The analogue of Hilbert's 1888 theorem for Even Symmetric Forms
aa r X i v : . [ m a t h . AG ] O c t THE ANALOGUE OF HILBERT’S 1888 THEOREM FOR EVENSYMMETRIC FORMS
CHARU GOEL, SALMA KUHLMANN, BRUCE REZNICKA bstract . Hilbert proved in 1888 that a positive semidefinite (psd) real form is asum of squares (sos) of real forms if and only if n = d = n , d ) = (3 , n is the number of variables and 2 d the degree of the form. We study theanalogue for even symmetric forms. We establish that an even symmetric n -ary2 d -ic psd form is sos if and only if n = d = n , d ) = ( n , n ≥ or( n , d ) = (3 ,
1. I ntroduction
A real form (homogeneous polynomial) f is called positive semidefinite (psd)if it takes only non-negative values and it is called a sum of squares (sos) if thereexist other forms h j so that f = h + · · · + h k . Let P n , d and Σ n , d denote the cone ofpsd and sos n -ary 2 d -ic forms (i.e. forms of degree 2 d in n variables) respectively.In 1888, Hilbert [9] gave a celebrated theorem that characterizes the pairs ( n , d )for which every n -ary 2 d -ic psd form can be written as a sos of forms. It states thatevery n -ary 2 d -ic psd form is sos if and only if n = d = n , d ) = (3 , Σ n , d ( P n , d for ( n , d ) = (4 , , (3 , f ( x , . . . , x n ) is called symmetric if f ( x σ (1) , . . . , x σ ( n ) ) = f ( x , . . . , x n ) for all σ ∈ S n . As an analogue of Hilbert’s approach, they reduced the problem to findingsymmetric psd not sos n -ary 2 d -ics for the pairs ( n , n ≥ and (3 , n ≥ even symmetric if it is symmetric and in each of its terms every vari-able has even degree. Let S P en , d and S Σ en , d denote the set of even symmetric psdand even symmetric sos n -ary 2 d -ic forms respectively. Set ∆ n , d : = S P en , d \ S Σ en , d .In this paper, we investigate the following question: Q ( S e ) : For what pairs ( n , d ) is ∆ n , d = ∅ ?The current answers to this question in the literature are ∆ n , d = ∅ if n = , d = , ( n , d ) = (3 ,
4) by Hilbert’s Theorem, ( n , d ) = (3 ,
8) due to Harris [7], and( n , d ) = ( n , n ≥ . The result ∆ n , = ∅ for n ≥ ∆ n , d , ∅ for( n , d ) = ( n , n ≥ due to Choi, Lam and Reznick [3], for ( n , d ) = (3 , , (4 , Mathematics Subject Classification.
Key words and phrases. positive polynomials, sums of squares, even symmetric forms. due to Harris [8] and for ( n , d ) = (3 ,
6) due to Robinson [10]. Robinson’s evensymmetric psd not sos ternary sextic is the form R ( x , y , z ) : = x + y + z − ( x y + y z + z x + x y + y z + z x ) + x y z . Thus the answer to Q ( S e ) in the literature can be summarized by the followingchart: deg \ var 2 3 4 5 . . . X X X X . . . X X X X . . . X × × × . . . X X × o o10 X × o o o12 X o o o o14 X o o o o ... ... o o o owhere, a tick ( X ) denotes a positive answer to Q ( S e ), a cross ( × ) denotes a negativeanswer to Q ( S e ), and a circle (o) denotes “undetermined”. Indeed to get a completeanswer to Q ( S e ), we need to investigate the question in these remaining cases,namely ( n ,
8) for n ≥
5, (3 , d ) for d ≥ n , d ) for n ≥ , d ≥ Main Theorem.
An even symmetric n -ary 2 d -ic psd form is sos if and only if n = d = n , d ) = ( n , n ≥ or ( n , d ) = (3 , × ”.The article is structured as follows. In Section 2, we develop the tools (Theorem2.3 and Theorem 2.4) we need to prove our Main Theorem. These tools allow us toreduce to certain basic cases, in the same spirit as Hilbert and Choi-Lam. In Section3 and Section 4 we resolve those basic cases by producing explicit examples for( n , d ); n ≥ , d = , ,
6. We conclude Section 4 by interpreting even symmetricpsd forms in terms of preorderings using our Main Theorem. Finally, for ease ofreference we summarize our examples in Section 5.2. R eduction to basic cases
The following Lemma will be used in Theorem 2.3.
