2.4 Intermediate cones between sos and psd cones
3.1.1 Zeroes and Test set of symmetric quartics
In this section we shall use x = (x1, . . . ,xn) to denote a generaln−tuple and
yrj. We will present some results from [9] including Corollary 3.11 that gives test set for symmetric quartics in n ≥ 4 variables. This corollary is the main result of this section and will be used to prove Proposition 3.12 in Section 3.1.2.
Timofte’s half degree principle [49] (seen as Theorems 2.15, 2.16) gave test sets for symmetric and even symmetricn−ary forms of degree 2d, as generalization of test set for even symmetricn−ary sextics given by Choi, Lam and Reznick in [10], and test set for even symmetric n−ary octics given by Harris in [20]. His results for symmetric forms were in fact a generalization of Corollary 3.11 given below from [9], since the work we present below from [9] was done much before [10] and [20].
Theorem 3.5. Supposen≥4, f ∈SPn,4andy=(y1, . . . ,yn)∈Z(f), whereyhas at least three distinct coordinatesyi,yj,yk. Then f is quadratic in M2 andM12, i.e.
[f]= [0,0,c,d,e].
For proving the above theorem we need a lemma, as given below.
Lemma 3.6. Supposeg(x1,x2,x3) ∈ P3,4 and (yi,yj,1) ∈Z(g) for 1 ≤ i , j ≤ 3 withy1 < y2 < y3. Thengis a perfect square and (u,v,1) ∈Z(g) for all (u,v) on the non-degenerate ellipse determined by (yi,yj).
Proof. Sinceg∈ P3,4,g=P
h2k (by Theorem 1.48) and (yi,yj,1)∈Z(hk) for each k; 1≤ i, j≤3 withy1 <y2 <y3.
Suppose (yi,yj,1) ∈ Z(h) andh(x1,x2,x3) = X
i≤j
αi jxixj, then (evaluating hat its zeroes and solving simultaneously) it is easy to see that
α11 =α12 =α22,
α13 =α23 =−(y1+y2+y3)α11, and α33 =(y1y2+y1y3+y2y3)α11. Hence
g(x1,x2,x3)=λh
x21+x22+[(y1+y2+y3)x3−(x1+x2)]2−(y21+y22+y23)x23i2
,and (u,v,1)∈Z(g) if (u,v) lies on the ellipseu2+v2+ y1+y2+y3−(u+v)2 =y21+y22+y23. This ellipse is not degenerate because its discriminant is non-zero (or since no three of the the points (yi,yj) fori, jare collinear).
3.7. Proof of Theorem 3.5.
Suppose [f]= [a,b,c,d,e], then f(x) = aM4+bM1M3+cM22+dM12M2+eM41, and since ∂Mr
∂xj
=rxr−1j we have
∂f
∂xj
(x1, . . . ,xn)=4ax3j+3bM1x2j+(4cM2+2dM12)xj+(bM3+2dM1M2+4eM13) Now suppose f y= 0. Since f is psd, ∂f
∂xj
y= 0 for all j. So by above equation the following holds for all j:
0= 4ay3j +3bN1y2j +(4cN2+2dN12)yj+(bN3+2dN1N2+4eN13) (3.6)
• Ify has four or more distinct coordinates, then the cubic in equation (3.6) above has at least four solutions, hence each coefficient vanishes. Therefore
3.1. n−ary quartics forn≥4 79 a = 0 and bN1 = 0, we will show that b = 0. For if for a contradiction b, 0, thenN1 = 0. So by equation (3.6): 4cN2+2dN12 =0, which implies cN2 = 0, so c = 0. But if a = c = 0, then (1,1,−2,0, . . . ,0) ∈ Z(p), and for this choice ofywe get N1 = 0,N2 = 6,N3 = −6. So by equation (3.6):
0=−6b, i.e.b=0. Soa=b=0 and f is a quadratic inM2 andM21.
• Now supposeyhas exactly three distinct coordinates and assume, without loss of generality, thaty1< y2< y3. Let
g(x1,x2,x3)= f x1,x2,(y1+y2+y3)x3−(x1+x2),y4x3, . . . ,ynx3. gneed not be symmetric, butg∈ Pn,4and (yi,yj,1)∈Z(g) for 1 ≤i, j≤3.
By Lemma 3.6 (just proved above), (u,v,1) ∈ Z(g) for all (u,v) on a non-degenerate ellipse, hence u,v,y1+y2+y3−(u+v),y4, . . . ,yn∈Z(f). By avoiding a finite set of lines, we can find (u,v) so thatu,v,y1+y2+y3−(u+v) are mutually distinct and distinct fromy4. Thus f has a zero with at least four distinct components and as in the above case, f is a quadratic in M2
andM12.
The above theorem suggests the following definition:
Definition 3.8. Forn ≥ 4, f ∈ SPn,4 is a dull formif it is quadratic in M2 and M21, i.e. if [f]= [0,0,c,d,e].
The above description of a dull form is motivated by Proposition 3.10 given below. We first note the following simple lemma on linear functions, which will be useful in proving Proposition 3.10:
Lemma 3.9. Supposeat+b≥ 0 for alltwithc≤t≤ d, wherec< d. Then (a,b) is a non-negative linear combination of (1,−c) and (−1,d).
Proof. Fort = λc+ (1− λ)d, at+b = λ(ac+b)+(1−λ)(ad +b), hence it is necessary and sufficient thatac+b≥ 0 andad+b≥ 0.
Any (a,b)=λ(1,−c)+µ(−1,d) for someλ, µ. Evaluation ofat+bat the end points givesλ, µ≥ 0, so the combination is a non-negative linear combination.
