
P
v A A A A
max m P Sing( ) . Simplepointsaresingularpointsofmultiplicity 2,andthesetofsuchpointsin
9 { P | ∈ A }
3 willbedenoted Sing2( ).
A A
0 In general, a hyperplane arrangement is supersolvable if its intersection lattice has a maximal chain of
4 modular elements ([19]). For the case of line arrangements P2, supersolvability is equivalent to the
0 A ⊂
existence ofaP Sing( )suchthatforanyotherQ Sing( ),thelineconnecting P andQbelongsto
.
1 ∈ A ∈ A
(inother words,there isanintersection pointP that“sees” alltheother intersection points through lines
0 A
5 of ). Such a point P will be called modular. Supersolvable hyperplane arrangements are an important
A
1 classofhyperplanearrangements,capturingagreatdealoftopological(overCtheyarealsoknownasfiber-
:
v typearrangements,see[11]),combinatorial, andhomologicalinformation(see[15]forlotsofinformation).
i Due to their highly combinatorial content, they are the best understood arrangements, yet there are some
X
questions andproblemsliketheonesweinvestigate here,whichwefeeldeservetobeanalyzed.
r
a
Thegoalsofthesenotesaretounderstand inaset-theoretical andcombinatorial contextthesingularities
of various types of supersolvable line arrangements; e.g. how many simple singularities such a divisor
can have (and in general, how many singularities are there all together), or what is the multiplicity of a
modular singularity. The driving forces behind our results are the following two classical problems about
configurationofpointsintherealplaneandthelinestheydetermine. Sinceanytwopointsdeterminealine,
under the classical duality (points lines) any two lines intersect at a point, so we are translating these
↔
problemsintoquestions aboutlinearrangements intheprojective plane.
The Dirac-Motzkin Conjecture states that if is a full rank real arrangement of n lines in P2, then, for
A
n sufficiently large, Sing ( ) n/2. This conjecture has been proven only recently by Ben Green and
2
| A | ≥
Terence Tao in [8], where they also solve another famous related problem (the Orchard Problem). They
2010MathematicsSubjectClassification. Primary:52C30;Secondary:52C35,05B35,06C10.
Keywordsandphrases. Dirac-Motzkinconjecture,Slopeproblem,supersolvablearrangements.
Authors’ addresses: Department of Mathematics, University of Idaho, Moscow, ID 83844, [email protected],
[email protected].
1
2 BENJAMINANZISANDS¸TEFANO.TOHAˇNEANU
presenttheuniqueclassofexamples,uptoprojectivetransformations, (called“Bo¨ro¨czkyexamples”,see[8,
Proposition 2.1 (i)]) for which the bound is attained (of course n must be even). On a side note, there are
only two known examples for which Sing ( ) < n/2 (see the brief history in Section 2); based on this,
2
| A |
Gru¨nbaum conjectured thatifn =7,13, then Sing ( ) n/2(see[10]).
2
6 | A | ≥
With a very short proof we show (see Theorem 2.4) that if P2 is a full rank real supersolvable
• A ⊂
linearrangement, then Sing ( ) |A|. Thoughitisnotstatedin[8],byasimplecalculation oneshould
| 2 A | ≥ 2
remark that the Bo¨ro¨czky examples are supersolvable line arrangements. Wenote here that our proof may
give insights towards proving Dirac-Motzkin conjecture for complex supersolvable line arrangements (see
thediscussions inSection3.2).
In [18], Scott proposed the following Slope Problem: Given n 3 points in the (real) plane, not all
≥
collinear, they determine atleast n 1 lines ofdistinct slopes. The firstto prove this conjecture wasPeter
−
Ungarin[23],usingabeautiful yetdifficultargument.
In Proposition 2.7 we show that the Slope problem is equivalent to showing that if is any full rank
• A
real supersolvable line arrangement in P2, then m( ) ( 1)/2. So providing an independent proof
A ≥ |A|−
(that wedon’t have at the moment) ofthe statement about the supersolvable line arrangements will lead to
analternativeproofoftheslopeproblem.
Atthe beginning ofSection 3theresults presented are valid over anyfieldof characteristic 0. Weprove
(Proposition 3.1)alowerbound for Sing ( ),when isasupersolvable line arrangement, that involves
2
| A | A
Sing( ), and we remark that the bound is attained also by the Bo¨ro¨czky examples. Next we give an
| A |
example which shows the upper bound for the degree of the reduced Jacobian scheme of a supersolvable
linearrangement obtained in[22,Proposition 3.1]issharp.
The remaining part of Section 3 is dedicated to analyzing the Dirac-Motzkin conjecture for complex
supersolvable arrangements (see Conjecture 3.2; basically the first bullet above with “real” replaced by
“complex”).
Lotsofourarguments arebased onthefollowing simpleobservation, whichisderivedfromthefactthat
anytwolinesinaprojectiveplanemustintersect: Let P2beanyarrangementofnlines,overanyfield.
