R ENDICONTI
del
S EMINARIO M ATEMATICO
della
U NIVERSITÀ DI P ADOVA
C. T. C. W ALL
Pencils of binary quartics
Rendiconti del Seminario Matematico della Università di Padova, tome 99 (1998), p. 197-217
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C. T. C. WALL (*)
1. - Introduction.
The
objective
of this article is to discuss the classification and invari- anttheory
ofpencils
ofbinary quartics.
The main motivation for this is the intrinsic interest of thequestion. Indeed,
it is now known(see [1])
that for very fewtypes
ofquantics (a single homogeneous equation)
isthe
ring
of invariants acomplete
intersection(for degree
d in r variableswe
require
either d ~2,
d =3 , r ~
4 or d ~6 , r
=2),
so it is of interest to document such further natural situations where-as here-thering
ofinvariants is a
complete
intersection.There are,
however,
two additional reasons forwriting
this paper at this time. One is that the results(or
at leastpart
ofthem)
are used in re-cent work of the author
[6] giving
analgorithm
to listpossible configura-
tions of
singularities
onquartic
surfaces inP 3 ( ~ ), given
that thesingu-
larities are isolated and not all
simple.
The other is that whereas at the time of
writing [5]
I considered thesequestions
but my notes wereincomplete
due to thedifficulty
of calcula- tions ofinvariants,
these can now beperformed relatively easily using computer algebra (my preference
is forMAPLE).
The paper opens with a section
discussing pencils
ofbinary quantics
in
general,
andestablishing
basic ideas andnotation,
inparticular
a sym- boldescibing
the ramification associated with each case. The varioustypes
ofpencil
ofbinary
cubics andquartics
are then listed andnamed,
thus
defining
a stratification and hoc. The map from strata topossible symbols
is notsurjective;
in thesedegrees
it isinjective.
The next section discusses the
application
ofgeometric
invariant the-ory, and determines the structure of the
ring
of invariants. Most of the invariants werealready given by
Salmon in[4],
but his list isincomplete.
(*) Indirizzo dell’A.:
Department
of Pure Mathematics,University
of Liver-pool, Liverpool
L693BX, GranBretagna.
The case of
pencils
ofbinary
cubics wasanalysed by
Newstead in[3],
asource I have found very useful.
Then we discuss the notion of
symmetry
of forms and ofpencils.
Wediscuss non-stable strata in a
separate section;
in the remainder of the paper we describe the stratification of the moduli space.The discussion of
symmetry
leads to a moresystematic
way to view the strata. Were-present
thelist,
and show how to characterise the cas- es on it.Finally
we calculate the invariants on theexamples
from our lists ofstrata,
and list thesymmetry types
that occur inspecial
cases.1. - General notions.
Consider the space V =
Vn
ofhomogeneous polynomials
ofdegree n
in the variables x and y. A
pencil of binary nics
is a 2-dimensional sub- space of V. We can define itby picking
any twolinearly independent
members
f = f ( x , y )
andg = g( x , y ),
andg).
Thus atypical
member of the
pencil
is a linear combinationpf +
qg. Weregard
two pen- cils asequal
if we have the samesubspace
of V and asisomorphic
ifthey
are
equivalent
under someautomorphism
inGL(V).
We can
distinguish
two mainsubcases, according
asf and g
do or donot have a common factor
(of degree
at least1).
If there is no commonfactor,
then for any linearexpression y - tx (or x)
there is a member of thepencil, unique
up to a scalarfactor,
which it divides: it is determinedby pf ( 1, t)
+t)
= 0(note
thatby hypothesis, f ( 1, t )
andt)
donot both
vanish). Equivalently,
thepoint of projective
space deter- minedby
the ratio(x : y)
determines aunique
ratio(p: q),
and hence apoint
ofprojective
space We can thusidentify
suchpencils
withrational maps
(of degree n)
fromP~(C)
toP~(C).
Note that weregard
these as
different projective
spaces, so will notattempt
to iterate such maps. Two ways ofobtaining
asingle polynomial
from apencil
are imme-diate. The Jacobian J is
defined,
asusual,
asIt thus is
homogeneous
ofdegree
2 n - 2 .Up
to a constantfactor,
it de-pends only
on thepencil.
