A proof of Hilbert's theorem on ternary quartic forms with the ladder technique

0.11 MB·

by  Jia Xu

A proof of Hilbert’s theorem on ternary quartic forms with the
ladder technique
JiaXu
ComputerScienceandTechnologyCollageofSouthwestUniversityforNationalities,Chengdu,China
YongYao
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ChengduInstituteofComputerApplications,ChineseAcademyofSciences,Chengdu,China
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a
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Abstract

1
Hilbert’stheoremstatesthateverypositivesemi-definiterealternaryquarticformcanbewritten
2
as a sum of squares of quadratic forms. In this paper, we give a constructive proof with the

] rudiments of real analysis and the ladder technique. In order to ensure that sum of squares
C
of quadratic forms can be practically constructed, the computation of real zero points is also
S discussed.
.
s Keywords: Hilbert’stheorem,positivesemi-definite,ternaryquarticforms,sumofsquares,the
c
[ laddertechnique

2010MSC: 11E25,14P05
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2 1. Introduction
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0 In 1888, David Hilbert published a paper (Hilbert, 1888) on the problem whether a posi-
.
1 tivesemi-definiterealpolynomialisinevitablyasumofsquaresofotherrealpolynomials. This
0 work is influential and inspires a lot of great works in researchers even today. In the excep-
7
tionalcase, Hilbertprovedthatapositivesemi-definiterealternaryquarticformcanbewritten
1
as a sum of squares of quadratic forms. Hilbert’s proof is very brief but rather hard, because
:
v some complicated mathematical tools are used (see comments in AlbrechtandClaus (2012)
i
X andChoiandLam (1977)). Furthermore,Hilbert’smethoddidnotlenditselftoareallypracti-
calconstruction. In1977,ChoiandLam(ChoiandLam,1977)showedagracefulelementary
r
a proof, in which only the rudiments of real analysis and the representation theorem of convex
setareutilized. However,thismethodisnotconstructiveeither. ComparedwithChoiandLam
(1977), though there are some similarities in the rudiments of real analysis, the proof of the
presentpaperhasthefollowingtwoconspicuousdifferences.First,wetaketheladdertechnique
✩Thisworkwaspartially supported byNational NatureScience Foundation (61273007, 61673016), SWUNCon-
structionProjectsforGraduateDegreePrograms(2017XWD-S0805),AdvanceResearchProgramofElectronicScience
andTechnologyNationalProgram(2017YYGZS16),InnovativeResearchTeamoftheEducationDepartmentofSichuan
Province(15TD0050),SichuanYouthScienceandTechnologyInnovationResearchTeam(2017TD0028).
Emailaddresses:[email protected](JiaXu),[email protected](YongYao)
PreprintsubmittedtoElsevier March22,2017
instead of the representation theorem of convex set. Furthermore, the method of our proof is
constructivewhereasthe one in ChoiandLam (1977) is not, since the representationtheorem
of convexset is an existence theorem. Thus, for a positive semi-definite ternary quartic form,
its explicitrepresentationasa sum ofsquarescan beconstructedstep by step accordingto the
methodaddressedinthispaper.Fromthisperspective,weemphasizethatthisisnotacompletely
newproofbut,rather,acompletelyconstructiveaccountoftheproofinChoiandLam (1977).
Therestofthispaperisorganizedasfollows. Insection2,fourlemmasarepresented. Each
ofthemislikeastepoftheladder,andthehighestone(Lemma4)achievesthegoal. Withthe
laddertechnique,theHilbert’stheoremisprovedeasily.Insection3,thecomputationofrealzero
pointsisdiscussed,whichprovidestheconcretedetailsfortheconstructionofsumofsquares.
2. ProofofHilbert’stheorem
Firstly,thefollowingarerequired. Aform f R[x,y,z]issaidtobepositivesemi-definite
∈
(psd)if
f(P) 0, P R3,
≥ ∀ ∈
and PSD4 representstheset of allpositivesemi-definiteternaryquarticforms. SOS standsfor
3
thesetofsumofsquaresofpolynomials.
