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S S hoe ..],which 2.2.1]), Theorem , [ C ˇ h otfundamental most The . sgvnb Schm¨udgen’s by given is ξ rcso hermitian of traces BCR98 3 1 stvtlest,Ra Null- Real ositivstellensatz, ξ , . . . , , g let ] PD01 K , S . Mar08 denote ussels i- y , 2 I. KLEP, S.ˇ SPENKO,ˇ AND J. VOLCIˇ Cˇ adapting the notion of a quadratic module and a preordering to Mn(R[ξ]), many of the results described 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 increasing synergy. 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 sums of hermitian squares [Hel02]. This principal result was followed by various Positivstellens¨atze in a free algebra [HM04, HKM12, KVV17], often with cleaner statements or stronger conclu- sions than their commutative counterparts. These dimension-free techniques are also applied to positivity in operator algebras [NT10, Oza16] and free probability [GS14]. Trace positivity of free polynomials presents the algebraic aspect of the renowned Connes’ embedding conjec- ture [KS08, Oza13]. In addition to convex optimization [BPT13], free positivity certificates frequently appear in quantum information theory [NC10] and control theory [BEFB94]. On the other hand, the dimension-dependent branch is less developed. Here the main tools come from 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 17th problem, 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 weights arise 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 and symmetric ∗ 1 g commuting variables Tr(w) for words w in xj,xj∗ satisfying Tr(w1w2) = Tr(w2w1) 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 (R)g is defined by ∈ ∈ n x X , x Xt and Tr(w) tr(w(X)). j 7→ j j∗ 7→ j 7→ Example. Consider f = 5 Tr(x x ) 2 Tr(x )(x + x ) T. We claim that f is positive 1 1∗ − 1 1 1∗ ∈ (semidefinite) on M (R). For X M (R) write 2 ∈ 2 t t t t t t H = X X ,H = XX X X,H = X2 2XX + 2X X (X )2. 1 − 2 − 3 − − If H1 is invertible, then one can check (see Example 6.5 for details) that
5 t 1 1 t t 1 1 t t f(X)= H H + H− H H H− + H− H H H− 2 1 1 2 1 2 2 1 2 1 3 3 1 and hence f(X) 0, so f 0 on M (R) by continuity. On the other hand, f is not positive on 2 M3(R): 2 0 0 4 0 0 2 0 0 4 0 0 2 0 0 − f 0 1 0 = 5Tr 0 1 0 I3 2 Tr 0 1 0 0 1 0 = 0 14 0 0. 0 0 1! 0 0 1! − 0 0 1! 0 0 1! 0 0 14! 6 In the rest of the paper we will develop a systematic theory for positivity of trace polynomials.
For S Sym T let ⊂ K = X M (R)g : s(X) 0 s S . S { ∈ n ∀ ∈ } If S is finite, then K is the semialgebraic set described by S. A set Q Sym T is a cyclic S ⊂ quadratic module if
1 Q, Q + Q Q, hQh∗ Q h T, Tr(Q) Q. ∈ ⊆ ⊆ ∀ ∈ ⊂ A cyclic quadratic module T is a cyclic preordering if T T is closed under multiplication. ∩ POSITIVE TRACE POLYNOMIALS 3
Proposition. If T T is a cyclic preordering, then f 0 for every f T. ⊂ |KT ∈ The converse of this simple proposition fails in general, but the next noncommutative version of the Krivine-Stengle Positivstellensatz uses cyclic preorderings to describe noncommutative polynomials positive semidefinite on a semialgebraic set KS.
Theorem B’. Let S f Sym T be finite and let T be the smallest cyclic preordering ∪ { } ⊂ containing S. Then f 0 if and only if |KS 2k (t f) g = (f + t ) g and (ft ) g = (t f) g 1 |Mn(R) 2 |Mn(R) 1 |Mn(R) 1 |Mn(R) for some k N and t ,t T. ∈ 1 2 ∈ See Theorem B below for an extended version in a slightly different language. The existence of trace identities on n n matrices suggests that the problem of positivity on n n matrices × × should be treated in an appropriate quotient of T, which we describe next. Let T be the trace ring of generic n n matrices, i.e., the R-algebra generated by n × generic matrices Ξ1,..., Ξg, their transposes and traces of their products. Here Ξj = (ξjı)ı is an n n matrix whose entries are independent commuting variables. The R-subalgebra × t GM T generated by Ξ , Ξ is called the ring of generic n n matrices and has a central n ⊂ n j j × role in the theory of polynomial identities [Pro76, Row80]. The ring of central quotients of GMn is the universal central simple algebra with orthogonal involution of degree n, denoted USAn. The ring Tn also has a geometric interpretation. Let the orthogonal group g On(R) act on Mn(R) by simultaneous conjugation. If Mn(R[ξ]) is identified with polynomial maps M (R)g M (R), then T is the ring of polynomial O (R)-concomitants in M (R[ξ]), n → n n n n i.e., equivariant maps with respect to the On(R)-action [Pro76]. If Cn, Tn,Zn are the centers of GM , T , USA , respectively, and is the “averaging” Reynolds operator for the O (R)-action, n n n R n then we have the following diagram.
R Cn
R
USAn Mn(R(ξ))
The elements of Tn are called trace polynomials and the elements of Tn are called pure trace polynomials. Since every evaluation of T at a tuple of n n matrices factors through × Tn, it suffices to prove our Positivstellens¨atze in the ring Tn. The purpose of this reduction is of course not to merely state Theorem B’ in a more compact form. Our proofs crucially rely on algebraic properties of Tn and their interaction with invariant and PI theory [Pro76, Row80]. The contribution of this paper is twofold. We prove the Krivine-Stengle, Schm¨udgen and Putinar Positivstellens¨atze for the trace ring of generic matrices in terms of cyclic quadratic modules and preorderings. We also prove Putinar’s Positivstellensatz for the ring of generic matrices (without traces). The proofs intertwine techniques from invariant theory, real algebraic 4 I. KLEP, S.ˇ SPENKO,ˇ AND J. VOLCIˇ Cˇ 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 about polynomial identities, positive involutions, and the rings GMn and Tn in Section 2, where we also prove some preliminary results that are used in the sequel. Section 3 deals with the question of Procesi and Schacher [PS76], which is a noncommutative version of Hilbert’s 17th problem for central simple -algebras. We say that a USA is totally ∗ ∈ n positive if a(X) 0 for every X M (R)g where a is defined. Then the universal Procesi- ∈ n Schacher conjecture states that totally positive elements in USAn are sums of hermitian squares in USAn. 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 USA3 that are not sums of hermitian squares in USA3.
The proof (see Theorem 3.2) relies on the central simple -algebra USA being split, i.e., ∗ 3 -isomorphic to M (Z ) with some orthogonal involution. After explicitly determining the in- ∗ 3 3 volution using quadratic forms (Proposition 3.4 of Tignol) and a transcendental basis of Z3 (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 USA3 that is not a sum of hermitian squares (Proposition 3.6). Section 4 first introduces cyclic quadratic modules and cyclic preorderings for the trace ring Tn, which are defined analogously as for T above. The main result in this section is the following version of the Krivine-Stengle Positivstellensatz for Tn.