The discriminant L1 is the usual discriminant of considered aspolynomial
in x and y. This ishomogeneous
ofdegree
2n - 2in p
and q.In the case of a
pencil
without commonfactor,
Jrepresents
the set ofbranch
points
of thecorresponding
rational map, and L1 the set ofimages
of these
points.
This statement is made more
precise
in thefollowing
lemma. Here wewrite to denote the
multiplicity
of x as factorof f :
this isequal
to rif f is
divisibleby x r
but notby x’’ + 1.
LEMMA 1.1.
(i) If the pencil
has no commonfactor,
thenv x(J) =
= r - 1
if
the member qg =h ,
say,of
thepencil
divisibleby
x hasv x (h)
= r, and = rif f
has n - r distinct linearfactors.
(ii) If the pencil (f, g)
has Jacobian J and discriminant L1 whilefor f ’
=xf , g ’
= xg we have J’ andL1’,
then J’ =X2 J (up
to a muLti-plicative constant). If
x divides bothf
and g then L1 ’ =0;
otherwise x di-vides qo g
for a unique (po: qo)
and 4 ’ =(pqo - (up to a constant).
PROOF. The assertions for J follow
by elementary
calculations. Ifv x ( f )
= r andv x (g)
= s > r then J isclearly
divisibleby x r + s -1
and wemay check that the coefficient 1 is non-zero.
Now suppose that
f
has n - r n distinct linearfactors;
writefi
fortheir
product,
so that both8f/8r
and8f/8y
are divisibleby f2 .
Now L1 is the determinant of the matrix of coefficients of the
(a
++ b = 2 n -
qg)lày.
This is a sumof 22, - 2
sim- ilar determinants each obtainedby omitting
either p or q in each row, hence each amultiple
ofIf t r we have at least 2 n - 1 - r rows formed
from f.
The corre-sponding polynomials,
ofdegree 2 n - 3 ,
are allmultiples
off2 ,
ofdegree
r, so form a vector space of dimension at most 2 n - 2 - r. The corre-
sponding
determinant thus vanishes. Thus L1 is divisibleby q r
First suppose the
pencil
has no common factor. Since ageneral
mem-ber does not then have a
repeated
root(by
Bertini’stheorem),
L1 does notvanish
identically.
We haveproved
that it is divisibleby
aproduct P,
say, of termscorresponding
to members of thepencil
withrepeated
factors.We will now show that P has the same
degree
2 n - 2 as L1: the desired result will follow.As described
above,
thepencil
defines a rational map fromP
toP~.
Let us calculate Euler characteristics 4 la Riemann-Hurwitz. Since
x(P 1 )
= 2 and thedegree
is n, the characteristic of the sourceP
isequal
to 2 n minus corrections due to the
branching
behaviour. A member of thepencil
withjust n - r
distinct factorscorresponds
to apoint
withjust
n - r distinct
pre-images,
and hencerequires
a correction of r. Since the characteristic of the source is2,
the sum of these correction terms mustequal
2 n - 2. This proves our claim.Now consider
f ’ = xf, g ’
= xg ; we may suppose thatf
is divisibleby
x,so
qo = 0.
WriteThen L1 is the determinant of
b = n - 2 }
with re-spect
to and L1’ is the determinant of. with
respect
towhere
It follows from
homogeneity
thatThus
elementary operations
transform the above list toThe first
part
of this list has determinant L1 withrespect
+ d = 2 n -
3},
so we must extract the coefficientsy 2n -1
in thelast two terms. We thus see that
(up
to a non-zeroscalar)
d ismultiplied by
the square of the coefficient ofy n -1
in,S,
orequivalently,
that ofy n
ing, which is zero if x also divides g, and is a non-zero
multiple of q
otherwise.
We use the above to define a
symbol
for anypencil
to describe itsbranching
behaviour. First suppose there is no common factor. For each root of L1(of multiplicity
r,say)
we list themultiplicities
of the factors(there
are n - r ofthem)
of thecorresponding
memberpf + qg
of thepencil, separated by
commas, and enclose this set inparentheses.
Do thisfor each root of
L1,
and enclose the whole in(square)
brackets.In the case of a
pencil
with common factorh,
say(fh, gh),
first wewrite the
symbol
of thepencil ( f, g),
Now for each linear factor ofh,
ofmultiplicity r,
if this hasalready
occurred as a factor of one of the mem-bers of the
pencil
with arepeated root,
we underline thecorresponding
number in the
symbol r
times. Ifnot,
we attach a new blankparenthesis
to the
symbol
and underline that instead. The definition isamply
illus-trated in the lists to follow.