Proposition 3.10. Suppose f is a psd dull form. Then f is a non-negative linear combination of (λM2 − M12)2,0 ≤ λ ≤ n, and M21(nM2 − M21) and is a sum of squares. Further, a dull form f is psd iff f(x1, . . . ,xn)≥ 0 for everyxwith at most two distinct coordinates.
Proof. For anyy,N12 = λN2with 0 ≤ λ≤ n by the Cauchy-Schwarz inequality.
If [f] = [0,0,c,d,e], then f y = N22(c+dλ+cλ2), so f is psd iffthe following holds:
fˆ(λ)=c+dλ+eλ2 ≥ 0 for 0≤ λ≤ n.
Letαbe the minimum of ˆf on [0,n]. Then we can write f(x)= αM22+q(x), where qis psd and ˆq(λ0) = 0 for someλ0 ∈[0,n]. (Note that M22 = 4
n2 hn
2M2−M122
+ M21(nM2− M21)i
satisfies the conclusion.)
If 0 < λ0 < n, then ˆq0(λ0) = 0 also holds so that ˆq(λ) = e(λ− λ0)2 and q(x) = e(λ0M2− M21)2.
Ifλ0 =0, then ˆq(λ)= λ(eλ+d) and (eλ+d)≥0 forλ∈[0,n]. By Lemma 3.9 (just proved above), (e,d) is a non-negative linear combination of (1,0) and (−1,n), so qˆ is a non-negative linear combination λ2 and λ(n− λ) and q is a non-negative linear combination ofM14and M21(nM2− M21).
Similarly ifλ0 = n,qis a non-negative linear combination ofM12(nM2− M21) and (nM2−M12)2. SincenM2− M21 =X
i<j
(xi−xj)2, a dull psd quartic f is a sos.
Finally, for y
t = (t,1, . . . ,1);−(n−1) ≤ t ≤ 1, λ increases from 0 ton. Thus, if f(x) ≥ 0 for all x with at most two distinct components, then f
yt
≥ 0 for
0≤λ≤nand f is psd.
Theorem 3.5 and Proposition 3.10 have the following remarkable corollary which allows us to determine directly whether a given symmetric quartic is psd or not only by checking its value at xwith two distinct components. This corollary is the main result of this section and will be used to prove Proposition 3.12 in Section 3.1.2.
Corollary 3.11. A symmetricn−ary quartic f is psd iff f(x)≥0 for everyx∈Rn with at most two distinct coordinates (ifn≥ 4), i.e.Λn,2= {x∈Rn|xi ∈ {r,s};r, s}is a test set for symmetricn−ary quartics.
3.1. n−ary quartics forn≥4 81 Proof. “⇒” Clearly, f ∈SFn,4psd implies f(x)≥0 for every x∈Λn,2 ⊂Rn.
“⇐” Suppose f(x) ≥ 0 for every x ∈ Rn with at most two distinct coordinates but f is not psd. Let f(x) ≥ −λ;λ > 0 for the unit sphere P
x2i = 1 with f y =
−λ,P
y2i = 1. Letq(x) = f(x)+λM22, thenq(x) ≥ 0 for the unit sphere and soq is psd by homogeneity andq(y)= 0. By hypothesis,yhas more than two distinct components, so by Theorem 3.5 qis dull and so f = q− λM22 is also dull. By Proposition 3.10, f would then be psd, a contradiction to supposition.
(Since the form considered above is symmetric, so after a suitable parametriza-tion, we need to check only psdness of a finite set of binary quartic forms. Some details of this parametrization were given in [9], but we will not discuss them here.)
3.1.2 Psd not sos symmetric n−ary quartics for n ≥ 5
In this section, we will construct explicit forms f ∈ SPn,4 \SΣn,4 for n ≥ 5, in Proposition 3.12 and Theorems 3.16, 3.17. This will complete the proof of Theorem 3.1 and hence the answer toQ(S).
Consider the following symmetric quarticLn(x) inn≥4 variables, which will be central to our discussion in this section, and is defined by
Ln(x1, . . . ,xn) :=m(n−m)X
i<j
(xi−xj)4− X
i<j
(xi−xj)22
,
wherem= hn 2 i.
We will see in Proposition 3.12 that the symmetric quarticLn(x) defined above is psd for all n. Then we will prove thatLn for oddn is not a sos (see Theorem 3.16), which will finish half of the work that we intend to present in this section since we can then take f := Ln ∈ SPn,4 \SΣn,4 forn = 2m+1 ≥ 5. In contrast we will show in Proposition 3.13 that Ln for even n is a sos and therefore L2m will no more be a candidate for psd not sos symmetric n−ary quartic for even n = 2m ≥ 6. So, we will define another psd symmetric quarticC2min 2m(≥ 4)
number of variables and prove that it is not sos (see Theorem 3.17). This will
which appeared on the 1971 International Mathematical Olympiad for high school students. The problem was to determine those nfor which An(x) is psd, and the answer is n = 3 or n = 5. Lax and Lax [25, p72] showed that A5(x), a psd symmetric quartic in five variables, is not a sos. By direct computation (and using a shorthand representation of a symmetric quartic form from equation (3.4)), we get (A5) = (1,−1,0,1,3), soA5 = 1
8L5. In general,Ln for oddnis not a sos (see Theorem 3.16).
Proposition 3.12. Ln is psd for alln.
Proof. In view of Corollary 3.11, it is enough to prove thatLn ≥ 0 on the test set Λn,2= {(r, . . . ,r
Proposition 3.13. Ifnis even, thenLnis sos.