Theorem B. Let S a Sym T be finite and T the cyclic preordering generated by S. ∪ { }⊂ n (1) a 0 if and only if at = t a = a2k + t for some t ,t T and k N. |KS 1 1 2 1 2 ∈ ∈ (2) a 0 if and only if at = t a =1+ t for some t ,t T. |KS ≻ 1 1 2 1 2 ∈ (3) a = 0 if and only if a2k T for some k N. |KS − ∈ ∈ See Theorem 4.13 for the proof, which decomposes into three parts. First we show that the finite set of constraints S Sym T can be replaced by a finite set S T (Corollary 4.4). ⊂ n ′ ⊂ n To prove this central reduction we use the fact that the positive semidefiniteness of a matrix can be characterized by symmetric polynomials in its eigenvalues and apply compactness of the real spectrum in the constructible topology [BCR98, Section 7.1]. In the second step we apply results on central simple algebras with involution and techniques from PI theory to prove the following extension theorem.
Theorem C. Let R R be a real closed field. Then an R-algebra homomorphism φ : Tn R ⊇ t → extends to an R-algebra homomorphism R[ξ] R if and only if φ(tr(hh )) 0 for all h T . → ≥ ∈ n On(R) Since Tn = R[ξ] , 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 Theorem C from these classical results; see Appendix B for a fuller discussion. Theorem C, proved as Theorem 4.8 below, is essential for relating evaluations of pure trace polynomials with orderings on Tn via Tarski’s transfer principle (Proposition 4.12). Finally, by combining the first two steps we obtain a reduction to the commutative situation, where we can apply an existing abstract version of the Krivine-Stengle Positivstellensatz [Mar08, Theorem 2.5.2]. Since trace polynomials are precisely On(R)-concomitants in Mn(R[ξ]), one might naively attempt to prove Theorem B by simply applying the Reynolds operator for the On(R)-action to analogous Positivstellens¨atze for matrix polynomials [Scm09, Cim12]. However, the Reynolds POSITIVE TRACE POLYNOMIALS 5 operator is not multiplicative and it does not preserve squares of trace polynomials, so in this manner 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, which encompass an algebraic notion of boundedness. Following the standard definition we say that a cyclic quadratic module Q T is archimedean if for every h T there exists ρ Q such ⊆ n ∈ n ∈ >0 that ρ hht Q. Then we prove Schm¨udgen’s Positivstellensatz for trace polynomials. − ∈ Theorem D. Let S a Sym T be finite and T be the cyclic preordering generated by S. ∪ { } ⊂ n If K is compact and a 0, then a T. S |KS ≻ ∈ 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 the commutative setting, where we use an abstract version of Schm¨udgen’s Positivstellensatz [Sce03]. Finally, we have the following version of Putinar’s Positivstellensatz for Tn and GMn, com- bining Theorems 5.5 and 5.7.
Theorem E. (a) Let Q Sym T be an archimedean cyclic quadratic module and a Sym T . If a ⊂ n ∈ n |KQ ≻ 0, then a Q. ∈ (b) Let Q Sym GM be an archimedean quadratic module and a Sym GM . If a 0, ⊂ n ∈ n |KQ ≻ then a Q. ∈ Theorem E is proved in a more functional-analytic way. We start by assuming a / Q and ∈ find an extreme separation of a and Q. Then we apply a Gelfand-Naimark-Segal construction towards finding a tuple of n n matrices in KQ at which a is not positive definite. For GM this × n is done using polynomial identities techniques, while for Tn we use Theorem C. As a consequence we have the following statement for noncommutative polynomials.
Corollary E’. Let Q Sym R
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 acknowledge fruitful Oberwolfach discussions with Cordian Riener and Markus Schweighofer.
2. Preliminaries In this section we collect some background material and preliminary results needed in the sequel. 6 I. KLEP, S.ˇ SPENKO,ˇ AND J. VOLCIˇ Cˇ
2.1. Polynomial and trace -identities. Throughout the paper let F be a field of character- ∗ istic 0. Let x = x ,...,x and x = x ,...,x be freely noncommuting variables, and let { 1 g} ∗ { 1∗ g∗}
m 2 λi = tr (A A )i j 1 2 j=1 X for i N. Now Newton’s identities imply ∈ k 1 i 1 k kck = ( 1) − tr (A1A2) ck i 2 − − Xi=1 for 1 k m and c = 1. ≤ ≤ 0 POSITIVE TRACE POLYNOMIALS 7
Now define fm T as ∈ m k m k f = ( 1) f ′ (x x ) − m − k · 1 2 Xk=0 with f0′ = 1 and k 1 i 1 k (2.1) fk′ = ( 1) − tr (x1x2) fk′ i 2k − − Xi=1 for 1 k m. The following lemmas will be important for distinguishing between different ≤ ≤ types of involutions of the first kind in the sequel.
Lemma 2.1. For every m N, f (x x ,x x ) is a -trace identity of (M (F ), t, tr). ∈ m 1 − 1∗ 2 − 2∗ ∗ 2m Proof. Observe that the set of pairs of t-antisymmetric A , A M (F ), such that A A has m 1 2 ∈ 2m 1 2 distinct eigenvalues, is Zariski dense in the set of all pairs of t-antisymmetric A , A M (F ). 1 2 ∈ 2m Hence the conclusion follows by the construction of fm.