A ⊂
Letℓ beanyline,andsuppose ithasexactlysintersection points: P ,...,P . Then
1 s
∈ A
s
(m 1) = n 1.
Pi − −
Xi=1
2. SUPERSOLVABLE LINE ARRANGEMENTS OVER THE REAL NUMBERS
Let beafullranksupersolvable linearrangement inP2. Suppose = n,anddenotethemultiplicity
A |A|
of by m := m( ). We begin by reproving [21, Lemma 2.1], which is a statement that is true when
A A
workingoveranyfield.
Lemma2.1. Let beasupersolvable linearrangement withamodularpointP. LetQ Sing( )notbe
A ∈ A
modular. Then
m(P, ) >m(Q, ).
A A
Therefore,anyintersection pointofmaximummultiplicity mismodular.
Proof. Lets = m(Q, )andletthelinespassing through Qbe ℓ ,...,ℓ . SinceQisnotmodular,
1 s
A { } ⊂ A
thereexistsapointP′ Sing( )withP′ / ℓi,i= 1,...,s. BecauseP ismodular,thereisalineℓPP′
∈ A ∈ ∈A
through P andP′. SinceP′ Sing( )thereisanother lineℓ through P′ andnotpassing through P.
∈ A ∈ A
ℓ intersects the ℓ ’s in s points, say P ,...,P Sing( ). Since P is modular, there are lines ℓ′
i { 1 s} ⊂ A i ∈ A
through P andP . Countingthenumberoflinesthrough P wegetm(P, ) s+1 > s = m(Q, ).
i
A ≥ A
LetD Sing( )besuchthatm(D, ) = m. IfDisnotmodular,letP beamodularpointof . Then
wehavem∈(P, )A> m(D, )= m,conAtradicting themaximalityofm. A (cid:3)
A A
ONTHEGEOMETRYOFREALORCOMPLEXSUPERSOLVABLELINEARRANGEMENTS 3
2.1. Dirac-Motzkin conjecture for real supersolvable arrangements. In [20], Sylvester proposed the
following problem: if is a full rank real line arrangement in P2, then Sing ( ) = . In 1944, Gallai
2
A A 6 ∅
solvedthisproblem (see[7]),whichisnowknownintheliterature asSylvester-Gallai Theorem.
DiracandMotzkin conjectured thatif isafullrankrealarrangement ofnlinesinP2,thenforn n
0
A ≥
one has Sing ( ) n/2. The existence of the absolute constant n is justified in part by the re-
2 0
| A | ≥
sults of Kelly and Moser ([14, Theorem 3.6]) and Csisma and Sawyer ([4, Theorem 2.15]) who proved
Sing ( ) 3n/7, and Sing ( ) 6n/13 if n > 7, respectively, as well as by the examples where
2 2
| A | ≥ | A | ≥
these two bounds are attained. For n = 7, the non-Fano arrangement ([14, Figure 3.1] or [4, Fig. 3]) has
Sing ( ) = 3, which clearly fails to be n/2; a more complicated example of Crowe and McKee ([3]
2
| A | ≥
or [4, Fig. 4]) is an arrangement of n = 13 lines with Sing ( ) = 6, so that again 6 (cid:3) n/2. Hence we
2
| A |
musthaven 14.
0
≥
The next lemma is crucial to the proof of the main result. The condition that R is our base field is
necessary fortheproof.
Lemma2.2. Let beafull ranksupersolvable reallinearrangement inP2. LetP Sing( )beapoint
A ∈ A
ofmaxmultiplicity(andhence,byLemma2.1,amodularpoint). Then,anylineof notpassingthroughP
A
hasatleastonesimplesingularity.
Proof. Definem = m(P, ). Afteralinear change ofvariables, wemayassume that P = [0,0,1], sothat
A
thelinespassing through P areparallel andvertical. Suppose forcontradiction thatthereexistsℓ with
∈ A
P / ℓ,andℓ Sing ( ) = . SinceP ismodular, wehaveℓ Sing( ) = P ,...,P whereeachP
2 1 m i
∈ ∩ A ∅ ∩ A { }
liesatthe intersection ofℓandalinethrough P, sayℓ′. Wemayassume thatthe P ’sareordered from left
i i
toright(i.e.fromleasthomogenized firstcoordinate tolargest).
Foreachi,letℓ beanotherlinethroughP differentfromℓandℓ′. Notethattheℓ aredistinct,since
i ∈ A i i
ℓistheuniquelinepassing throughtheP .
i
Leta be the slope of ℓ for each i. Note that each a is finite, since P = [0,0,1]. Wemay assume that
i i i
a 0.