Lemma 2.1.
For n ≥
3, the even symmetric real forms p n ( x , . . . , x n ) = n X j = x j − X ≤ i < j ≤ n x i x j ; q n ( x , . . . , x n ) = n X j = x j + X ≤ i , j ≤ n x i x j − X ≤ i < j < k ≤ n x i x j x k are irreducible over R . Proof.
First observe that if a form g has a factorization g ( x , . . . , x n ) = u Y r = f r ( x , . . . , x n ) , HE ANALOGUE OF HILBERT’S 1888 THEOREM FOR EVEN SYMMETRIC FORMS 3 then the same holds when x k + = · · · = x n =
0, hence it su ffi ces to show that p and q are irreducible over R . Second, observe that if (in addition) g is even andsymmetric, then for all σ ∈ S n and choices of sign, f r ( ± x σ , . . . , ± n x σ n ) is also afactor of g . We call distinct (non-proportional) forms of this kind cousins of f r . If(in addition) f r is irreducible, deg f r = d and deg g = n , then f r can have at most n / d cousins.If p is reducible, then it has a factor of degree ≤
2. Suppose that p has a linearfactor α x + α x + α x . Upon setting x =
0, we see that α x + α x | x − x x + x = ( x + x )( x − x )(2 x + x )(2 x − x ) , so α /α ∈ {± / , ± } . Similarly, α /α ∈ {± / , ± } , so α /α ∈ {± / , ± , ± } ,which contradicts α /α ∈ {± / , ± } . It follows that p has no linear factors.Suppose p has a quadratic (irreducible) factor f = α x + α x + α x + . . . . Ifit is not true that α = α = α , then by permuting variables, f has at least 3 > / α = α = α , and by scaling we may assume the common valueis 2. The binary quartic 4 x − x x + x has six quadratic factors, found bytaking pairs of linear factors as above. Of these, the ones in which α = α are2 x ± x x + x . It follows that, more generally, the coe ffi cient of x i x j is ± f ( x , x , x ) = x + x + x + ± (5 x x ) + ± (5 x x ) + ± (5 x x ) . Regardless of the initial choice of signs, making the single sign changes x i
7→ − x i for i = , , f has 4 cousins, which again is too many. Therefore, wemay conclude that p is irreducible.We turn to q and first observe that q ( x , x , x ) = ( x + x + x ) − x x x . Suppose now that q is reducible, so it has at least one factor of degree ≤
3, and let f be such a factor of q . Once again, we set x = f ( x , x , | q ( x , x , = ( x + x ) . Since x + x is irreducible over R , we conclude that deg f = f ( x , x ) = λ ( x + x ). Writing f ( x , x , x ) = α x + α x + α x + X ≤ i < j ≤ β i j x i x j , we see from the foregoing that α = α and β =
0. By setting x = x = α i ’s are equal and β i j =
0, so f is a multiple of x + x + x .But since x x x is not a multiple of x + x + x , f cannot divide q , completingthe proof. (cid:3) Lemma 2.2.
Let f be a psd not sos n -ary 2 d -ic form and p an irreducible indefiniteform of degree r in R [ x , . . . , x n ]. Then the n -ary (2 d + r )-ic form p f is also psdnot sos. Proof.
See [6, Lemma 2.1]. (cid:3)
Theorem 2.3. ( Degree Jumping Principle ) Suppose f ∈ ∆ n , d for n ≥
3, then1. for any integer r ≥
2, the form f p an q bn ∈ ∆ n , d + r , where r = a + b ; a , b ∈ Z + ,and p n , q n are as defined in Lemma 2.1;2. ( x . . . x n ) f ∈ ∆ n , d + n . C. GOEL, S. KUHLMANN, B. REZNICK
Proof.
1. For r ∈ Z + , r ≥
2, there exists non-negative a , b ∈ Z such that r = a + b .Since f p an q bn is a product of even symmetric forms, it is even and symmetric;since it is a product of psd forms, it is psd. Thus we have f p an q bn ∈ S P en , d + r .Since p n and q n are indefinite and irreducible forms by Lemma 2.1, we get f p n ∈ ∆ n , d + and f q n ∈ ∆ n , d + by Lemma 2.2. Finally, by repeating thisargument we get f p an q bn ∈ ∆ n , d + r .2. Taking p = x i in turn for each 1 ≤ i ≤ n , the assertion follows by Lemma 2.2. (cid:3) Theorem 2.4. (Reduction to Basic Cases) If ∆ n , d , ∅ for ( n , n ≥ , ( n , n ≥ and( n , n ≥ , then ∆ n , d , ∅ for ( n , d ) n ≥ , d ≥ . Proof.