For a
pencil
with no commonfactor,
we caninterpret
thesymbol topologically.
We have a branched n-foldcovering
of a2-sphere;
at eachbranch
point,
thesymbol provides
apartition
of n which describes the nature of thebranching.
Themonodromy given by encircling
thepoint
once is a
permutation
of the n branches whosecycle type
isgiven by
thispartition.
Globally,
if wepuncture
thesphere
at the k branchpoints
we obtainan unramified
covering.
The fundamental group of thepunctured sphere
is
generated by
theclasses {ai}
ofloops circling
thesepoints,
which aresubject
to the sole relation a 1 a 2 ...a k = 1(so generate
a free group ofrank k -
1).
This actstransitively by permutations
on a set ofcardinality
n; the fundamental group of the cover is a stabiliser
subgroup.
As we have noted in the above
proof,
the numerical conditions ensurethat when we
compactify
thiscovering by filling
in thepunctures
we ob- tain a surface of genus 0.2. - Enumeration.
We
begin by enumerating types
ofpencil
ofdegree n S
3. Note thatpencils
with a common factor are enumeratedby removing
thefactor,
and then
considering
the differentplaces
in which it can be inserted in thesymbol.
The enumeration with n = 3 isgiven
in[3];
see also[5],
where a notation is introduced and normal forms discussed more
fully (though
theemphasis
there is on the realcase).
The first column in Table 1
gives
the namesassigned
in[5]
to the various cases with n = 3. The fourth column is the dimension of the stra- tum(as
a subset of the Grassmannian of 2-dimensionalsubspaces
of then + 1-dimensional vector space
V).
The column headed Jgives
the multi-plicities
of the linear factors of J. We observe that in each case the num-ber of such factors
equals
the dimension. Thecorresponding
column ford would be the same
except
for case(e),
where L1 = 0 .TABLE 1.
TABLE 2.
We now
begin
the same forpencils
ofquartics.
First we consider cas-es with a common
factor,
which we list in Table 2: asbefore,
we accom-plish
thisby inserting
a common factor in the cases in Table 1.The
precise
conditions on theparameters defining
these strata willbe discussed below. An
exception
occurs in case(h),
where if a = 0 thepencil
contains(X2
+y 2 )2 ,
and d hastype ( 2 , 2 , 2 ).
We willdistinguish
this case and denote it
by (ho).
We next enumerate the cases with no common factor. The list is sum-
marized in Table 3. In order to
display
all thedata,
each caseoccupies
two lines.
To obtain the list in Table
3,
we must arguedirectly.
First suppose thepencil
contains a fourth power,say f = X4.
We may take the coeffi- cient ofy 4
in g as1; perform
a substitutiony ’
= y + kx to remove the termXy 3
and add amultiple of f to g
to reduce g to the form ++ +
y 4. (In
future we will omitessentially
trivialarguments
of thistype.)
We now find(it
is easyusing MAPLE)
that d =q 3 D,
where D is aTABLE 3.
cubic
(not
divisibleby q)
with discriminanta 2 ( 27 a 2 + 8 b 3 )3.
We thushave cases
(A),
where a = b =0, (B)
with a =0 ,
b ~0, (C)
with a =1,
b == -3/2,
and(D)
where none of these occur. A more convenient normal form for(C)
isgiven
in Table 3.Next we enumerate cases where the
pencil
contains no fourth power but has two members each withonly
2 distinctprime
factors. First sup- pose both of these havemultiplicities ( 2 , 2 ).
Then we cang =
(X2
+2 axy
+y 2 )2.
Here we have L1= ~ro 2 q 2 D
where q divides Donly
if
a 2
= 1(which
is excluded since thepencil
contains no fourthpower)
and D has a
repeated
root if andonly
if a = 0. We thus have cases(E)
with a = 0 and(F)
with a # 0.Next suppose we have
multiplicities ( 3 , 1 )
and(2, 2),
so we can takef = and g
=(x 2 + 2 axy
+y 2 )2 .
Thenagain
L1 withneither p
nor q
dividing D;
the discriminantof D is ( a 2 + 3 )3 (up
to aconstant).