Proof. Ifnis even,n= 2m, then
3.1. n−ary quartics forn≥4 83
For proving thatLnfor oddnis not a sos, we first give the following definition and thereafter prove a lemma:
Definition 3.14. A subsetS ⊆ Rnof the formS ={x|xi ∈ {0,1} ∀i=1, . . . ,n}is
Thenhis identically zero.
Proof. Fix distinct i, j,k and let S such that |S| = m−1, be a set of indices not
Subtracting above two equations gives:
ak+X
Thusaik =ajk(from equations (3.7), (3.8)). 0. Then we get a symmetric quarticC2m(x) in 2mnumber of variables form≥ 2, defined by
C2m(x1, . . . ,x2m) := L2m+1(x1, . . . ,x2m,0).
3.1. n−ary quartics forn≥4 85 TriviallyC2m(x1, . . . ,x2m)∈SP2m,4, in addition we prove that it is not a sos as follows:
Theorem 3.17. Form≥ 2,C2m(x1, . . . ,x2m) is not a sos.
Proof. IfC2m =X
t
h2t, thenC2m(x)=0⇒eachht(x)=0, for anyx∈Rn.
In particular,C2m(x) = 0 when xhasm or (m+1) 1’s and mor (m−1) 0’s. So, ht(x)=0 for xwithmor (m+1) 1’s andmor (m−1) 0’s respectively.
Write
ht(x)=
n
X
i=1
aix2i +
n
X
i<j
ai jxixj
(ai j =ajiif needed).
Then by Lemma 3.15 above, we getht =0. Hence,C2mis not a sos.
From Proposition 3.12 and Theorems 3.16, 3.17, it follows that:
Corollary 3.18. 1. For allm≥2,L2m+1 ∈SP2m+1,4\SΣ2m+1,4, and 2. For allm≥ 3,C2m∈SP2m,4\SΣ2m,4.
This finishes the construction of explicit psd not sos symmetricn−ary quartic forms for alln ≥ 5, thereby completing the proof of Theorem 3.1 and hence the answer toQ(S).
In Chapter 4 we will actually prove a stronger version (see Theorems 4.12, 4.15) of the above Corollary 3.18 by showing that L2m+1(x21, ...,x22m+1) for m ≥ 2 andC2m(x21, . . . ,x22m) form≥3 are not sos, from which it follows thatL2m+1(x) for m ≥ 2 andC2m(x) for m ≥ 3 are not sos (because if L2m+1(x) orC2m(x) or both were sos, then by substitutingxi → x2i ∀i= 1, . . . ,n they still have to be sos, but it is not the case).
Chapter 4
Even symmetric forms
The inclusion SΣen,2d ⊆ SPen,2d holds for all n,d, since every sos form is clearly psd. In this chapter we will investigate the converse, i.e. the following question:
Q(Se) : For what pairs (n,2d) will SPen,2d ⊆ SΣen,2d? (4.1) In Section 4.1, we will construct explicit forms f ∈ SPen,2d\SΣen,2d for the pairs (n,2d) = (3,12), (n,8)n≥5 (see Propositions 4.9, 4.12, 4.15) and give a degree jumping principle (see Theorem 4.5) to find psd not sos even symmetricn−ary forms of degree 2d + 8,2d + 12,2d + 16, . . . etc. and 2d + 2n from given psd not sos even symmetricn−ary 2d−ic form. We will deduce that for the pairs (n,2d) = (n,6)n≥3,(n,8)n≥4,(3,2d)d≥5, and (n,2d)n≥4,d≥7, the answer to Q(Se) is negative.
Q(Se) is an extension of Hilbert’s original question (Q), mentioned before as equa-tion (1.2), to the special case when the form considered is in addiequa-tion even sym-metric.
The following proposition gives a partial answer toQ(Se) based on the results already known in the literature:
Proposition 4.1. 1. SPen,2d =SΣen,2d ifn=2,d =1,(n,2d)= (n,4)n≥3,(3,8).
2. SPen,2d ) SΣen,2dfor (n,2d)=(n,6)n≥3,(3,10),(4,8).
87
For part 1 of the above proposition, we give a proof forn= 2,d =1,(n,2d)= (n,4)n≥3 and an idea of the proof for (n,2d) = (3,8), as below. For part 2, we give explicit examples of forms f ∈ SPen,2d \ SΣen,2d for the pairs (n,2d) = (n,6)n≥3,(3,10) and (4,8) as below.
1. SPen,2d =SΣen,2d ifn=2,d= 1,(n,2d)=(n,4)n≥3,(3,8).
(a) Forn= 2,d =1,(n,2d)= (3,4):
From Hilbert’s Theorem 1.48 we know that a psdn−ary form of degree 2d is a sos ifn = 2,2d = 2, and (n,2d) = (3,4). So this must hold in particular for even symmetric forms also.
(b) For (n,2d)= (n,4);n≥4:
The vector space of even symmetricn−ary quartics is two-dimensional (as seen in Section 1.1.5). One basis is n
n
X
j=1
x4j,X
j<k
x2jx2ko
. We will consider a more useful basis {f,g}, where f = X
j<k
(x2j − x2k)2, g = X
j<k
x2jxk2. Clearly, f andgare sos.
If p is any even symmetric quartic, then there exist uniquely deter-mined realsaandbsuch that p= a f +bg.
Letu=(1,0, ..,0) andv=(1, ...,1). Then
f(u)= n−1,g(u)= 0, f(v)= 0,g(v)= n(n−1)/2.
If pis psd, then 0≤ p(u)=(n−1)aand 0≤ p(v)=n(n−1)b/2.