For n =
3, the basic examples are R ( x , y , z ) ∈ ∆ , (by Robinson [10]),several examples in ∆ , (by Harris [7]) and p R ( x , y , z ) ∈ ∆ , (by Theorem 2.3(1)). Every even integer ≥
12 can be written as 6 + k , + k or 14 + k , k ≥ ∆ , d is non-empty for 2 d ≥
6, 2 d , n ≥ ∆ n , , ∅ (by Choi, Lam, Reznick [3]). We shall show in Sections 3and 4 that ∆ n , , ∆ n , , ∆ n , are non-empty. Every even integer ≥
14 can be writtenas 6 + k , + k , + k or 12 + k and so, given our claimed examples, by Theorem2.3, ∆ n , d is non-empty for n ≥
4, 2 d ≥ (cid:3) In order to find psd not sos even symmetric n -ary octics, psd not sos even sym-metric n -ary decics and psd not sos even symmetric n -ary dodecics for n ≥
4, wefirst recall the following theorems which will be particularly useful in proving themain results of Sections 3 and 4.
Theorem 2.5.
Suppose p = r X i = h i is an even sos form. Then we may write p = s X j = q j , where each form q j is even. In particular, q j ( x ) = X c j ( α ) x α , where thesum is taken over α ’s in one congruence class mod 2 component-wise. Proof.
See [3, Theorem 4.1]. (cid:3)
Theorem 2.6.
A symmetric n -ary quartic f is psd if and only if f ( x ) ≥ x ∈ R n with at most two distinct coordinates (if n ≥ Proof.
This was originally proved in [2]; see [5, Corollary 3.11], [7, Section 2]. (cid:3)
Theorem 2.7. (i) For odd 2 m + ≥
5, the symmetric 2 m + L m + ( x ) : = m ( m + X i < j ( x i − x j ) − (cid:18) X i < j ( x i − x j ) (cid:19) is psd not sos.(ii) For 2 m ≥
4, the symmetric 2 m -ary quartic C m ( x , . . . , x m ) : = L m + ( x , . . . , x m , Proof.
See [6, Theorems 2.8, 2.9]. (cid:3)
Theorem 2.8.
For an integer r ≥
1, let M r = M r ( x , . . . , x n ) : = x r + · · · + x rn . Forreals a , b , c , the sextic p = aM + bM M + cM is psd if and only if at + bt + c ≥ t ∈ { , , . . . n } and sos if and only if at + bt + c ≥ t ∈ { } ∪ [2 , n ]. HE ANALOGUE OF HILBERT’S 1888 THEOREM FOR EVEN SYMMETRIC FORMS 5
Proof.
See [3, Theorems 3.7, 4.25]. (cid:3)
Observation 2.9.
Let v t denote any n -tuple with t components equal to 1 and n − t components equal to zero. Then M r ( v t ) = t , so p ( v t ) = t ( at + bt + c ). It will beuseful in the proofs of Theorems 3.1, 4.1 and 4.4 to let v t ( a , . . . , a t ) denote theparticular v t with 1’s in positions a , . . . , a t .3. P sd not sos even symmetric n - ary octics for n ≥ m ≥ G m + ( x , . . . , x m + ) : = L m + ( x , . . . , x m + ) ∈ S P e m + , ; D m ( x , . . . , x m ) : = G m + ( x , . . . , x m , ∈ S P e m , . We showed in [6] that G m + ( x ) = x ∈ R m + which are a permutationof m + r ’s and m s ’s for ( r , s ) ∈ R , so that D m ( x ) =
0, projectively, at any v m or v m + . Theorem 3.1.
For m ≥ D m ∈ ∆ m , and G m + ∈ ∆ m + , . Proof.