We thus have cases(G)
witha 2
= - 3 and(H)
otherwise. Ifboth f and g
havemultiplicities (3, 1)
we can take andwhere as there is no common
factor, ad( ad - bc ) ~
0 and as there is no fourth power, bc ~ 0. Weagain
have L1= ~ 2 q 2 D
with neither p nor q di-viding D ;
the discriminant of D is(ad - bc)( 8 ad
+bc)( 4 a,d - bc).
Thecase 8 ad + bc = 0
again yields
the stratum(G);
otherwise we have case(I )
with 4 ad = bc and case(J)
where the above discriminant does not vanish.There remain three cases:
(K) where f has
factormultiplicities ( 2 , 2 )
but each other member of the
pencil
has at least 3 distinct linearfactors;
(L) similarly
with( 3 , 1 )
and thegeneric
case(M).
In all cases in Table 2 and Table
3,
the dimension of the stratum isequal
to the number of distinct linear factors ofL1, except
in cases( a ), ( b ), ( d ), (g )
where there is a commonrepeated
factor and L1= 0;
in these cases, we may count the factors of J instead.Apart
fromthese,
thefactors of L1
correspond
to those of Jexcept
whenever thesymbol
con-tains
( 2 , 2 ) (in
Table2, only
cases(m)
and(p)).
A more
interesting
remark is that thesymbol [(2, 2)(2, 2)(3, 1)]
does not appear. This could be
predicted
from thetopological interpreta-
tion. For if such a cover
existed,
we could findpermutations
of a set of 4objects,
ofrespective cycle types (2, 2) (2, 2)
and(3, 1),
whoseproduct
is the
identity.
But this ismanifestly impossible.
3. - Invariant
theory.
To
apply
invarianttheory
in itssimplest form,
we considergeneric quartics (or
indeednics) f,
g, and the action ofGL2
xGL2
on thepolyno-
mial
ring
in the coefficientsof f and
g, where one copy ofGL2
actsby
lin-ear substitutions in x and y and the other
by
linear substitutionsin f and
g . There are several ways to construct invariants of apencil
ofbinary
quartics.
One may considerpf + qg
as aquartic
in x and y and take its in- variants : its self-transvectantI(p, q)
and its catalecticantJ(p, q).
Wethen have the discriminant DI
of I ( p , q),
which hasweight
2(in
the coef- ficientsof f
and also in those ofg);
the resultant DR ofI(p, q )
andJ(p, q),
which hasweight 6;
and the discriminant DJ ofJ(p , q),
whichalso has
weight
6. We may also form the Hessian determinantH(p, q )
ofJ(p, q )
and the transvectant DHq )
andI(p, q),
which hasweight
4.Alternatively
we may start with the Jacobian J and construct its in- variants as abinary
sextic(our
version of these -including
thealgorithm
to
compute
them - is taken from[2], p.156).
We denote theseby JI2 , JI4 , JI6 ,
andJI15 ,
where the suffixgives
theweight
in each case. We re-call that the first four of these are
independent,
while the square ofJI15
may be
expressed
as apolynomial
in them. Moreprecisely,
we call an in-variant 0 even if
~( - f ) _ ~( f ),
and odd if thesign
ischanged.
Thenthe even invariants form a
polynomial ring
inJI2 , JI4 , JI6
andJI10;
eachodd invariant is divisible
by JI15 ,
withquotient
an even invariant.(We
could also consider the invariants for L1 but here theweights
would be
tripled,
and in any case as L1 =41(p, q )3
+27J(p, q )2
many of these can be recovered from the above covariants ofI(p, q )
andJ(p, q) ).
We also have the resultant
Res( f , g ) of f and
g . We choose thefollowing
as our basic invariants:These are
essentially
the same as the invariantsgiven by
Salmon[4],
Arts 213-217.
There are various conventions as to the
precise
definition of invari-ants,
so that each one has a somewhat variable scalar factor. Our conven-tion
is,
ineffect,
definedby
thecomputer
programme we have used. Itcan be inferred from the
following examples
whichgive
the invariants for normal forms for thetop
dimensional strata(K), (L), (M).
Most othercases may be obtained
by substituting
in these. Insubsequent
calcula-tions we write
conventionally C ,~
to denote anunimportant
scalar factor(usually involving
alarge
power of2).
Write F for the
generic quartic
+
ey 4.
