Positive trace polynomials and the universal Procesi-Schacher conjecture
aa r X i v : . [ m a t h . R A ] M a y POSITIVE TRACE POLYNOMIALS ANDTHE UNIVERSAL PROCESI-SCHACHER CONJECTURE
IGOR KLEP , ˇSPELA ˇSPENKO , AND JURIJ VOL ˇCI ˇC Abstract.
Positivstellens¨atze are fundamental results in real algebraic geometry providing al-gebraic certificates for positivity of polynomials on semialgebraic sets. In this article Positivstel-lens¨atze for trace polynomials positive on semialgebraic sets of n × n matrices are provided. AKrivine-Stengle-type Positivstellensatz is proved characterizing trace polynomials nonnegativeon a general semialgebraic set K using weighted sums of hermitian squares with denominators.The weights in these certificates are obtained from generators of K and traces of hermitiansquares. For compact semialgebraic sets K Schm¨udgen- and Putinar-type Positivstellens¨atzeare obtained: every trace polynomial positive on K has a sum of hermitian squares decomposi-tion with weights and without denominators. The methods employed are inspired by invarianttheory, classical real algebraic geometry and functional analysis.Procesi and Schacher in 1976 developed a theory of orderings and positivity on central simplealgebras with involution and posed a Hilbert’s 17th problem for a universal central simple algebraof degree n : is every totally positive element a sum of hermitian squares? They gave an affir-mative answer for n = 2. In this paper a negative answer for n = 3 is presented. Consequently,including traces of hermitian squares as weights in the Positivstellens¨atze is indispensable. Introduction
Positivstellens¨atze are pillars of modern real algebraic geometry [BCR98, PD01, Mar08,Sce09]. A Positivstellensatz is an algebraic certificate for a real polynomial to be positive on aset described by polynomial inequalities. For a finite set S ⊂ R [ ξ ] = R [ ξ , . . . , ξ g ] let K S denotethe semialgebraic set of points α ∈ R g for which s ( α ) ≥ s ∈ S . The most fundamentalresult here is the Krivine-Stengle Positivstellensatz (see e.g. [Mar08, Theorem 2.2.1]), whichcharacterizes polynomials that are positive on K S as weighted sums of squares with denomina-tors, where weights are products of elements in S . This theorem is the real analog of Hilbert’sNullstellensatz and a far-reaching generalization of Artin’s solution to Hilbert’s 17th problem. Ifthe set K S is compact, a simpler description of strict positivity on K S is given by Schm¨udgen’sPositivstellensatz [Scm91]. If moreover S generates an archimedean quadratic module, thenPutinar’s Positivstellensatz [Put93] presents an even simpler form of strictly positive polynomi-als on K S . The latter leads to a variety of applications of real algebraic geometry via semidefiniteprogramming [WSV12, BPT13] in several areas of applied mathematics and engineering. By Date : October 23, 2018.2010
Mathematics Subject Classification.
Primary 16R30, 13J30; Secondary 16W10, 14P10.
Key words and phrases.
Trace ring, generic matrices, positive trace polynomial, Positivstellensatz, Real Null-stellensatz, Procesi-Schacher conjecture. Supported by the Marsden Fund Council of the Royal Society of New Zealand. Partially supported by theSlovenian Research Agency grants P1-0222, L1-6722 and J1-8132. The second author is a FWO [PEGASUS] Marie Sk lodowska-Curie fellow at the Free University of Brussels(funded by the European Union Horizon 2020 research and innovation programme under the Marie Sk lodowska-Curie grant agreement No 665501 with the Research Foundation Flanders (FWO)). During part of this work shewas also a postdoc with Sue Sierra at the University of Edinburgh. Supported by the University of Auckland Doctoral Scholarship and the Deutsche Forschungsgemeinschaft(DFG) Grant No. SCHW 1723/1-1. adapting the notion of a quadratic module and a preordering to M n ( R [ ξ ]), many of the resultsdescribed above extend to matrix polynomials [GR74, SH06, Cim12].Positivstellens¨atze are also key in noncommutative real algebraic geometry [dOHMP09, Scm09,Oza13], where the theory essentially divides into two parts between which there is increasingsynergy. The dimension-free branch started with Helton’s theorem characterizing free non-commutative polynomials, which are positive semidefinite on all matrices of all sizes, as sumsof hermitian squares [Hel02]. This principal result was followed by various Positivstellens¨atzein a free algebra [HM04, HKM12, KVV17], often with cleaner statements or stronger conclu-sions than their commutative counterparts. These dimension-free techniques are also appliedto positivity in operator algebras [NT10, Oza16] and free probability [GS14]. Trace positivityof free polynomials presents the algebraic aspect of the renowned Connes’ embedding conjec-ture [KS08, Oza13]. In addition to convex optimization [BPT13], free positivity certificatesfrequently appear in quantum information theory [NC10] and control theory [BEFB94]. Onthe other hand, the dimension-dependent branch is less developed. Here the main tools comefrom the theory of quadratic forms, polynomial identities and central simple algebras with in-volution [KMRT98, Row80, AU15]. A fundamental result in this context, a Hilbert’s 17thproblem, was solved by Procesi and Schacher [PS76]: totally positive elements in a central sim-ple algebra with a positive involution are weighted sums of hermitian squares, and the weightsarise as traces of hermitian squares. Analogous conclusions hold for trace positive polynomials[Kle11]. The basic problem here is whether the traces of hermitian squares are actually needed;cf. [KU10, AU+, SS12].We next outline the contributions of this paper. Let T be the free trace ring , i.e., the R -algebra with involution ∗ generated by noncommuting variables x , . . . , x g and symmetriccommuting variables Tr( w ) for words w in x j , x ∗ j satisfying Tr( w w ) = Tr( w w ) and Tr( w ∗ ) =Tr( w ). Let Sym T be the subspace of symmetric elements, T the center of T and Tr : T → T the natural T -linear map. For a fixed n ∈ N , the evaluation of T at X ∈ M n ( R ) g is defined by x j X j , x ∗ j X t j and Tr( w ) tr( w ( X )). Example.
Consider f = 5 Tr( x x ∗ ) − x )( x + x ∗ ) ∈ T . We claim that f is positive(semidefinite) on M ( R ). For X ∈ M ( R ) write H = X − X t , H = XX t − X t X, H = X − XX t + 2 X t X − ( X t ) . If H is invertible, then one can check (see Example 6.5 for details) that f ( X ) = 52 H H t + 12 H − H H t H − t + 12 H − H H t H − t and hence f ( X ) (cid:23)
0, so f (cid:23) ( R ) by continuity. On the other hand, f is not positive onM ( R ): f (cid:16) !(cid:17) = 5 Tr (cid:16) !(cid:17) I − (cid:16) !(cid:17) ! = − ! . In the rest of the paper we will develop a systematic theory for positivity of trace polynomials.For S ⊂ Sym T let K S = { X ∈ M n ( R ) g : s ( X ) (cid:23) ∀ s ∈ S } . If S is finite, then K S is the semialgebraic set described by S . A set Q ⊂ Sym T is a cyclicquadratic module if1 ∈ Q , Q + Q ⊆ Q , h Q h ∗ ⊆ Q ∀ h ∈ T , Tr( Q ) ⊂ Q . A cyclic quadratic module T is a cyclic preordering if T ∩ T is closed under multiplication. OSITIVE TRACE POLYNOMIALS 3
Proposition. If T ⊂ T is a cyclic preordering, then f | K T (cid:23) for every f ∈ T . The converse of this simple proposition fails in general, but the next noncommutative versionof the Krivine-Stengle Positivstellensatz uses cyclic preorderings to describe noncommutativepolynomials positive semidefinite on a semialgebraic set K S . Theorem B’.
Let S ∪ { f } ⊂ Sym T be finite and let T be the smallest cyclic preorderingcontaining S . Then f | K S (cid:23) if and only if ( t f ) | M n ( R ) g = ( f k + t ) | M n ( R ) g and ( f t ) | M n ( R ) g = ( t f ) | M n ( R ) g for some k ∈ N and t , t ∈ T . See Theorem B below for an extended version in a slightly different language. The existenceof trace identities on n × n matrices suggests that the problem of positivity on n × n matricesshould be treated in an appropriate quotient of T , which we describe next.Let T n be the trace ring of generic n × n matrices , i.e., the R -algebra generated bygeneric matrices Ξ , . . . , Ξ g , their transposes and traces of their products. Here Ξ j = ( ξ jı ) ı is an n × n matrix whose entries are independent commuting variables. The R -subalgebraGM n ⊂ T n generated by Ξ j , Ξ t j is called the ring of generic n × n matrices and has a centralrole in the theory of polynomial identities [Pro76, Row80]. The ring of central quotients ofGM n is the universal central simple algebra with orthogonal involution of degree n ,denoted USA n . The ring T n also has a geometric interpretation. Let the orthogonal groupO n ( R ) act on M n ( R ) g by simultaneous conjugation. If M n ( R [ ξ ]) is identified with polynomialmaps M n ( R ) g → M n ( R ), then T n is the ring of polynomial O n ( R )-concomitants in M n ( R [ ξ ]),i.e., equivariant maps with respect to the O n ( R )-action [Pro76]. If C n , T n , Z n are the centers ofGM n , T n , USA n , respectively, and R is the “averaging” Reynolds operator for the O n ( R )-action,then we have the following diagram. R C n R < x , x ∗ > GM n T T n R [ ξ ] T T n M n ( R [ ξ ]) Z n R ( ξ )USA n M n ( R ( ξ )) R RR R
The elements of T n are called trace polynomials and the elements of T n are called puretrace polynomials . Since every evaluation of T at a tuple of n × n matrices factors through T n , it suffices to prove our Positivstellens¨atze in the ring T n . The purpose of this reduction isof course not to merely state Theorem B’ in a more compact form. Our proofs crucially rely onalgebraic properties of T n and their interaction with invariant and PI theory [Pro76, Row80].The contribution of this paper is twofold. We prove the Krivine-Stengle, Schm¨udgen andPutinar Positivstellens¨atze for the trace ring of generic matrices in terms of cyclic quadraticmodules and preorderings. We also prove Putinar’s Positivstellensatz for the ring of genericmatrices (without traces). The proofs intertwine techniques from invariant theory, real algebraic I. KLEP, ˇS. ˇSPENKO, AND J. VOLˇCIˇC geometry, PI theory and functional analysis. Our second main result is a counterexample to the(universal) Procesi-Schacher conjecture.1.1.
Main results and reader’s guide.
After this introduction we recall known facts aboutpolynomial identities, positive involutions, and the rings GM n and T n in Section 2, where wealso prove some preliminary results that are used in the sequel.Section 3 deals with the question of Procesi and Schacher [PS76], which is a noncommutativeversion of Hilbert’s 17th problem for central simple ∗ -algebras. We say that a ∈ USA n is totallypositive if a ( X ) (cid:23) X ∈ M n ( R ) g where a is defined. Then the universal Procesi-Schacher conjecture states that totally positive elements in USA n are sums of hermitian squaresin USA n . While this is true for n = 2 [PS76, KU10], we show that the conjecture fails for n = 3. Theorem A.
There exist totally positive elements in
USA that are not sums of hermitiansquares in USA . The proof (see Theorem 3.2) relies on the central simple ∗ -algebra USA being split, i.e., ∗ -isomorphic to M ( Z ) with some orthogonal involution. After explicitly determining the in-volution using quadratic forms (Proposition 3.4 of Tignol) and a transcendental basis of Z (Lemma 3.5), we use Prestel’s theory of semiorderings [PD01] to produce an example of a to-tally positive element (a trace of a hermitian square) in USA that is not a sum of hermitiansquares (Proposition 3.6).Section 4 first introduces cyclic quadratic modules and cyclic preorderings for the trace ring T n , which are defined analogously as for T above. The main result in this section is the followingversion of the Krivine-Stengle Positivstellensatz for T n . Theorem B.
Let S ∪ { a } ⊂ Sym T n be finite and T the cyclic preordering generated by S . (1) a | K S (cid:23) if and only if at = t a = a k + t for some t , t ∈ T and k ∈ N . (2) a | K S ≻ if and only if at = t a = 1 + t for some t , t ∈ T . (3) a | K S = 0 if and only if − a k ∈ T for some k ∈ N . See Theorem 4.13 for the proof, which decomposes into three parts. First we show that thefinite set of constraints S ⊂ Sym T n can be replaced by a finite set S ′ ⊂ T n (Corollary 4.4).To prove this central reduction we use the fact that the positive semidefiniteness of a matrixcan be characterized by symmetric polynomials in its eigenvalues and apply compactness of thereal spectrum in the constructible topology [BCR98, Section 7.1]. In the second step we applyresults on central simple algebras with involution and techniques from PI theory to prove thefollowing extension theorem. Theorem C.