Thusa,b≥0 and hence pis sos.
(c) For (n,2d)= (3,8):
The vector space of even symmetric ternary octics is four-dimensional (as seen in Section 1.1.5). Any even symmetric ternary octic form
p(x,y,z) can be written uniquely as:
p(x,y,z) := [α, β, γ, δ]=α
3
Xx8+β
6
Xx6y2+γ
3
Xx4y4+δ
3
Xx4y2z2,
89
where each summation
k
Xdenotes the sum taken over all thek per-mutations of variables which yield distinct expressions (for example X3
x8 = x8+y8+z8).
Harris showed in [20, p221]
that any psd even symmetric ternary octic form is sos, by demonstrating that the following elements and families of elements inSΣe3,8comprise all the extremal forms ofSPen,8:
(i)A(x,y,z)=[0,1,−2,0]=
3
X(x2−y2)2y2;
(ii)B(x,y,z)=[0,0,1,−1]= 1 2
3
X(x2−y2)2z4;
(iii)C(x,y,z)=[0,0,0,1]= X3
x4y2z2; (iv) Db(x,y,z)=[1,2b,b2+2,2b2+2b]
= x4+y4+z4+b(x2y2+ x2z2+y2z2)2,forb<−1;
(v)Eu(x,y,z)=[1,−(u+1),u2+2u,−u2+1]
= 1 6
3
X 2x4−y4−z4−(u+1)(x2y2+x2z2−2y2z2)2,foru≥ 0, and then using the fact that any element of SPe3,8 is a finite non-negative linear combination of its extremal elements (i.e. A,B,C,Dt,Eu).
2. To show:SPen,2d ) SΣen,2dfor (n,2d)=(n,6)n≥3,(3,10),(4,8).
Proof. We divide this into subparts depending on the order of their appear-ance in the literature:
(a) For (n,2d)= (n,6);n≥3:
Forn≥3;t ∈Z,2≤t≤ n−1, Choi , Lam and Reznick in [10, p572]
gave the following psd not sos even symmetricn−ary sextic forms:
ft(x1, . . . ,xn) := (t2−t)
whereR(x,y,z) is the Robinson’s even symmetric ternary sextic.
Thus the forms f2, . . . , fn−1were actually generalizations of the Robin-son’s form to the case of any arbitrary number (≥3) of variables.
(b) For (n,2d)= (3,10):
For 0 ≤ u ≤ 2, u , 1, the following even symmetric ternary decic forms are psd but not sos see [20, p239]
X denotes the sum taken over all thek per-mutations of variables which yield distinct expressions (for example
3
Xx6y2z2 = x6y2z2+ x2y6z2+x2y2z6).
(c) For (n,2d)= (4,8)
For 0 ≤ v ≤ 5, v , 1, the following even symmetric quaternary octic forms are psd but not sos see [21, p81]
denotes the sum taken over all thek per-mutations of variables which yield distinct expressions (for example
6
Xx4y4= x4y4+x4z4+x4w4+y4z4+y4w4+z4w4).
91 (For more examples of psd not sos even symmetric ternary decics and
quaternary octics see [20] or [21].)
The above partially known answer to Q(Se) given in Proposition 4.1 can be summarized by the following chart:
deg\var 2 3 4 5 . . .
2 X X X X . . . 4 X X X X . . .
6 X × × × . . .
8 X X × ? ?
10 X × ? ? ?
12 X ? ? ? ?
... ... ? ? ? ?
where, a tick (X) denotes a positive answer toQ(Se), a cross (×) denotes a negative answer toQ(Se), and a (?) denotes an unknown answer toQ(Se).
To get a complete answer to Q(Se) it is interesting to look at the following remaining cases in which it is not known whetherSPen,2d ⊆SΣen,2dor not:
1. (3,2d) ford≥6, 2. (n,8) forn≥5, and 3. (n,2d) forn≥ 4,d ≥5.
In the coming section we will deal with these cases one by one and prove some results that will answerQ(Se) in more detail, and will bring us very close to a complete answer ofQ(Se).
4.1 Version of Hilbert’s 1888 theorem
In this section we will construct explicit forms f ∈ SPen,2d \SΣen,2d for the pairs (n,2d) = (3,12), (n,8)n≥5 and show that for the pairs (n,2d) = (3,2d)d≥6 and (n,2d)n≥4,d≥7, the answer toQ(Se) is negative. For the pairs (n,2d) forn ≥ 4 and d= 5,6 we are still working on getting an answer toQ(Se).
Our answer toQ(Se) can be summarized by the following chart, giving a ver-sion of Hilbert’s 1888 Theorem (i.e. Theorem 1.48) for even symmetric forms (see Theorem 4.16):
deg\var 2 3 4 5 6 . . .
2 X X X X X . . . 4 X X X X X . . .
6 X × × × × . . .
8 X X × × × . . .
10 X × ? ? ? ?
12 X × ? ? ? ?
14 X × × × × . . .
... ... ... ... ... ... ...
where, a tick (X) denotes a positive answer to Q(Se), a cross (×)\(×) denotes a negative known\our answer respectively toQ(Se), and a (?) denotes an unknown answer toQ(Se).
We first need the following lemmas for proving that SPen,2d ) SΣen,2d for the pairs (n,2d)=(3,2d)d≥6,(n,8)n≥5, and (n,2d)n≥4,d≥7:
Lemma 4.2. Forθ ∈1 3,1
, define Fθ = 1
1−θ (x4+y4+z4)−θ(x2+y2+z2)2
= (x4+y4+z4)− 2θ
1−θ(x2y2+ x2z2+y2z2).