Then the invariants for thepencil (F, X4) (generically
in case(D))
are
for the
pencil (F, (if
we set a = e = 1 this is the normal form forcase
(L))
we haveand for the
pencil (F, x z y 2 ) (here
we may set a = e = 1 togive
the nor-mal form for case
(K) )
we haveThe two latter cases are, of course,
independent
of b and crespectively,
since these do not affect the
pencil. Similarly
in the invariants for thegeneric
case(F, x 4 - x 2 y 2 ~,
the coefficients a and conly
appear in the combination a + c.Taking
c = 0(for
the normal form for(M)
we also have e =1),
we haveAs a final
example,
for thegiven
normal form for case(q),
the invariantsare
In our enumeration of
strata,
we defined 3 codimension 1 condi- tions :there is a common
factor,
the
pencil
contains aperfect
square,the
pencil
contains aquartic
divisibleby
a cube.The invariant
I6
is defined as the resultant DR of andJ, q.
Itthus vanishes if and
only
if these have a common factor. This means that for some elementpf + qg
in thepencil,
both its invariants I and Jvanish,
so it has a factor of
multiplicity
at least 3. Hence the third condition is in-variantly
characterisedby
thevanishing
ofI6.
The invariant
I4
is defined as the resultantof f and
g. It thus vanishes if andonly
iff and g
have a common factor: the first condition above.These conclusions were obtained a
priors. By
direct calculation wealso find that the second condition holds if and
only if J22
=I4.
These arethe considerations which
guided
our choice of the invariants to take asbasic. The conditions
Is = 0 , I4 = 0
andI4 = J2
are illustrated on the aboveexamples.
PROPOSITION 3.1. The even invariants
of J
areexpressed
in termsof
the aboveby
thefollowing formulae:
The odd invariant
JI15
is aproduct C * FIg FI6 of
irreduciblefactors,
where
and
FI9
is obtainedby substituting
A =I2 ,
B =3 J2 ,
C =81 I4
and D ==
64 I6
inPROOF. The result was established
by
direct calculationusing
MAPLE. The calculations were
performed
as follows. It isenough
to con-sider the
generic
case~F, x 4 - x 2 y 2 ~
consideredabove,
with c = 0. Notethat
14
has a factor e .Substituting
e = 0yields
We may solve these
equations
asand hence define substitutions
(with
new variablesi2 , j2 , i6 )
Now for any invariant (P of J
(or
indeed ofL1)
weproceed
as follows.Write X for 0
computed
on the aboveexample.
Substitute e = 0 to obtainAi .
Now make the above substitutions for c~,b,
d inAl
andsimplify:
thisyields
apolynomial Bl
ini2, j2
andi6. Substituting
thecomputed
valuesof
I2 , J2 , I6
fori2 , j2 , i6 in B1 gives
anexpression Cl .
Now divideAl - C1 by
thecomputed
value of14
andsimplify
to obtain anexpression A2 (the
fact that this division ispossible
is a crucial check on thecalculation).
Re-peat
the entireprocedure, obtaining
in turnB2 , C2 , A3
and so on. Sincethe
degree
is lowered at eachstep,
theprocedure
terminates.The desired formula is now
given by
the final valueDn
whereDr
is in-ductively
definedby Do = 0 , D~. = D~. _
withI2
fori2 ,
etc. Thisprocedure
iseasily programmed,
andrunning
the programmegives
theresults above.
We can also run the programme on other
invariants-e.g.
those ofthe
binary
sextic L1(after
thefirst,
wejust
obtaincomplicated
expres-sions) and,
moreinterestingly,
thediscriminants, giving
conforming
to the considerations of Section 1.THEOREM 3.1. The
ring of
invariantsof
apencil o, f binary quartics
is
generated by I2 , J2 , I4 , 16
and Theonly
syzygy is thatexpressing
the square
of
termsof
the rest.The invariant
FI9
was overlookedby
Salmon. We shall show below how tointerpret
it.PROOF. We will model our
proof
on thecorresponding argument given by
Newstead[3]
forpencils
of cubics. Webegin
with the vectorspace V of
quartics, containing f
and g. A 2-dimensionalsubspace
of V isdetermined
by
its 10 Plfckercoordinates,
orequivalently by
the exteriorproduct fA g
E/~2 ( T~.