Lemma 2.2. For every n,m N and d N 2N there exist s-antisymmetric A , A M (F ) ∈ ∈ \ 1 2 ∈ 2n such that d d f A⊕ A⊕ = 0. m 1 2 6 Proof. Every t-symmetric matrix S M (F ) can be written as S = ( SJ)J, where J = 0 1 ∈ 2n − 1 0 and SJ, J are s-antisymmetric matrices. Hence it suffices to prove that f (S d) = 0 holds− for − m ⊕ 6 2n 1 − S = diag(1,..., 1, 0) M (F ). ∈ 2n Since tr((S d)k)= d(2n 1) is odd, we canz }| use{ (2.1) and induction on k to show that ⊕ − d ℓ ℓ k!f ′ (S⊕ ) : ℓ Z : ℓ Z k ∈ 2k ∈ \ 2k 1 ∈ − for 1 k m. In particular we have f (S d) = 0 and thus f (S d) = 0. ≤ ≤ m′ ⊕ 6 m ⊕ 6 For n N and m 2N let J (n, t) denote the set of polynomial -identities of (M (F ), t) ∈ ∈ ∗ n and let J (m, s) denote the set of polynomial -identities of (M (F ), s). By [Row80, Corollary ∗ m 2.5.12 and Remark 2.5.13] we have J (m, s) J (n, t) if and only if 2n m. ⊆ ≤ Proposition 2.3. Let n N and m 2N. Then J (n, t) J (m, s) if and only if 2m n. ∈ ∈ ⊆ ≤ Proof. ( ) Let ⇒ cm = sgn πxπ(1)xm+1xπ(2)xm+2 x2m 1xπ(m) · · · − π Sym ∈X m be the mth Capelli polynomial [Row80, Section 1.2]. If is a central simple F -algebra and A a , . . . , a , then a , . . . , a is linearly dependent over F if and only if 1 m ∈A { 1 m} cg(a1, . . . , ag, b1, . . . , bm 1) = 0 bi − ∀ ∈A by [Row80, Theorem 1.4.34]. Now assume n< 2m. If A , A M (F ) are t-antisymmetric, then the set 1 2 ∈ n n/2 +1 A1A2,..., (A1A2)⌊ ⌋ is linearly dependent. Indeed, forn an even n this holds directlyo by Lemma 2.1, while for an odd n we use the fact that A1A2 is singular and then apply Lemma 2.1 for n+1. On the other hand, since every t-symmetric matrix in Mm(F ) is a product of two s-antisymmetric matrices, there 8 I. KLEP, S.ˇ SPENKO,ˇ AND J. VOLCIˇ Cˇ exist s-antisymmetric A , A M (F ) such that 1,..., (A A )m 1 is linearly independent. 1 2 ∈ m { 1 2 − } Since n + 1 m, ⌊ 2 ⌋ ≤ m c (x x∗)(x x∗),..., ((x x∗)(x x∗)) ,x ,...,x m 1 − 1 2 − 2 1 − 1 2 − 2 3 m+1 is a -identity of M (F ) endowed with t but is not a -identity of M (F ) endowed with s. ∗ n ∗ m ( ) By [Row80, Corollary 2.3.32] we can assume that F is algebraically closed; let i F be ⇐ ∈ such that i2 = 1. Since m 2N, (M (F ), s) -embeds into (M (F ), t) via − ∈ m ∗ 2m a + d i(a d) c b i(b + c) − − a b 1 i(d a) a + d i(b + c) b c . M (F ), s) ֒ (M (F ), t), − − ) m → 2m c d 7→ 2 b c i(b + c) a + d i(a d) − − − i(b + c) c b i(d a) a + d − − −
Remark 2.4. The same reasoning as in the proof of Proposition 2.3 also implies that elements of J (n, t) are polynomial *-identities of Mm(F ) with an involution of the second kind if and only if 2m n. Recall that an involution on M (F ) is of the second kind if it induces an ≤ m automorphism of order two on F . 2.2. Generic matrices and the trace ring. For g,n N let ∈ ξ = ξ : 1 j g, 1 ı, n { jı ≤ ≤ ≤ ≤ } be a set of commuting indeterminates. We recall the terminology from Section 1. Let Ξ = (ξ ) M (R[ξ]) j jı ı ∈ n be n n generic matrices and let GMn Mn(R[ξ]) be the R-algebra generated by Ξj and × t ⊂ their transposes Ξ . Furthermore, let T M (R[ξ]) be the R-algebra generated by GM and j n ⊂ n n traces of elements in GM . This algebra is called the trace ring of n n generic matrices (see n × e.g. [Pro76, Section 7]) and inherits the transpose involution t and trace tr from Mn(R[ξ]). Let C T R[ξ] be the centers of GM and T , respectively. The elements of T are called n ⊂ n ⊂ n n n trace polynomials and the elements of Tn are called pure trace polynomials. There is another, more invariant-theoretic description of the trace ring. Define the following g action of the orthogonal group On(R) on Mn(R) : t t (2.2) (X ,...,X )u := (uX u , . . . , uX u ), X M (R), u O (R) 1 g 1 g j ∈ n ∈ n g and consider R[ξ] as the coordinate ring of Mn(R) . By [Pro76, Theorems 7.1 and 7.2], Tn is the ring of On(R)-invariants in R[ξ] and Tn is the ring of On(R)-concomitants in Mn(R[ξ]), i.e., elements f Mn(R[ξ]) satisfying ∈ t f(Xu)= uf(X)u for all X M (R)g and u O (R). ∈ n ∈ n We list a few important properties of GMn and Tn that will be used frequently in the sequel. t t (a) Let J (n, , tr) T denote the set of trace -identities of (Mn(R), , tr). Then GMn ∼= ⊂ t ∗ t R
2.2.1. Reynolds operator. This subsection is to recall some basic properties of the Reynolds operator [DK02, Subsection 2.2.1]. Let G be an algebraic group and X an affine G-variety. The Reynolds operator : F [X] F [X]G is a linear map with the properties: R → (1) (f)= f for f F [X]G, R ∈ (2) is a G-module homomorphism; i.e., (f u)= (f) for f F [X], u G. R R R ∈ ∈ The Reynolds operator is hence a G-invariant projection onto the space of the invariants. The Reynolds operator exists if G is linearly reductive and is then unique (see e.g. [DK02, Theorem 2.2.5]). 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 , , RM RN resp., and f G a G-module homomorphism f restricted to M G. Then f = f G . This easily RN RM follows by the uniqueness of the Reynolds operator. The Reynolds operator is thus functorial. g In our case On(R) acts on Mn(R) by simultaneous conjugation as in (2.2). Since Mn(R[ξ]) can be identified with polynomial maps M (R)g M (R), we have the Reynolds operator n → n : M (R[ξ]) T with respect to the action (2.2). Since O (R) is a compact Lie group, Rn n → n n can be given by the averaging integral formula (with respect to the normalized left Haar Rn measure µ on On(R))
(2.3) (f)= f u dµ(u). Rn ZOn(R) Consequently is a trace-intertwining T -module homomorphism, i.e., Rn n (2.4) (hf)= h (f), (fh)= (f)h, tr( (f)) = (tr(f)) Rn Rn Rn Rn Rn Rn for all h T and f M (R[ξ]). In Appendix A we present algebraic ways of computing . ∈ n ∈ n Rn 2.3. Positive involutions and totally positive elements. Let be a central simple algebra A with involution τ and -center F (that is, F is the subfield of -invariant elements in the center of ∗ ∗ ). The F -space of τ-symmetric elements in is denoted Sym . Following the terminology of A A A [PS76] and [KU10], an ordering of F is a -ordering if tr (aaτ ) 0 for every a . In this ≥ ∗ A ≥ ∈A case we also say that τ is positive with respect to such an ordering. An element a Sym is ∈ A positive in a given -ordering if the hermitian form x tr(xτ ax) on is positive semidefinite. ∗ 7→ A Finally, a Sym is totally positive if it is positive with respect to every -ordering. ∈ A ∗ Let α ,...,α F be the elements appearing in a diagonalization of the form x tr(xxτ ) 1 n ∈ 7→ on . By [PS76, Theorem 5.4], a symmetric s is totally positive if and only if it has a A ∈ A weighted sum of hermitian squares representation
I τ (2.5) s = α hI,ihI,i, I 0,1 n i ∈{X} X I I1 In where α = α1 αn and hI,i . · · · ∈A t Let Ω T be the preordering generated by tr(hh ) for h T , i.e., the set of all sums n ⊂ n ∈ n of products of tr(hht) (note that c2 = tr(( c )2) Ω for every c T , so Ω is really a √n ∈ n ∈ n n preordering). Further, let
t Ωn = ωihihi : ωi Ωn, hi Tn . ( ∈ ∈ ) Xi Note that Ωn = tr(Ωn).