Let R ⊇ R be a real closed field. Then an R -algebra homomorphism φ : T n → R extends to an R -algebra homomorphism R [ ξ ] → R if and only if φ (tr( hh t )) ≥ for all h ∈ T n . Since T n = R [ ξ ] O n ( R ) , this statement resembles variants of the Procesi-Schwarz theorem [PS85,Theorem 0.10] (cf. [CKS09, Br¨o98]). Nevertheless, it does not seem possible to deduce TheoremC from these classical results; see Appendix B for a fuller discussion. Theorem C, proved asTheorem 4.8 below, is essential for relating evaluations of pure trace polynomials with orderingson T n via Tarski’s transfer principle (Proposition 4.12). Finally, by combining the first two stepswe obtain a reduction to the commutative situation, where we can apply an existing abstractversion of the Krivine-Stengle Positivstellensatz [Mar08, Theorem 2.5.2].Since trace polynomials are precisely O n ( R )-concomitants in M n ( R [ ξ ]), one might naivelyattempt to prove Theorem B by simply applying the Reynolds operator for the O n ( R )-action toanalogous Positivstellens¨atze for matrix polynomials [Scm09, Cim12]. However, the Reynolds OSITIVE TRACE POLYNOMIALS 5 operator is not multiplicative and it does not preserve squares of trace polynomials, so in thismanner one obtains only weak and inadequate versions of Theorem B.In Section 5 we refine the strict positivity certificate (2) of Theorem B in the case of com-pact semialgebraic sets. We start by introducing archimedean cyclic quadratic modules, whichencompass an algebraic notion of boundedness. Following the standard definition we say that acyclic quadratic module Q ⊆ T n is archimedean if for every h ∈ T n there exists ρ ∈ Q > suchthat ρ − hh t ∈ Q . Then we prove Schm¨udgen’s Positivstellensatz for trace polynomials. Theorem D.
Let S ∪ { a } ⊂ Sym T n be finite and T be the cyclic preordering generated by S .If K S is compact and a | K S ≻ , then a ∈ T . In the proof (see Theorem 5.3) we apply techniques similar to those in the proof of Theorem B.That is, we replace S by finitely many central constraints and apply Theorem C to reduce to thecommutative setting, where we use an abstract version of Schm¨udgen’s Positivstellensatz [Sce03].Finally, we have the following version of Putinar’s Positivstellensatz for T n and GM n , com-bining Theorems 5.5 and 5.7. Theorem E. (a)
Let Q ⊂ Sym T n be an archimedean cyclic quadratic module and a ∈ Sym T n . If a | K Q ≻ , then a ∈ Q . (b) Let Q ⊂ Sym GM n be an archimedean quadratic module and a ∈ Sym GM n . If a | K Q ≻ ,then a ∈ Q . Theorem E is proved in a more functional-analytic way. We start by assuming a / ∈ Q andfind an extreme separation of a and Q . Then we apply a Gelfand-Naimark-Segal constructiontowards finding a tuple of n × n matrices in K Q at which a is not positive definite. For GM n thisis done using polynomial identities techniques, while for T n we use Theorem C. As a consequencewe have the following statement for noncommutative polynomials. Corollary E’.
Let Q ⊂ Sym R < x , x ∗ > be an archimedean quadratic module and a ∈ R < x , x ∗ > .If a | K Q ≻ , then a = q + f for some q ∈ Q and f ∈ R < x , x ∗ > satisfying f | M n ( R ) g = 0 .Proof. If π : R < x , x ∗ > → GM n is the canonical ∗ -homomorphism, then π ( Q ) is an archimedeanmodule in GM n and hence π ( a ) ∈ π ( Q ) by Theorem E(b). Corollary E’ now follows becausethe kernel of π consists precisely of the polynomial identities for n × n matrices. (cid:3) The paper concludes with Section 6 containing examples and counterexamples. In AppendixA where we present algebraic constructions of the Reynolds operator for the action of O n ( R )on polynomials and matrix polynomials as alternatives to the integration over the orthogonalgroup, which is of interest in mathematical physics [C´S06]. Appendix B explains why theProcesi-Schwarz theorem cannot be used to obtain the extension theorem C. Acknowledgments.
The authors thank Jean-Pierre Tignol for sharing his expertise and gen-erously allowing us to include his ideas that led to the counterexample for the universal Procesi-Schacher conjecture, and James Pascoe for his thoughtful suggestions. We also acknowledgefruitful Oberwolfach discussions with Cordian Riener and Markus Schweighofer.2.
Preliminaries
In this section we collect some background material and preliminary results needed in thesequel.
I. KLEP, ˇS. ˇSPENKO, AND J. VOLˇCIˇC
Polynomial and trace ∗ -identities. Throughout the paper let F be a field of character-istic 0. Let x = { x , . . . , x g } and x ∗ = { x ∗ , . . . , x ∗ g } be freely noncommuting variables, and let < x , x ∗ > be the free monoid generated by x j , x ∗ j . The free algebra F < x , x ∗ > is then endowedwith the unique involution of the first kind determined by x j x ∗ j . If A is an F -algebra withinvolution τ and f = f ( x , . . . , x g , x ∗ , . . . , x ∗ g ) ∈ F < x , x ∗ > is such that f ( a , . . . , a g , a τ , . . . , a τg ) = 0for all a j ∈ A , then f is a polynomial ∗ -identity of ( A , τ ).Let ∼ be the equivalence relation on < x , x ∗ > generated by w w ∼ w w , w ∼ w ∗ for w , w ∈ < x , x ∗ > . Let Tr( w ) be the equivalence class for w ∈ < x , x ∗ > . Then we definethe free trace ring with involution T = T ⊗ F F < x , x ∗ > , where T is the free commutative F -algebra generated by Tr( w ) for w ∈ < x , x ∗ >/ ∼ . Note that Tr(1) ∈ T is one of the generatorsof T and not a real scalar. If A is an F -algebra, then an F -linear map χ : A → F satisfying χ ( ab ) = χ ( ba ) for a, b ∈ A is called an F -trace on A . If f = X i α i Tr( w i ) · · · Tr( w iℓ i ) w i , α i ∈ F, w ij ∈ < x , x ∗ > satisfies X i α i χ ( w i ( a )) · · · χ ( w iℓ i ( a )) w i ( a ) = 0for every tuple a of elements in A , then f is a trace ∗ -identity of ( A , τ, χ ).2.1.1. A particular trace ∗ -identity. For n ∈ N let t denote the transpose involution on M n ( F )and let s denote the symplectic involution on M n ( F ): (cid:18) a bc d (cid:19) s = (cid:18) d t − b t − c t a t (cid:19) . Let tr : M n ( F ) → F be the usual trace. Finally, for X ∈ M n ( F ) let X ⊕ d ∈ M dn ( F ) denote theblock-diagonal matrix with d diagonal blocks all equal to X .Fix m ∈ N . For a t -antisymmetric A ∈ M m ( F ) let pf( A ) ∈ F be its Pfaffian, pf( A ) = det( A ).Suppose that A , A ∈ M m ( F ) are t -antisymmetric and A is invertible. Now consider f = pf( A ) pf( tA − − A ) ∈ F [ t ] . Then f is the characteristic polynomial of A A , so ± f is monic of degree m and the coefficientsof f are polynomials in the entries of A , A by Gauss’ lemma. Also, as in the proof of theCayley-Hamilton theorem we see that f ( A A ) = 0.Hence for every t -antisymmetric A , A ∈ M m ( F ) there exists f = t m + P k ( − k c k t m − k ∈ F [ t ] such that f ( A A ) = 0. If A A has distinct eigenvalues λ , . . . , λ m , then their blocks inthe Jordan decomposition of A A have multiplicity 2 and2 m X j =1 λ ij = tr (cid:0) ( A A ) i (cid:1) for i ∈ N . Now Newton’s identities imply kc k = k X i =1
12 ( − i − tr (cid:16) ( A A ) k (cid:17) c k − i for 1 ≤ k ≤ m and c = 1. OSITIVE TRACE POLYNOMIALS 7
Now define f m ∈ T as f m = m X k =0 ( − k f ′ k · ( x x ) m − k with f ′ = 1 and(2.1) f ′ k = k X i =1 k ( − i − tr (cid:16) ( x x ) k (cid:17) f ′ k − i for 1 ≤ k ≤ m . The following lemmas will be important for distinguishing between differenttypes of involutions of the first kind in the sequel. Lemma 2.1.
For every m ∈ N , f m ( x − x ∗ , x − x ∗ ) is a ∗ -trace identity of (M m ( F ) , t , tr) .Proof. Observe that the set of pairs of t -antisymmetric A , A ∈ M m ( F ), such that A A has m distinct eigenvalues, is Zariski dense in the set of all pairs of t -antisymmetric A , A ∈ M m ( F ).Hence the conclusion follows by the construction of f m . (cid:3) Lemma 2.2.
For every n, m ∈ N and d ∈ N \ N there exist s -antisymmetric A , A ∈ M n ( F ) such that f m (cid:16) A ⊕ d A ⊕ d (cid:17) = 0 . Proof.
Every t -symmetric matrix S ∈ M n ( F ) can be written as S = ( − SJ ) J , where J = (cid:0) − (cid:1) and − SJ, J are s -antisymmetric matrices. Hence it suffices to prove that f m ( S ⊕ d ) = 0 holds for S = diag( n − z }| { , . . . , , ∈ M n ( F ) . Since tr(( S ⊕ d ) k ) = d (2 n −
1) is odd, we can use (2.1) and induction on k to show that k ! f ′ k ( S ⊕ d ) ∈ (cid:26) ℓ k : ℓ ∈ Z (cid:27) \ (cid:26) ℓ k − : ℓ ∈ Z (cid:27) for 1 ≤ k ≤ m . In particular we have f ′ m ( S ⊕ d ) = 0 and thus f m ( S ⊕ d ) = 0. (cid:3) For n ∈ N and m ∈ N let J ( n, t ) denote the set of polynomial ∗ -identities of (M n ( F ) , t )and let J ( m, s ) denote the set of polynomial ∗ -identities of (M m ( F ) , s ). By [Row80, Corollary2.5.12 and Remark 2.5.13] we have J ( m, s ) ⊆ J ( n, t ) if and only if 2 n ≤ m . Proposition 2.3.
Let n ∈ N and m ∈ N . Then J ( n, t ) ⊆ J ( m, s ) if and only if m ≤ n .Proof. ( ⇒ ) Let c m = X π ∈ Sym m sgn πx π (1) x m +1 x π (2) x m +2 · · · x m − x π ( m ) be the m th Capelli polynomial [Row80, Section 1.2]. If A is a central simple F -algebra and a , . . . , a m ∈ A , then { a , . . . , a m } is linearly dependent over F if and only if c g ( a , . . . , a g , b , . . . , b m − ) = 0 ∀ b i ∈ A by [Row80, Theorem 1.4.34].Now assume n < m . If A , A ∈ M n ( F ) are t -antisymmetric, then the set n A A , . . . , ( A A ) ⌊ n/ ⌋ +1 o is linearly dependent. Indeed, for an even n this holds directly by Lemma 2.1, while for an odd n we use the fact that A A is singular and then apply Lemma 2.1 for n + 1. On the other hand,since every t -symmetric matrix in M m ( F ) is a product of two s -antisymmetric matrices, there I. KLEP, ˇS. ˇSPENKO, AND J. VOLˇCIˇC exist s -antisymmetric A , A ∈ M m ( F ) such that { , . . . , ( A A ) m − } is linearly independent.Since ⌊ n ⌋ + 1 ≤ m , c m (cid:16) ( x − x ∗ )( x − x ∗ ) , . . . , (( x − x ∗ )( x − x ∗ )) m , x , . . . , x m +1 (cid:17) is a ∗ -identity of M n ( F ) endowed with t but is not a ∗ -identity of M m ( F ) endowed with s .( ⇐ ) By [Row80, Corollary 2.3.32] we can assume that F is algebraically closed; let i ∈ F besuch that i = −
1. Since m ∈ N , (M m ( F ) , s ) ∗ -embeds into (M m ( F ) , t ) via(M m ( F ) , s ) ֒ → (M m ( F ) , t ) , (cid:18) a bc d (cid:19) a + d i ( a − d ) c − b i ( b + c ) i ( d − a ) a + d i ( b + c ) b − cb − c − i ( b + c ) a + d i ( a − d ) − i ( b + c ) c − b i ( d − a ) a + d . (cid:3) Remark . The same reasoning as in the proof of Proposition 2.3 also implies that elementsof J ( n, t ) are polynomial *-identities of M m ( F ) with an involution of the second kind if andonly if 2 m ≤ n . Recall that an involution on M m ( F ) is of the second kind if it induces anautomorphism of order two on F .2.2. Generic matrices and the trace ring.