We observe that D m ( v ) >
0; in fact, it is equal to m ( m + m ) − (2 m ) = m ( m − ffi cient of x i in D m is positive. Suppose D m = P h t .Then x i must appear with non-zero coe ffi cient in at least one h t . Since we mayassume that h t is even (using Theorem 2.5), we must have h t = n X i = a i x i + X ≤ i < j ≤ n b i , j x i x j . Since D m ( v m ) = D m ( v m + ) =
0, it follows that h t ( v m ) = h t ( v m + ) =
0, andthis holds for all permutations of v m and v m + . Our goal is to show that theseequations imply that h t =
0, which will contradict the assumption that D m is sos.By symmetry, it su ffi ces to prove that a i = i .To this end, let y (1) = v m (1 , . . . , m − , m − , y (2) = v m (1 , . . . , m − , m ) and y (3) = v m + (1 , . . . , m − , m − , m ). Then0 = h t ( y (1) ) = m − X i = a i + a m − + X ≤ i < j ≤ m − b i , j + m − X i = b i , m − ;0 = h t ( y (2) ) = m − X i = a i + a m + X ≤ i < j ≤ m − b i , j + m − X i = b i , m ;0 = h t ( y (3) ) = m − X i = a i + a m − + a m + X ≤ i < j ≤ m − b i , j + m − X i = b i , m − + m − X i = b i , m + b m − , m . Taking the first equation plus the second minus the third yields m − X i = a i + X ≤ i < j ≤ m − b i , j = b m − , m . Since m ≥ m − < m −
2; thus, the same argument implies that m − X i = a i + X ≤ i < j ≤ m − b i , j = b m − , m . C. GOEL, S. KUHLMANN, B. REZNICK
That is, the coe ffi cient of x m − x m in h t equals the coe ffi cient of x m − x m , and soby symmetry, for all distinct i , j , k , ℓ , the coe ffi cient of x i x j equals the coe ffi cientof x i x k , which equals the coe ffi cient of x k x ℓ . Thus, for all i , j , b i , j = u for some u . Subtracting the first from the second equation gives now a m − = a m , and sofor all i , a i = v for some v . Finally, our previous equations imply that0 = mv + m ! u = ( m + v + m + ! u = = ⇒ − v = m − · u = m · u = ⇒ u = = ⇒ v = . In other words, h t =
0, establishing the contradiction.Suppose now that G m + were sos. Then G m + = r X t = h t = ⇒ D m = r X t = h t ( x , . . . , x m , , a contradiction. Thus G m + is not sos. (cid:3) Remark 3.2.
It was asserted in [3] that the psd even symmetric n -ary octic M (cid:16) M − (2 k + M M + k ( k + M (cid:17) is not sos, provided 2 ≤ k ≤ n −
2. We prove this below for k = n ≥ Theorem 3.3.
For n ≥ T n ( x , . . . , x n ) = M (cid:16) M − M M + M (cid:17) ∈ ∆ n , . Proof.
Note that T n is psd by Theorem 2.8. Suppose T n ( x , . . . , x n ) = m X r = h r ( x , . . . , x n ) . Then, T n ( v ) = T n ( v ) = T n ( v ) >
0. In particular, the terms x j must appearon the right hand side. As in the proof of Theorem 3.1, these terms must appear in n X k = a k x k + X ≤ j < k ≤ n b jk x j x k , which must vanish at every v and every v . In particular, for i < j < k , we have a i + a j + b i j = , a i + a k + b ik = , a j + a k + b jk = , a i + a j + a k + b i j + b ik + b jk = . It easily follows that a i + a j + a k =
0. Now assume i , j , k are distinct, but not nec-essarily increasing. Since n ≥
4, there is an unused index ℓ and we may concludethat a i + a j + a ℓ =
0. Hence a k = a ℓ . Since these are arbitrary, we conclude that a m is independent of m , and since a i + a j + a k =
0, it follows that each a m =
0, acontradiction. (cid:3)
Remark 3.4.
For n = M ( M − M M + M ) = M R is sos, see [10], orequation (7.4) in [3]. HE ANALOGUE OF HILBERT’S 1888 THEOREM FOR EVEN SYMMETRIC FORMS 7
4. P sd not sos even symmetric n - ary decics and dodecics for n ≥ Theorem 4.1.
For n ≥ P n ( x , . . . , x n ) = ( nM − M )( M − M M + M ) ∈ ∆ n , . Proof.