4.1. Version ofHilbert’s1888theorem 93
ThenFθ is irreducible if and only ifθ, 1 2.
Proof. See [20, p214].
Lemma 4.3. If 2t= 4,6, andn≥ 3, then ht(x1, . . . ,xn) :=
n
X
i=1
x2ti −10X
i,j
x2t−2i x2j
is an indefinite irreducible even symmetricn−ary form of degree 2t.
Proof. Fort= 2,3;htis clearly indefinite (sinceht(1,0, . . . ,0)=1 and ht(1, . . . ,1)<<0). Alsoh2,h3are irreducible as below:
1. We have h2 =
n
X
i=1
x4i −10X
i,j
x2ix2j. Suppose for a contradiction h2 is re-ducible, and let h2 = f g; where f,g are non-constant forms (by Remark 1.13) such that deg(f)≥ 1, deg(g)≥1, deg(f)+deg(g)=4.
Settingx4 = x5 =. . .= xn= 0,we get:
h2(x1,x2,x3,0, . . . ,0)=h2(x1,x2,x3)
= f(x1,x2,x3,0, . . . ,0)g(x1,x2,x3,0, . . . ,0)
⇒h2 has a factorization as a product of two (or more) non-constant forms, which is a contradiction, sinceh2(x,y,z) = x4+y4 +z4−20(x2y2+ x2z2+ y2z2)= F10
11(x,y,z) is irreducible by above lemma with θ= 10 11. 2. We have h3 =
n
X
i=1
x6i −10X
i,j
x4ix2j. Suppose for a contradiction h3 is re-ducible, and let h3 = f g; where f,g are non-constant forms (by Remark 1.13) such that deg(f)≥ 1, deg(g)≥1, deg(f)+deg(g)=6.
Settingx4 = x5 =. . .= xn= 0,we get:
h3(x1,x2,x3,0, . . . ,0)=h3(x1,x2,x3)
= f(x1,x2,x3,0, . . . ,0)g(x1,x2,x3,0, . . . ,0)
⇒h3 has a factorization as a product of two (or more) non-constant forms, which is not possible sinceh3(x,y,z)= x6+y6+z6−10(x4y2+x4z2+x2y4+ y4z2+x2z4+y2z4) cannot have a linear, irreducible quadratic or irreducible cubic factor as shown below in (a), (b) and (c) respectively. For simplicity sayh3 =h.
(a) hcannot have a linear factor:
Supposel(x,y,z)= ax+by+cz|h, then a, 0,b,0,c,0
(because ifa=0, i.e. if
h(x,y,z)= (by+cz)q(x,y,z); deg(q)= 5, (4.2) then there will be nox6term on the R.H.S. of equation (4.2) and hence inh, which is a contradiction. Similarlyb,0,c, 0).
So by Remark 1.16, we have:
l(x,y,z)=ax+by+cz|h l(x,y,−z)=ax+by−cz|h l(x,−y,z)=ax−by+cz|h l(x,−y,−z)= ax−by−cz|h.
Now
i. if|a|,|b| then again by Remark 1.16:
l(y,x,z)= bx+ay+cz|h l(y,x,−z)=bx+ay−cz|h l(−y,x,z)=bx−ay+cz|h l(−y,x,−z)= bx−ay−cz|h.
So we have 8 distinct linear factors ofh, namely
(ax+by+cz),(ax+by−cz),(ax−by+cz),(ax−by−cz),(bx+ay+cz), (bx+ay−cz),(bx−ay+cz),and (bx−ay−cz),
which is a contradiction since deg(h)= 6.
ii. if|a|=|b| then|a|=|b|=|c|. So, (x+y+z),(x+y−z),(x−y+z), and (x−y−z) are linear factors ofh.
4.1. Version ofHilbert’s1888theorem 95 Let T(x,y,z) :=(x+y+z)(x+y−z)(x−y+z)(x−y−z).
If h(x,y,z)=T(x,y,z)q(x,y,z),
for some quadratic formq(x,y,z),thenT(1,2,1)=0⇒ h(1,2,1)= 0, which gives a contradiction sinceh(1,2,1)=−354, 0.
Hencehcannot have a linear factor.
(b) hcannot have an irreducible quadratic factor:
Supposeq(x,y,z) = a11x2 +a22y2+a33z2+a12xy+a13xz+a23yz |h, then either of the following holds:
i. ai j = 0 for alli< j, or ii. one ofai j , 0 fori< j, or iii. two ofai j ,0 fori< j, or
iv. ai j , 0 for alli< j.
i. ifa12 =a13 =a23 =0 then q(x,y,z) = a11x2 +a22y2 +a33z3, and either of the following holds:
• ifa11,a22,a33are all distinct then by Remark 1.16:
q1:= q(x,y,z)= a11x2+a22y2+a33z2 q2:= q(x,z,y)= a11x2+a33y2+a22z2 q3:= q(y,x,z)= a22x2+a11y2+a33z2 q4:= q(z,x,y)= a22x2+a33y2+a11z2 q5:= q(y,z,x)= a33x2+a11y2+a22z2 q6:= q(z,y,x)= a33x2+a22y2+a11z2
are 6 irreducible quadratic factors of h. If at least 4 among them are pairwise non scalar multiples, then their product di-vides himplying deg(h) ≥ 8, which is a contradiction since deg(h) = 6. Otherwise, if for somei, j,k,l ∈ {1, . . . ,6}, qi is a non-zero scalar multiple of qj and qk is a non-zero scalar multiple ofql, say w.l.o.g.q1 = λq6, then we get:
λa22 =a22 (4.3)
λa11 =a33, λa33= a11; (4.4)
equation (4.3) impliesλ = 1 or a22 = 0; but λ = 1 implies a11 = a33 (by equation (4.4)), which is not possible (since a11,a33must be distinct), soa22= 0. Now, equation (4.4) im-pliesλ = ±1 ora11 = a33 = 0; so only possibility is λ = −1 (since λ , 1 and a11,a33 must be distinct), so a33 = −a11. Thus we have a22 = 0,a33 = −a11, but then p(x,y,z) = a116(x2 −z2)(x2 −y2)(y2−z2) = q1q2q3 = −q4q5q6 does not divideh, sincep(1,1,0)= 0 buth(1,1,0)=−18, 0.