As
representation
ofGL2 (acting
on x andy), A 2 (V) splits
as directsum of
representations
ofdegrees
7 and 3. The Jacobian J is obtainedby projecting f A
g onto the 7-dimensionalcomponent.
Asprojective variety,
the Grassmannian has dimension 6 and we are
projecting
onto a 6-di-mensional linear space. This
projection
isgenerically smooth,
as we seeby
asimple
calculation in local coordinates. Since the Grassmannian va-riety
hasdegree 5,
this is also thedegree
of theprojection.
It follows that the field of invariants of J is an extension ofdegree
5 of the field of invariants of thepencil.
The field of invariants of J is an extension of
degree
2 of the fieldgenerated by
theindependent
invariantsJI2 , JI4 , JI6
andJIIO .
We haveexpressed
these aspolynomials
in our basic invariantsI2 , J2 , I4 , I6 . Comparing
thedegrees
of these invariants we see that we have a field extension ofdegree (2
x 4 x 6 x10 ) ~( 2
x 2 x 4 x6)
= 5. Moreover thequadratic
extensiongiven by JI15 corresponds
to thequadratic
extensiongiven by FIg.
Since our invariantsgenerate
a field over which the invari- ants of Jgenerate
an extension ofdegree 5,
it follows that we do indeed have the field of all invariants.It remains to show that this field contains no
polynomials
which arenot
expressible
aspolynomials
in thegiven
invariants.Suppose
itdoes,
and that
A ,
B arepolynomials,
with no non-trivial commonfactor,
suchthat
when evaluated for a
generic pair (f, g)
ofbinary quartics,
reduces to apolynomial
in the coefficients. We may suppose without loss ofgenerality
that A and B are
homogeneous,
and that we have acounter-example
ofminimal
degree.
We recall that
substituting gives
Since the first three of these are
algebraically independent, substituting
in B will
yield
zeroidentically only
ifI6
divides B.But,
in the casewhen f and g
aregeneric, I6
is irreducible(it
is soalready
in Case(K)
forexample),
so sinceA/B
becomespolynomial
in this case,I6
must also di-vide A: a contradiction.
Thus
A(3d2,
4ce -d2, 64ace3, 0)/B(3d2,
4ce -d2, 64ae3, 0)
hasnon-vanishing denominator,
so is ahomogeneous polynomial
in a, c, dand e: say
P( a ,
c ,d , e).
It follows thatsince this holds on
substituting I2 = 3 d 2 , J2 = 4 c - d 2 , I4 = a .
Since theleft hand side is a rational
function, I2
canonly
appear on theright
hand side via its square. Hence
the right
hand side is apolynomial C(I2, J2~ I4).
Now set
Then A ’ /B has the same
properties
asA/B,
and vanishes for the aboveexample.
But thisexample
is thegeneric
element of its stratum. Thus vanishes wheneverI6
= 0 . Asabove,
it follows thatthis,
and henceA ’,
isformally
divisibleby I6 : A ’ = I6A ".
But now A "/B has the sameproperties
and lowerdegree, contradicting
the inductivehypothe-
sis.
PROOF. Since we have established our invariants
by comparison
withthe
ring
of invariants of thejacobian J,
it is of some interest to consider the induced map of moduli spaces in more detail. We confine ourselves tofinding
its branch locus. This isgiven by
thevanishing
of(after
a brief calculation inMAPLE).
Thevanishing
locus of this function meets our strata in a rathercomplicated
manner: we will notinvestigate
this further.
4. - Pencils with
symmetry.
,As a
preliminary
todiscussing symmetry
ofpencils,
it is convenient to note the situation forbinary sextics,
or moregenerally,
nics. We definethe
symmetry
group of a form F to be thatsubgroup
ofGL2
which trans-forms F to a
multiple
ofitself,
or(better)
itsimage
inPGL2.
For the
n th
power x n of asingle
linearform,
we obtain the full group of uppertriangular
matrices. For a monomialx r y n - r
we have the group ofdiagonal matrices, augmented,
if r = n - r,by
thetransposition
of thetwo factors.
In all other cases, the
subgroup
ofPGL2
preserves a finite set of at least 3points
on theprojective line,
hence is itself finite. It may thus be taken to be contained in a maximalcompact subgroup P U2 =
ofThis group is thus
cyclic, dihedral,
or one of the 3polyhedral
groups. Let us write
Cn
for acyclic
group of order n;D2n
for a dihedralgroup of order
2 n;
andTet, Oct,
Icos for therespective polyhedral
groups.