1
≥
Consider the case that all the a have the same sign. We may assume that all a > 0. Let Q be the
i i
intersection point ofℓ and ℓ . IfQlies “above” ℓ, then Qnecessarily lies tothe“right” of ℓ and hence,
1 m m
since a 0, to the “right” of any of the ℓ′. Similarly, if Q lies “below” ℓ, then Q necessarily lies to the
m ≥ i
“left” ofℓ andhence, since a 0, tothe“left” ofanyoftheℓ′. Ineither case, Qisnot onalinethrough
1 1 ≥ i
P,contradicting thatP ismodular.
Intheremaining case, wemayassume thata > 0. Letj bethe least index inwhicha < 0,and letQ
m j
betheintersection pointofℓ andℓ . ThenQmustlie“between” ℓ′ andℓ′,andsocannotlieonaline
j−1 j j−1 j
through P. Thisagaincontradicts thatP ismodular. (cid:3)
Corollary 2.3. Let beafullranksupersolvable reallinearrangement inP2. Then
A
Sing ( ) +m( ) .
2
| A | A ≥ |A|
Proof. Thereare exactly m( )lines of not passing through P, and, by Lemma2.2, each ofthese
|A|− A A
must have a simple singularity of on it. Because P is modular, such a simple point can occur only
A
as the intersection of a line of through P and of a line of not through P. Therefore Sing ( )
2
m( ). A A | A | ≥(cid:3)
|A|− A
Theorem2.4. Let beafullranksupersolvable reallinearrangement inP2. Then
A
Sing ( ) |A|.
2
| A | ≥ 2
Proof. Letn := ,andletm := m( ). LetP Sing( )withm(P, ) = m. Then, byLemma2.1,P
|A| A ∈ A A
isamodularpoint.
Case1. Supposem n/2. Thenn m n/2,andfromCorollary2.3wehave Sing ( ) n/2.
2
≤ − ≥ | A | ≥
4 BENJAMINANZISANDS¸TEFANO.TOHAˇNEANU
Case 2. Suppose m > n/2. In general, any P2 with each line containing a simple singularity has
A ⊂
Sing ( ) /2. Suppose that there is ℓ with no simple singularity on it. From Lemma 2.2 we
2
| A | ≥ |A| ∈ A
necessarily havethatP ℓ.
∈
Letℓ′ withP / ℓ′. Let Q = ℓ ℓ′. Thenm(Q, ) 3.
∈ A ∈ { } ∩ A ≥
Supposeℓ′ Sing( ) = Q ,...,Q ,Q ,withQ = Q,andm(Q , ) = 2(fromLemma2.2).
1 m−1 m m 1
∩ A { } A
Denoten := m(Q , ),i = 1,...,m. Itisclearthat
i i
A
(1) (n 1)+(n 1)+ +(n 1) = n 1.
1 2 m
− − ··· − −
1
| {z }
Case2.1. Suppose3n/4 m > n/2. Ifn 3,foralli= 2,...,m,then,from(1)above
i
≥ ≥
n 2 2(m 1)
− ≥ −
giving n 2m; contradiction.d Hence we can assume n = 2 as well, so that ℓ′ has at least two simple
2
≥
singularities onit. Thisleadsto
Sing ( ) 2(n m) 2n 3n/2 = n/2.
2
| A | ≥ − ≥ −
Case2.2. Suppose 5n/6 m > 3n/4. From the previous case wecan assume n = n = 2. Suppose
1 2
≥
n 3,foralli = 3,...,m. Then,from(1),
i
≥
n 3 2(m 2)
− ≥ −
giving (n +1)/2 m. This contradicts m > 3n/4 and the fact that n 3 ( has rank 3). So we can
≥ ≥ A
assumen = 2aswell,whichmeansthatℓ′ hasatleastthreesimplesingularities onit. Thisleadsto
3
Sing ( ) 3(n m) 3n 5n/2 = n/2.
2
| A | ≥ − ≥ −
Case 2.u-1. Suppose 2u−1 n m > 2u−3 n, where u is some integer > 3. From the inductive
2u · ≥ 2u−2 ·
hypothesis (since m > 2u−3 n > 2(u−1)−3 n)we can assume n = = n = 2. Suppose n 3,
2u−2 · 2(u−1)−2 · 1 ··· u−1 i ≥
foralli = u,...,m. Then,from(1),
n u 2(m u+1)
− ≥ −
giving(n+u 2)/2 m.
Wehave n −1 m≥, from the full rank hypothesis. Then n 1 > 2u−3 n, which gives n > 2u 2.
− ≥ − 2u−2 · −
Theinequality (n+u 2)/2 m obtained before, together withm > 2u−3 n,leads tou 1 > n. But
− ≥ 2u−2 · −
we previously obtained n > 2u 2, a contradiction. Hence we can assume n = 2 as well, which means
u
−
thatℓ′ hasatleastusimplesingularities onit. Thisleadsto
Sing ( ) u(n m) un (2u 1)n/2 = n/2.