Lemma 2.5. Let f SymM (R[ξ]). If f(X) 0 for all X M (R)g, then (f)= c 2q for ∈ n ∈ n Rn − some q Ω and c T 0 . ∈ n ∈ n \ { } 10 I. KLEP, S.ˇ SPENKO,ˇ AND J. VOLCIˇ Cˇ
Proof. By the integral formula (2.3) it is clear that (f)(X) 0 for all X M (R)g. Hence Rn ∈ n (f) is a totally positive element in USA by [KU10, Lemma 5.3], so Rn n I t n(f)= α hI,ihI,i R 2 I 0,1 n i ∈{X} X t for some h USA and a diagonalization α ,...,α 2 over Z of the form x tr(xx ) on I,i ∈ n h 1 n i n 7→ USA . Hence α = tr(h˜ h˜t ) for some h˜ USA . Since USA is the ring of central quotients n k k k k ∈ n n of T , there exist q Ω and c T such that (f)= c 2q. n ∈ n ∈ n Rn − As demonstrated in Example 6.2, the denominator in Lemma 2.5 is in general indispensable even if f is a hermitian square or f R[ξ]. For more information about images of squares under ∈ Reynolds operators for reductive groups acting on real affine varieties see [CKS09].
Remark 2.6. In particular, the linear operator does not map squares in R[ξ] into Ω or Rn n hermitian squares in Mn(R[ξ]) into Ωn. Hence our Positivstellens¨atze in the sequel cannot simply be deduced from their commutative or matrix counterparts by averaging with . Furthermore, Rn even if one were content with using totally positive polynomials (which by Lemma 2.5 are of the form c 2q for q Ω and c T 0 ) instead of Ω , one could still not derive our results − ∈ n ∈ n \ { } n since is not multiplicative. Rn 3. Counterexample to the 3 3 universal Procesi-Schacher conjecture × By [PS76, Corollary 5.5] every totally positive element in USA2 is a sum of hermitian squares, i.e., of the form c 2 h ht for c C and h GM . Indeed, by (2.5) it suffices to show that − i i i ∈ 2 i ∈ 2 tr(aat) is a sum of hermitian squares. Since USA is a division ring, we have P 2 t t 1 1 t tr(aa )= a a + (det(a)a− )(det(a)a− ) for every a USA 0 by the Cayley-Hamilton theorem. In their 1976 paper [PS76], Procesi ∈ 2 \{ } and Schacher asked if the same holds true for n> 2:
Conjecture 3.1 (The universal Procesi-Schacher conjecture). Let n 2. Then every totally ≥ positive element in USAn is a sum of hermitian squares. By (2.5), Conjecture 3.1 is equivalent to the following: every trace of a hermitian square in USAn is a sum of hermitian squares in USAn. In this section we show that Conjecture 3.1 fails for n = 3:
Theorem 3.2. There exist totally positive elements in USA3 that are not sums of hermitian squares in USA3.
In the first step of the proof we identify the split central simple algebra USA3 as a matrix algebra M3(F ) for a rational function field F , endowed with an involution of the orthogonal type. 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 V ′ ∈ F such that q = q θ. Quadratic forms q and q are similar if αq and q are equivalent for some ′ ◦ ′ ′ α F 0 . Every quadratic form is equivalent to a diagonal quadratic form, which is denoted ∈ \ { } α ,...,α for α F . h 1 ni i ∈ First we fix g = 1 and write Ξ = Ξ1. Since USA3 is an odd degree central simple algebra with involution of the first kind, USA is split by [KMRT98, Corollary 2.8]. Let us fix a - 3 ∗ representation USA3 = EndZ3 V , where V is a 3-dimensional vector space over Z3, the center of POSITIVE TRACE POLYNOMIALS 11
USA . By [KMRT98, Proposition 2.1], there exists a symmetric bilinear form b : V V Z 3 × → 3 such that t b(xu, v)= b(u, x v) for all u, v V and x End V , where t denotes the involution on End V originating from ∈ ∈ Z3 Z3 USA . Let q : V Z given by q(u)= b(u, u) be the associated quadratic form. 3 → 3 Lemma 3.3. Let a End V be t-antisymmetric with tr(a2) = 0. Define e = 1 2 tr(a2) 1a2. ∈ Z3 6 − − 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 1 tr(a2). − 2 Proof. Since a and at = a have the same trace and determinant, we have tr(a) = det(a) = 0, − so by the Cayley-Hamilton theorem it follows that 1 (3.1) a3 tr(a2)a = 0. − 2 4 1 2 2 Hence a = 2 tr(a )a and it is straightforward to check that e is a symmetric idempotent. It has 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 antisymmetric operator with determinant 1 tr(a2), and the determinant of the restriction of q to ker e is − 2 the square class of the determinant of any nonzero antisymmetric operator; see [KMRT98, Proposition 7.3].
Proposition 3.4. For i = 1, 2 let a End V be t-antisymmetric with tr(a2) = 0, and let i ∈ Z3 i 6 e = 1 2 tr(a2) 1a2. If e e = e e = 0, then q is similar to 1, 1 tr(a2), 1 tr(a2) . i − i − i 1 2 2 1 h − 2 1 − 2 2 i Proof. Let e = 1 e e and V = im e for i = 1, 2, 3; we have dim V = 1 for each i. Moreover, 3 − 1 − 2 i i i if u V and v V for i = j, then ∈ i ∈ j 6 b(u, v)= b(eiu, ejv)= b(u, eiejv) = 0. Therefore V = V V V . We have ker e = V V and ker e = V V , and Lemma 1 ⊥ 2 ⊥ 3 1 2 ⊥ 3 2 1 ⊥ 3 3.3 shows that the determinant of the restriction of q to ker e is 1 tr(a2) for i = 1, 2. If i − 2 i α Z 0 is such that the restriction of q to V is equivalent to α , then it follows that ∈ 3 \ { } 3 h i the restriction of q to V is equivalent to α tr(a2) for i = 1, 2. Hence q is equivalent to i h− 2 i i α, α tr(a2), α tr(a2) . h − 2 1 − 2 2 i Now let t t a =Ξ Ξ , a = e Ξ(1 e ) (1 e )Ξ e , 1 − 2 1 − 1 − − 1 1 2 1 2 2 1 2 e = 1 2 tr(a )− a , e = 1 2 tr(a )− a , 1 − 1 1 2 − 2 2 1 1 β = tr(a2), β = tr(a2). 1 −2 1 2 −2 2 Since a , a are nonzero, it follows from a3 1 tr(a2)a = 0 that β = 0 for i = 1, 2. It is also 1 2 i − 2 i i i 6 easy to check that e1e2 = e2e1 = 0, so the conclusion of Proposition 3.4 holds. Hence we can choose a basis of V in such a way that
