R ENDICONTI
del
S EMINARIO M ATEMATICO
della
U NIVERSITÀ DI P ADOVA
E ZIO S TAGNARO
A new construction of a surface of degree 5 having 31 nodes
Rendiconti del Seminario Matematico della Università di Padova, tome 69 (1983), p. 27-35
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of
asurface of degree 5 having 31 nodes.
EZIO STAGNARO
(*)
SUNTO - È
noto un soloesempio
disuperficie
di ordine 5 di avente 31 nodi (=punti doppi conici);
esso 6 dovuto a E. G.Togliatti ( cfr. [3]).
Recen-temente A. Beauville ha dimostrato che 31 6 il massimo numero di nodi che una
superficie
di ordine 5 diP’ c puo
avere. Qui,seguendo
i nostrimetodi
(cfr.
[6],[7]),
si dh una nuova costruzione di unasuperficie
diordine 5
di ,
con 31 nodi. La nostrasuperficie
6 definita da unpolinomio
omogeneo
biquadratico
in,Xo : cxX’ 0
-+- -f- y;lX,,8, Y
E~2~ ~3],
ed 6 costruita considerando la curva di diramazione
(#2- ay)2y
delpiano Xo
nellaproiezione
dellasuperficie
dalpunto
(1, 0, 0, 0). Il risultato vale anche se la caratteristica di k 6 p > 0,purche
Zn # 2, 3, 5 di-verso
dagli
eventuali valori che danno nodi coincidenti sullasuperficie.
Introduction.
The
largest
number of conical doublepoints, shortly nodes,
that itwas
possible
toassign
to a surfaceF5
ofdegree 5
in Ccomplex field,
is 31. This result is due to E. G.Togliatti (cfr. [3])
who con-structs
F5
as branch locus of a cubichypersurface .g3
inhaving
15nodes,
in theprojection
from ageneric
line of.H3
over aP~.
The bestlimitation for the maximum number of nodes for a surface of
degree
5(*)
Indirizzo dell’A.: Istituto di MatematicaApplicata,
Via Belzoni, 7 -35100 Padova
(Italy).
Lavoro
eseguito
nell’ambito delGruppo
Nazionale per le StruttureAlge-
briche e Geometriche e loro
Applicazioni.
28
in
Pg
was 34. Such a limit is due to A. B. Basset(cfr. [1], [2];
for aproof
of Basset’s limitation cfr.[8]).
Then it is natural to search for-examples
of surfaces ofdegree 5
with a number of nodes > 31.Despite
several
attempts
I never was able to find a surface ofdegree 5
with 32nodes.
Only recently
I understood that the motivation of this failurewas not in the
method,
as Ibelieved,
but in the non existence of such surface. In fact A. Beauvilleproved (cfr. [9])
that a surface ofdegree 5
in
Pg
cannot have a number of nodes >31;
so the surface ofTogliatti
has the maximum number of nodes.
Since
Togliatti’s example
is theunique example well-known,
y inthe
complex
case, of a surfaceFs
with 31 nodes and since 31 isjust
the maximum for surfaces of
degree 5,
it seems to me useful to showthat also with our method
(substantially
contained in[6]
and[7])
we are able to
construct,
notonly
a surface ofdegree 5
with 31nodes,
ybut also to check that such a construction works even if the characteri- stic of k
is p
>0, except
for a finite number of values of p.The idea of our construction is the
following.
LetFs
be a surface~ of
degree 5
definedby
ahomogeneous polynomial
in the variables.X~o , Xi, X2, biquadratic
in.Xo ,
and let us
project ~5
from thepoint ( 1, 0, 0, 0 )
over theplane Xo .
The branch locus on the
plane Xo
is(f32
-ay)2y.