2
| A | ≥ − ≥ − −
(cid:3)
Example2.5. Form 3,theBo¨ro¨czkyconfiguration ofpointsis
≥
2πj 2πj
X := [cos ,sin ,1] : 0 j < m
2m
{ m m ≤ }
Λ1
| {z }
πj πj
[ sin ,cos ,0] :0 j < m .
∪{− m m ≤ }
Λ2
| {z }
Let P2 betherealarrangement ofthe2mlinesdualtothepointsofX .
2m 2m
A ⊂
ONTHEGEOMETRYOFREALORCOMPLEXSUPERSOLVABLELINEARRANGEMENTS 5
The calculations done in the proof of [8, Proposition 2.1 (i)] show that is supersolvable. The dual
2m
A
linestothepointsofΛ allpassthroughthepointP := [0,0,1],andanytwolinesdualtotwodistinctpoints
2
[cos 2πj,sin 2πj,1],[cos 2πj′,sin 2πj′,1] Λ intersect inapointthatbelongs tothelinewithequation
m m m m ∈ 1
π(j +j′) π(j +j′)
sin x+cos y = 0.
− m · m ·
Since0 j,j′ m 1,then0< j +j′ 2m 3.
≤ ≤ − ≤ −
Ifj +j′ m 1,thentheabovelineisin ,andalsopassesthroughP.
2m
≤ − A
Ifj +j′ m,thenconsider k := j +j′ m,whichsatisfies0 k m 3 < m. Trigidentities
≥ − ≤ ≤ −
π(j +j′) πk π(j +j′) πk
sin = sin ,cos = cos
m − m m − m
givealsothatthelineofequation listedaboveisin .
2m
A
Everythingputtogether showsthatP isamodularpoint(ofmaximummultiplicity m),so issuper-
2m
A
solvableandattains theboundofourTheorem2.4.
Atlarge,[8,Theorem2.2]showstheuniquenessoftheBo¨ro¨czkyexamples;theproofisquitechallenging.
Forourspecificcaseofsupersolvable arrangements theproofismoreintuitiveandwebelievesimpler.
Theorem 2.6. Let be a supersolvable, real line arrangement with m := m( ) 3 and = 2m.
A A ≥ |A|
Then, Sing ( ) = mifandonlyif and arecombinatorially equivalent(i.e.,theyhaveisomorphic
2 2m
| A | A A
Orlik-Solomon algebras).
Proof. Wefirstanalyzethegeometryof .
2m
Let L := V(cos 2πjx + sin2πjy +Az) for j = 0,...,m 1 and L′ := V(sin πkx cos πky) for
j m m − k m − m
k = 0,...,m 1bethelinesof ;observethatthemodularpointP := [0,0,1]liesonalltheL′’s. We
− A2m k
havethefollowingcircuits(i.e.minimaldependent sets):
{Lj,Lj′,L′j+j′},j < j′ and{L′k,L′k′,L′k′′},k < k′ < k′′,
where j +j′ denotes the reminder of the division of j +j′ by m. Note that if j +j′ j +j′′ (modm)
≡
withj′,j′′ 0,...,m 1 ,thenm (j′ j′′)andhencej′ = j′′ (since j′ j′′ m 1). Therefore,the
∈ { − } | − | − | ≤ −
intersection pointsonlinesnotpassingthrough P areeithersimplepointsortriplepoints.
Claim1. ThereexistsexactlyonesimplepointoneachL ,namelytheintersection ofL andL′ .
j j
2j
SupposethereexistsLj′ thatpassesthroughthesameintersectionpoint. Thenwemusthavej +j′ = 2j,
leadingtom (j′ j). Soj′ = j.
| −
As a consequence we obtain that if m is even, then half of the lines through P have exactly two simple
points and half have no simple points, and if m is odd, then each line through P has exactly one simple
point.
Let be a real supersolvable line arrangement with m( ) = m consisting of n = 2m lines. Suppose
A A
that Sing ( ) = m. Toshowthat and arecombinatorially equivalentwefollowroughlythesame
2 2m
| A | A A
ideasasGreenandTao,withthehopethatforsupersolvablearrangementstheargumentismoretransparent.
IfforthemtheCaley-BacharachTheoremwasthekeyingredientintheproof,forusasimpleplanegeometry
problem doesthetrick(seetheproofofClaim2below).
LetQ Sing( )withm(Q, ) = m. Then,byLemma2.1,Qisamodular point. LetM′,...,M′
∈ A A 0 m−1
bethelinesof passingthroughQ. Also,denotebyM ,...,M theremainingn m = mlinesof .
0 m−1
A − A
ByLemma2.2, each line M has at least one simple point on it which is the intersection of this M and
i i
one of the M′. Since we have m simple points in total and m lines not passing through Q, each M has
k i
exactlyonesimplepointonit. IfanM hasapointwith4ormorelinesof throughit,then,fromequation
i
A
(1)intheproof ofTheorem 2.4,weobtain 2(m 1) < n 2, contradicting that2m = n. Soalltheother
− −
pointsonM havemultiplicity 3.
i
6 BENJAMINANZISANDS¸TEFANO.TOHAˇNEANU
LetusconsidersomearbitrarylineM′ throughQ. Supposeithasusimplepointsandvtriplepoints.