t 1 τ (3.2) x = diag(1, β1, β2)− x diag(1, β1, β2) for all x End V , where τ is the transpose in M (Z ) = End V with respect to the chosen ∈ Z3 3 3 Z3 basis of V . 12 I. KLEP, S.ˇ SPENKO,ˇ AND J. VOLCIˇ Cˇ
The field Z3 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 basis for Z3 over R. Denote 1 t 1 t 1 s = (Ξ+Ξ ), a = (Ξ Ξ ), s = s tr(s) 2 2 − 0 − 3 and tr(a2)2 tr(s2) 6 tr(s a2)2 α = tr(s), α = 0 − 0 , 1 4 tr(a2)2 tr(s2) 4 tr(a2) tr(s2a2) 2 tr(s a2)2 0 − 0 − 0 tr(a2)3 tr(s3) + 6 tr(s a2)3 (3.3) α = tr(a2), α = 0 0 , 2 5 tr(a2)2 tr(s2) 6 tr(s a2)2 0 − 0 tr(as a2s2) α = tr(s a2), α = 0 0 . 3 0 6 tr(a2)2 tr(s2) 6 tr(s a2)2 0 − 0 Lemma 3.5. The elements α1,...,α6 are algebraically independent over R, Z3 = R(α1,...,α6), and 3 2 2 2 1 288α2α4α6 (3α3α4 + 2α4α5 + 9α3) (3.4) β1 = α2, β2 = − 2 . −2 9α2(α4 + 1) Proof. Using a computer algebra system one can verify that the determinant of the Jacobian matrix Jα1,...,α6 is nonzero, so by the Jacobian criterion α1,...,α6 are algebraically independent over R. Likewise, (3.4) is checked by a computer algebra system. A minimal set of generators of pure trace polynomials in two 3 3 generic matrices without × involution is given in [ADS06, Section 1] or in the proof of [LV88, Proposition 7]. Replacing the first generic matrix by s and the second generic matrix by a we obtain the following generators of T3: 2 3 2 2 2 2 2 2 (3.5) tr(s), tr(s0), tr(s0), tr(a ), tr(s0a ), tr(s0a ), tr(as0a s0). 2 From (3.3) we can directly see that (3.5) are rational functions in α1,...,α6, tr(s0) ; for example, 2 2 2 3 3 α5(α2 tr(s0) 6α3) 6α3 tr(s0)= −3 − . α2 Then we use a computer algebra system to verify that 2 2 2α4 6α3 tr(s0)= β2 + 2 α4 + 1 α2 is a rational function in α1,...,α6 and hence Z3 = R(α1,...,α6). Proposition 3.6. β β Z is totally positive in USA but is not a sum of hermitian squares 1 2 ∈ 3 3 in USA3. t Proof. Since β1β2 = tr(hh ) for 0 0 0 h = 0 0 β , 2 0 0 0 β β is totally positive in USA . Now suppose β β = r rt for r USA . If r = (ρ ) , 1 2 3 1 2 i i i i ∈ 3 i iı ı then the (1, 1)-entry of r rt equals i i i P 2 1 2 1 2 P ρi11 + β1− ρi12 + β2− ρi13 i X and therefore 2 2 2 ρi11 ρi12 ρi13 (3.6) 1 = β1β2 + β2 + β1 . β1β2 β1β2 β1β2 Xi Xi Xi POSITIVE TRACE POLYNOMIALS 13
By [PD01, Exercise 5.5.3 and Lemma 5.1.8] there exists a semiordering Q R(α ,...,α ) ⊂ 1 6 satisfying α , α , α α Q and p Q R[α ,...,α ] if and only if the term of the highest 2 − 4 − 2 4 ∈ ∈ ∩ 1 6 degree in p belongs to Q. These assumptions on Q yield β , β , β β Q, so (3.6) implies − 1 − 2 − 1 2 ∈ 1 Q, a contradiction. − ∈ Proof of Theorem 3.2. To be more precise we write USA3,g for USA3 generated by g generic matrices Ξ . Proposition 3.6 proves Theorem 3.2 for g = 1. Now let g N be arbitrary; note j ∈ that USA naturally -embeds into USA . Let s USA be a totally positive element that 3,1 ∗ 3,g ∈ 3,1 is not a sum of hermitian squares in USA3,1. Suppose that s is a sum of hermitian squares in USA3,g, i.e., 2 t s = c− hihi Xj for some c, h GM with c central. Since the sets of polynomial -identities of GM and i ∈ 3,g ∗ 3,1 GM coincide, it is easy to see that there exists a -homomorphism φ : GM GM 3,g ∗ 3,g → 3,1 satisfying φ(Ξ )=Ξ and φ(c) = 0. Then 1 1 6 2 t s = φ(c)− φ(hi)φ(hi) Xj is a sum of hermitian squares in USA3,1, a contradiction.
We do not know if Conjecture 3.1 holds for n = 4, where USA4 is a division biquaternion algebra by [Pro76, Theorem 20.1] and hence does not split. In [AU+] the authors use signatures of hermitian forms to distinguish between sums of hermitian squares and general totally positive elements.
4. The Krivine-Stengle Positivstellensatz for trace polynomials In this section we prove the Krivine-Stengle Positivstellensatz representing trace polynomials positive on semialgebraic sets in terms of weighted sums of hermitian squares with denominators.
4.1. Cyclic quadratic modules and preorderings. For a finite S SymM (R[ξ]) let ⊂ n K = X M (R)g : s(X) 0 s S S { ∈ n ∀ ∈ } be the semialgebraic set described by S. A set Q Sym T is a cyclic quadratic module ⊆ n if t 1 Q, Q + Q Q, hQh Q h T , tr(Q) Q. ∈ ⊆ ⊆ ∀ ∈ n ⊂ A cyclic quadratic module T Sym Tn is a cyclic preordering if T Tn is closed under multi- ⊆Qtr Ttr ∩ plication. For S Sym Tn let S and S denote the cyclic quadratic module and preordering, ⊂ tr tr respectively, generated by S. For example, Q = T = Ωn. ∅ ∅ Lemma 4.1. Let S Sym T . ⊆ n (1) If Q is a cyclic quadratic module, then tr(Q)= Q Tn. Qtr ∩ (2) Elements of S are precisely sums of t t q1, h1s1h1, tr(h2s2h2)q2
for qi Ωn, hi Tn and si S. Ttr ∈Qtr ∈ ∈ (3) S = S′ , where
t S′ = S tr(hisihi): hi Tn,si S . ∪ ( ∈ ∈ ) Yi Proof. Straightforward. 14 I. KLEP, S.ˇ SPENKO,ˇ AND J. VOLCIˇ Cˇ
Our main result of this subsection is a reduction to central generators for cyclic quadratic modules, see Corollary 4.4. It will be used several times in the sequel. In its proof we need the following lemma.