• if two ofa11,a22,a33are same w.l.o.g. let a11 = r; a22 = a33 = s for r , s, then (rx2 + sy2 + sz2),(sx2 + ry2 + sz2), and (sx2+sy2+rz2) are 3 distinct quadratic factors ofh.
So,his a scalar multiple of
p= (rx2+sy2+ sz2)(sx2+ry2+ sz2)(sx2+sy2+rz2), sayh= λp, then comparing the coefficients ofλpwith those ofh = P3
i=1x6i −10P
i,jx4ix2j we get an insolvable system of equations inr, s:
λrs2 =1;λ(rs2+sr2+s3)=−10;λ(r3+2s3+3rs2)= 0.
So this case doesn’t arise.
• ifa11 =a22 =a33 thenq= x2+y2+z2 Replacingz2by−(x2+y2), we get h= (x2+y2) 11(x4+y4)+7x2y2
, which is positive definite, a contradiction, so nothing to prove.
ii. if one ofai j , 0 fori< j suppose w.l.o.g. a12, 0 anda13 =a23 = 0, then by Remark 1.16:
q(x,y,z)= a11x2+a22y2+a33z3+a12xy, q(x,−y,z)=a11x2+a22y2+a33z3−a12xy, q(z,y,x)= a33x2+a22y2+a11z3+a12yz, and q(z,−y,x)=a33x2+a22y2+a11z3−a12yz
4.1. Version ofHilbert’s1888theorem 97 are 4 distinct quadratic factors ofh. This would imply deg(h)≥8, which is a contradiction since deg(h)= 6.
iii. if two ofai j ,0 fori< j suppose w.l.o.g. a12,a13 , 0, then q(x,y,z) and (by Remark 1.16)q(x,−y,z),q(x,y,−z),q(x,−y,−z) are 4 distinct quadratic factors ofh. This would imply deg(h)≥8, which is a contradiction since deg(h)= 6.
iv. ifai j , 0 for alli< j similarly as in case (iii) above.
Hencehcannot have an irreducible quadratic factor.
(c) hcannot have an irreducible cubic factor:
Supposec(x,y,z)= αx3+βy3+γz3+c1(x,y,z)|h, wherec1(x,y,z) := a210x2y+a201x2z+a120xy2+a021y2z+a102xz2+a012yz2+a111xyz.
Then α, β, γ, 0
because if
h=c(x,y,z) α0x3+β0y3+γ0z3+c01(x,y,z), (4.5) wherec01(x,y,z) is a cubic form defined likec1(x,y,z), then expanding the R.H.S. of equation (4.5) and comparing its coefficients with those ofh=P3
i=1x6i −10P
i,jx4ix2j we get αα0 =1, ββ0 =1, γγ0= 1 . So,c(x,y,z) and (by Remark 1.16)
c(x,−y,z)=αx3−βy3+γz3+c1(x,−y,z), c(x,y,−z)=αx3+βy3−γz3+c1(x,y,−z), c(x,−y,−z)= αx3−βy3−γz3+c1(x,−y,−z),
are 4 distinct cubic factors ofh. This would imply deg(h)≥12, which is a contradiction since deg(h)= 6.
Lemma 4.4. Ifr∈Z+;r ≥2, then there existsa,b∈Z+such thatr=2a+3b.
Proof. Ifris even, then takea= r
2,b=0.
Ifris odd, thenr≥ 3. So,r−3 is even. So,r =2a+3b;a∈Z+,b=1.
Now we present as follows in Theorem 4.5, two amazing techniques that can be used to find psd not sos even symmetric forms of degree 2d+8,2d+12,2d+ 16, . . .etc. and 2d+2nfrom a given psd not sos even symmetric form of degree 2d. We will call it a Degree Jumping Principle, since the degree of the newly constructed form will be
=(degree of the previous form)+jump of degree 4r(for integerr≥ 2) and 2n respectively.
Theorem 4.5. Degree Jumping Principle:
If f ∈SPen,2d\SΣen,2d, (n≥3), then
1. for any integer r ≥ 2, the form f h2a2 h2b3 ∈ SPen,2d+4r\SΣen,2d+4r, where r = 2a+3b;a,b∈Z+.
2. (x1. . .xn)2f ∈SPen,2d+2n\SΣen,2d+2n.
Proof. 1. By above lemma, for r ∈ Z+;r ≥ 2, ∃non-negativea,b ∈ Z such thatr= 2a+3b.
Consider the form
f h2a2 h2b3
of degree 2d+4rinnvariables, whereht(x) :=
n
X
i=1
x2ti −10X
i,j
x2t−2i x2j for t=2,3 as defined in Lemma 4.3.
Since f h2a2 h2b3 is a product of even symmetric forms, it is even and sym-metric. Also it is a product of psd forms so it is still psd. Thus we have
f h2a2 h2b3 ∈SPen,2d+4r.