∈ A
Then,thesameequation(1)gives
m 1+u+2v = n 1,
− −
leadingtou+2v = n m = m.
−
Claim2. u 2.
≤
Supposeu 3. PickM ,M ,M through3oftheusimplepointsonM′. IfM M M = Q′ ,then
1 2 3 1 2 3
≥ ∩ ∩ { }
since Q is modular, then there is the line connecting Q′ and Q that leads to m(Q′, ) 4. Contradiction.
A ≥
SoM M = Q ,M M = Q ,M M = Q ,withthesethreepointsdistinct. SinceQ
1 2 1,2 1 3 1,3 2 3 2,3
∩ { } ∩ { } ∩ { }
ismodular, itmust connect to these three points through somelines M′,M′,M′ respectively. Intersecting
3 2 1
M and M′, for i = 1,2,3, we obtain Q . The lines M ,M ,M already have their simple points, so
i i i 1 2 3
m(Q , ) = 3,i = 1,2,3. Also, Q isalready connected to each Q , sothe extra line through each Q that
i i i
A
addsuptothetotalmultiplicity of3,mustcomefromalinenotthroughQanddifferentthanM ,M ,M .
1 2 3
The geometry of the real plane cannot allow for the points Q to be collinear (see figure below): we
i
havethe triangle (Q Q Q )and apoint Qnotonanyofitsedges M ,M ,M . LetQ = QQ
1,2 1,3 2,3 1 2 3 1 2,3
△ ∩
M ,Q = QQ M ,Q = QQ M . ThenQ ,Q ,Q arenotcollinear.
1 2 1,3 2 3 1,2 3 1 2 3
∩ ∩
Suppose one such extra line contains Q and Q . Then it intersects M in a “new” point Q′. The extra
1 2 3 3
line passing through Q should not contain either Q , nor Q because itwill make their multiplicity bump
3 1 2
to4. Sothisextralinepassing through Q intersects thelinesM andM intwo“new”points Q′ andQ′.
3 1 2 1 2
In the case when each Q comes with its own extra line, we still obtain three “new” points Q′, Q′, Q′ on
i 1 2 3
M ,M ,M , respectively. Now with these three “new” non-collinear (by the same geometry argument as
1 2 3
above)pointsreplacing thethreepointsQ ,wecancontinue theargument repeatedly untilweexhaustthe
i,j
linesthrough Q,obtaining acontradiction.
Claim2tellsusthatifmisodd,thenu = 1,andhenceeachlinethroughQhasexactlyonesimplepoint,
and if m is even, then u = 0or u = 2, and hence, keeping in mind that there are exactly msimple points,
halfofthelinesthroughQhaveexactlytwosimplepointsandhalfhavenosimplepoint.
Toconclude the proof, we show that the Orlik-Solomon algebras are isomorphic via an isomorphism of
NBC-bases (see [15] for definitions and more information, but more specifically [17]). The abbreviation
NBC stands for non broken circuit (in [17, Definition 2.1] it is called basic), and it means the following:
ONTHEGEOMETRYOFREALORCOMPLEXSUPERSOLVABLELINEARRANGEMENTS 7
Suppose one picks an ordering of the hyperplanes of an arrangement = H ,...,H . A broken ciruit
1 n
A { }
(underthischosenordering)isasubsetS AsuchthatthereisH withH smaller(withrespecttothe
⊆ ∈ A
ordering) thanallelementsofS andsuchthatS H isacircuit. AnNBCisanindependent subsetof
∪{ } A
thatdoesnotcontain abrokencircuit.
issupersolvable, andsuppose weorderthelinesofthisarrangement as
2m
A
L′ < L′ < < L′ < L <L < < L .
0 1 ··· m−1 0 1 ··· m−1
Then[17,Theorem3.22](ortheoriginal[2,Theorem2.8])saysthattheOrlik-Solomonalgebra,OS( ),
2m
A
isquadratic. Thequadratic NBCelementsinitsbasisconsisting ofthefollowing:
(L′,L′): 1 i m 1 (L ,L′ ) :0 j m 1 .
{ 0 i ≤ ≤ − }∪{ j 2j ≤ ≤ − }
Similarly, issupersolvable, andifweorderitslinesas
A
M′ < M′ < < M′ < M < M < < M
0 1 ··· m−1 0 1 ··· m−1
thenOS( )isquadratic, withthequadratic NBCelementsinitsbasisconsisting ofthefollowing:
A
(M′,M′): 1 i m 1 (M ,M′ ) :0 j m 1 ,
{ 0 i ≤ ≤ − }∪{ j δ(j) ≤ ≤ − }
whereM′ isthelinepassing throughQsuchthatM M′ istheuniquesimplepointonM .