First recall that nM − M = n n X k = x k − n X k = x k = X i < j ( x i − x j ) is psd by Cauchy-Schwarz. The zero set is ( ± , . . . , ± t ( at + bt + c ) gives the valueof the sextic aM + bM M + cM at an n -tuple v t with t n − t t ( t − t − ≥
0, this criterion is satisfied, and the second factor is also psd withzeros at v and v .It follows that P n is psd and its coe ffi cient of x is ( n − − + >
0. Weshow that P n is not sos by showing that in any sos expression, x cannot occur.Using Theorem 2.5, we see that if P n = P h r and x occurs in h r , then h r = ax + x n X k = b k x k + x n X k = c k x k + x X ≤ j < k < n d jk x j x k . Since P n ( v (1 , j )) = P n ( v (1 , j , k )) = j , k , 2 ≤ j < k ≤ n , it follows that0 = h r ( v (1 , j )) = h r ( v (1 , j , k )) =
0, and we have the equations0 = a + b j + c j , = a + b j + b k + c j + c k + d jk = ( a + b j + c j ) + ( a + b k + c k ) + d jk − a . From these equations, we may conclude that for all 2 ≤ j < k ≤ n , b k + c k = − a , d jk = a . Finally, P n ( v n ) =
0, so h r ( v n ) =
0; that is,0 = a + n X k = ( b k + c k ) + X ≤ j < k < n d jk = a − ( n − + n − !! = a · ( n − n − . Thus, a = x occurs in no h r . This gives the contradiction. (cid:3) Remark 4.2.
When n = P n is sos: P = (3 M − M )( M − M M + M ) = x + y + z − x y − x z − y z ) R ( x , y , z ) = x ( x − y ) ( x − z ) + y ( y − x ) ( y − z ) + z ( z − x ) ( z − y ) ) . Remark 4.3.
We have also shown that for m ≥ M G m + ∈ ∆ m + , . We shalldiscuss M G m + and M D m in a future publication. Theorem 4.4.
For n ≥ Q n ( x , . . . , x n ) = ( M − M M + M )( M − M M + M ) ∈ ∆ n , . Proof.
Since ( t − t − ≥ t − t − ≥
0, both factors in Q n are psdby Theorem 2.8. The first has zeros at every v and v and the second has zeros atevery v and v . But note that neither has a zero at v . In fact, the coe ffi cient of x in Q n is (1 − + − + > C. GOEL, S. KUHLMANN, B. REZNICK
Suppose Q n is sos and Q n = P f k . As before, assume the f k ’s are even (usingTheorem 2.5). Then f k ( v t ) = v t with t = , ,
4. Since Q n ( v ) >
0, theremust be an f k containing x i , which will be itself even. To this end, suppose f k = n X i = α i x i + X ≤ i , j ≤ n β i j x i x j + X ≤ i < j < k ≤ n γ i jk x i x j x k . For i < j , let µ i j = β i j + β ji . By evaluating at v ( i , j ), we see that0 = α i + α j + β i j + β ji = α i + α j + µ i j = ⇒ µ i j = − α i − α j . By evaluating at v ( i , j , k ), we have0 = α i + α j + α k + µ i j + µ ik + µ jk + γ i jk = ( α i + α j + α k ) − α i + α j + α k ) + γ i jk = ⇒ γ i jk = ( α i + α j + α k ) . Finally, by evaluating at v ( i , j , k , ℓ ), we have0 = α i + α j + α k + α ℓ + µ i j + µ ik + µ jk + µ i ℓ + µ j ℓ + µ k ℓ + γ i jk + γ i j ℓ + γ ik ℓ + γ jk ℓ = ( α i + α j + α k + α ℓ )(1 − + = ⇒ α i + α j + α k + α ℓ = . In other words, the sum of any four distinct α r ’s is 0. Since n ≥
5, there exists m ∈{ , . . . , n } di ff erent from i , j , k , ℓ and we have α i + α j + α k + α m =
0. Thus α ℓ = α m ,and since the choice of ℓ and m was arbitrary, we conclude that α = · · · = α n = α ,so that 4 α = ffi cient of x i in f k must be zero, a contradiction. (cid:3) Remark 4.5.
We have been unable to determine whether Q and Q are sos. Theorem 4.6.