n i Lemma 4.2. Let R be an ordered field, λ , . . . , λ R and p = λ for i N. If λ 0 < 0 1 n ∈ i j=1 j ∈ j for some 1 j n, then there exists f Q[p ,...,p ][ζ] such that ≤ 0 ≤ ∈ 1 n P n 2 (4.1) f(λj) λj < 0. Xj=1 n 1 i Proof. Denote E = Q(p ,...,p ) and F = Q(λ , . . . , λ ). For every f = − α ζ F [ζ] we 1 n 1 n i=0 i ∈ have P n 2 i+i′+1 i+i′+1 (4.2) f(λj) λj = αiαi′ λj = λj αiαi′ = pi+i′+1αiαi′ . ′ ′ ′ Xj=1 Xj Xi,i Xi,i Xj Xi,i Note that pi E for every i N and define P Mn(E) by Pij = pi+j 1. If λj0 < 0, then there ∈ ∈ ∈ − clearly exists f0 F [ζ] of degree n 1 such that f0(λj0 ) = 0 and f0(λj)=0 for λj = λj0 . Then ∈ − 6 6 t f0 satisfies (4.1), so P is not positive semidefinite as a matrix over F by (4.2). Since P = QDQ for some Q GLn(E) and diagonal D Mn(E), we conclude that P is not positive semidefinite ∈ ∈ t n t as a matrix over E, so there exists v = (β0,...,βn 1) E such that v Pv < 0. By (4.2), n 1 i − ∈ f = − β ζ E[ζ] satisfies (4.1). After clearing the denominators of the coefficients of f 1 i=0 i ∈ 1 we obtain f Q[p ,...,p ][ζ] satisfying (4.1). P ∈ 1 n The proof of the next proposition requires some well-known notions and facts from real algebra that we recall now. Let Λ be a commutative unital ring. Then P Λ is a ordering if P is ⊂ closed under addition and multiplication, P P = Λ and P P is a prime ideal in Λ. Note ∪− ∩− that every ordering in Λ gives rise to a ring homomorphism from Λ into a real closed field and vice 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 of { ∈ ∈ } \ ∈ the constructible topology [BCR98, Section 7.1] (also called patch topology [Mar08, Section 2.4]). By [BCR98, Proposition 1.1.12] or [Mar08, Theorem 2.4.1], Sper Λ endowed with this topology 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 let T be the ring of polynomials in s and tr(si) ∈ n O ⊂ n for i N with rational coefficients, and set ∈ Qtr S = tr(hsh): h tr s . { ∈O}⊂ { } Then there exists a finite subset S0 S such that K s = KS0 . ⊂ { } Proof. First we prove that for every real closed field R we have (4.3) X M (R)g : s(X) 0 = X M (R)g : c(X) 0 . { ∈ n } { ∈ n ≥ } c S \∈ The inclusion is obvious. Let X M (R)g be such that s(X) is not positive semidefinite. ⊆ ∈ n Since R is real closed and pure trace polynomials are On(R)-invariant, we can assume that i s(X) = diag(λ1, . . . , λn) is diagonal and λj < 0 for some j. If pi = tr(s(X) ), then by Lemma 4.2 there exists a polynomial f Q[p ,...,p ][ζ] such that ∈ 1 n n 2 f(λi) λi < 0. Xi=1 If h is such that h(X)= f(s(X)), then tr(h(X)s(X)h(X)) < 0. Hence in (4.3) holds. ∈ O ⊇ POSITIVE TRACE POLYNOMIALS 15
Let σ = tr( js) T for 1 j n, where js denotes the jth exterior power of s; hence σ j ∧ ∈ n ≤ ≤ ∧ j are signed coefficients of the characteristic polynomial for s and X M (R)g : s(X) 0 = X M (R)g : σ (X) 0,...,σ (X) 0 { ∈ n } { ∈ n 1 ≥ n ≥ } 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 n (4.4) K(σj)= K(c) j=1 c S \ \∈ by the correspondence between homomorphisms from R[ξ] to real closed fields and orderings in R[ξ]. Since the complement of the left-hand side of (4.4) is compact in the constructible topology, there exists a finite subset S S such that 0 ⊂ n K(σj)= K(c) j=1 c S0 \ \∈ and consequently K s = K σ1,...,σn = KS0 . { } { } Corollary 4.4. For every finite set S Sym T there exists a finite set S tr(Qtr) such that ⊂ n ′ ⊂ S KS = KS′ . Proof. Let S = s ,...,s . By Proposition 4.3 there exist finite sets S Qtr T with 1 ℓ i si n { } ⊂ { } ∩ KSi = K si . If S′ = S1 Sℓ, then { } ∪···∪ Qtr Qtr S′ si Tn S Tn ⊂ { } ∩ ⊂ ∩ [i and
′ KS = KSi = K si = KS. { } \i \i Corollary 4.5. For every cyclic quadratic module Q Sym T we have KQ = K Q . ⊆ n tr( ) Proof. Direct consequence of Corollary 4.4. 4.2. An extension theorem. The main result in this subsection, Theorem 4.8, characterizes homomorphisms from pure trace polynomials Tn to a real closed field R which arise via point evaluations ξ α R. jı 7→ jı ∈ We start with some additional terminology. Let F be a field and let be a finite-dimensional A simple F -algebra with center C. If tr is the reduced trace of as a central simple algebra and A A trC/F is the trace of the field extension C/F , then F tr = trC/F tr : F A ◦ A A→ is called the reduced F -trace of [DPRR05, Section 4]. A Proposition 4.6. Let R R be a real closed field and a finite-dimensional semisimple ⊇ A 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 such that Φ(T ) R and ∗ n →A n ⊆ is generated by Φ(T ) over R. Then there exists a trace preserving -embedding of into A n ∗ A (Mn(R), t, tr).