δ(j) j ∩ δ(j) j
FromClaim1,itisclearthatthetwoOrlik-Solomon algebrasareisomorphic. (cid:3)
2.2. TheSlope problem. Given n 3 points in the (real) plane, not all collinear, they determine at least
≥
n 1lines distinct slopes. Thebound can beachieved only ifn 5isodd. In[12]4infinite families and
− ≥
102sporadic examplesareshowntosatisfytheequality inthebound(herewecitedfrom[1,Chapter11]).
Theconnectionwithsupersolvable linearrangementsisthefollowing: ConsiderP ,...,P R2,n 3
1 n
∈ ≥
not all collinear. We consider these points in P(R2), by adding one extra (homogeneous) coordinate equal
to 1. Two lines in R2 are parallel (respectively, identical) and have the same slope, if, when embedded in
P(R2),theyintersectoneanother atthelineatinfinity(ofequationz = 0).
Next let D ,...,D to be all the points lying at the line of infinity, obtained by intersecting this line at
1 w
infinity with all the lines determined by P ,...,P . Then, w is the number of distinct slopes of all these
1 n
lines. TheSlopeproblemsaysthat
w n 1.
≥ −
Dualizingtheaboveconstruction, weobtainthefollowing:
P ,...,P becomethelinesℓ ,...,ℓ P2,andD ,...,D becomethelinesδ ,...,δ P2.
1 n 1 n 1 w 1 w
• ∈ ∈
The line at infinity (of equation z = 0) becomes the point [0,0,1] P2, which belongs to every
• ∈
δ ,i = 1,...,w.
i
P andP determinethelineℓifandonlyifℓ andℓ intersectinthepointdualtoℓ. Furthermore,since
i j i j
•
ℓintersects thelineatinfinityatsomeD ,thepointdualtoℓbelongstothelineδ .
k k
Consider the line arrangement = ℓ ,...,ℓ ,δ ,...,δ P2. Then, the three bullets above
PD 1 n 1 w
A { } ⊂
(especially thelasttwo)implythat issupersolvable (withamodularpointP := [0,0,1]).
PD mod
A
Wehavem(P ) = wandn = w,therefore
mod PD
|A |−
1
PD
w n 1ifandonlyifm(P ) |A |− .
mod
≥ − ≥ 2
Thisstatement leadedustothefollowing
Proposition 2.7. Thefollowingareequivalent:
(1) TheSlopeproblem.
(2) If isafullrankrealsupersolvable linearrangement inP2,then
A
1
m( ) |A|− .
A ≥ 2
8 BENJAMINANZISANDS¸TEFANO.TOHAˇNEANU
Proof. First let us note that if a supersolvable line arrangement has two modular points Q and Q , of
1 2
A
distinctmultiplicitiesm(Q ) >m(Q ),thenm(Q )+m(Q )= +1,meaningthatanylineof must
1 2 1 2
|A| A
pass through Q or Q . If there were aline ℓ not passing through Q and Q , then from the modularity of
1 2 1 2
these two points, the number of points on ℓ is precisely m(Q ). However, it must also be m(Q ), giving
1 2
thatm(Q )= m(Q ),acontradiction.
1 2
If we are in the situation of the observation above, then m( ) m(Q ) > ( +1)/2, hence (2) in
1
A ≥ |A|
the statement is satisfied immediately. Also, if the line arrangement has m(P ) < m( ), then
PD mod PD
A A
ℓ ,...,ℓ all pass through the modular point of of maximum multiplicity m( ), and therefore
1 n PD PD
A A
the points P ,...,P we started with are collinear, contradicting the conditions of (1). So m(P ) =
1 n mod
m( ).
PD
A
With the two observations above, the equivalence is evident. That (2) implies (1) is immediate, and for
(1) implies (2) we may pick a (modular) point of maximum multiplicity and change variables so that this
pointisP = [0,0,1]. Thenthelinearrangement becomesanarrangement oftheform . (cid:3)
mod PD
A
3. SUPERSOLVABLE LINE ARRANGEMENTS OVER ANY FIELD OF CHARACTERISTIC 0
3.1. The degree of the reduced Jacobian scheme. In this section we assume K to be any field of char-
acteristic 0. Let P2 be a full rank supersolvable arrangement of n lines. Let m := m( ) be the
A ⊂ K A
multiplicity of ,andsuppose m 3. LetP Sing( )withm(P, ) = m;therefore byLemma2.1,P
A ≥ ∈ A A
ismodular.
In R := K[x,y,z], let f be the defining polynomial of , let g be the product of linear forms defining
A
themlinesof passingthroughP,andlethbetheproductofthelinearformsdefiningthen mlinesof
A −
notpassing throughP. Withoutlossofgenerality wemayassume
A
f = g h.