Sinceh2andh3are indefinite and irreducible forms by Lemma 4.3, so taking p = h2 or h3 in Lemma 3.2 we get f h22 ∈ SPen,2d+8 \ SΣen,2d+8 and f h23 ∈ SPen,2d+12\SΣen,2d+12. Repeating this argument we get f h2a2 h2b3 <SΣen,2d+4r. 2. Taking p= xi in turn for each 1≤i≤n, we are done by Lemma 3.2.
4.1. Version ofHilbert’s1888theorem 99 Proposition 4.6. If we can find psd not sos even symmetricn−ary 2d−ic forms for the following pairs:
1. (n,2d)=(n,8) forn≥ 5, and 2. (n,2d) forn≥ 4,d =5,6.
then the complete answer toQ(Se) will be:
SPen,2d ⊆ SΣen,2dif and only ifn=2,d=1,(n,2d)=(n,4)n≥3,(3,8).
Proof. We proved in Proposition 4.1 that
SPen,2d ⊆ SΣen,2difn=2,d=1,(n,2d)=(n,4)n≥3,(3,8).
For all the remaining cases we break the discussion into the following two cases:
1. Forn=3:
(a) for 2d= 6 : there exists f ∈SPe3,6\SΣe3,6, as seen in Proposition 4.1 (b) for 2d= 8:SPe3,8= SΣe3,8
(c) for 2d =10: there exists f ∈SPe3,10\SΣe3,10, as seen in Proposition 4.1 (d) for 2d≥ 12: using Theorem 4.5 (part 1 and 2 respectively):
• p∈SPe3,6\SΣe3,6 ⇒ ∃ p0 ∈SPe3,14\SΣe3,14, and
• f ∈SPe3,6\SΣe3,6; g∈SPe3,10\SΣe3,10; h∈SPe3,14\SΣe3,14
(jump of deg 6)⇓ ⇓ ⇓
∃ f1 ∈SPe3,12\SΣe3,12;g1 ∈SPe3,16\SΣe3,16;h1∈SPe3,20\SΣe3,20
(jump of deg 6)⇓ ⇓ ⇓
∃ f2 ∈SPe3,18\SΣe3,18;g2 ∈SPe3,22\SΣe3,22;h2∈SPe3,26\SΣe3,26
... ... ...
2. Forn≥4:
(a) for 2d= 6 : there exists f ∈SPen,6\SΣen,6, as seen in Proposition 4.1 (b) for 2d= 8:
• for n = 4, there exists f ∈ SPe4,8\SΣe4,8, as seen in Proposition 4.1
• forn≥5, there exists f ∈SPen,8\SΣen,8by hypothesis (we will do this in Section 4.1.2)
(c) for 2d=10: there exists f ∈SPen,10\SΣen,10 by hypothesis (d) for 2d=12: there exists f ∈SPen,12\SΣen,12 by hypothesis (e) for 2d≥14: using (part 1 of) Theorem 4.5:
f ∈SPen,6\SΣen,6; g∈SPen,8\SΣen,8
(jump of degree 8) ⇓ ⇓
∃ f1∈SPen,14\SΣen,14; g1 ∈SPen,16\SΣen,16
(jump of degree 12)⇓ ⇓
∃ f2∈SPen,18\SΣen,18; g2 ∈SPen,20\SΣen,20
... ...
So we get the following as an immediate corollary to Proposition:
Corollary 4.7. SPen,2d ) SΣen,2d for (n,2d)=(3,2d)d≥6,(n,2d)n≥4,d≥7.
In addition to the degree jumping techniques presented above in Theorem 4.5 we also note the following result that can be used to find psd not sos even sym-metric ternary forms of degree 6,10,14,18. . .etc.:
Lemma 4.8. Ifµis odd integer,
Hµ(x,y,z) :=hµ(x2,y2,z2)∈SPe3,2µ+4\SΣe3,2µ+4, whereh(x,y,z)= xµ(x−y)(x−z)+yµ(y−z)(y−x)+zµ(z−x)(z−y).
Proof. See [8, p6].
4.1. Version ofHilbert’s1888theorem 101
4.1.1 Psd not sos even symmetric ternary dodecics
We give some explicit examples of psd not sos even symmetric ternary dodecics as below.
Proposition 4.9. The forms R1 := R(xy,xz,yz) and R2 := (xyz)2R(x,y,z) ∈ SPe3,12 \SΣe3,12, where R(x,y,z) = x6 + y6 + z6 − (x4y2 + y4z2 + z4x2 + x2y4 + y2z4+z2x4)+3x2y2z2is the psd not sos Robinson’s form.
Proof. (i) ClearlyR1∈SPe3,12. SupposeR1(x,y,z)=P
fk2(x,y,z), then R1(xy,xz,yz)=X
fk2(xy,xz,yz) Letgk(x,y,z) := fk(xy,xz,yz), then
R1(xy,xz,yz)=R(x2yz,xy2z,xyz2)= (xyz)6R(x,y,z)=P
g2k(x,y,z)
⇒ x3y3z3|gk
⇒ gk = x3y3z3hk(x,y,z)
⇒ R= P
h2k(x,y,z), a contradiction.
(ii)R2 ∈SPe3,12\SΣe3,12follows directly from (part 2 of) Theorem 4.5.
Proposition 4.10. The formH(x,y,z) :=h(x2,y2,z2)∈SPe3,12\SΣe3,12, forh(x,y,z)= P3
x6−P6
x5y+P3
x4yz, where each summationPk
denotes the sum taken over all thekpermutations of variables which yield distinct expressions.
Proof. See [8, p6].