GivenP ,P R3 (0,0,0) satisfying f(P ) = f(P ) = 0,wecall P and P aresameif
1 2 1 2 1 2
∈ \{ }
thereisanon-zeronumberλsuchthatP = λP . Thatis, wediscussthezerosof f inthereal
1 2
projectivespaceP2(R). Theset ofzerosinP2(R)is denotedbyZ(f), and Z(f) standsforthe
| |
numberofelementsofZ(f). TheunitsphereiswrittenasS2 = (x,y,z) x2+y2+z2 =1 .
{ | }
Next,wewillpresentfourlemmas.
Lemma1. Ifthepsdquarticform f R[x,y,z]hasnozeroinP2(R)(i.e. Z(f) =0),thenthere
isaquadraticformg R[x,y,z]suc∈hthat f g2 ispositivesemi-definite| and|hasatleastone
zeroinP2(R)(i.e. Z(f∈ g2) 1). −
| − |≥
Proof. Since f is continuous on the unit sphere S2, f inevitably attains a minimum λ. It is
obviousthatλ>0and
f λ(x2+y2+z2)2 0.
− ≥
Letg= √λ(x2+y2+z2),andthen f g2ispositivesemi-definiteand Z(f g2) 1.
− | − |≥
Lemma 2. If the psd quartic form f R[x,y,z] has only one zero in P2(R) (i.e. Z(f) = 1),
then there are quadratic forms g ,g ∈ R[x,y,z] and non-negativereal numbers a|and|b with
1 2
∈
a2+b2 ,0suchthat f ag2 bg2ispositivesemi-definiteand Z(f ag2 bg2) 2.
− 1− 2 | − 1− 2 |≥
Proof. Bycoordinatestransformation,wecanassumethat f(1,0,0) = 0,and f canbewritten
as
f = x2p(y,z)+2xq(y,z)+r(y,z), (1)
where p 0, r 0,and pr q2 0. Therearethreecasesof ptobediscussed.
≥ ≥ − ≥
Case(i): p=0(0form).Itisclearlythatqisnecessarilya0form.Then
f =r(y,z).
2
Since f hasonlyonezero(1,0,0),itholdsthatr(y,z) > 0forall(y,z) S1 = (y,z) R2 y2+
∈ { ∈ |
z2 = 1 .Furthermore, f iscontinuousontheunitcircleS1,sothat f hastheminimumλonS1.
}
Itisobviouslythatλ>0and
f λ(y2+z2)2 0.
− ≥
Letg= √λ(y2+z2). Thus f g2ispositivesemi-definiteand Z(f g2) 2.
− | − |≥
Case (ii): The rank of p is 1. By coordinate transformation, assume that p = y2. The
condition pr q2 0impliesthatqisdividedbyy(thecoefficientofz3inqis0).Thatis,there
− ≥
isq satisfyingq=yq . Substitutingitinto(1)onehas
1 1
f = x2y2+2xyq +r=(xy+q )2+r q2.
1 1 − 1
Noticethatr q2 ( 0)isnota0form,otherwise f wouldhaveinfinitezerosonS2. Thereare
− 1 ≥
twosituationstobeconsidered.
(a) r q2hasazero (y ,z )on S1,thenithasatleast2zeros(1,0,0)and(0,y ,z )on S2.
− 1 0 0 0 0
Hence f (xy+q )2ispositivesemi-definiteand Z(f (xy+q )2) 2.
1 1
− | − |≥
(b) r q2 hasnozeroonS1,then f (xy+q )2 yieldscase(i),sothatthereisaquadratic
− 1 − 1
formgsuchthat f (xy+q )2 g2ispositivesemi-definiteand f (xy+q )2 g2 2.
1 1
− − | − − |≥
Case(iii):Therankofpis2. Bycoordinatetransformation,pmaytaketheformp=y2+z2.
Consequently,weclaimthatpr q2 >0onS1.Otherwiseifpr q2 =0for(y ,z ),(0,0),
− − |(y0,z0) 0 0
then( q(y ,z )/p(y ,z ),y ,z )isazeroof f, conflictingwiththefactthat f onlyhasthezero
0 0 0 0 0 0
−
(1,0,0).Thus pr q2attainstheminimumλ>0onS1. Asaresult,
−
(pr q2) λ(y2+z2)3 0,
− − ≥
withequalityat(y ,z ) S1. Hence
0 0
∈
f λ(y2+z2)2
−
isstillpositivesemi-definitebydiscriminant,andhasatleasttwozeros:(1,0,0)and(−pq((yy00,z,z00)),y0,z0).