For n ≥ R n ( x , . . . , x n ) = · ( M − M M + M )( M − M M + M ) = X ≤ i < j < k ≤ n x i x j x k n X i = x i − X ≤ i , j ≤ n x i x j + X ≤ i < j < k ≤ n x i x j x k ∈ ∆ n , . Proof.
Since ( t − t − ≥ t − t − ≥
0, both factors in R n are psd byTheorem 2.8. Moreover, the first factor implies that(4.1) R n ( t , u , , . . . , = t , u , and at all n -tuples which are permutations of ( t , u , , . . . , t ,(4.2) R n ( t , , , , . . . , = t · · ((2 + t ) − + t )(2 + t ) + + t )) = t ( t − ;this also holds by symmetry at any permutation of n − t .We first remark that if n = R n ( x , y , z ) = x y z R ( x , y , z ); since R is not sos, thesame holds for R multiplied by a product of squared linear factors. For n ≥
4, morework is necessary.Suppose R n = P r h r , so that deg h r =
6. Suppose as usual that each h r iseven (using Theorem 2.5). It follows from equation (4.1) that for any such h r , h r ( t , u , , . . . , =
0, for all ( t , u ). If, say, the terms in h r involving only x − k x k are P k = a k x − k x k , then P a k t − k u k = t , u , which implies that a k = h r . HE ANALOGUE OF HILBERT’S 1888 THEOREM FOR EVEN SYMMETRIC FORMS 9
For equation (4.2), let φ r ( t ) = h r ( t , , , , . . . , w X r = φ r ( t ) = t ( t − . Evaluation at t = , , − φ r ( t ) = t ( t − ψ r ( t ), so that w X r = ψ r ( t ) = t , which in turn implies that ψ r ( t ) = λ r t for some real λ r . To recapitulate, we have(4.3) h r ( t , , , , . . . , = λ r t ( t − , and similar equations hold for all permutations of the variables.Since x x x appears in r n with coe ffi cient 1, it also appears in P h r with coef-ficient 1. Since no monomial occurs in any h r with only two variables, it followsthat x x x must appear in at least one h r , and since h r is even, all terms in h r must be x x times an even quartic monomial. Further, we already know that x x , x x , x x do not occur. Thus h r ( x , . . . , x n ) = x x n X j = ( a j x x j + b j x x j + c j x j ) + X ≤ j < k ≤ n d jk x j x k . Thus, h r ( t , , , , . . . , = t ( a t + b + c ) , h r (1 , t , , , . . . , = t ( a + b t + c ) . In view of equation (4.3), both of these cubics must be identically zero, hence a = b + c = b = a + c =
0, and so, in particular, c =
0. This means that x x x does not appear in any h r , establishing the contradiction. (cid:3) Proof of the Main Theorem.
Combine Theorems 2.4, 3.1, 4.1, 4.4 and 4.6. (cid:3)
We now present an application of the Main Theorem to the interpretation ofeven symmetric psd forms in terms of preorderings. We briefly recall the necessarybackground. Let S = { g , . . . , g s } ⊆ R [ x ], and let T S : = X e ,..., e s ∈{ , } σ e g e . . . g e s s | σ e ∈ Σ R [ x ] , e = ( e , . . . , e s ) be the associated finitely generated quadratic preordering, and K S : = { x ∈ R [ x ] | g ( x ) ≥ , . . . , g s ( x ) ≥ } be the associated basic closed semi-algebraic set.We recall the following result which follows from [11, Proposition 6.1]: Proposition 4.7.
Let S be a finite subset of R [ x ], such that dim( K S ) ≥
3. Thenthere exists a g ∈ R [ x ] s.t. g ≥ K S but g < T S .In the concluding Remark 4.9, we investigate when can the form g of Proposition4.7 be chosen to be symmetric. We set S ′ : = { x , . . . , x n } and K S ′ = R n + (thepositive orthant). We need the following relation between the preordering T S ′ andeven sos forms, as verified in [4, Lemma 1]: Lemma 4.8.
Let g ∈ R [ x ]. Then g ( x , . . . , x n ) is sos if and only if g ∈ T S ′ . Remark 4.9.