If there is a
symmetry
of orderk,
we mayconjugate
it todiagonal
form. This transforms
f
to theproduct
of a monomialby
apolynomial
inx k and y k;
this still leaves the freedom tomultiply x
and yby
constants.Excluding
the case whenf
is amonomial,
it follows that k S n.Now we fix n = 6. For
f
not amonomial,
we have thefollowing
cases:
The case k = 6 is given by taking b = 0 in f3; the case k = 4 by b = d = 0 in f2
orby
b = 0 in g2.For f3
and g2, if 0 we can reduce to a = c = 1 and have a furthersymmetry interchanging
x and y,giving
a dihedral group.In fact all cases
admitting
the four groupD4
can be reduced to g2.Case f2
is characterisedby
thevanishing
of the invariantI15.
Theonly
case of a group neither
cyclic
nor dihedral isgiven by y 4 ),
repre-senting
the vertices of aregular
octahedron and thusadmitting
the octa-hedral group.
Now we turn to the case of
pencils.
Here we consider theproduct GL2
xGL2
of the groups of linear substitutions on the variables( x , y)
and on the forms
(~ , q ),
and consider the action of this group onpairs
(p, q ~. Since, by hypothesis, p and q
arelinearly independent,
the sta-TABLE 4.
biliser intersects 1 x
GL2 trivially,
soprojects isomorphically
into thefirst factor. Since the
subgroup ( x ,
y , ~ ,q ) ~ ( ax ,
ay ,a -4p , a - 4 q )
al-ways acts
trivially,
we mayproject
further toPGL2 .
Theresulting
groupis,
of course, asubgroup
of theautomorphism
group of the Jacobian of p and q, which we havejust
described.In
particular,
if J hasjust
two linearfactors,
we may takeboth p
andq to be monomials: for
pencils
ofquartics,
this arises in cases(a), (b), (d), ( e ), ( i ), (A).
Here the group is infinite: in case(a)
we have thetriangular matrices,
in the rest thediagonal
ones, withinterchange
pos- sible for(e)
and(A).
Otherwise we have a finite group, which may be
regarded
as a sub-group of If a
cyclic
group of order m actsby symmetries,
we may di-agonalise
the group tomultiply x by m th
roots ofunity,
say,leaving
y in-variant,
and look for this tomultiply
each of pand q by
a constant. Thisimplies
that each of pand q
isequal
to a monomialmultiplied by
apoly-
nomial in x m and
For
pencils
ofquartics,
with the case when both pand q
are monomi-als
excluded,
thisimplies
m ~4 ,
and leads to the list in Table 4. Here thecases
(S4), (t4) = (E), (t3 )
areunique
up toisomorphism;
the1-parameter
families
( s3 )
and( t2 ) (which
isessentially
the same as(F) )
contain somespecial
cases, listed in Section 6 below. The2-parameter family (s2 )
has anumber of
interesting specialisations,
which are also discussed in detail in Section 6.5. - Non-stable
pencils.
In the next section we will use our invariants to describe strata in the moduli space. As this
ignores
non-stable cases, we consider these morefully
now.Applying geometric
invarianttheory (in
its naiveform)
to oursituation,
we find 3 casesg)
not semistable:(i) g = 0 ;
1(ii) x
dividesf
divides g;(iii) x 2 divides f and x 3
divides g ; and 3 casesg)
not stable:(iv) x 4
divides g ;’
(v) x
dividesf
andx 3
divides g ;(vi) X2
dividesf
divides g.However, (i)
does notcorrespond
to apencil;
and if(vi) holds,
thensome other member of the
pencil
is divisibleby x 3 ,
so we reduce to case(iii).
The correct list ofpossibilities
is thus thefollowing.
A non-semistable
pencil
is either oftype (ii),
which occurs if andonly
if the
pencil
contains a4th
power and has a common factor: thisgives
cas-es
(a), (d), ( i ), ( 1~ );
or oftype (iii),
which occurs if andonly
if there is a re-peated
commonfactor, viz,
for cases(a), (b), (g).