Asingularity
onF,
is
projected
in asingularity
on~2
- ocy or on y andconversely
asingu- larity
notbelonging
to a or asingularity
on y is theprojec-
tion of two or one
singularities
on Then toimpose
alarge
numberof isolated
singularities
onFs
it isequivalent
toimpose
alarge
numberof
singularities
on~2
- ay or on y. Now in order toimpose singula-
rities on
fl2 -
ay we consider a curve 6having
alarge
number of sin-gularities
and we show that it ispossible
to write 6 in the formfl2-
ay.We choose as 6 a
curve split
in three conics 0 ==~~ ~~ ~3
and more-over we want the
quintic
y tosplit
into aquartic 99
and a line r. Inother
words,
want to write thefollowing equality (identity)
For this we remark that if
(1)
holds the sextic818283
is tritan-.gent
to the two lines a and r. But this condition is also sufficiente in -order that(1) holds;
inparticular if 3,
istangent
to aand r,
we havewhere
is a linear form(i
=1, 2, 3)
andconsequently
which
gives (1)
withAt this
point
incorrespondence
to the choosenpolynomials
0153,~8 ~
y = rcp the surface
ocX" 0 + 2BX20 + y
has(at least)
28 nodes. Next if we were able toimpose
three nodes on thequartic cp
which appears in(2),
then thecorresponding
surface would have 31 nodes. Unfor-tunatly
we were not able to find aquartic
with three nodes. To over- come thisdifficulty
we add and substract in(2)
the term+
+
where I is anarbitrary
linear form. Weget
Putting now # =
y =r(p
+ arl2 +2ll Z2 1, l)
thecorrespond- ing
surfaceFs
hasagain
28 nodes and among thequartics y
= 99+
+ arl2 + we shall be able to find one, forparticular
values ofthe
parameters,
y with three nodes and therefore a surfaceFs
with31 nodes.
In the n. 1 of this paper we recall some results
connecting
the sin-gularities
onT’S
with thesingularities
on the branchlocus;
in the n. 2we
give
thepolynomials 8i,
ce,fl,
y and we check thatthey
leadto a surface
Fs
with 31nodes; finally
in the n. 3 we show the way followed toimpose
on thequartic y
three nodes.1. The results of this number are due
substantially
to D. Galla-rati
( [4], [5 ). They
are written here with a sketch of theproof
forreasons of
completeness
and for convenience of the reader. The ex-position
is like that in[7]
to which we refer for further details.Let be the surface of
degree 5
inPjJ , k algebraically
closed field of characteristicp ~ 2, 3, 5,
definedby
thehomogeneous polynomial
30
We
project F5
from thepoint (1, 0, 0, 0)
over theplane Xo (such
aprojection
is not defined in(1, 0, 0, 0 ) ) ;
the branch locus on.~o
is(P 2 - oey) 2 y.
LEMMA 1.
If
A =(Yo,
y1, y2 , with yo =1=0,
is asingular point
on
F5,
then theprojection
A’=( o,
Yl, Y2, issingular point
on thecurve
fl2
- ay on.X’o; conversely if
A’ issingular
onfJ2 -
ay andif oc(A’) =
=
I
=1=0,
then the twopoints
toyl , y2 , distinct
if ~8(A’ )
=1=0,
aresingular
onFs.
2) If a(A’ )
=1= 0 moreprecisely
we have: A is a node onFs if
andonly if
A’ is a node( =
doublepoint
with distincttangents)
ocy.Now we consider the case Yo = 0.
3)
B =(0,
Yl, Y2,Y3)
issingular
orc~’5 if
andonly if
it is sin-gular
on y ;if
=1= 0 then B is a node onFs if
andonly if
it is a nodeon y.
PROOF. The statement
1)
follows from theequality
Namely
we have ++ y)
=+
conse-quently
if A issingular
on.Fb ,
since characteristic of k is ~ 2 and yo ~0,
we have that
ocX’ 0 -E- ~
vanishes on A. Then from(4) ~., equivalently A’,
is
singular
on#2
- «y.Conversely
from thehypotheses
we have thatthe
right-hand
side of(4)
issingular
inÅ2,
hence also the left- handside;
but therefore.~’5
issingular
inA i , i = 1, 2 .
The
proof
of2)
followsby considering
thetangent
coneFA
in Ato
.F5
andshowing
that in theprojection
from(1, 0, 0, 0)
the branchlocus of
FA
isjust
thetangent
cone in A’to B2
- ay(remark
that«(.A’)~o ~(l,o,o,o)~rA).