·
Let := V(h) P2; this is the line arrangement consisting of all the lines of not passing through
h
A ⊂ A
P. The following equivalent statement is immediate: Q Sing( ) if and only if Q Sing( ) and
h
∈ A ∈ A
m(Q, ) 3. Therefore, taking into account that P is not a simple point of (m 3), we have the
A ≥ A ≥
formula
(2) Sing ( ) = Sing( ) Sing( ) 1.
2 h
| A | | A |−| A |−
Thisisthe mainreason whythe study of Sing( ) goes together withthe study of Sing ( ). In fact
2
| A | | A |
wehavethefollowingimmediateresult:
Proposition 3.1. Let be afull rank supersolvable arrangement of nlines, with maxmultiplicity m 3.
A ≥
Then
Sing ( ) 2Sing( ) m(n m) 2.
2
| A | ≥ | A |− − −
Equality holds if and only if is generic (i.e., Sing ( ) = Sing( )), and in this case Sing ( ) =
h 2 h h 2
A A A | A |
(n m)(2m n+1).
− −
Proof. Let P Sing( ) be such that m(P, ) = m. Then P is modular. Let ℓ be an arbitrary line
∈ A A ∈ A
passingthroughP. Supposetherearessimplepointsonℓ,andtmultiplepointsonℓ,distinctfromP,with
multiplicities n ,...,n 3. Equation(1)intheproofofTheorem2.4gives:
1 t
≥
(m 1)+s+(n 1)+ +(n 1) = n 1.
1 t
− − ··· − −
Asn 3leadsto
i
≥
n m s 2t.
− − ≥
SummingoverallthelinesthroughP implies
m(n m) Sing ( ) 2Sing( ).
2 h
− −| A | ≥ | A |
Formula (2) above proves the assertion, and the “if and only if” statement. If is generic, then
h
Sing( ) = n−m ,andweimmediatelyobtaintheclaimedformula. A (cid:3)
| Ah | 2
(cid:0) (cid:1)
ONTHEGEOMETRYOFREALORCOMPLEXSUPERSOLVABLELINEARRANGEMENTS 9
It is worth noting that for the arrangement(s) in Theorem 2.6, Sing( ) = m+1 + 1. To see this
| A | 2
calculate u+v foreach line through P,sum these and add 1toaccount for P. Th(cid:0)en fo(cid:1)rmula (2), together
with Sing ( ) = m,gives
2
| A |
m
Sing( ) = ,
h
| A | (cid:18)2(cid:19)
meaningthat isgeneric (as = n m = m).
h h
A |A | −
3.1.1. An interesting example. For any homogeneous polynomial ∆ R, let J = ∆ ,∆ ,∆ R
∆ x y z
∈ h i ⊂
be the Jacobian ideal of ∆. Since ∆ is a homogeneous polynomial, J defines the scheme of the singular
∆
locus ofthedivisor V(∆) P2. Asingular point showsupinJ withacertain multiplicity (known inthe
∆
⊂
literature as the Tjurina number), but we are interested only in the number of singular points, which is the
degreeof√J (thedefiningidealofthereduced Jacobian scheme).
∆
Thehomological information ofJ , where f isthe defining polynomial ofasupersolvable line arrange-
f
ment,isverywellunderstood. Forexample,thefirstsyzygiesmoduleofJ RisafreeR-moduleofrank
f
⊂
2,withbasiselementshavingdegreesm 1andn m(see[11],or[15]).
− −
By [22, Theorem 2.2], there exists (α,β,γ) a syzygy on J with α,β,γ forming a regular sequence.
f
{ }
The degree of this syzygy is d (equal to m 1 or n m). Then, if m < n 1, by [22, Proposition 3.1],
− − −
Sing( ) d2+d+1.
| A | ≤
When [22] appeared, there was a question as to if the bound can be attained. The next example shows
that it can. We came across this example while trying to prove Proposition 2.7(2) without using the Slope
problem. Wehavesofarbeenunabletofindsuchaproof.
Considerthelinearrangement withdefiningpolynomialf = xyz(x y)(x z)(y z)(x+y z)(x
A − − − − −
y+z)(x y z).
− −
Wehaven = 9andm = 4. Thisistheprojectivepicture withtheborderbeingthelineatinfinityz = 0.
Themarkedpointsaremodularpointsofmaximummultiplicity.
Thecalculations belowweredonewith[9].
P = [1,1,0] Sing( ) with m(P, ) = 4; the lines of equation z = 0,x y = 0,x y + z =
∈ A A − −
0,x y z = 0allpassthroughP. Infact,sincez(x y)(x y+z)(x y z) J ,by[21,Theorem
f
− − − − − − ∈
2.2],P isamodularpointand issupersolvable. p
A
Calculations showthat
1 11 1 1 1 1
( x2z xyz+ y2z+ xz2+ yz2 z3,
10 − 30 10 12 12 − 20
1 1 1 11 1 1
x2y+ xy2 y3 xyz+ y2z+ yz2,
5 6 − 10 − 15 6 5
3 1 3 1 11 3
x3+ x2y+ xy2+ x2z xyz+ xz2)
−10 2 5 2 − 5 5
10 BENJAMINANZISANDS¸TEFANO.TOHAˇNEANU
is a syzygy on J (i.e., on some linear combination of the partial derivatives of f). Its entries generate an
f
idealofheight3,andhencetheyformaregularsequence. Thedegreeofthissyzygyis3= m 1.