A semi-stable
pencil
which is not stable is either oftype (iv),
whichoccurs when there is a fourth power, hence for cases
(A), (B), (C), (D);
orof
type (v),
whichimplies,
but is notequivalent to, having
a common fac-tor : it
comprises
cases(e), ( f ), ( L )
and(o)
All invariants vanish for the non-stable
pencils;
for each of the semi- stable groups(iv), (v),
the invariants take aunique
value(up
toscaling by constants)
for the whole group. Such a group istypified by
theunique
case
representing
a closed orbit: this is(A )
fortype (iv)
and(e)
fortype (v).
In the next section we will refer to thecorresponding points
inmoduli space as
(A ) +
and( e ) + .
We now go
through
these casesconsidering first,
theprecise
condi-tions on the normal form in Tables 2 and
3,
andsecond,
what is the sym-metry
group in each case.The first
question
isquickly
answered in most cases, sinceonly
forcases
(D)
and( o )
does theproposed
normal form contain aparameter.
For each of
these,
our reduction to normal form fixedand y
up to scalar factors: x as therepeated factor;
y since in the( o )
case, thepencil
con-tains
x 3 y,
in the(D)
case we eliminated the term inxy 3 .
Now theonly
scalars which leave the form invariant involve
multiplying
x(or y ) by
acube root of
unity:
thismultiplies
bby
such a root. Thus in both casesb 3
is the
parameter determining
thepencil
up toisomorphism.
It follows that the
y( y 3
+ +ex 3 ) ~
hastype (A)
ifd=e=o, type (B)
ife=o, type (C) 8d3+e2=0
and other- wisetype (D),
withparameter c~/e~ 20131/8.
TheX(y3 +
+ 3dxy2 + ex3)> has type (e) if d = e = 0, type (f) if e = 0, d # 0, type (l) if
e #
0, d 3
= e and otherwisetype (o),
withparameter
1.The same
argument
as gave the normal form shows also that for each of(D)
and(o)
ageneric pencil
has no non-trivialautomorphism;
the caseb = 0 admits a group of order 3.
We have
already
discussed theautomorphism
group in the cases when eachof f and g
is a monomial. Here the group isinfinite,
so the caseis
necessarily
non-stable. These are the cases( ac ), ( b ), ( d ), ( e ), ( i ), (A )
For the
remaining
cases we recall the list ofmultiplicities
of linearfactors of J :
In cases
( f ), ( L ), ( C),
J has no non-trivialautomorphism,
hence nor hasthe
pencil.
In cases(g)
and( k )
we obtain a group of order 2(interchange
x and y, resp.
change
asign).
In case(B),
the sameargument
as for(D)
shows that x and y are determined up to
scalars;
it is now clear that theautomorphism
group has order 2.6. - Stratification of the moduli space.
If f and
g aresubjected
to a linear transformation with determinant6,
an invariant ofweight
r ismultiplied by
6 r. When we calculate the in- variants(for
a case which is at leastsemi-stable)
we shallregard
them asdefining
thepoint
inweighted projective
spaceP( 1, 1, 2, 3)
with coordi- nates(I2, J2, I4, I6):
the results below aregiven
with this understand-ing.
Note that if 3 is takenas -1,
these four invariants areunchanged
but
FI9 changes sign.
SinceFI9
is determined up tosign by
the other in-variants,
its value does notyield
additionalinformation,
and we omit itfrom the tabulations to follow.
We next tabulate invariants for all cases
giving
aunique point
inmoduli space. As well as the
1-point
strata of Tables 2 and3,
we list otherpoints picked
outby symmetry
considerations. Thepreflx S
in the last two cases denotes an element of a stratum with extrasymmetry.
We observe that
(s4 )
and(t3 )
are the twopoints
at which the moduli spaceP( 1, 1, 2, 3)
is not smooth.It will
clarify
our discussion of 1-dimensional strata if we first discusscases with
symmetry.
Thegeneric symmetric
case(S2)
is characterisedby
theequation FI9
=0,
for a direct calculation shows thatFI9
vanisheson
it,
and isirreducible,
so its zero locus also is.Special
cases of thismay be obtained
by substituting
inFI92
andfactorising.
This leads to 6TABLE 5.
cases, as follows.
The factors
defining (S4)
and(S6 )
each occursquared
when we sub-stitute in We can at once
identify
three of these cases: we have(,S2 )
=(J), (S4 ) _ (h), (S6 ) _ (F).
For if we take new normal forms for TABLE 6.TABLE 7.