Proof. Let ∼= 1 ℓ be the decomposition in simple factors and let nk be the degree A A ×···×A 1, 1 of for 1 k ℓ. Moreover, let C = R[i] be the algebraic closure of R and H = ( − − ) the Ak ≤ ≤ R division quaternion algebra over R. By [PS76, Theorem 1.2], each of is -isomorphic to one Ak ∗ of the following: 16 I. KLEP, S.ˇ SPENKO,ˇ AND J. VOLCIˇ Cˇ
(I) Mnk (R) with the transpose involution;
(II) Mnk (C) with the conjugate-transpose involution;
(III) Mnk/2(H) with the symplectic involution. Without loss of generality assume that there are 1 ℓ ℓ ℓ such that is of type (I) for ≤ 1 ≤ 2 ≤ Ank k ℓ , of type (II) for ℓ < k ℓ , and of type (III) for ℓ < k. By [DPRR05, Theorem 4.2] ≤ 1 1 ≤ 2 2 there exist d ,...,d N such that 1 ℓ ∈ R (4.5) χ ak = dk tr k (ak) ! A Xk Xk for a . We claim that d 2N for every k>ℓ . Let n = n and fix k>ℓ . By Lemma k ∈Ak k ∈ 2 ′ ⌈ 2 ⌉ 2 2.1, f = fn′ (x1 x1∗,x2 x2∗) is a -trace identity for Tn. Therefore f is also a -trace identity − − ∗ R∗ for by the assumptions on Φ. Hence f is a -trace identity for ( k,τk, dk tr ), where τk is A ∗ A · k the restriction of the involution on . Since -trace identities are preserved by scalarA extensions A ∗ and R s s C R k,τk, dk tr = C R Mnk/2(H), , dk (trH tr) = (Mnk (C), , dk tr) , ⊗ A · Ak ∼ ⊗ · ◦ ∼ · f is a -trace identity for (M (C), s, d tr). Now Lemma 2.2 implies d 2N. ∗ nk k · k ∈ Thus we have d (4.6) n = d n + 2d n + 4 k n k k k k 2 k k ℓ1 ℓ1 Lemma 4.7. Let Λ be a Noetherian domain with char Λ = 2, M a finitely generated Λ-module, 6 K a field, φ : Λ K a ring homomorphism, and b : M M Λ a symmetric Λ-bilinear form. → × → Let π : M K M be the natural homomorphism. Then there exist u , . . . , u M such that → ⊗φ 1 ℓ ∈ π(u ),...,π(u ) is a K-basis of K M and φ(b(u , u ′ )) = 0 for i = i . { 1 ℓ } ⊗φ i i 6 ′ Proof. Let ℓ = dim (K M). We prove the statement by induction on ℓ. The case ℓ = 1 is K ⊗φ trivial. Now assume that statement holds for ℓ 1 and suppose K M is of dimension ℓ. − ⊗φ If φ b = 0, we are done. Otherwise there exists u M ker π with b(u , u ) / ker φ. Indeed, ◦ 1 ∈ \ 1 1 ∈ if φ(b(u, u)) = 0 for all u M ker π, then φ(b(u, u)) = 0 for all u M, so by ∈ \ ∈ 2b(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 M such that ∈ 2 ℓ ∈ π(u ),π(v ) ...,π(v ) is a K-basis of K M. For 2 i ℓ let { 1 2 ℓ } ⊗φ ≤ ≤ v′ = b(u , u )v b(u , v )u i 1 1 i − 1 i 1 and let M be the Λ-module generated by v . Note that b(u , v) = 0 for all v M and ′ i′ 1 ∈ ′ dim (K M )= ℓ 1 since φ(b(u , u )) is invertible in K. Hence we can apply the induction K ⊗φ ′ − 1 1 hypothesis to obtain u , . . . , u M such that π(u ),...,π(u ) is a K-basis of K M and 2 ℓ ∈ ′ { 1 ℓ } ⊗φ φ(b(u , u ′ )) = 0 for all i = i . i i 6 ′ Theorem 4.8. Let R R be a real closed field. Then an R-algebra homomorphism φ : T R ⊇ n → extends to an R-algebra homomorphism ϕ : R[ξ] R if and only if φ(Ωn) R 0. → ⊆ ≥ Proof. The implication ( ) is obvious, so we prove ( ). In the terminology of [DPRR05, ⇒ ⇐ Subsection 2.3], Tn is an n–Cayley-Hamilton algebra. Since Tn is finitely spanned over Tn, = R T is a finite-dimensional R-algebra which inherits an involution τ and an R- A′ ⊗φ n ′ trace χ : R from T . By [DPRR05, Subsection 2.3] is again an n–Cayley-Hamilton ′ A′ → n A′ algebra. Let be the Jacobson radical of ′. Since ′ is finite-dimensional, elements of are J A A ′ J characterized as generators of nilpotent ideals. Hence clearly τ . Moreover, if f , then J ⊆J ∈J χ (f) = 0 by applying [DPRR05, Proposition 3.2] to the scalar extension of by the algebraic ′ A′ closure of R and [Lam91, Theorem 5.17]. Therefore = / is a finite-dimensional semisimple R-algebra with involution τ and an A A′ J R-trace χ : R. IfΦ: T is the canonical -homomorphism, then A→ n →A ∗ (4.7) χ Φ=Φ tr . ◦ ◦ We claim that τ(aaτ ) 0 for every a . Indeed, if π : T R T is the canonical ≥ ∈ A n → ⊗φ n -homomorphism, then by Lemma 4.7 there exist a finite set u of symmetric elements in T ∗ { i}i n and a finite set v of antisymmetric elements in T such that π(u ) π(v ) form an { j}j n { i }i ∪ { j }j R-basis of R T and ⊗φ n φ(tr(uiui′ )) = φ(tr(vjvj′ )) = φ(tr(uivj)) = 0 for all i = i and j = j . If 6 ′ 6 ′ a = α Φ(u )+ β Φ(v ), α , β R, i i j j i j ∈ Xi Xj then t t χ(aaτ )= α2φ(tr(u u )) + β2φ(tr(v v )) 0 i i i j j j ≥ Xi Xj by (4.7). By Wedderburn’s structure theorem we have = A A1 ×···×Aℓ 18 I. KLEP, S.ˇ SPENKO,ˇ AND J. VOLCIˇ Cˇ for some finite-dimensional simple R-algebras . Moreover, by [DPRR05, Theorem 4.2] there Ak exist d ,...,d N such that 1 ℓ ∈ R (4.8) χ ak = dk tr k (ak) ! A Xk Xk for a . k ∈Ak Next we show that τ is split, i.e., ( )τ for 1 k ℓ. Since every involution Ak ⊆ Ak ≤ ≤ preserves centrally primitive idempotents [Lam91, Section 22], for every k there exists k′ such τ τ that ( ) ′ . Suppose that τ is not split and without loss of generality assume ( ) . Ak ⊆Ak A1 ⊆A2 Let e and e be the identity elements, respectively. Then 1 ∈A1 2 ∈A2 χ (e e )(e e )τ = χ (e e )(e e ) 1 − 2 1 − 2 1 − 2 2 − 1 = χ( e e ) − 1 − 2 R R = d1 tr 1 (e1) d2 tr 2 (e2) < 0, − A − A a contradiction. Let τk be the restriction of τ on k. By (4.8) and the previous paragraph it follows that R τk τk A tr (aa ) 0 and hence tr (aa ) 0 for every a k, so τk is a positive involution. k ≥ Ak ≥ ∈A ATherefore the assumptions of Proposition 4.6 are met and we obtain a trace preserving - ∗ homomorphism Ψ : T M (R) extending φ. Now we define ϕ : R[ξ] R by n → n → ϕ(ξjı) = Ψ(Ξj)ı. Remark 4.9. The condition φ(Ωn) R 0 in Theorem 4.8 is clearly necessary since tr(hh∗) is ⊆ ≥ a nonzero sum of squares in R[ξ] for every nonzero h Tn. Moreover, it is not vacuous. For ∈s example, let τ be the involution on H defined by uτ = iu i 1 for u H, where s is the standard − ∈ symplectic involution on H. Then τ is of orthogonal type and we have a trace preserving - ∗ epimorphism Φ : T2 H defined by Φ(Ξ1)= i and Φ(Ξ2)= j. Since Φ(T2)= R, the restriction → t yields a homomorphism φ : T R and φ(tr(Ξ Ξ )) = 1. 