−
By [22, Proposition 3.1], Sing( ) 32 + 3 + 1 = 13, and calculations show that in fact we have
| A | ≤
equality.
3.2. Dirac-Motzkin conjectureforsupersolvablearrangements overC. [5,Theorem 2]showsthatany
complex smooth cubic plane curve has exactly nine inflection points, and Theorem 3 in the same paper
proves that any two of these points are collinear with a third (an image of such a curve can be found by
searching for“Hesseconfiguration”). ThisshowsthatSylvester’soriginalproblem doesnothaveasolution
overC.
Suchaconfigurationofpoints(whereanytwopointsinthesetarecollinearwithathirdintheset)iscalled
Sylvester-Gallaiconfiguration(SGC).Serreproposedthefollowingproblem: AnySCGinCnwithn 3is
≥
containedina(2-dimensional) plane. TheproblemwassolvedbyKelly,forn = 3(see[13,Theorem]),and
foranyn 3,byElkies-Pretorius-Swanepoel (see[6,Theorem 2];theyreduced theproblem tothecaseof
≥
n = 3,thoughtheyuseadifferentmethodthanKelly).
In [24, Example 4.1] it is shown that the line arrangement dual to these nine points (called the Hessian
arrangement) is a free arrangement with exponents (1,4,4), but it is not supersolvable. Alternatively, we
can see this by using the formula in Proposition 3.1. For, we have n = 9,m = 3 and Sing ( ) = 0; if
2
| A |
were supersolvable, the formula would give Sing( ) 10. However, has at least 12 singularities
A | A | ≤ A
corresponding tothe12linesdetermined bythe9inflectionpoints.
Weconjecture thefollowing
Conjecture3.2. Let P2 beafullrankcomplexsupersolvable arrangement ofnlines. Then
A ⊂
n
Sing ( ) .
2
| A | ≥ 2
Theorem 2.4 relies on the arrangement being real only in the use of Lemma 2.2; everything else is
independent of the base field and relies on the observation wemade at the end of Introduction. The key in
provingConjecture 3.2lieswithinprovingaversionofLemma2.2validoverC.
Suppose P2 is a full rank supersolvable complex arrangement of n lines. Let P Sing( ) be a
A ⊂ ∈ A
pointofmaximummultiplicitym := m( ). Suppose(forcontradiction) thatthereisalineP / ℓ with
A ∈ ∈ A
no simple points on it. If there is Q ℓ Sing( ), with m(Q , ) 4, by removing any line through
i i
∈ ∩ A A ≥
Q not passing through P and distinct from ℓ,theresulting complex arrangement isstill supersolvable, has
i
n 1lines,andstillhasthelineℓwithnosimplepointsonit. Bythisobservation,thefollowingconjectured
−
resultwillprovebycontradiction thecomplexversionofLemma2.2.
Conjecture3.3. Let P2beafullranksupersolvablearrangementofnlinesandwithsingularmodular
A ⊂
point P of maximum multiplicity m 3. If there exists a line ℓ not passing through P and only with
≥ ∈ A
triplesingularities onit,then isnotrealizable overC.
A
Becauseℓdoesnotpass through P,itmusthaveexactly msingularities onit: P ,...,P . Theformula
1 m
(m 1) = n 1,together withm = 3,givesthatn = 2m+1.
Pi − − Pi
PAfterachangeofvariables, wemayassumethatP = [0,0,1]andℓ = V(z).
Suppose ℓ = V(a x+b y),i = 1,...,m are the lines of through P, with a b a b = 0, for all
i i i i j j i
A − 6
i = j. Suppose each point P is the intersection of ℓ , ℓ, and another line ℓ′ of not passing through P.
6 i i i A
Thenwecanassumeℓ′ = V(a x+b y+z),i = 1,...,m.
i i i
The supersolvable condition translates to the following: for i = j, the point of intersection of ℓ′ and ℓ′
6 i j
must lie on a line through P. Thisis equivalent tothe following determinantal condition: for any 1 i <
≤
j m,thereexistsk(i,j) 1,...,m i,j suchthat
≤ ∈{ }−{ }
a b 1
i i
(cid:12) aj bj 1 (cid:12) = 0.
(cid:12) (cid:12)
(cid:12) a b 0 (cid:12)
k(i,j) k(i,j)
(cid:12) (cid:12)
(cid:12) (cid:12)
(cid:12) (cid:12)