For g, n ∈ N let ξ = { ξ jı : 1 ≤ j ≤ g, ≤ ı, ≤ n } be a set of commuting indeterminates. We recall the terminology from Section 1. LetΞ j = ( ξ jı ) ı ∈ M n ( R [ ξ ])be n × n generic matrices and let GM n ⊂ M n ( R [ ξ ]) be the R -algebra generated by Ξ j andtheir transposes Ξ t j . Furthermore, let T n ⊂ M n ( R [ ξ ]) be the R -algebra generated by GM n andtraces of elements in GM n . This algebra is called the trace ring of n × n generic matrices (seee.g. [Pro76, Section 7]) and inherits the transpose involution t and trace tr from M n ( R [ ξ ]). Let C n ⊂ T n ⊂ R [ ξ ] be the centers of GM n and T n , respectively. The elements of T n are called trace polynomials and the elements of T n are called pure trace polynomials .There is another, more invariant-theoretic description of the trace ring. Define the followingaction of the orthogonal group O n ( R ) on M n ( R ) g :(2.2) ( X , . . . , X g ) u := ( uX u t , . . . , uX g u t ) , X j ∈ M n ( R ) , u ∈ O n ( R )and consider R [ ξ ] as the coordinate ring of M n ( R ) g . By [Pro76, Theorems 7.1 and 7.2], T n isthe ring of O n ( R )-invariants in R [ ξ ] and T n is the ring of O n ( R )-concomitants in M n ( R [ ξ ]), i.e.,elements f ∈ M n ( R [ ξ ]) satisfying f ( X u ) = uf ( X ) u t for all X ∈ M n ( R ) g and u ∈ O n ( R ).We list a few important properties of GM n and T n that will be used frequently in the sequel.(a) Let J ( n, t , tr) ⊂ T denote the set of trace ∗ -identities of (M n ( R ) , t , tr). Then GM n ∼ = R < x , x ∗ > / J ( n, t ) by [Row80, Remark 3.2.31] and T n ∼ = T / J ( n, t , tr) by [Pro76,Theorem 8.4].(b) By [Pro76, Theorem 20.1], the ring of central quotients of GM n is a central simple algebraof degree n , which is also the ring of rational O n ( R )-concomitants in M n ( R ( ξ )). It iscalled the universal central simple algebra with orthogonal involution of degree n . We denote it by USA n and its center by Z n . Note that USA n is also the ring ofcentral quotients of T n .(c) By [Pro76, Theorem 7.3], T n is a finitely generated R -algebra and T n is finitely spannedover T n . In particular, T n and T n are Noetherian rings. OSITIVE TRACE POLYNOMIALS 9
Reynolds operator.
This subsection is to recall some basic properties of the Reynoldsoperator [DK02, Subsection 2.2.1]. Let G be an algebraic group and X an affine G -variety. The Reynolds operator R : F [ X ] → F [ X ] G is a linear map with the properties:(1) R ( f ) = f for f ∈ F [ X ] G ,(2) R is a G -module homomorphism; i.e., R ( f u ) = R ( f ) for f ∈ F [ X ] , u ∈ G .The Reynolds operator is hence a G -invariant projection onto the space of the invariants. TheReynolds operator exists if G is linearly reductive and is then unique (see e.g. [DK02, Theorem2.2.5]).Let M, N be G -modules and f : M → N a G -module homomorphism. Denote by M G , N G the modules of invariants of M , N , resp., the corresponding Reynolds operators by R M , R N ,resp., and f G a G -module homomorphism f restricted to M G . Then R N f = f G R M . This easilyfollows by the uniqueness of the Reynolds operator. The Reynolds operator is thus functorial.In our case O n ( R ) acts on M n ( R ) g by simultaneous conjugation as in (2.2). Since M n ( R [ ξ ])can be identified with polynomial maps M n ( R ) g → M n ( R ), we have the Reynolds operator R n : M n ( R [ ξ ]) → T n with respect to the action (2.2). Since O n ( R ) is a compact Lie group, R n can be given by the averaging integral formula (with respect to the normalized left Haarmeasure µ on O n ( R ))(2.3) R n ( f ) = Z O n ( R ) f u dµ ( u ) . Consequently R n is a trace-intertwining T n -module homomorphism, i.e.,(2.4) R n ( hf ) = h R n ( f ) , R n ( f h ) = R n ( f ) h, tr( R n ( f )) = R n (tr( f ))for all h ∈ T n and f ∈ M n ( R [ ξ ]). In Appendix A we present algebraic ways of computing R n .2.3. Positive involutions and totally positive elements.
Let A be a central simple algebrawith involution τ and ∗ -center F (that is, F is the subfield of ∗ -invariant elements in the center of A ). The F -space of τ -symmetric elements in A is denoted Sym A . Following the terminology of[PS76] and [KU10], an ordering ≥ of F is a ∗ -ordering if tr A ( aa τ ) ≥ a ∈ A . In thiscase we also say that τ is positive with respect to such an ordering. An element a ∈ Sym A is positive in a given ∗ -ordering if the hermitian form x tr( x τ ax ) on A is positive semidefinite.Finally, a ∈ Sym A is totally positive if it is positive with respect to every ∗ -ordering.Let α , . . . , α n ∈ F be the elements appearing in a diagonalization of the form x tr( xx τ )on A . By [PS76, Theorem 5.4], a symmetric s ∈ A is totally positive if and only if it has aweighted sum of hermitian squares representation(2.5) s = X I ∈{ , } n α I X i h I,i h τI,i , where α I = α I · · · α I n n and h I,i ∈ A .Let Ω n ⊂ T n be the preordering generated by tr( hh t ) for h ∈ T n , i.e., the set of all sumsof products of tr( hh t ) (note that c = tr(( c √ n ) ) ∈ Ω n for every c ∈ T n , so Ω n is really apreordering). Further, let Ω n = (X i ω i h i h t i : ω i ∈ Ω n , h i ∈ T n ) . Note that Ω n = tr(Ω n ). Lemma 2.5.
Let f ∈ Sym M n ( R [ ξ ]) . If f ( X ) (cid:23) for all X ∈ M n ( R ) g , then R n ( f ) = c − q forsome q ∈ Ω n and c ∈ T n \ { } . Proof.
By the integral formula (2.3) it is clear that R n ( f )( X ) (cid:23) X ∈ M n ( R ) g . Hence R n ( f ) is a totally positive element in USA n by [KU10, Lemma 5.3], so R n ( f ) = X I ∈{ , } n α I X i h I,i h t I,i for some h I,i ∈ USA n and a diagonalization h α , . . . , α n i over Z n of the form x tr( xx t ) onUSA n . Hence α k = tr(˜ h k ˜ h t k ) for some ˜ h k ∈ USA n . Since USA n is the ring of central quotientsof T n , there exist q ∈ Ω n and c ∈ T n such that R n ( f ) = c − q . (cid:3) As demonstrated in Example 6.2, the denominator in Lemma 2.5 is in general indispensableeven if f is a hermitian square or f ∈ R [ ξ ]. For more information about images of squares underReynolds operators for reductive groups acting on real affine varieties see [CKS09]. Remark . In particular, the linear operator R n does not map squares in R [ ξ ] into Ω n orhermitian squares in M n ( R [ ξ ]) into Ω n . Hence our Positivstellens¨atze in the sequel cannot simplybe deduced from their commutative or matrix counterparts by averaging with R n . Furthermore,even if one were content with using totally positive polynomials (which by Lemma 2.5 are ofthe form c − q for q ∈ Ω n and c ∈ T n \ { } ) instead of Ω n , one could still not derive our resultssince R n is not multiplicative.3. Counterexample to the × universal Procesi-Schacher conjecture By [PS76, Corollary 5.5] every totally positive element in USA is a sum of hermitian squares,i.e., of the form c − P i h i h t i for c ∈ C and h i ∈ GM . Indeed, by (2.5) it suffices to show thattr( aa t ) is a sum of hermitian squares. Since USA is a division ring, we havetr( aa t ) = a t a + (det( a ) a − )(det( a ) a − ) t for every a ∈ USA \{ } by the Cayley-Hamilton theorem. In their 1976 paper [PS76], Procesiand Schacher asked if the same holds true for n > Conjecture 3.1 (The universal Procesi-Schacher conjecture) . Let n ≥ . Then every totallypositive element in USA n is a sum of hermitian squares. By (2.5), Conjecture 3.1 is equivalent to the following: every trace of a hermitian square inUSA n is a sum of hermitian squares in USA n . In this section we show that Conjecture 3.1 failsfor n = 3: Theorem 3.2.
There exist totally positive elements in
USA that are not sums of hermitiansquares in USA . In the first step of the proof we identify the split central simple algebra USA as a matrixalgebra M ( F ) for a rational function field F , endowed with an involution of the orthogonaltype. For the constructive proof of Theorem 3.2 we then use Prestel’s theory of semiorderings[PD01].We recall some terminology of quadratic forms from [KMRT98]. Let F be a field and V an n -dimensional vector space. Quadratic forms q and q ′ are equivalent if there exists θ ∈ GL F V such that q ′ = q ◦ θ . Quadratic forms q and q ′ are similar if αq and q ′ are equivalent for some α ∈ F \ { } . Every quadratic form is equivalent to a diagonal quadratic form, which is denoted h α , . . . , α n i for α i ∈ F .First we fix g = 1 and write Ξ = Ξ . Since USA is an odd degree central simple algebrawith involution of the first kind, USA is split by [KMRT98, Corollary 2.8]. Let us fix a ∗ -representation USA = End Z V , where V is a 3-dimensional vector space over Z , the center of OSITIVE TRACE POLYNOMIALS 11
USA . By [KMRT98, Proposition 2.1], there exists a symmetric bilinear form b : V × V → Z such that b ( xu, v ) = b ( u, x t v )for all u, v ∈ V and x ∈ End Z V , where t denotes the involution on End Z V originating fromUSA . Let q : V → Z given by q ( u ) = b ( u, u ) be the associated quadratic form. Lemma 3.3.
Let a ∈ End Z V be t -antisymmetric with tr( a ) = 0 . Define e = 1 − a ) − a .Then e is a symmetric idempotent of rank 1 such that V = im e ⊥ ker e . Moreover, im a = ker e and ker a = im e , and the determinant of the restriction of q to ker e is − tr( a ) .Proof. Since a and a t = − a have the same trace and determinant, we have tr( a ) = det( a ) = 0,so by the Cayley-Hamilton theorem it follows that(3.1) a −
12 tr( a ) a = 0 . Hence a = tr( a ) a and it is straightforward to check that e is a symmetric idempotent. Ithas rank 1 because tr( e ) = 1, and the decomposition V = im e ⊕ ker e is orthogonal because e is symmetric. The equation (3.1) also yields ea = ae = 0, hence im a ⊆ ker e and im e ⊆ ker a .Since the rank of every antisymmetric matrix is even, we have im a = ker e and im e = ker a .To prove the last statement, observe that the restriction of a to ker e is an antisymmetricoperator with determinant − tr( a ), and the determinant of the restriction of q to ker e isthe square class of the determinant of any nonzero antisymmetric operator; see [KMRT98,Proposition 7.3]. (cid:3) Proposition 3.4.
For i = 1 , let a i ∈ End Z V be t -antisymmetric with tr( a i ) = 0 , and let e i = 1 − a i ) − a i . If e e = e e = 0 , then q is similar to h , − tr( a ) , − tr( a ) i .Proof. Let e = 1 − e − e and V i = im e i for i = 1 , ,
3; we have dim V i = 1 for each i . Moreover,if u ∈ V i and v ∈ V j for i = j , then b ( u, v ) = b ( e i u, e j v ) = b ( u, e i e j v ) = 0 . Therefore V = V ⊥ V ⊥ V . We have ker e = V ⊥ V and ker e = V ⊥ V , and Lemma3.3 shows that the determinant of the restriction of q to ker e i is − tr( a i ) for i = 1 ,
2. If α ∈ Z \ { } is such that the restriction of q to V is equivalent to h α i , then it follows thatthe restriction of q to V i is equivalent to h− α tr( a i ) i for i = 1 ,
2. Hence q is equivalent to h α, − α tr( a ) , − α tr( a ) i . (cid:3) Now let a = Ξ − Ξ t , a = e Ξ(1 − e ) − (1 − e )Ξ t e ,e = 1 − a ) − a , e = 1 − a ) − a ,β = −
12 tr( a ) , β = −
12 tr( a ) . Since a , a are nonzero, it follows from a i − tr( a i ) a i = 0 that β i = 0 for i = 1 ,
2. It is alsoeasy to check that e e = e e = 0, so the conclusion of Proposition 3.4 holds. Hence we canchoose a basis of V in such a way that(3.2) x t = diag(1 , β , β ) − x τ diag(1 , β , β )for all x ∈ End Z V , where τ is the transpose in M ( Z ) = End Z V with respect to the chosenbasis of V . The field Z is rational over R by [Sal02, Theorem 1.2] and of transcendental degree 6 by[BS88, Theorem 1.11]. Inspired by [For79, Section 3] we present an explicit transcendental basisfor Z over R . Denote s = 12 (Ξ + Ξ t ) , a = 12 (Ξ − Ξ t ) , s = s −
13 tr( s )and(3.3) α = tr( s ) , α = tr( a ) tr( s ) − s a ) tr( a ) tr( s ) − a ) tr( s a ) − s a ) ,α = tr( a ) , α = tr( a ) tr( s ) + 6 tr( s a ) tr( a ) tr( s ) − s a ) ,α = tr( s a ) , α = tr( as a s )tr( a ) tr( s ) − s a ) . Lemma 3.5.