Itfollowsfromcase(i)tocase(iii)thatLemma2holds.
Lemma 3. Ifthepsd quarticform f R[x,y,z]hasonlytwo zerosin P2(R)(i.e. Z(f) = 2),
thentherearequadraticformsg ,g ,g∈ R[x,y,z]andnon-negativerealnumbers|a,b,|cwith
1 2 3
∈
a2+b2+c2 ,0suchthat f ag2 bg2 cg2ispositivesemi-definiteand Z(f ag2 bg2 cg2) 3.
− 1− 2− 3 | − 1− 2− 3 |≥
Proof. Changecoordinatessothat f(1,0,0)= f(0,1,0)=0. Write
f = x2p(y,z)+2xzq(y,z)+z2r(y,z), (2)
where p, qandrarequadraticformswith p 0, r 0,and pr q2 0. Nexttheproofwillbe
≥ ≥ − ≥
splitintotwocases.
Case(i): Ifatleastoneof pandrhasazeroonS1,thenlet(y ,z )bethezeroofrwithout
0 0
lossofgenerality.Substitutingitinto(2),oneyields
f(x,y ,z )= x2p(y ,z )+2xz q(y ,z ).
0 0 0 0 0 0 0
Then(0,y ,z )isalsoazeroof f, hencey = 1,z = 0. Substitutingthemintoquadraticform
0 0 0 0
r(y,z),itisreducedtotheform
r(y,z)=tz2, t>0.
3
Frompr q2 =tpz2 q2 0,qmustbedividedbyz. Thusq= √tzq ,whereq isalinearform.
1 1
− − ≥
Therefore,
pr q2 =tz2(p q2) 0= p q2 0,
− − 1 ≥ ⇒ − 1 ≥
and
f = x2p+2xz2√tq +z2(tz2)=(√tz2+xq )2+x2(p q2).
1 1 − 1
Then f (√tz2+xq )2 = x2(p q2)ispositivesemi-definiteandhasinfinitezeros(0,y,z).Hence
− 1 − 1
Z(f (√tz2+xq )2) >3.
1
| − |
Case(ii):IfneitherpnorrissupposedtohaveazeroonS1(bothpandrarestrictlypositive
onS1),thendiscusstwopossibilitiesofthediscriminantpr q2.
(a)Thediscriminantpr q2hasazero(y ,z ) S1,and−letλ= q(y0,z0). Thenonehas
− 0 0 ∈ −p(y0,z0)
f (x,y,z)= f(x+λz,y,z)= x2p+2xz(q+λp)+z2(r+2λq+λ2p).
1
Since(y ,z )isazeroofr+2λq+λ2p, f isreducedtothecase(i).
0 0 1
(b)Thediscriminantpr q2hasnozeroonS1. Since pandrarestrictlypositiveonS1,the
−
function
pr q2
−
p(y2+z2)
is also strictly positive on S1, and its minimum λ > 0. Thus pr q2 λp(y2 +z2), and f
− ≥ −
λz2(y2+z2)ispositivesemi-definite.Ifr λ(y2+z2)haszeros,then f λz2(y2+z2)isreduced
− −
tothecase(i). Incontrast,ifr λ(y2+z2)hasnozeroonS1,then f λz2(y2+z2)isreducedto
− −
thecase(ii)(a).
Becausewehaveconsideredalltwocases,wecanconcludethatLemma3holds.
Lemma4. Ifthenumberofzerosofthepsdquarticform f R[x,y,z]inP2(R)ismorethan3
∈
(i.e. Z(f) 3),then f isasumofsquaresofquadraticforms.
| |≥
Proof. Theproofwillbesplitintotwocases.
Case(i): Iftherearethreezerosnotonthesameline,thenbyarrangingcoordinates,wemay
assumethat
f(1,0,0)= f(0,1,0)= f(0,0,1)=0.
Thisimpliesthatthedegreeofeachvariableislessthanorequalto2. Write
f =ax2y2+by2z2+cz2x2+a x2yz+b y2xz+c z2xy, a,b,c,a ,b ,c R.