4.1.2 Psd not sos even symmetric n−ary octics for n ≥ 5
For finding psd not sos even symmetric n−ary octics forn ≥ 5 we first note the following lemma which will be particularly useful in proving the main results of this section.
Lemma 4.11. Suppose p =
r
X
i=1
h2i is an even sos form. Then we may write p =
s
X
j=1
q2j, where each formq2j is even. In particular,qj(x) = X
cj(α)xα, where the sum is taken overα’s in one congruence class mod 2 componentwise.
Proof. See [10, Theorem 4.1].
Proposition 4.12. The form
B(x1, . . . ,x5) := A5(x21, . . . ,x25)∈SPe5,8\SΣe5,8, whereA5(x) :=
X5
i=1
Y
j,i
(xi− xj) is a symmetric psd not sos 5−ary quartic form.
(A5 = 1
8L5was discussed already in Section 3.1.2).
Proof. ClearlyB is psd, sinceA5 is psd. Suppose B = P
h2t. Since Bis an even form, by Lemma 4.11 eachh2t is even. Sohtcan be only of the following 3 types:
1. ht =ei jklxixjxkxl, or 2. ht = xixj
d1x2i +d2x2j +X
k,i,j
dkx2k , or
3. ht =
5
X
i=1
cix4i +X
i<j
ci jx2ix2j. SinceΛ5,2 =
x∈R5 | ∃r, s∈R: xi ∈ {r,s} is a test set for symmetric quartics (by Corollary 3.11), so A5 ≥ 0 on Λ5,2. In particular, A5(x) = 0 on the points x∈ {(r,r,r,s,s) and its permutations, (r,r,r,r,r);r , s ∈R}. ThereforeB= 0 on such points xand hence forr, s∈R, eachht = 0 on the points (r,r,r,s,s), their permutations and (r,r,r,r,r). Now
1. ifht = ei jklxixjxkxl, then
ht(1,1,1,1,1)=0⇒ei jkl =0, which impliesht = 0.
4.1. Version ofHilbert’s1888theorem 103 2. ifht = xixj
d1xi2+d2x2j +X
k,i,j
dkx2k
W.l.o.g. takeht = x1x2
d1x21+d2x22+ X5
k=3
dkx2k , then
ht(1,1,0,0,0)=0⇒ d1+d2= 0 (4.6) ht(1,1,1,0,0)= 0⇒d1+d2+d3 =0 (4.7) ht(1,1,0,1,0)= 0⇒d1+d2+d4 =0 (4.8) ht(1,1,0,0,1)= 0⇒d1+d2+d5 =0 (4.9) ht(1,2,1,1,2)= 0⇒2(d1+4d2+d3+d4+4d5)=0 (4.10) Using equation (4.6) in equations (4.7), (4.8) and (4.9), we get d3 = d4 = d5 =0. Using this in equation (4.10), we get 2(d1+4d2)= 0, and then again using equation (4.6), we getd1= d2= 0. Soht = 0.
3. ifht =
5
X
i=1
cix4i +X
i<j
ci jx2ix2j, then ht(1,1,0,0,0)=0⇒c1+c2+c12 =0.
Similarly atxwithxi = xj = 1;xk =0 fork, i, j;i, j,k∈ {1, . . . ,5}:
ht(x)= 0⇒ci+cj+ci j =0⇒ci j = −(ci+cj) (4.11) Evaluating ht at x with xi = xj = xk = 1;xl = 0 for l , i, j,k; i, j,k,l ∈ {1, . . . ,5}:
ht(x)=0⇒ ci +cj+ck+ci j+cik+cjk =0
⇒
(equation (4.11))|{z}
ci+cj+ck−(ci+cj)−(ci+ck)−(cj+ck)=0
⇒ci+cj+ck =0;i, j,k∈ {1, . . . ,5}distinct (4.12) Similarly,
ci+cj+cl = 0;i, j,k,l∈ {1, . . . ,5}distinct (4.13) Equations (4.12) and (4.13)⇒cl = ck;l,k∈ {1, . . . ,5}distinct,
i.e.cl =ck for arbitraryl,k. This⇒c1 =c2 =. . . =c5
⇒
(equation (4.12))|{z}
3ci = 0⇒ci =0∀i
⇒
(equation (4.11))|{z}
ci j = 0∀i< j So eachht =0.
For finding psd not sos even symmetricn−ary octics forn ≥ 6 we first prove two lemmas, as below, that will be used in proving Theorem 4.15 following im-mediately after these:
Lemma 4.13. Forn≥ 6, ifh(x1, . . . ,xn)= x1x2
d1x21+d2x22+
n
X
k=3
dkx2k
is a quartic form that vanishes on all 0/1 points with mor (m+1) 1’s, where m = hn
2 i, i.e.
h(x)=0 for allxwithmor (m+1) 1’s and
(m+1) orm0’s (resp.) for oddn=2m+1;
mor (m−1) 0’s (resp.) for evenn= 2m.
Thenh=d1x1x2(x21−x22).
Proof. Pickr ∈ {3, . . . ,n} and choose A ⊆ {3, . . . ,n} such that |A| = m−2 and r< A. Then|A∪ {r}| = m−1. So,h= 0 onx, where the 1’s on xoccur precisely onA∪ {1,2}, A∪ {1,2,r}. So we have:
onA∪ {1,2}: d1+d2+X
k∈A
dk =0 (4.14)
onA∪ {1,2,r}: d1+d2+X
k∈A
dk+dr = 0
Subtracting above two equations gives: dr = 0. Since r was arbitrary, we can
Subtracting above two equations gives: dr = 0. Since r was arbitrary, we can