The elements α , . . . , α are algebraically independent over R , Z = R ( α , . . . , α ) ,and (3.4) β = − α , β = 288 α α α − (3 α α + 2 α α + 9 α ) α ( α + 1) . Proof.
Using a computer algebra system one can verify that the determinant of the Jacobianmatrix J α ,...,α is nonzero, so by the Jacobian criterion α , . . . , α are algebraically independentover R . Likewise, (3.4) is checked by a computer algebra system.A minimal set of generators of pure trace polynomials in two 3 × s and the second generic matrix by a we obtain the following generatorsof T :(3.5) tr( s ) , tr( s ) , tr( s ) , tr( a ) , tr( s a ) , tr( s a ) , tr( as a s ) . From (3.3) we can directly see that (3.5) are rational functions in α , . . . , α , tr( s ) ; for example,tr( s ) = α ( α tr( s ) − α ) − α α . Then we use a computer algebra system to verify thattr( s ) = 2 α α + 1 β + 6 α α is a rational function in α , . . . , α and hence Z = R ( α , . . . , α ). (cid:3) Proposition 3.6. β β ∈ Z is totally positive in USA but is not a sum of hermitian squaresin USA .Proof. Since β β = tr( hh t ) for h = β ,β β is totally positive in USA . Now suppose β β = P i r i r t i for r i ∈ USA . If r i = ( ρ iı ) ı ,then the (1 , P i r i r t i equals X i (cid:0) ρ i + β − ρ i + β − ρ i (cid:1) and therefore(3.6) 1 = β β X i (cid:18) ρ i β β (cid:19) + β X i (cid:18) ρ i β β (cid:19) + β X i (cid:18) ρ i β β (cid:19) . OSITIVE TRACE POLYNOMIALS 13
By [PD01, Exercise 5.5.3 and Lemma 5.1.8] there exists a semiordering Q ⊂ R ( α , . . . , α )satisfying α , − α , − α α ∈ Q and p ∈ Q ∩ R [ α , . . . , α ] if and only if the term of the highestdegree in p belongs to Q . These assumptions on Q yield − β , − β , − β β ∈ Q , so (3.6) implies − ∈ Q , a contradiction. (cid:3) Proof of Theorem 3.2.
To be more precise we write USA ,g for USA generated by g genericmatrices Ξ j . Proposition 3.6 proves Theorem 3.2 for g = 1. Now let g ∈ N be arbitrary; notethat USA , naturally ∗ -embeds into USA ,g . Let s ∈ USA , be a totally positive element thatis not a sum of hermitian squares in USA , . Suppose that s is a sum of hermitian squares inUSA ,g , i.e., s = c − X j h i h t i for some c, h i ∈ GM ,g with c central. Since the sets of polynomial ∗ -identities of GM , andGM ,g coincide, it is easy to see that there exists a ∗ -homomorphism φ : GM ,g → GM , satisfying φ (Ξ ) = Ξ and φ ( c ) = 0. Then s = φ ( c ) − X j φ ( h i ) φ ( h i ) t is a sum of hermitian squares in USA , , a contradiction. (cid:3) We do not know if Conjecture 3.1 holds for n = 4, where USA is a division biquaternionalgebra by [Pro76, Theorem 20.1] and hence does not split. In [AU+] the authors use signaturesof hermitian forms to distinguish between sums of hermitian squares and general totally positiveelements.4. The Krivine-Stengle Positivstellensatz for trace polynomials
In this section we prove the Krivine-Stengle Positivstellensatz representing trace polynomialspositive on semialgebraic sets in terms of weighted sums of hermitian squares with denominators.4.1.
Cyclic quadratic modules and preorderings.
For a finite S ⊂ Sym M n ( R [ ξ ]) let K S = { X ∈ M n ( R ) g : s ( X ) (cid:23) ∀ s ∈ S } be the semialgebraic set described by S . A set Q ⊆ Sym T n is a cyclic quadratic module if 1 ∈ Q , Q + Q ⊆ Q , h Q h t ⊆ Q ∀ h ∈ T n , tr( Q ) ⊂ Q . A cyclic quadratic module T ⊆ Sym T n is a cyclic preordering if T ∩ T n is closed under multi-plication. For S ⊂ Sym T n let Q tr S and T tr S denote the cyclic quadratic module and preordering,respectively, generated by S . For example, Q tr ∅ = T tr ∅ = Ω n . Lemma 4.1.
Let S ⊆ Sym T n . (1) If Q is a cyclic quadratic module, then tr( Q ) = Q ∩ T n . (2) Elements of Q tr S are precisely sums of q , h s h t , tr( h s h t ) q for q i ∈ Ω n , h i ∈ T n and s i ∈ S . (3) T tr S = Q tr S ′ , where S ′ = S ∪ (Y i tr( h i s i h t i ) : h i ∈ T n , s i ∈ S ) . Proof.
Straightforward. (cid:3)
Our main result of this subsection is a reduction to central generators for cyclic quadraticmodules, see Corollary 4.4. It will be used several times in the sequel. In its proof we need thefollowing lemma.
Lemma 4.2.
Let R be an ordered field, λ , . . . , λ n ∈ R and p i = P nj =1 λ ij for i ∈ N . If λ j < for some ≤ j ≤ n , then there exists f ∈ Q [ p , . . . , p n ][ ζ ] such that (4.1) n X j =1 f ( λ j ) λ j < . Proof.
Denote E = Q ( p , . . . , p n ) and F = Q ( λ , . . . , λ n ). For every f = P n − i =0 α i ζ i ∈ F [ ζ ] wehave(4.2) n X j =1 f ( λ j ) λ j = X j X i,i ′ α i α i ′ λ i + i ′ +1 j = X i,i ′ X j λ i + i ′ +1 j α i α i ′ = X i,i ′ p i + i ′ +1 α i α i ′ . Note that p i ∈ E for every i ∈ N and define P ∈ M n ( E ) by P ij = p i + j − . If λ j <
0, then thereclearly exists f ∈ F [ ζ ] of degree n − f ( λ j ) = 0 and f ( λ j ) = 0 for λ j = λ j . Then f satisfies (4.1), so P is not positive semidefinite as a matrix over F by (4.2). Since P = QDQ t for some Q ∈ GL n ( E ) and diagonal D ∈ M n ( E ), we conclude that P is not positive semidefiniteas a matrix over E , so there exists v = ( β , . . . , β n − ) t ∈ E n such that v t P v <
0. By (4.2), f = P n − i =0 β i ζ i ∈ E [ ζ ] satisfies (4.1). After clearing the denominators of the coefficients of f we obtain f ∈ Q [ p , . . . , p n ][ ζ ] satisfying (4.1). (cid:3) The proof of the next proposition requires some well-known notions and facts from real algebrathat we recall now. Let Λ be a commutative unital ring. Then P ⊂ Λ is a ordering if P isclosed under addition and multiplication, P ∪ − P = Λ and P ∩ − P is a prime ideal in Λ. Notethat every ordering in Λ gives rise to a ring homomorphism from Λ into a real closed field andvice versa. The set of all orderings is the real spectrum of Λ, denoted Sper Λ. For a ∈ Λ let K ( a ) = { P ∈ Sper Λ : a ∈ P } . Then the sets K ( a ) and Sper Λ \ K ( a ) for a ∈ Λ form a subbasis ofthe constructible topology [BCR98, Section 7.1] (also called patch topology [Mar08, Section2.4]). By [BCR98, Proposition 1.1.12] or [Mar08, Theorem 2.4.1], Sper Λ endowed with thistopology is a compact Hausdorff space. In particular, since the sets K ( a ) are closed in Sper Λ,they are also compact. Proposition 4.3.
For every s ∈ Sym T n let O ⊂ T n be the ring of polynomials in s and tr( s i ) for i ∈ N with rational coefficients, and set S = { tr( hsh ) : h ∈ O} ⊂ tr (cid:16) Q tr { s } (cid:17) . Then there exists a finite subset S ⊂ S such that K { s } = K S .Proof. First we prove that for every real closed field R we have(4.3) { X ∈ M n ( R ) g : s ( X ) (cid:23) } = \ c ∈ S { X ∈ M n ( R ) g : c ( X ) ≥ } . The inclusion ⊆ is obvious. Let X ∈ M n ( R ) g be such that s ( X ) is not positive semidefinite.Since R is real closed and pure trace polynomials are O n ( R )-invariant, we can assume that s ( X ) = diag( λ , . . . , λ n ) is diagonal and λ j < j . If p i = tr( s ( X ) i ), then by Lemma4.2 there exists a polynomial f ∈ Q [ p , . . . , p n ][ ζ ] such that n X i =1 f ( λ i ) λ i < . If h ∈ O is such that h ( X ) = f ( s ( X )), then tr( h ( X ) s ( X ) h ( X )) <
0. Hence ⊇ in (4.3) holds. OSITIVE TRACE POLYNOMIALS 15
Let σ j = tr( ∧ j s ) ∈ T n for 1 ≤ j ≤ n , where ∧ j s denotes the j th exterior power of s ; hence σ j are signed coefficients of the characteristic polynomial for s and { X ∈ M n ( R ) g : s ( X ) (cid:23) } = { X ∈ M n ( R ) g : σ ( X ) ≥ , . . . , σ n ( X ) ≥ } for all real closed fields R . In terms of Sper R [ ξ ] and the notation introduced before the propo-sition, (4.3) can be stated as(4.4) n \ j =1 K ( σ j ) = \ c ∈ S K ( c )by the correspondence between homomorphisms from R [ ξ ] to real closed fields and orderingsin R [ ξ ]. Since the complement of the left-hand side of (4.4) is compact in the constructibletopology, there exists a finite subset S ⊂ S such that n \ j =1 K ( σ j ) = \ c ∈ S K ( c )and consequently K { s } = K { σ ,...,σ n } = K S . (cid:3) Corollary 4.4.
For every finite set S ⊂ Sym T n there exists a finite set S ′ ⊂ tr( Q tr S ) such that K S = K S ′ .Proof. Let S = { s , . . . , s ℓ } . By Proposition 4.3 there exist finite sets S i ⊂ Q tr { s i } ∩ T n with K S i = K { s i } . If S ′ = S ∪ · · · ∪ S ℓ , then S ′ ⊂ [ i Q tr { s i } ∩ T n ⊂ Q tr S ∩ T n and K S ′ = \ i K S i = \ i K { s i } = K S . (cid:3) Corollary 4.5.
For every cyclic quadratic module Q ⊆ Sym T n we have K Q = K tr( Q ) .Proof. Direct consequence of Corollary 4.4. (cid:3)
An extension theorem.
The main result in this subsection, Theorem 4.8, characterizeshomomorphisms from pure trace polynomials T n to a real closed field R which arise via pointevaluations ξ jı α jı ∈ R .We start with some additional terminology. Let F be a field and let A be a finite-dimensionalsimple F -algebra with center C . If tr A is the reduced trace of A as a central simple algebra andtr C/F is the trace of the field extension
C/F , thentr F A = tr C/F ◦ tr A : A → F is called the reduced F -trace of A [DPRR05, Section 4]. Proposition 4.6.