1 1 1 1 1 1
∈
Do substitution xy = X,yz = Y,zx = Z, and f canbe transformedinto a positivesemi-definite
forminX,Y,Z.Itisobviousthat f isasumofsquaresofquadraticforms.
Case(ii):Ifallofthezerosareonthesameline,thenitcanbesplitintotwosubcases.
(a) f hasinfinitezeros. Then f haslinearfactorsbyBe´zout’stheorem. Since f is positive
semi-definite,thedegreeofthelinearfactorsiseven.Hencetheyare
l2h , l4,
2
wherelislinearformandh ispositivesemi-definitequadraticform(maybedegenerate). Itis
2
obviousthatallofthemaresumofsquaresofquadraticforms.
4
(b) f hasfinitezeros.weclaimthatthiscasewillneverhappen.Thereasonisasfollows.
Arrangecoordinatessothat f(1,0,0)= f(0,1,0)= f(x ,y ,0)=0with x y ,0. Write
0 0 0 0
f = x2p(y,z)+2xzq(y,z)+z2r(y,z), (3)
where p,qandrarequadraticformswith p 0,r 0,pr q2 0. Therefore,
≥ ≥ − ≥
f(x ,y ,0)= x2p(y ,0)=0.
0 0 0 0
Furthermore,x ,0yields p(y ,0)=0,where pisaquadraticformonlyinyandz. Let
0 0
p=ay2+2byz+cz2,a,b,c R.
∈
Consequently p(y ,0) = ay2 = 0,anda = 0byy , 0. Thuszdivides f by(3),andallofthe
0 0 0
pointsonthelinez=0arezerosof f. Thiscontradictsthepremisethat f hasfinitezerosonthe
linez=0.
Withtheabove4lemmas,Hilbert’stheoremwillbeprovedwithachart.
Theorem1(Hilbert). Apositivesemi-definiteternaryquarticformovertherealscanbewritten
asasumofsquaresofquadraticforms.
Proof. Thefollowingchartwillpresenttheprocessofproof.
Z(f) 3(SOS,Lemma4)
| |≥
ր ⇑
Z(f) 2 Lemma3
| |≥ k
ր ⇑ ց k
Z(f) 1 Lemma2 Z(f) =2
| |≥ k | |
ր ⇑ ց k
PSD4 Lemma1 Z(f) =1
3 k | |
ց k
Z(f) =0
| |
ThesetPSD4isdividedintotwodisjointsubsets,thatis,thesubsetwith Z(f) =0andtheother
3 | |
one with Z(f) 1. Accordingto Lemma 1, the former can be transformedinto the latter by
| | ≥
minusa squareofa quadraticform. Furthermore,the subsetwith Z(f) 1 canbe dealtwith
| | ≥
in the same way according to Lemma 2 and so forth. Finally the theorem holds according to
Lemma4.
Remark1. Theorem1 is theweak editionofHilbert’stheorem. The strongeditionis thatthe
positivesemi-definiteternaryquarticformsaresumofnomorethanthreesquaresofquadratic
forms. TheproofofthestrongeditioncanseeAlbrechtandClaus (2012),Powersetal. (2004)
andScheiderer (2010).
Remark2. Therealzerosareassumedtobeknownintheaboveproof. Toachievethegoalof
reallypracticalconstruction,westillneedsomemethodinsolvingtherealzerosofternaryquar-
ticforms. ThisproblemwillbeconqueredinSection3,andwewillreallyobtainthepractically
constructiveproofofHilbert’stheorem.
Nextwewillpresentanexample.
5
Example1. (Cirtoaje,2006)Computethesumofsquaresofthefollowingternaryquarticform.
f = 4(x4+y4+z4)+21(xy+yz+zx)2 10(x2+y2+z2)(xy+yz+zx)
−
37xyz(x+y+z).
−
Solution:Since f has4zerosintheprojectivespace. Thereare
(1,1,1), (3,2,2), (2,3,2), (2,2,3).
Selectthreeofthemandconstructthefollowingmatrix(eachcolumnisazeroof f)
1 3 2
A=1 2 3.
1 2 2
Dolineartransformation
x x¯
y= Ay¯.