The first
part
of3)
follows from theequality
++ y
=+ X’o --~- y
and the second one from2. We consider the
polynomials
The
key
to the existence of the surfaceFs
with 31 nodes is the follow-ing equality
whosechecking
isonly
aquestion
of calculationPutting
=II l2la
+arl, y
= ry(5)
can be writtenWith
respect
to the curves onXo
definedby
the abovepolynomials,
the
following
statementshold, except possibly
finite manypositive
values of the characteristic of k:
(a)
the curve~2
- ay,split
in the three conics~~, ~~, ~3 (c f r. (6)),
has 12
(distinct)
nodes in the commonpoints
to twoconics;
(b)
the curve y,split
in r and in 1p, has 7(distinct)
nodes: 4 in thecommon
points
to r andto 1p
and 3 in the three nodes on 1p:(1, 0, 0 ), (0, 1, 0), (0, 0, 1);
(c)
the nodes in(b)
are distinctf rom
those in(ac) ; (d)
the curves a and~8
do not passthrough
any node.The
possibly exceptional positive
values of the characteristic of 7~ are those whichgive
coincidentnodes ;
we are not interested in such values.Before
proving (a), (b), (c)
and(d),
we show how from them theexistence of a surface
Fs,
ofdegree 5,
with 31 nodes follows.32
THEOREM. Let oc,
f3,
y be thepolynomials de f ined
in this number.If
the characteristicof
k isdifferent from 2, 3, 5
andmaybe from
otherpositive values,
thesurface Fs, of degree 5, de f ined by
thehomogeneous polynomial
has 31
(distinct)
nodes andF5
has no othersingularity
outside the nodes.PROOF. From
1)
and2)
of the lemma of the n. 1 and from(a)
and(d) .F5
has 24(distinct)
nodes incorrespondece
to the 12 nodeson p 2
- ay.From
1)
and3)
of the lemma and from(b)
and(d) F5
has 7 nodes inthe nodes on y on
Xo .
From(c)
all the nodes are distinct. From the firstpart
of1)
and(3)
of the lemma.F5
has no othersingularity
out-side the 24 + 7 = 31 nodes.
Q.E.D.
Now we prove
(a), (b), (c)
and(d).
PROOF OF
(c~). (a)
can beproved directly calculating
the commonpoints
to twoconics 3;, j
=1, 2,
3:ð1
nð3
= commonpoints
to the lines36(11 ~
+ð2
= commonpoints
to the lines36(11 ~ i3) X2
+where i2
= -1,
andchecking
that all thesepoints
are distint in charac-teristic zero, hence in
positive
characteristicexcept, maybe,
finitemany values.
PROOF OF
(b).
It isenough
to check that the 4 commonpoints
to rand
to y
are distinct in characteristic zero; this can be done with the discriminantof r and y
withrespect
to or otherwisedrawing
on the realplane
the curve V andseeing
that rmeets y
in 4distint real
points.
PROOF OF
(c).
The threepoints (1, 0, 0), (0, 1, 0), (0, 0, 1)
areclearly
distinct from the nodes on
~1 ~2 ~3
because none of thembelongs j = 1, 2, 3 ;
moreover theremaining
nodes on ry are distinct from those on for otherwise if apoint
Pbelongs to r, ~g, 3, (i
#j),
then P
belongs to r, li, 1j
which isimpossible.
PROOF OF
(d).
a does not passthrough
any node P on~1~3?
forotherwise .P
belongs
to a,li, li (i ~ j )
which isimpossible.
Since otmeets r in the
point (1, 1, 0),
it follows that « does not passthrough
any node on
ry. Also B
does not passthrough
any node onðlð2ð3
otherwise if
Q
is such anode,
from theequality ~1 ~2 ~3
=f32 -
ocy,being Q singular
on~2
- ay and onf32, Q
would besingular
on ay ==since Q
does notbelong
to a, thenQ
is a node on ry,against (c).
Final-ly f3
does not passthrough
any node on ry otherwise if .R is such anode,
R is
singular
onf32
and on y, hence issingular on @2
- ay,again contradicting (c).
3. In this number we want to show
shortly
the way followed toimpose
three doublepoints
on thequartic y
= g+
arl2 +2l1 l2 Z3 t
ofthe
introduction,
in such waythat y
does notsplit
in a conic countedtwice,
otherwise incorrespondence
onF’5
we have a doubleconic,
sonon isolated
singularities.