Let R ⊇ R be a real closed field and A a finite-dimensional semisimple R -algebra with an R -trace χ and a split involution, which is positive on every simple factor.Assume there exists a trace preserving ∗ -homomorphism Φ : T n → A such that Φ( T n ) ⊆ R and A is generated by Φ( T n ) over R . Then there exists a trace preserving ∗ -embedding of A into (M n ( R ) , t , tr) .Proof. Let
A ∼ = A × · · · × A ℓ be the decomposition in simple factors and let n k be the degreeof A k for 1 ≤ k ≤ ℓ . Moreover, let C = R [ i ] be the algebraic closure of R and H = ( − , − R ) thedivision quaternion algebra over R . By [PS76, Theorem 1.2], each of A k is ∗ -isomorphic to oneof the following: (I) M n k ( R ) with the transpose involution;(II) M n k ( C ) with the conjugate-transpose involution;(III) M n k / ( H ) with the symplectic involution.Without loss of generality assume that there are 1 ≤ ℓ ≤ ℓ ≤ ℓ such that A n k is of type (I) for k ≤ ℓ , of type (II) for ℓ < k ≤ ℓ , and of type (III) for ℓ < k . By [DPRR05, Theorem 4.2]there exist d , . . . , d ℓ ∈ N such that(4.5) χ X k a k ! = X k d k tr R A k ( a k )for a k ∈ A k . We claim that d k ∈ N for every k > ℓ . Let n ′ = ⌈ n ⌉ and fix k > ℓ . By Lemma2.1, f = f n ′ ( x − x ∗ , x − x ∗ ) is a ∗ -trace identity for T n . Therefore f is also a ∗ -trace identityfor A by the assumptions on Φ. Hence f is a ∗ -trace identity for ( A k , τ k , d k · tr R A k ), where τ k isthe restriction of the involution on A . Since ∗ -trace identities are preserved by scalar extensionsand C ⊗ R (cid:0) A k , τ k , d k · tr R A k (cid:1) ∼ = C ⊗ R (cid:0) M n k / ( H ) , s , d k · (tr H ◦ tr) (cid:1) ∼ = (M n k ( C ) , s , d k · tr) ,f is a ∗ -trace identity for (M n k ( C ) , s , d k · tr). Now Lemma 2 . d k ∈ N .Thus we have(4.6) n = X k ≤ ℓ d k n k + X ℓ Lemma 4.7. Let Λ be a Noetherian domain with char Λ = 2 , M a finitely generated Λ -module, K a field, φ : Λ → K a ring homomorphism, and b : M × M → Λ a symmetric Λ -bilinear form.Let π : M → K ⊗ φ M be the natural homomorphism. Then there exist u , . . . , u ℓ ∈ M such that { π ( u ) , . . . , π ( u ℓ ) } is a K -basis of K ⊗ φ M and φ ( b ( u i , u i ′ )) = 0 for i = i ′ .Proof. Let ℓ = dim K ( K ⊗ φ M ). We prove the statement by induction on ℓ . The case ℓ = 1 istrivial. Now assume that statement holds for ℓ − K ⊗ φ M is of dimension ℓ .If φ ◦ b = 0, we are done. Otherwise there exists u ∈ M \ ker π with b ( u , u ) / ∈ ker φ . Indeed,if φ ( b ( u, u )) = 0 for all u ∈ M \ ker π , then φ ( b ( u, u )) = 0 for all u ∈ M , so by2 b ( u, v ) = b ( u + v, u + v ) − b ( u, u ) − b ( v, v )it follows that φ ( b ( u, v )) = 0 for every u, v ∈ M . Clearly there exist v , . . . , v ℓ ∈ M such that { π ( u ) , π ( v ) . . . , π ( v ℓ ) } is a K -basis of K ⊗ φ M . For 2 ≤ i ≤ ℓ let v ′ i = b ( u , u ) v i − b ( u , v i ) u and let M ′ be the Λ-module generated by v ′ i . Note that b ( u , v ) = 0 for all v ∈ M ′ anddim K ( K ⊗ φ M ′ ) = ℓ − φ ( b ( u , u )) is invertible in K . Hence we can apply the inductionhypothesis to obtain u , . . . , u ℓ ∈ M ′ such that { π ( u ) , . . . , π ( u ℓ ) } is a K -basis of K ⊗ φ M and φ ( b ( u i , u i ′ )) = 0 for all i = i ′ . (cid:3) Theorem 4.8. Let R ⊇ R be a real closed field. Then an R -algebra homomorphism φ : T n → R extends to an R -algebra homomorphism ϕ : R [ ξ ] → R if and only if φ (Ω n ) ⊆ R ≥ .Proof. The implication ( ⇒ ) is obvious, so we prove ( ⇐ ). In the terminology of [DPRR05,Subsection 2.3], T n is an n –Cayley-Hamilton algebra. Since T n is finitely spanned over T n , A ′ = R ⊗ φ T n is a finite-dimensional R -algebra which inherits an involution τ ′ and an R -trace χ ′ : A ′ → R from T n . By [DPRR05, Subsection 2.3] A ′ is again an n –Cayley-Hamiltonalgebra. Let J be the Jacobson radical of A ′ . Since A ′ is finite-dimensional, elements of J arecharacterized as generators of nilpotent ideals. Hence clearly J τ ′ ⊆ J . Moreover, if f ∈ J , then χ ′ ( f ) = 0 by applying [DPRR05, Proposition 3.2] to the scalar extension of A ′ by the algebraicclosure of R and [Lam91, Theorem 5.17].Therefore A = A ′ / J is a finite-dimensional semisimple R -algebra with involution τ and an R -trace χ : A → R . If Φ : T n → A is the canonical ∗ -homomorphism, then(4.7) χ ◦ Φ = Φ ◦ tr . We claim that τ ( aa τ ) ≥ a ∈ A . Indeed, if π : T n → R ⊗ φ T n is the canonical ∗ -homomorphism, then by Lemma 4.7 there exist a finite set { u i } i of symmetric elements in T n and a finite set { v j } j of antisymmetric elements in T n such that { π ( u i ) } i ∪ { π ( v j ) } j form an R -basis of R ⊗ φ T n and φ (tr( u i u i ′ )) = φ (tr( v j v j ′ )) = φ (tr( u i v j )) = 0for all i = i ′ and j = j ′ . If a = X i α i Φ( u i ) + X j β j Φ( v j ) , α i , β j ∈ R, then χ ( aa τ ) = X i α i φ (tr( u i u t i )) + X j β j φ (tr( v j v t j )) ≥ A = A × · · · × A ℓ for some finite-dimensional simple R -algebras A k . Moreover, by [DPRR05, Theorem 4.2] thereexist d , . . . , d ℓ ∈ N such that(4.8) χ X k a k ! = X k d k tr R A k ( a k )for a k ∈ A k .Next we show that τ is split, i.e., ( A k ) τ ⊆ A k for 1 ≤ k ≤ ℓ . Since every involutionpreserves centrally primitive idempotents [Lam91, Section 22], for every k there exists k ′ suchthat ( A k ) τ ⊆ A k ′ . Suppose that τ is not split and without loss of generality assume ( A ) τ ⊆ A .Let e ∈ A and e ∈ A be the identity elements, respectively. Then χ (cid:0) ( e − e )( e − e ) τ (cid:1) = χ (cid:0) ( e − e )( e − e ) (cid:1) = χ ( − e − e )= − d tr R A ( e ) − d tr R A ( e ) < , a contradiction.Let τ k be the restriction of τ on A k . By (4.8) and the previous paragraph it follows thattr R A k ( aa τ k ) ≥ A k ( aa τ k ) ≥ a ∈ A k , so τ k is a positive involution.Therefore the assumptions of Proposition 4.6 are met and we obtain a trace preserving ∗ -homomorphism Ψ : T n → M n ( R ) extending φ . Now we define ϕ : R [ ξ ] → R by ϕ ( ξ jı ) = Ψ(Ξ j ) ı . (cid:3) Remark . The condition φ (Ω n ) ⊆ R ≥ in Theorem 4.8 is clearly necessary since tr( hh ∗ ) isa nonzero sum of squares in R [ ξ ] for every nonzero h ∈ T n . Moreover, it is not vacuous. Forexample, let τ be the involution on H defined by u τ = iu s i − for u ∈ H , where s is the standardsymplectic involution on H . Then τ is of orthogonal type and we have a trace preserving ∗ -epimorphism Φ : T → H defined by Φ(Ξ ) = i and Φ(Ξ ) = j . Since Φ( T ) = R , the restrictionyields a homomorphism φ : T → R and φ (tr(Ξ Ξ t )) = − Corollary 4.10. Let φ : T n = R [ ξ ] O n ( R ) → R be an R -algebra homomorphism into a real closedfield R ⊇ R . Then the following are equivalent: (i) φ extends to an R -algebra homomorphism ϕ : R [ ξ ] → R ; (ii) φ ( R [ ξ ] O n ( R ) ∩ P R [ ξ ] ) ⊆ R ≥ ; (iii) φ (tr( hh t )) ∈ R ≥ for all h ∈ T n .Remark . At first glance one might ponder whether Theorem 4.8 could be derived from theProcesi-Schwarz theorem [PS85]. In Appendix B we explain why this does not seem to be thecase.4.3. Stellens¨atze. We are now ready to give the main result of this section, the Krivine-StenglePositivstellensatz for trace polynomials a that are positive (semidefinite) on K S , see Theorem4.13. In the proof we use Corollary 4.4 to reduce the problem to the commutative ring T n [ a ].Before applying the abstract Positivstellensatz for commutative rings, we need the relationbetween orderings and matrix evaluations of trace polynomials that is given in Proposition4.12 below, which is a crucial consequence of the extension Theorem 4.8 and Tarski’s transferprinciple. Proposition 4.12. Let S ⊂ T n be finite, a ∈ Sym T n and P an ordering in T n [ a ] containing S ∪ Ω n . (1) a | K S (cid:23) implies a ∈ P . (2) a | K S ≻ implies a ∈ P \ − P . OSITIVE TRACE POLYNOMIALS 19 (3) a | K S = 0 implies a ∈ P ∩ − P .Proof. Let P be an ordering in T n [ a ] containing S and let σ j = tr( ∧ j a ) ∈ T n for 1 ≤ j ≤ n .(1) The restriction of P to T n gives rise to a real closed field R and a homomorphism φ : T n → R satisfying φ ( S ∪ Ω n ) ⊆ R ≥ . By Theorem 4.8 we extend it to a homomorphism ϕ : R [ ξ ] → R .Suppose that ϕ ( σ j ) < j ; in other words, there exist α ∈ R gn such that σ j ( α ) < s ′ ( α ) ≥ s ′ ∈ S . By Tarski’s transfer principle [Mar08, Theorem 1.4.2] there exists α ′ ∈ R gn such that σ j ( α ′ ) < s ′ ( α ′ ) ≥ s ′ ∈ S . But this contradicts σ j | K S ≥ s | K S (cid:23) 0. Hence φ ( σ j ) = ϕ ( σ j ) ≥ j , so σ j ∈ P for all j . Bythe Cayley-Hamilton theorem we have(4.9) ( − a ) n + n X j =1 σ j ( − a ) n − j = 0 . Suppose a / ∈ P . Then − a ∈ P , so (4.9) implies ( − a ) n ∈ P ∩ − P . Therefore a ∈ P ∩ − P , acontradiction.(2) Because a | K S ≻ σ j | K S > 0, we obtain σ j ∈ P \− P for all j by the same reasoningas in (1). If a / ∈ P \ − P , then − a ∈ P , so (4.9) implies σ n ∈ P ∩ − P , a contradiction.(3) If a | K S = 0, then a | K S (cid:23) − a | K S (cid:23) 0, so a ∈ P ∩ − P by (1). (cid:3) Theorem 4.13 (Krivine-Stengle Positivstellensatz for trace polynomials) . Let S ∪{ a } ⊂ Sym T n be finite. (1) a | K S (cid:23) if and only if at = t a = a k + t for some t , t ∈ T tr S and k ∈ N . (2) a | K S ≻ if and only if at = t a = 1 + t for some t , t ∈ T tr S . (3) a | K S = 0 if and only if − a k ∈ T tr S for some k ∈ N .Proof. The directions ( ⇐ ) are straightforward. For the implications ( ⇒ ), by Corollary 4.4 wecan assume that S ⊂ T n . Let T be the preordering in T n [ a ] generated by S ∪ Ω n . Note that T ⊂ T tr S since S ⊂ T n . If a | K S (cid:23) 0, then a ∈ P for every ordering P of T n [ a ] containing T by Proposition 4.12. Therefore t a = a k + t for some t , t ∈ T and k ∈ N by the abstractPositivstellensatz [Mar08, Theorem 2.5.2], so (1) is holds. (2) and (3) are proved analogously. (cid:3) Remark . In general we cannot choose a central t in Theorem 4.13; see Example 6.4. Remark . A clean Krivine-Stengle Positivstellensatz for generic matrices clearly does notexist for n = 3 due to Theorem 3.2. Moreover, in Example 6.3 we show that even for n = 2,where Conjecture 3.1 holds, the traceless equivalent of Theorem 4.13 for GM n fails. Corollary 4.16. If S ⊂ Sym T n is finite, then K S = ∅ if and only if − ∈ T tr S .Proof. If K S = ∅ , then − | K S ≻ 0, so by Theorem 4.13 there exist t , t ∈ T tr S such that( − t = 1 + t , so − t + t ∈ T tr S . The converse is trivial. (cid:3) Corollary 4.17. Let s ∈ Sym T n . Then s ( A ) for all A ∈ M n ( R ) g if and only if − X i ω i Y j tr( h ij sh t ij ) for some ω i ∈ Ω n and h ij ∈ T n .Proof. Follows by Corollary 4.16 and Lemma 4.1. (cid:3) Corollary 