Bycomputing,wehave z z¯
f(A[x¯,y¯,z¯]T) = x¯2y¯2+49y¯2z¯2+z¯2x¯2 x¯2y¯z¯+7y¯2z¯x¯+7z¯2y¯x¯
−
1
= (x¯y¯ z¯x¯)2+(x¯y¯+7y¯z¯)2+(z¯x¯+7y¯z¯)2 . (4)
2 −
(cid:16) (cid:17)
ComputetheinversematrixofA,
2 2 5
− −
A−1 = 1 0 1.
 0 1 −−1
Let
x¯ x 2x 2y+5z
− −
y¯= A−1y= x z ,
andsubstituteitinto(4),andthenz¯thesumzofsquaresxo−f−fyisasfollows.
1
f = ( 2x2+5xz+2y2 5yz)2.
2 − −
X
3. Computationofzeros
Inthissection,wewilldiscusstheproblemofcomputingzeros. Lemma4presentsakeyof
constructingsumofsquares,thatis,findingoutatleastthreerealzerosofapositivesemi-definite
ternaryquarticform.Nextwewillproveaproposition(Lemma5)thatcansolvetheproblemof
computingthe realzerosof positivesemi-definite ternaryforms. The fundamentalidea comes
fromYangandXia (2000);XiaandYang (2016).
Lemma 5. Given positive semi-definite form f R[x,y,z], let f be the derivative of f with
∈ x′
respect to x, andres(f, f ,x) be the resultantof f and f with respectto x. Then the equation
x′ x′
f =0 andtheequations f =0, res(f, f ,x)=0 areequivalentintheprojectivespaceP2(R).
{ } { x′ }
6
Proof. ItisobviousthatZ(f) Z(f, res(f, f ,x)).NextwewillproveZ(f) Z(f, res(f, f ,x)).
⊇ x′ ⊆ x′
It is easy to get that the real zero (x ,y ,z ) S2 of positive semi-definite form f need to
0 0 0
∈
satisfythefollowingequationsforsolvingstationarypoints.
f =0,
x′
 fy′ =0, (5)
 fz′ =0.
Thisisbecauseofthefactthatifatleastoneofthefollowingvalues
f , f , f
x′|(x0,y0,z0) y′|(x0,y0,z0) z′|(x0,y0,z0)
is not zero, then f can be approximated by linear functions on a neighborhood of the point
(x ,y ,z ). So f isnotnonnegativeonaneighborhoodofthepoint(x ,y ,z ). Thiscontradicts
0 0 0 0 0 0
thepremisethat f ispositivesemi-definite.Thusx satisfiesthefollowingequations.
0
f(x,y ,z )=0,
0 0 (6)
( fx′(x,y0,z0)=0.
Thenres(f, f ,x) = 0accordingtothefundamentalpropertyofresultant. Thatis, Z(f)
x′ |(y0,z0) ⊆
Z(f,res(f, f ,x)).
x′
Remark1. In Lemma 5 f cannot include square factors, otherwise res(f, f ,x) would iden-
x′
tically equal to 0. Thus the equations f = 0,res(f, f ,x) = 0 in P2(R) is zero dimension,
{ x′ }
and we can solve the aboveequationswith variousefficientmethodssuch as methodof resul-
tant (Kapuretal. , 1994; CannyandManocha, 1993), rational single variable present method
(Rouillier,1999),etc.
Remark2. ThegeneralversionofLemma5(successiveresultantmethod(YangandXia,2000;
XiaandYang, 2016)) also ensure the minimum of Lemma 1, Lemma 2 and Lemma 3 can be
computedaccurately, and then ensure the aboveproofof Hilbert’s theorem is completelycon-
structive.However,thefollowingexample(example3)showsthatthepracticalcomputationmay
beverycomplexandfinallycomestonothingbecauseofhightimecomplexity.
Nextwewillpresenttwocomputingexamples.
Example2. (VasileCirtoaje)(Cirtoaje,2006)Computerealzerosof
f =(x2+y2+z2)2 3(x3y+y3z+z3x).
−
Solution:Firstlycompute
f =4(x2+y2+z2)x 9x2y 3z3.
x′ − −
Thencomputetheresultantres(f, f ,x),
x′
res(f, f ,x)=9(13y4 18y3z y2z2 6yz3+13z4)( z+y)2(y3 5y2z+6yz2 z3)2.
x′ − − − − − −
ByGrammatrixmethod(Choietal. ,1995),oneyields
h = (13y4 18y3z y2z2 6yz3+13z4)
− − −
9 7 2 51 3
= 13(y2 yz z2)2+286( yz z2)2+ z4.