The more naturalthing
is toimpose
tothe y
the three
singular points (1, 0, 0 ), (0, 1, 0 ) , (0, 0, 1).
For this we mustannihilate the coefficients of
.X4,
=1, 2, 3 7 i # j,
in the ~.We
get
analgebraic system
of 9equations.
Since ispratically impos-
sible to resolve this
system,
theproblem
is to makepositions
among theparameters
in such a way as to resolve thesystem
and not losethe non
trivial
solution. For this weimpose
onthe y
to besymmetric
in the affine
plane
to the lineXi - ~Y2,
ychoosing
A,
,u,ai , b i ,
c; , a,(the parameters a,
c aresuperflous
but usefulto
get semplifications
in thecalculations).
With this choice we have
34
The coefficients to annihilate are now five instead of nine.
Anyhow
we
get
verycomplicated equations.
Agood simplification
isgot
onputting a2
=0;
even if with this choice we lose aparameter
we shallsee that it is
possible
to have non trivial solutions. We remark that with otherchoices,
for instancebl
=0,
we haveonly
the trivial solu-tion split
in a conic counted twice.Choosing
then a2 = 0 and anni-hilating
the coefficient ofX i
and of inthe V,
we may write-
Annihilating
the coefficient of inthe 1jJ,
we havewhere it is
possible
toreplace bl
and Cl with the valuegiven by (7).
We have still to annihilate the last two coefficient : those of and
X§ in
y. At thispoint
we cantry
someattempts
withparticular
numerical values of the
parameters. Namely choosing a
= ~, == al === aa == 1, c = -1,
« -f- >> _ c J~ from(7)
and(8)
we havebl = 4,
cl
= 3, c2 = (2b,
+2)/3; annihilating
the coefficient of and~Y’3
we have
p2
=(b2
+7)/9, (b2
-2)2(b2
+2)
= 0. We cannot chooseb2
= 2 otherwise31
and62
passthrough (o, 0, 1) against (c),
n. 2. Con-sequently
we haveb2 = - 2,
hence c, = -2/3, p2
=5/9.
Thereforethe
polynomials
a, are those of n. 2 and the verifications made there assurethat,
with ourchoice,
we have resolved theproblem.
BIBLIOGRAPHY
[1] A. B.
BASSET,
The maximum numberof
doublepoints
on asurface,
Nature,73 (1906), p. 246.
[2] A. B.
BASSET,
On thesingularities of surfaces,
Quart. Journ., 38 (1907),pp. 63-83.
[3] E. G. TOGLIATTI, Una notevole
superficie
del 5° ordine con solipunti doppi
isolati, Festschrift R. Fueter, Zurich (1940), pp. 127-132.[4] D. GALLARATI, Intorno a certe
superficie algebriche
aventi un elevato nu-mero di
punti singolari
isolati, Rend. Acc. Naz. Lincei, Serie VIII, 11, fasc. 6 (1951), pp. 344-347.[5] D. GALLARATI, Sulle
superficie
delquinto
ordine dotate dipunti tripli,
Rend. Acc. Naz. Lincei, Serie VIII, fasc. 1 (1952), pp. 70-75.
[6] E. STAGNARO,
Sopra
certesuperficie algebriche
consingolarità
isolate, Le Matematiche, Catania, 23 (1968), pp. 338-343.[7] E. STAGNARO, Sul massimo numero di
punti doppi
isolati di una super-ficie algebrica
diP3,
Rend. Sem. Mat. Padova, 59 (1978), pp. 179-198.[8] E. STAGNARO, On the maximum number
of
isolatedsingularities of algebraic
curves and
surfaces
and on Basset’s limitations, Seminar Notes, Bhaska-racharya
Pratishthana Institute, Pune(India)
(1979-1980).[9] A. BEAUVILLE, Sur le nombre maximum de
points
doubles d’unesurface
dans P3
(03BC(5)
=31), Algebraic Geometry Angers
1979,Sijthoff
and Noor-dhoff (1980), pp. 207-215.
Manoscritto pervenuto in redazione il 21 settembre 1981.