4.18 (Real Nullstellensatz for trace polynomials) . Let J ⊂ T n be an ideal andassume tr( J ) ⊆ J . For h ∈ T n the following are equivalent: (i) for every X ∈ M n ( R ) g , u ( X ) = 0 for every u ∈ J implies h ( X ) = 0 ; (ii) there exists k ∈ N such that − ( h t h ) k ∈ Ω n + J . Proof. The implication (2) ⇒ (1) is clear. Conversely, T n is Noetherian, so J is (as a left ideal)generated by some u , . . . , u ℓ ∈ T n . Let S = {− u t u , . . . , − u t ℓ u ℓ } ; then (1) is equivalent to h t h | K S = 0. Hence (1) ⇒ (2) follows by Theorem 4.13(3) and T tr S ⊆ Ω n + J . (cid:3) Corollary 4.19. Let S ⊂ T n . Then { A ∈ M n ( R ) g : s ( A ) = 0 ∀ s ∈ S } = ∅ if and only if − ω + X i tr( h i s i ) for some ω ∈ Ω n , h i ∈ T n and s i ∈ S . We mention that Hilbert’s Nullstellensatz for n × n generic matrices over an algebraicallyclosed field is given by Amitsur in [Ami57, Theorem 1].5. Positivstellens¨atze for compact semialgebraic sets In this section we give certificates for positivity on compact semialgebraic sets. We prove aversion of Schm¨udgen’s theorem [Scm91] for trace polynomials (Theorem 5.3). We also presenta version of Putinar’s theorem [Put93] for trace polynomials (Theorem 5.7) and for genericmatrices (Theorem 5.5).5.1. Archimedean (cyclic) quadratic modules. A cyclic quadratic module Q ⊂ T n is archimedean if for every h ∈ T n there exists ρ ∈ Q > such that ρ − hh t ∈ Q . Equivalently, forevery s ∈ Sym T n there exists ε ∈ Q > such that 1 ± εs ∈ Q .For a cyclic quadratic module Q let H Q be the set of elements h ∈ T n such that ρ − hh t ∈ Q for some ρ ∈ Q > . It is clear that Q is archimedean if and only if H Q = T n . Proposition 5.1. H Q is a trace ∗ -subalgebra over R in T n .Proof. H Q is a ∗ -subalgebra over R by [Vid59]. Let h ∈ H Q . Then s = h + h t ∈ H Q and let ρ ∈ Q > be such that ρ − s ∈ Q . Then ρ ± s = 12 ρ (cid:0) ( ρ ± s ) + ( ρ − s ) (cid:1) ∈ Q and consequently nρ ± tr( s ) ∈ Q . Therefore( nρ ) − tr( s ) = 12 nρ (cid:0) ( ρ − s )( ρ + s )( ρ − s ) + ( ρ + s )( ρ − s )( ρ + s ) (cid:1) ∈ Q , so tr( h ) = tr( s ) ∈ H Q . (cid:3) Corollary 5.2. A cyclic quadratic module Q is archimedean if and only if there exists ρ ∈ Q > such that ρ − P j Ξ j Ξ t j ∈ Q .Proof. ( ⇒ ) is trivial. Conversely, ρ − P j Ξ j Ξ t j ∈ Q implies Ξ j ∈ H Q for 1 ≤ j ≤ n , so H Q = T n since T n is generated by Ξ j as a trace ∗ -subalgebra over R by Proposition 5.1. (cid:3) It is easy to see that K Q is compact if Q is archimedean. The converse fails already with n = 1 ([Mar08, Section 7.3] or [PD01, Example 6.3.1]). If K Q is compact, say k X k ≤ N for all X ∈ K Q , then we can add N − P j Ξ j Ξ t j to Q to make it archimedean without changing K Q . OSITIVE TRACE POLYNOMIALS 21 Schm¨udgen’s Positivstellensatz for trace polynomials. In this subsection we prove aversion of Schm¨udgen’s Positivstellensatz for trace polynomials a that are positive on a compactsemialgebraic set K S . The proof is a two-step commutative reduction. Firstly, the constraints S are replaced with central ones by Corollary 4.4. Then the abstract version of Schm¨udgen’sPositivstellensatz is used in the commutative ring T n [ a ]. Theorem 5.3 (Schm¨udgen’s Positivstellensatz for trace polynomials) . Let S ∪ { a } ⊂ Sym T n be finite. If K S is compact and a | K S ≻ , then a ∈ T tr S .Proof. First we apply Corollary 4.4 to reduce to the case S ⊂ T n . Let T be the preordering in T n [ a ] generated by S ∪ Ω n . Note that K S = K T and T ⊂ T tr S .Let b ∈ T n [ a ] be arbitrary. Since K S is compact, there exists β ∈ R ≥ such that β ± b | K S (cid:23) β ± b ∈ P for every ordering P in T n [ a ] containing T ⊃ S ∪ Ω n by Proposition 4.12. In theterminology of [Sce03], T is weakly archimedean. Since T n [ a ] is a finitely generated R -algebra, T is an archimedean preordering in T n [ a ] by the abstract version of Schm¨udgen’s Positivstellensatz[Sce03, Theorem 3.6]. Similarly, Proposition 4.12 implies a ∈ P \ − P for every ordering P in T n [ a ] containing T ⊃ S ∪ Ω n , so a ∈ T by [Sce03, Proposition 3.3] or [Mon98, Theorem 4.3]. (cid:3) Corollary 5.4. Let S ⊂ Sym T n be finite. Then T tr S is archimedean if and only if K S is compact. Putinar’s Positivstellensatz for generic matrices. Our next theorem is a Putinar-type Positivstellensatz for generic matrices on compact semialgebraic sets, which requires afunctional analytic proof. While the proof generally follows a standard outline (using a sepa-ration argument followed by a Gelfand-Naimark-Segal construction), several modifications areneeded. For instance, the separation is taken to be extreme in a convex sense, and polynomialidentities techniques are applied to produce n × n matrices.A set Q ⊆ Sym GM n is a quadratic module if1 ∈ Q , Q + Q ⊆ Q , h Q h t ⊆ Q ∀ h ∈ GM n . We say that Q is archimedean if for every h ∈ GM n there exists ρ ∈ Q > such that ρ − aa t ∈ Q .As in Corollary 5.2 we see that a quadratic module is archimedean if and only if it contains ρ − P j Ξ j Ξ t j for some ρ ∈ Q > . Theorem 5.5 (Putinar’s Positivstellensatz for generic matrices) . Let Q ⊂ Sym GM n be anarchimedean quadratic module and a ∈ Sym GM n . If a | K Q ≻ , then a ∈ Q .Proof. Assume a ∈ Sym GM n \ Q . We proceed in several steps. Step 1: Separation.Consider Q as a convex cone in the vector space Sym GM n over R . Since Q is archimedean,for every s ∈ Sym T n there exists ε ∈ Q > such that 1 ± εs ∈ Q , which in terms of [Bar02,Definition III.1.6] means that 1 is an algebraic interior point of the cone Q in Sym T n . By theEidelheit-Kakutani separation theorem [Bar02, Corollary III.1.7] there exists a nonzero R -linearfunctional L : Sym GM n → R satisfying L ( Q ) ⊆ R ≥ and L ( a ) ≤ 0. Moreover, L (1) > Q is archimedean, so after rescaling we can assume L (1) = 1. Let L : GM n → R bethe symmetric extension of L , i.e., L ( f ) = L ( f + f t ) for f ∈ GM n . Step 2: Extreme separation.Now consider the set C of all linear functionals L ′ : GM n → R satisfying L ′ ( Q ) ⊆ R ≥ and L ′ (1) = 1. This set is nonempty because L ∈ C . Endow GM n with the norm k p k = max {k p ( X ) k : X ∈ M n ( R ) g , k X k ≤ } . By the Banach-Alaoglu theorem [Bar02, Theorem III.2.9], the convex set C is weak*-compact.Thus by the Krein-Milman theorem [Bar02, Theorem III.4.1] we may assume that our separatingfunctional L is an extreme point of C . Step 3: GNS construction.On GM n we define a semi-scalar product h p, q i = L ( q t p ). By the Cauchy-Schwarz inequality forsemi-scalar products, N = (cid:8) q ∈ GM n | L ( q t q ) = 0 (cid:9) is a linear subspace of GM n . Hence(5.1) h p, q i = L ( q t p )is a scalar product on GM n / N , where p = p + N denotes the residue class of p ∈ GM n inGM n / N . Let H denote the completion of GM n / N with respect to this scalar product. Since1 6∈ N , H is non-trivial.Next we show that N is a left ideal of GM n . Let p, q ∈ GM n . Since Q is archimedean, thereexists ε > − εp t p ∈ Q and therefore(5.2) 0 ≤ L ( q t (1 − εp t p ) q ) ≤ L ( q t q ) . Hence q ∈ N implies pq ∈ N .Because N is a left ideal, we can define linear maps M p : GM n / N → GM n / N , q pq for p ∈ GM n . By (5.2), M p is bounded and thus extends to a bounded operator ˆ M p on H . Step 4: Irreducible representation of GM n .The map π : GM n → B ( H ) , p ˆ M p is clearly a ∗ -representation, where B ( H ) is endowed with the adjoint involution ∗ . Observe that η = 1 ∈ H is a cyclic vector for π by construction and(5.3) L ( p ) = h π ( p ) η, η i . Write A = π (GM n ). We claim that the self-adjoint elements in the commutant A ′ of A in B ( H )are precisely real scalar operators. Let P ∈ A ′ be self-adjoint. By the spectral theorem, P decomposes into real scalar multiples of projections belonging to { P } ′′ ⊆ A ′ . So it suffices toassume that P is a projection. By way of contradiction suppose that P / ∈ { , } ; since η is cyclicfor π , we have P η = 0 and (1 − P ) η = 0. Hence we can define linear functionals L j on GM n by L ( p ) = h π ( p ) P η, P η ik P η k and L ( p ) = h π ( p )(1 − P ) η, (1 − P ) η ik (1 − P ) η k for all p ∈ GM n . One checks that L is a convex combination of L and L . Since also L j ∈ C ,we obtain L = L = L by the extreme property of L . Let λ = k P η k ; then (5.3) implies h π ( p ) η, λη i = λ h π ( p ) η, η i = h π ( p ) P η, P η i = h P π ( p ) η, P η i = h π ( p ) η, P η i for all p ∈ GM n . Therefore P η = λη since η is a cyclic vector for π . Then λ ∈ { , } since P isa projection, a contradiction.Next we show that π is an irreducible representation. Suppose that U ⊆ H is a closed π -invariant subspace and P : H → U the orthogonal projection. If p ∈ GM n and p t = ± p , then π ( p ) P = P π ( p ) P = ± ( P π ( p ) P ) ∗ = ± ( π ( p ) P ) ∗ = P π ( P ) . Consequently P ∈ A ′ and hence P ∈ R . Since P is an orthogonal projection, we have P ∈ { , } ,so π is irreducible. Step 5: Transition to n × n matrices.We claim that A is a prime algebra. Indeed, suppose a A b = 0 for a, b ∈ A . If b = 0, then thereis a u ∈ H with bu = 0. Since π is irreducible, the vector space V = A bu is dense in H . Now aV = 0 implies aH = 0, i.e., a = 0. OSITIVE TRACE POLYNOMIALS 23 Since the ∗ -center of A equals R , we have π ( C n ) = R , so A is generated by π (Ξ j ) for 1 ≤ j ≤ g as an R -algebra with involution. Let f ∈ R < x , x ∗ > be a polynomial ∗ -identity of (M n ( R ) , t ).Then f ( p , . . . , p k , p t , . . . , p t k ) = 0 for all p i ∈ GM n . Therefore f ( π ( p ) , . . . , π ( p k ) , π ( p ) ∗ , . . . , π ( p k ) ∗ ) = π ( f ( p , . . . , p k , p t , . . . , p t k )) = 0 , so f is a polynomial ∗ -identity for A . By the ∗ -version of Posner’s theorem [Row73, Theorem 2]it follows that A is central simple algebra of degree n ′ ≤ n with involution ∗ and with ∗ -center R . Furthermore, the involution on A is positive since it is a restriction of the adjoint involutionon B ( H ). By [PS76, Theorem 1.2], A is ∗ -isomorphic to one of(M n ′ ( R ) , t ) , (M n ′ ( C ) , ∗ ) , (M n ′ / ( H ) , s ) . Since A = Φ(GM n ), A satisfies all polynomial ∗ -identities of (M n ( R ) , t ). If ( A , ∗ ) ∼ = (M n ′ / ( H ) , s ),then Proposition 2.3 implies n ′ ≤ n . Similarly, ( A , ∗ ) ∼ = (M n ′ ( C ) , ∗ ) implies 2 n ′ ≤ n by Remark2.4. Hence in all cases there exists a ∗ -embedding of ( A , ∗ ) into (M n ( R ) , t ), so we can assumethat X j := ˆ M Ξ j ∈ M n ( R ), ∗ = t and η ∈ R d . Since L ( Q ) ⊆ R ≥ , (5.3) implies that q ( X ) ispositive semidefinite for all q ∈ Q , so X ∈ K Q . Step 6: Conclusion.By (5.1) we have 0 ≥ L ( a ) = h a, i = h a ( X, X t ) η, η i . Therefore a is not positive definite at X ∈ K Q . (cid:3) Putinar’s theorem for trace polynomials. Our final result in this section is Putinar’sPositivstellensatz for trace polynomials, Theorem 5.7. Our proof combines functional analytictechniques from the proof of Theorem 5.5 with an algebraic commutative reduction. Lemma 5.6. Let Q ⊂ Sym T n be an archimedean cyclic quadratic module and c ∈ T n . If c | K Q > , then c ∈ Q .Proof. Assume c ∈ Sym T n \ Q . Steps 1–4 in the proof of Theorem 5.5 work if we replace GM n with T n . Hence we obtain a Hilbert space H , a ∗ -representation π : T n → B ( H ) with π ( T n ) = R ,and a cyclic unit vector η ∈ H for π such that the linear functional L : T n → R , p 7→ h π ( p ) η, η i satisfies L ( Q ) ⊆ R ≥ and L ( c ) ≤ 0. Let φ = π | T n : T n → R . By the proof of Theorem 4.8, φ extends to a trace preserving ∗ -homomorphism Ψ : T n → M n ( R ). Let X j = Ψ(Ξ j ) ∈ M n ( R ) for j = 1 , . . . , g . Because π ( T n ) = R , we have L | T n = π | T n and therefore tr( q ( X )) = φ (tr( q )) ≥ q ∈ Q , so X ∈ K Q by Corollary 4.5 and c ( X ) = φ ( c ) ≤ (cid:3) Theorem 5.7 (Putinar’s theorem for trace polynomials) . Let Q ⊂ Sym T n be an archimedeancyclic quadratic module and a ∈ Sym T n . If a | K Q ≻ , then a ∈ Q .Proof. Let σ j = tr( ∧ j a ) and assume a | K Q ≻ 0. Since K Q is compact, there exists ε > σ j − ε ) | K Q > ≤ j ≤ n . By Lemma 5.6 we have σ j − ε ∈ Q for all j . Let c , . . . , c N be the generators of T n as an R -algebra. Since Q is archimedean, there exist ρ , . . . , ρ N ∈ Q > such that ρ i − c i ∈ Q . Write S = (cid:8) σ j − ε, ρ i − c i : 1 ≤ j ≤ n, ≤ i ≤ N (cid:9) and let Q be the quadratic module in T n [ a ] generated by S ∪ Ω n . Clearly we have Q ⊂ Q and a | K Q > 0. Let H Q ⊆ T n [ a ] be the subring of bounded elements with respect to Q , i.e., b ∈ T n [ a ]such that ρ − b ∈ Q for some ρ ∈ Q > . Because c i generate T n , we have T n ⊆ H Q . Since H Q isintegrally closed in T n [ a ] by [Bru79, Section 6.3] or [Scw03, Theorem 5.3], we also have a ∈ H Q .Hence Q is archimedean. If P is an ordering in T n [ a ] containing S ∪ Ω n , then a ∈ P \ − P by Proposition 4.12. Therefore a ∈ Q ⊂ Q by Jacobi’s representation theorem [Mar08, Theorem5.4.4]. (cid:3) Examples In this section we collect some examples and counterexamples pertaining to the results pre-sented above. Example 6.1. Proposition 4.3 states that for every s ∈ Sym T n there exists a finite set S ⊂ tr( Q tr { s } ) such that K { s } = K S . Let us give a concrete example of such a set for n = 3. Let σ j = tr( ∧ j s ); using the Cayley-Hamilton theorem and the relations between σ j and tr( s i ) it iseasy to check that tr( s ) = σ , tr (cid:0) ( s − σ ) s ( s − σ ) (cid:1) = σ σ + 3 σ , tr (cid:0) ( s − σ s + σ ) s ( s − σ s + σ ) (cid:1) = σ σ , tr (cid:0) ( s − σ − s ( s − σ − (cid:1) = σ + 4 σ + 3 σ + σ σ . Denote these elements by c , c , c , c ∈ tr( Q tr { s } ). We claim that K { s } = K { c ,c ,c ,c } . It sufficesto prove the inclusion ⊇ .To simplify the notation let s = s t ∈ M ( R ). If s (cid:15) 0, then σ j < j . If σ < c < 0. Hence assume σ ≥ 0. If σ , σ are of opposite sign, then c < 0. If σ > σ , σ ≤ 0, then c < 0. Finally, if σ = 0 and one of σ , σ equals 0, then σ < c < σ < c < Example 6.2. The denominator in Lemma 2.5 is unavoidable even if f is a hermitian squareor f ∈ R [ ξ ]. For example, let n = 4 and a = Ξ − Ξ t . Then det( a ) is a square in R [ ξ ]. Supposedet( a ) ∈ Ω , i.e., det( a ) = X i tr( h i h t i ) · · · tr( h im i h t im i )for some h ij ∈ T n . Because a is independent of Ξ + Ξ t and Ξ j for j > 1, we can assume that h ij are polynomials in a and tr( a k ) for k ∈ N . Moreover, det( a ) is homogeneous of degree 4 withrespect to the entries of a , so h ij are of degree at most 2 with respect to a . Finally, tr( a ) = 0,so we conclude that det( a ) = X i tr (cid:0) ( α i a + β i tr( a )) (cid:1) , α i , β i ∈ R . If R is a real closed field containing R ( ξ ), then there exist λ, µ ∈ R that are algebraicallyindependent over R such that a = − λ λ − µ µ after an orthogonal basis change. Therefore( λµ ) = 2 X i (cid:0) ( α i λ + 2 β i ( λ + µ )) + ( α i µ + 2 β i ( λ + µ )) (cid:1) , which is clearly a contradiction. OSITIVE TRACE POLYNOMIALS 25 Example 6.3. Next we show that a traceless equivalent of Theorem 4.13 fails for GM . Aquadratic module T ⊆ Sym GM n is a preordering if T ∩ C n is closed under multiplication.For S ⊂ Sym GM n let Q S and T S denote the quadratic module and preordering, respectively,generated by S . For example, T ∅ = Q ∅ is the set of sums of hermitian squares in GM n .Fix n = 2 and let s = Ξ + Ξ t , a = Ξ − Ξ t , f = [ s , a ][ s, a ] . Since f = tr( s )[ s, a ] , it is clear that f | K { s } (cid:23) 0. We will show that there do not exist t , t ∈ T S and k ∈ N such that(6.1) f t = t f = f k + t . It clearly suffices to assume g = 1. If R is a real closed field containing R ( ξ ), then afterdiagonalizing s we may assume that s = (cid:18) λ λ (cid:19) , a = (cid:18) − µµ (cid:19) for some λ , λ , µ ∈ R that are algebraically independent over R . Let h ∈ GM be homogeneousof degree ( d, e ) with respect to ( s, a ). Then it is not hard to check that(6.2) hsh t = µ e (cid:18) λ ˜ h ( λ , λ ) λ ˜ h ( λ , λ ) (cid:19) for some homogeneous polynomial ˜ h ∈ R [ y , y ] of degree d . Therefore X i h i sh t i ∈ C ⇒ h i = 0 ∀ i, so we deduce that(6.3) T { s } = ( T ∅ ∩ C ) · Q { s } . Now suppose that (6.1) holds for some t , t ∈ T S and k ∈ N . Since f is homogeneous ofdegree (5 , 2) with respect to ( s, a ), t , t can be taken homogeneous as well. Then t is of degree(10 k, k ) and t is of degree (10 k − , k − t and t areodd and even, respectively, so by (6.3) we conclude that t ∈ T ∅ and t is of the form P i h i sh t i for h i ∈ GM . Hence (6.2) implies t = µ k − (cid:18) λ P i ˜ h i ( λ , λ ) λ P i ˜ h i ( λ , λ ) (cid:19) for some homogeneous ˜ h i ∈ R [ y , y ]. Now f = ( λ + λ )( λ − λ ) µ implies(6.4) f t = ( λ + λ ) µ k ( λ − λ ) (cid:18) λ P i ˜ h i ( λ , λ ) λ P i ˜ h i ( λ , λ ) (cid:19) . The nonempty set { X ∈ M ( R ) : det( s ( X )) < s ( X )) > } is open in the Euclidean topology, so by (6.4) there exists X ∈ M ( R ) such that ( st )( X ) isnonzero and indefinite. However, this contradicts f t = f k + t ∈ T ∅ . Example 6.4. Here we show that the element t in Theorem 4.13 cannot be chosen central ingeneral. Let n = 2, s = (Ξ + Ξ t ) and S = { tr( s ) , det( s ) } . Suppose t s = s k + t for some k ∈ N and t , t ∈ T tr S and t ∈ T . LetΨ : T → M ( R [ ζ ]) , Ξ (cid:18) ζ 00 1 (cid:19) . Since ( ζ + 1) = ζ + ( ζ + 1) + ζ , we conclude that Ψ( t ) belongs to the commutativepreordering generated by ζ and Ψ( t ) belongs to the matricial preordering generated by ζ inM ( R [ ζ ]). But then Ψ( t )Ψ( s ) = Ψ( s ) k + Ψ( t ) contradicts [Cim12, Example 4]. Example 6.5. Let f = 5 Tr(Ξ Ξ t ) − )(Ξ + Ξ t ) ∈ T . We will show that f is totallypositive and write it as a sum of hermitian squares in USA . Write Ξ = Ξ = ( ξ ı ) ı and let u = (cid:0) η η (cid:1) , v = (cid:0) ξ η ξ η ξ η ξ η ξ η ξ η ξ η ξ y (cid:1) . Then uf u t can be viewed as a quadratic form in v and uf u t = vG α v t , G α = α α − α − α − α − α − − − α − − α − − α − − α − − 20 0 0 0 − α − α − α − α − α α for α ∈ R . Observe that G α is positive semidefinite if and only if − ≤ α ≤ − . Hence f isindeed totally positive and a sum of hermitian squares in M ( R [ ξ ]). By diagonalizing G α at α = − we obtain f = 52 ( ξ − ξ ) + 12 ˜ H ˜ H t + 12 ˜ H ˜ H t , where ˜ H = (cid:18) ξ + ξ ξ − ξ ξ − ξ ξ + ξ (cid:19) , ˜ H = (cid:18) ξ + ξ ) ξ − ξ ξ − ξ ξ + ξ ) (cid:19) . Note that while ˜ H ˜ H t , ˜ H ˜ H t ∈ Sym T , we can compute R ( ˜ H ) = R ( ˜ H ) = 0, so ˜ H , ˜ H / ∈ T .However, if we set H = Ξ − Ξ t , H = ΞΞ t − Ξ t Ξ , H = Ξ − t + 2Ξ t Ξ − (Ξ t ) , then H H t = ( ξ − ξ ) , H H t = ( ξ − ξ ) ˜ H ˜ H t , H H t = ( ξ − ξ ) ˜ H ˜ H t and so f = 52 H H t + 12 H − H H t H − t + 12 H − H H t H − t . A. Constructions of the Reynolds operator In this appendix we describe a few more ways of constructing R n for the action of O n ( R ) onM n ( R [ ξ ]) defined in Subsection 2.2. We refer to [Stu08] for algorithms for finite group actions.Let R ′ n : R [ ξ ] → T n be the restriction of R n : M n ( R [ ξ ]) → T n , i.e., the Reynolds operator forthe action of O n ( R ) on R [ ξ ] given by f u = f ( u Ξ u t , . . . , u Ξ g u t )for f ∈ R [ ξ ] and u ∈ O n ( R ). OSITIVE TRACE POLYNOMIALS 27 A.1. Computing R ′ n . We start by describing two ways of obtaining R ′ n .A.1.1. First method. We follow [DK02, Section 4.5.2] to present an algorithm for computing R ′ n . We define a linear map c ∈ R [O n ( R )] ∗ : c ( r ) = dd t dd s n X i,j =1 r (cid:0) (1 + se ij )(1 + te ij ) (cid:1) − r (cid:0) (1 + se ij )(1 + te ji ) (cid:1)(cid:12)(cid:12)(cid:12)(cid:12) s = t =0 , where e ij , 1 ≤ i, j ≤ n , denote the standard matrix units in M n ( R ). (In fact, c equals theCasimir operator of the Lie algebra o n of skew symmetric matrices of the group O n ( R ) up to ascalar multiple.) For example, if n = 2, then c ( u u ) = dd t dd s (cid:16)(cid:0) (1 + se )(1 + te ) (cid:1) (cid:0) (1 + se )(1 + te ) (cid:1) −− (cid:0) (1 + se )(1 + te ) (cid:1) (cid:0) (1 + se )(1 + te ) (cid:1) ++ (cid:0) (1 + se )(1 + te ) (cid:1) (cid:0) (1 + se )(1 + te ) (cid:1) −− (cid:0) (1 + se )(1 + te ) (cid:1) (cid:0) (1 + se )(1 + te ) (cid:1) (cid:17)(cid:12)(cid:12)(cid:12)(cid:12) s = t =0 = dd t dd s (1 − st + 1 − st ) (cid:12)(cid:12)(cid:12)(cid:12) s = t =0 = − . For f ∈ R [ ξ ], u = ( u ı ) ı ∈ O n ( R ), write f u as P i f i µ i , where f i are linearly independentpolynomials in the variables ξ jı and µ i are polynomials in the variables u ı . Define˜ c ( f ) = X f i c ( µ i ) . Find the monic polynomial p of smallest degree such that p (˜ c )( f ) = 0. If p (0) = 0, set R ′′ n ( f ) = f .If p (0) = 0, write p ( t ) = tq ( t ) and define R ′′ n ( f ) = q (0) − q (˜ c ( f )). By [DK02, Proposition 4.5.17], R ′′ n defines the Reynolds operator for the action of SO n ( R ) on R [ ξ ]. Since O n ( R ) / SO n ( R ) ∼ = Z / Z , setting R ′ n ( f ) = ( R ′′ n ( f ) + R ′′ n ( f ) v ), where v is an arbitrary element in O n ( R ) \ SO n ( R ),we obtain the Reynolds operator for the action of O n ( R ) on R [ ξ ].A.1.2. Second method. Here we mention another way of computing the Reynolds operator R ′ n in terms of an integral. This approach is based on