− 13 − 13 13 − 286 22
7
ThushisstrictlypositiveonS1. Hence
res(f, f ,x)=0 ( z+y)(y3 5y2z+6yz2 z3)=0.
x′ ⇐⇒ − − −
Byusingrealrootisolationalgorithm(XiaandYang,2016),weknowthattheequationres(f, f ,x)
x′
=0hasfourrealzerosintheprojectivespaceP2(R):
(1,1), (α ,1), (α ,1), (α ,1),
1 2 3
whereα <α <α arethreepositiverealzerosoft3 5t2+6t 1=0. Substitutingtheabove
1 2 3
− −
four points into the polynomial f, and we have four real zeros of f = 0 in the real projective
spaceP2(R). Thereare
1 1 1
(1,1,1), ( ,α ,1), ( ,α ,1), ( ,α ,1). (7)
α 1 α 2 α 3
2 3 1
Consequently f canbewrittenasasumofsquaresaccordingtoLemma4.
Example3. (C. Scheiderer)(Scheiderer, 2016) g = x4 +y4 +z4 + xy3 + xz3 +yz3 3x2yz
− −
4xy2z+2x2y2.
InScheiderer (2016)thisformisproventohavethefollowingcharacters:gcannotbewritten
as sum of squares of quadratic forms with rational coefficients. That is to say the answer of
Sturmfels’squestion(Scheiderer,2016)isnegative.Itmeansthatitisverydifficulttowritegas
sumofsquaresforthetoolsthatdependonnumericalcomputationsuchasSOSTOOLS(Parrilo,
2003).
Nextwecomputethezerosofgintherealprojectivespace.
Firstlywecompute
g =4x3+4xy2 6xyz+y3 4zy2+z3.
′x − −
Thencomputetheresultant
res(g,g ,x) = 229y12 1904y11z+5896y10z2+1376y9z3 12176y8z4
′x − −
+6432y7z5+8630y6z6 9472y5z7+952y4z8+3232y3z9
−
96y2z10+336yz11+229z12.
−
Byusingrealrootisolationalgorithm,weknowthatres(g,g ,x)onlyhasonetrivialzero(0,0),
′x
but(1,0,0)isnotthezeroofg. HenceZ(g)= (emptyset),andgisstrictlypositiveonS2.
∅
ByLemma1,oneyieldsthatthereisapositiveconstanttsuchthat
g =g t(x2+y2+z2)2
t
−
isstillpositivesemi-definiteandhasatleasta zerointherealprojectivespaceP2(R). Further-
more,byusingsuccessiveresultantalgorithm(YangandXia,2000;XiaandYang,2016),one
can compute t accurately. t, in the interval [51/512,103/1024]and the approximation being
0.10009018,istherealzeroofthefollowingequationwithdegree12.
1540909743009169408x12 13437733654176464896x11
−
+51805978528683065344x10 116396366581901484032x9
−
+168975565335348900096x8 165910705322168135008x7
−
+111957978056509355125x6 51652982930080321180x5
−
+15876922302830413280x4 3088008227838928440x3
−
+347409936566531728x2 19347901948050048x
−
[phone]=0.
8
Itisverydifficulttocomputetherealzerosofg forhightimecomplexity.
t
InScheiderer (2016)gcanbewrittenas
1 1 y2 2xz
g= (2x2+βy2 yz+(2+ )z2)2 β(2xy + +βyz z2)2 ,
4 − β − − β β − !
whereβisanegativezeroofequationt4 t+1=0.
−
4. Conclusion
Inthispaper,wegiveaproofofHilbert’stheorembyfourlemmasandtheladdertechnique.
Accordingtotheproof,catchingatleastthreerealzerosofapositivesemi-definiteternaryquartic
formisnecessaryforconstructingitssumofsquares. Consequently,wepresentthemethodof
locating the zeros based on the property that zeros of a positive semi-definite form satisfy the
equationsforstationarypoints. Howevertherepresentationofsumofsquaresusingthismethod
maybecomplexandhowtobuildaclearrepresentationisanewquestion.
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