Let f be an even symmetric n -ary form of degree 2 d , and g be the n -ary symmetric form of degree d such that g ( x , . . . , x n ) = f ( x , . . . , x n ). Clearly, f is psd if and only if g is nonnegative on R n + . Moreover, by Lemma 4.8, f ∈ Σ R [ x ] if and only if g ∈ T S ′ . Applying our Main Theorem, we get that for n ≥ d ≥ n , d ) , (3 , symmetric n -ary d -ic form g such that g ≥ R n + but g < T S ′ . 5. A glossary of forms For easy reference, we list the examples discussed in this paper. L m + ( x , . . . , x m + ) : = m ( m + X i < j ( x i − x j ) − (cid:18) X i < j ( x i − x j ) (cid:19) , (Theorem 2 . C m ( x , . . . , x m ) = L m + ( x , . . . , x m , , (Theorem 2 . M r ( x , . . . , x n ) = x r + · · · + x rn , (Theorem 2 . G m + ( x , . . . , x m + ) = L m + ( x , . . . , x m + ) , (Section 3); D m ( x , . . . , x m ) = G m + ( x , . . . , x m , , (Section 3); T n ( x , . . . , x n ) = M (cid:16) M − M M + M (cid:17) , (Theorem 3 . P n ( x , . . . , x n ) = ( nM − M )( M − M M + M ) , (Theorem 4 . Q n ( x , . . . , x n ) = ( M − M M + M )( M − M M + M ) , (Theorem 4 . R n ( x , . . . , x n ) = · ( M − M M + M )( M − M M + M ) , (Theorem 4 . . A cknowledgements This paper contains results from the Ph.D. Thesis of Charu Goel [5], supervisedby Salma Kuhlmann. The authors are thankful to the Zukunftskolleg of the Uni-versity of Konstanz for a mentorship program that supported the visit of BruceReznick to Konstanz and the visit of Charu Goel to Urbana in the summer of 2015.The first author acknowledges support from the Gleichstellungsrat of the Univer-sity of Konstanz via a Postdoc Br¨uckenstipendium in 2015 and the third author isthankful for the support of Simons Collaboration Grant 280987.R eferences [1] M.D. Choi, T.Y. Lam, An old question of Hilbert, Proc. Conf. quadratic forms, Kingston 1976,Queen’s Pap. Pure Appl. Math. 46 (1977), 385-405.[2] M.D. Choi, T.Y. Lam and B. Reznick, Symmetric quartic forms, unpublished, 1980.[3] M.D. Choi, T.Y. Lam and B. Reznick, Even Symmetric Sextics, Math. Z. 195 (1987), 559-580.[4] P. E. Frenkel and P. Horv´ath, Minkowski’s Inequality and Sums of Squares, Cent. Eur. J. Math.12 (2014), no. 3, 510-516.[5] C. Goel, Extension of Hilbert’s 1888 Theorem to Even Symmetric Forms, Dissertation, Univer-sity of Konstanz, 2014.[6] C. Goel, S. Kuhlmann, B. Reznick, On the Choi-Lam Analogue of Hilbert’s 1888 Theorem forsymmetric forms, Linear Algebra Appl. 496 (2016), 114-120.[7] W. R. Harris, Real Even Symmetric Ternary Forms, J. Algebra 222, no. 1 (1999), 204-245.[8] W. R. Harris, Real Even Symmetric Forms, Thesis, University of Illinois at Urbana-Champaign,1992.
HE ANALOGUE OF HILBERT’S 1888 THEOREM FOR EVEN SYMMETRIC FORMS 11 [9] D. Hilbert, ¨Uber die Darstellung definiter Formen als Summe von Formenquadraten, Math. Ann.,32 (1888), 342-350; Ges. Abh. 2, 154-161, Springer, Berlin, reprinted by Chelsea, New York, 1981.[10] R.M. Robinson, Some definite polynomials which are not sums of squares of real polynomials;Selected questions of algebra and logic, Acad. Sci. USSR (1973), 264-282, Abstracts in NoticesAmer. Math. Soc. 16 (1969), p. 554.[11] C. Scheiderer, Sums of squares of regular functions on real algebraic varieties, Trans. Amer.Math. Soc. 352 (2000), 1039-1069.D epartment of M athematics and S tatistics , U niversity of K onstanz , U niversit ¨ atsstrasse onstanz , G ermany E-mail address : [email protected] D epartment of M athematics and S tatistics , U niversity of K onstanz , U niversit ¨ atsstrasse onstanz , G ermany E-mail address : [email protected] D epartment of M athematics , U niversity of I llinois at U rbana -C hampaign , U rbana , IL 61801 E-mail address ::