the way the invariants of O n ( R ) for the actionon R [ ξ ] were described by Procesi in [Pro76]. Let f ∈ R [ ξ ]. We first multihomogenize f as afunction f : M n ( R ) g → R , then multilinearize its homogeneous components f i and view f i as anelement f i ∈ (M n ( R ) ⊗ d i ) ∗ for d i = deg( f i ). Since M n ( R ) ∼ = V ∗ ⊗ V for a n -dimensional vectorspace V on which O n ( R ) acts naturally, and V ∗ is isomorphic as a O n ( R )-module to V , f i canbe seen as an element ˜ f i ∈ V ⊗ d i . The monomial ξ ı · · · ξ dı d d corresponds to the element e ı ⊗ e ⊗ · · · ⊗ e ı d ⊗ e d , where e i , 1 ≤ i ≤ n , is an orthonormal basis of V . Then we can compute R ′ n ( f i ) by integratingthe function u ( ˜ f i ) u over O n ( R ). To obtain R ′ n ( f i ) we need to restitute R ′ n ( f i ) and multiplythe result by a suitable integer. Finally, R ′ n ( f ) = P R ′ n ( f i ).A.2. From R ′ n to R n . Once we have R ′ n , we can compute R n as follows. A.2.1. First method. Let f ∈ M n ( R [ ξ ]). We can assume that f is independent of Ξ g . Let uscompute R ′ n (tr( f Ξ g )). Since this is an invariant, linear in Ξ g , it has the form R ′ n (tr( f Ξ g )) = tr (cid:16) f Ξ g (cid:17) for some f ∈ T n . Here we used the fact that tr( h Ξ t g ) = tr( h t Ξ g ) for h ∈ T n . We define R n ( f ) = f . We claim that R n : M n ( R [ ξ ]) → T n is the Reynolds operator. We have R n ( f ) = f for f ∈ T n since R ′ n (tr( f Ξ g )) = tr( f Ξ g ) as tr( f Ξ g ) ∈ T n . For u ∈ O n ( R ), f ∈ M n ( R [ ξ ]) we havetr(( f u )Ξ g ) = tr(( f Ξ g ) u ) = tr( f Ξ g ) u , where the first equality follows as Ξ g is an invariant and the second one since tr is linear. Thus, R n ( f u ) = R n ( f ), and R n is the Reynolds operator.A.2.2. Second method. If we have R ′ n , then the Reynolds operator can also be computed byexpressing an element f ∈ M n ( R [ ξ ]) as an R ( ξ )-linear combination of Ξ i Ξ t j , 0 ≤ i, j ≤ n − n ( R ( ξ )) as there exists X ∈ M n ( R ) suchthat X i X t j , 0 ≤ i, j ≤ n − 1, are linearly independent. We denote Ξ i Ξ t j , 0 ≤ i, j ≤ n − 1, by y , . . . , y n . Let c be a n -normal (i.e., multilinear and alternating in the first n -variables) centralpolynomial of M n ( R ) in 2 n − y , . . . , y n areindependent we can find y n +1 , . . . , y n − ∈ GM n such that0 = c ( y , . . . , y n , y n +1 , . . . , y n − ) = z ∈ C n . If g ≥ n , we can take y n +1 = Ξ , . . . , y n − = Ξ n . Then f can be written as follows(A.1) f = n X i =1 ( − i − z − c ( f, y , . . . , y i − , y i +1 , . . . , y d ) y i . Let z i ( f ) = R ′ n ( c ( f, y , . . . , y i − , y i +1 , . . . , y d )). We define R n ( f ) = n X i =1 ( − i +1 z − z i ( f ) y i . If f ∈ T n , then the coefficients in the expression (A.1) are already in T n , so in this case R n ( f ) = f . Note that R ′ n ( z i ( f u )) = R ′ n ( z i ( f )) for u ∈ O n ( R ) and f ∈ M n ( R [ ξ ]). Therefore R n ( f u ) = R n ( f ) and R n : M n ( R [ ξ ]) → T n is the Reynolds operator.B. How not to prove the extension Theorem 4.8 One might attempt to prove Theorem 4.8 using geometric invariant theory of Lie groups[PS85, Br¨o98, CKS09]. Here we explain why this approach fails.Let G be a compact Lie group with an orthogonal representation on W = R N . The invariantring R [ W ] G is a finitely generated R -algebra; let p , . . . , p m be its generators. Let ( · , · ) denote a G -invariant inner product on W and its dual on W ∗ . Since the differentials d p i : W → W ∗ are G -equivariant, we have (d p i , d p j ) ∈ R [ W ] G . Finally let H = (cid:0) (d p i , d p j ) (cid:1) i,j ∈ M m ( R [ W ] G ) . The following theorem is a reformulation of the celebrated Procesi-Schwarz theorem [PS85,Theorem 0.10] and is essentially due to Schrijver [Scr+] (see also [Scr08]). We thank M.Schweighofer for drawing our attention to Schrijver’s work. Theorem B.1 (Procesi-Schwarz) . Let φ : R [ W ] G → R be an R -algebra homomorphism into areal closed field R ⊇ R . Then the following are equivalent: (i) φ extends to an R -algebra homomorphism ϕ : R [ W ] → R ; OSITIVE TRACE POLYNOMIALS 29 (ii) φ ( R [ W ] G ∩ P R [ W ] ) ⊆ R ≥ ; (iii) φ ( H ) ∈ M m ( R ) is positive semidefinite.Proof. While (i) ⇒ (ii) and (i) ⇒ (iii) are straightforward, (iii) ⇒ (ii) is involved and proved in[PS85, Theorem 0.10] and [CKS09, Subsection 2.7]. Hence we are left with (ii) ⇒ (i). Withoutloss of generality we can assume that φ ( R [ W ] G ) generates R as a field.First we observe that if R : R [ W ] → R [ W ] G is the Reynolds operator for the action of G ,then R ( P R [ W ] ) ⊆ P R [ W ] . Note that since G acts linearly on W , the action of G on R [ W ]does not increase the degree of polynomials. Since G is compact, R is given by the integrationformula R ( f ) = R G f g dµ ( g ), where µ is the normalized left Haar measure. If f ∈ R [ W ] is ofdegree d , then R ( f ) is a limit of sums of squares of degree 2 d . Using Carath´eodory’s theorem[Bar02, Theorem I.2.3] it is easy to see that the cone of sums of squares in R [ W ] of degree atmost 2 d is closed in the space of polynomials of degree at most 2 d (cf. [Mar08, Section 4.1]).Hence we conclude that R ( f ) is indeed a sum of squares in R [ W ].Now let T ⊆ R [ W ] be the preordering generated by φ − ( R ≥ ). We claim that − / ∈ T .Otherwise − s + P i ≥ s i t i for some s i ∈ P R [ W ] and t i ∈ φ − ( R ≥ ). By applying R weget(B.1) − R ( s ) + X i ≥ R ( s i ) t i . By the above observation we have R ( s i ) ∈ R [ W ] G ∩ P R [ W ] , so (B.1) implies − φ ( − 1) = φ ( R ( s )) + X i ≥ φ ( R ( s i )) φ ( t i ) ≥ , a contradiction. Therefore we can extend T to an ordering P ⊂ R [ W ], which gives rise to ahomomorphism ϕ : R [ W ] → R , where R is the real closure of the ordered field of fractions of R [ W ] / ( P ∩ − P ). Since ker φ ⊆ T ∩ − T ⊆ P ∩ − P = ker ϕ , we see that ϕ extends φ and hence R ⊆ R . By the Artin-Lang homomorphism theorem[BCR98, Theorem 4.1.2] there exists a homomorphism ϕ : R [ W ] → R satisfying ker ϕ = ker ϕ .Thus ϕ extends φ . (cid:3) Theorem B.1 can be used to prove the following weakened version of Theorem 4.8. Corollary B.2. An R -algebra homomorphism φ : T n → R extends to an R -algebra homomor-phism ϕ : R [ ξ ] → R if and only if φ (Ω n ) ⊆ R ≥ . Let us outline the proof. Let p , . . . , p m be generators of the R -algebra T n = R [ ξ ] O n ( R ) . Theirdifferentials d p i : M n ( R ) g → (M n ( R ) g ) ∗ are O n ( R )-equivariant maps. Since we can identify(M n ( R ) g ) ∗ with (M n ( R ) ∗ ) g , and the O n ( R )-equivariant polynomial maps M n ( R ) g → M n ( R ) areprecisely trace polynomials [Pro76, Theorems 7.1 and 7.2], we have d p i ∈ T gn . On M n ( R ) g thereis an O n ( R )-invariant inner product( X, Y ) = tr X j X j Y t j . Finally, let H = ((d p i , d p j )) i,j ∈ M m ( T n ).Let φ : T n → R be an R -algebra homomorphism. Then Theorem B.1 implies that φ extends toan R -algebra homomorphism ϕ : R [ ξ ] → R if and only if φ ( H ) ∈ M m ( R ) is positive semidefinite.To prove the non-trivial direction in Corollary B.2 it therefore suffices to show the following. Lemma B.3. If φ (Ω n ) ⊆ R ≥ , then α t φ ( H ) α ≥ for all α ∈ R m . Proof. Denote h ij = (d p i ) j ∈ T n . If α = ( α i ) i ∈ R m , then α t Hα = X i ,i α i α i (d p i , d p i )= X i ,i α i α i tr X j h i j h t i j = X j tr X i ,i α i h i j α i h t i j = X j tr X i α i h ij ! X i α i h ij ! t ! . Therefore φ (tr( hh t )) ≥ h ∈ T n implies that φ ( H ) is positive semidefinite. (cid:3) From the proof of Lemma B.3 we see that to prove Theorem 4.8 using the Procesi-Schwarztheorem, one would need to extend the chain of equivalences in Theorem B.1 with the condition(iii’) α t φ ( H ) α ≥ α ∈ R m .However, (iii’) ⇒ (iii) fails in our context. Example B.4. Let O ( R ) act on M ( R ) by conjugation, i.e., n = 2 and g = 1 in the setting ofthis paper. If Ξ = Ξ (2) is a generic 2 × y = tr(Ξ) , y = tr(Ξ ) , y = tr(ΞΞ t ) , then R [ ξ ] O ( R ) = R [ y , y , y ] (see e.g. [ADS06]; algebraic independence follows from the Jaco-bian criterion). For this choice of generators we have H = y y y y y y y y . Let R be the real closure of the rational function field R ( ε ) endowed with the ordering 0 < ε < α for every α ∈ R > . Consider the R -algebra homomorphism φ : R [ y , y , y ] → R, y ε , y , y ε (cid:16) p ε − ε (cid:17) . For α = ( α , α , α ) ∈ R we have(B.2) α t φ ( H ) α α + εα ( α + α ) + ε (1 + √ ε − ε )4 ( α + α )= (cid:18) α + ε ( α + α )2 (cid:19) + ε ( √ ε − ε )4 (cid:18) α − α √ ε − ε (cid:19) − ε α . If α , α ∈ R and α = 0 or α = − α , then (cid:18) α + ε ( α + α )2 (cid:19) > ε α ;and if α = − α ∈ R \ { } , then ε (cid:16)p ε − ε (cid:17) (cid:18) α + α √ ε − ε (cid:19) = ε (cid:16) p ε (cid:17) α > ε α . Therefore α t φ ( H ) α > α ∈ R \ { } . On the other hand, from (B.2) it is clear that wecan choose α ∈ R such that α t φ ( H ) α = − ε < 0, so φ ( H ) is not positive semidefinite. OSITIVE TRACE POLYNOMIALS 31 References [Ami57] S. A. Amitsur: A generalization of Hilbert’s Nullstellensatz , Proc. Amer. Math. Soc. 8 (1957) 649–656.[ADS06] H. Aslaksen, V. Drensky, L. 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Igor Klep, Department of Mathematics, University of Auckland E-mail address : [email protected] ˇSpela ˇSpenko, Departement Wiskunde, Vrije Universiteit Brussel E-mail address : [email protected] Jurij Volˇciˇc, Department of Mathematics, Ben-Gurion University of the Negev E-mail address : [email protected] OSITIVE TRACE POLYNOMIALS 33 NOT FOR PUBLICATION Contents 1. Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.1. Main results and reader’s guide . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4Acknowledgments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52. Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52.1. Polynomial and trace ∗ -identities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62.1.1. A particular trace ∗ -identity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62.2. Generic matrices and the trace ring . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82.2.1. Reynolds operator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92.3. Positive involutions and totally positive elements . . . . . . . . . . . . . . . . . . . . . . . . . . . 93. Counterexample to the 3 × R ′ n . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27A.1.1. First method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27A.1.2. Second method. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27A.2. From R ′ n to R nn