2 → 2 2 − Corollary 4.10. Let φ : T = R[ξ]On(R) R be an R-algebra homomorphism into a real closed n → field R R. Then the following are equivalent: ⊇ (i) φ extends to an R-algebra homomorphism ϕ : R[ξ] R; On(R) 2 → (ii) φ(R[ξ] R[ξ] ) R 0; t ∩ ⊆ ≥ (iii) φ(tr(hh )) R 0 for all h Tn. ∈ P≥ ∈ Remark 4.11. At first glance one might ponder whether Theorem 4.8 could be derived from the Procesi-Schwarz theorem [PS85]. In Appendix B we explain why this does not seem to be the case. 4.3. Stellens¨atze. We are now ready to give the main result of this section, the Krivine-Stengle Positivstellensatz for trace polynomials a that are positive (semidefinite) on KS, see Theorem 4.13. In the proof we use Corollary 4.4 to reduce the problem to the commutative ring Tn[a]. Before applying the abstract Positivstellensatz for commutative rings, we need the relation between orderings and matrix evaluations of trace polynomials that is given in Proposition 4.12 below, which is a crucial consequence of the extension Theorem 4.8 and Tarski’s transfer principle. Proposition 4.12. Let S T be finite, a Sym T and P an ordering in T [a] containing ⊂ n ∈ n n S Ω . ∪ n (1) a 0 implies a P . |KS ∈ (2) a 0 implies a P P . |KS ≻ ∈ \− POSITIVE TRACE POLYNOMIALS 19 (3) a = 0 implies a P P . |KS ∈ ∩ − Proof. Let P be an ordering in T [a] containing S and let σ = tr( ja) T for 1 j n. n j ∧ ∈ n ≤ ≤ (1) The restriction of P to T gives rise to a real closed field R and a homomorphism φ : T n n → R satisfying φ(S Ωn) R 0. By Theorem 4.8 we extend it to a homomorphism ϕ : R[ξ] R. ∪ ⊆ ≥ 2 → Suppose that ϕ(σ ) < 0 for some j; in other words, there exist α Rgn such that σ (α) < 0 j ∈ j and s′(α) 0 for all s′ S. By Tarski’s transfer principle [Mar08, Theorem 1.4.2] there exists 2 ≥ ∈ α Rgn such that σ (α ) < 0 and s (α ) 0 for all s S. But this contradicts σ 0, ′ ∈ j ′ ′ ′ ≥ ′ ∈ j|KS ≥ which is a consequence of s 0. Hence φ(σ )= ϕ(σ ) 0 for all j, so σ P for all j. By |KS j j ≥ j ∈ the Cayley-Hamilton theorem we have n n n j (4.9) ( a) + σ ( a) − = 0. − j − Xj=1 Suppose a / P . Then a P , so (4.9) implies ( a)n P P . Therefore a P P , a ∈ − ∈ − ∈ ∩− ∈ ∩− contradiction. (2) Because a 0 implies σ > 0, we obtain σ P P for all j by the same reasoning |KS ≻ j|KS j ∈ \− as in (1). If a / P P , then a P , so (4.9) implies σ P P , a contradiction. ∈ \− − ∈ n ∈ ∩− (3) If a = 0, then a 0 and a 0, so a P P by (1). |KS |KS − |KS ∈ ∩− Theorem 4.13 (Krivine-Stengle Positivstellensatz for trace polynomials). Let S a Sym T ∪{ }⊂ n be finite. (1) a 0 if and only if at = t a = a2k + t for some t ,t Ttr and k N. |KS 1 1 2 1 2 ∈ S ∈ (2) a 0 if and only if at = t a =1+ t for some t ,t Ttr. |KS ≻ 1 1 2 1 2 ∈ S (3) a = 0 if and only if a2k Ttr for some k N. |KS − ∈ S ∈ Proof. The directions ( ) are straightforward. For the implications ( ), by Corollary 4.4 we ⇐ ⇒ can assume that S T . Let T be the preordering in T [a] generated by S Ω . Note that ⊂ n n ∪ n T Ttr since S T . If a 0, then a P for every ordering P of T [a] containing T ⊂ S ⊂ n |KS ∈ n by Proposition 4.12. Therefore t a = a2k + t for some t ,t T and k N by the abstract 1 2 1 2 ∈ ∈ Positivstellensatz [Mar08, Theorem 2.5.2], so (1) is holds. (2) and (3) are proved analogously. Remark 4.14. In general we cannot choose a central t1 in Theorem 4.13; see Example 6.4. Remark 4.15. A clean Krivine-Stengle Positivstellensatz for generic matrices clearly does not exist 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 GMn fails. Corollary 4.16. If S Sym T is finite, then K = if and only if 1 Ttr. ⊂ n S ∅ − ∈ S Proof. If K = , then 1 0, so by Theorem 4.13 there exist t ,t Ttr such that S ∅ − |KS ≻ 1 2 ∈ S ( 1)t =1+ t , so 1= t + t Ttr. The converse is trivial. − 1 2 − 1 2 ∈ S Corollary 4.17. Let s Sym T . Then s(A) 0 for all A M (R)g if and only if ∈ n 6 ∈ n t 1= ω tr(h sh ) − i ij ij Xi Yj for some ω Ω and h T . i ∈ n ij ∈ n Proof. Follows by Corollary 4.16 and Lemma 4.1. Corollary 4.18 (Real Nullstellensatz for trace polynomials). Let T be an ideal and J ⊂ n assume tr( ) . For h T the following are equivalent: J ⊆J ∈ n g (i) for every X Mn(R) , u(X) = 0 for every u implies h(X) = 0; ∈ t ∈J (ii) there exists k N such that (h h)k Ω + . ∈ − ∈ n J 20 I. KLEP, S.ˇ SPENKO,ˇ AND J. VOLCIˇ Cˇ Proof. The implication (2) (1) is clear. Conversely, Tn is Noetherian, so is (as a left ideal) ⇒ t t J generated by some u , . . . , u T . Let S = u u ,..., u u ; then (1) is equivalent to 1 ℓ ∈ n {− 1 1 − ℓ ℓ} hth = 0. Hence (1) (2) follows by Theorem 4.13(3) and Ttr Ω + . |KS ⇒ S ⊆ n J Corollary 4.19. Let S T . Then ⊂ n A M (R)g : s(A) = 0 s S = { ∈ n ∀ ∈ } ∅ if and only if 1= ω + tr(h s ) − i i Xi for some ω Ω , h T and s S. ∈ n i ∈ n i ∈ We mention that Hilbert’s Nullstellensatz for n n generic matrices over an algebraically × closed field is given by Amitsur in [Ami57, Theorem 1]. 5. Positivstellensatze¨ for compact semialgebraic sets In this section we give certificates for positivity on compact semialgebraic sets. We prove a version of Schm¨udgen’s theorem [Scm91] for trace polynomials (Theorem 5.3). We also present a version of Putinar’s theorem [Put93] for trace polynomials (Theorem 5.7) and for generic matrices (Theorem 5.5). 5.1. Archimedean (cyclic) quadratic modules. A cyclic quadratic module Q Tn is t ⊂ archimedean if for every h T there exists ρ Q such that ρ hh Q. Equivalently, for ∈ n ∈ >0 − ∈ every s Sym Tn there exists ε Q>0 such that 1 εs Q. ∈ ∈ ± ∈ t For a cyclic quadratic module Q let HQ be the set of elements h T such that ρ hh Q ∈ n − ∈ for some ρ Q . It is clear that Q is archimedean if and only if HQ = T . ∈ >0 n Proposition 5.1. HQ is a trace -subalgebra over R in T . ∗ n t Proof. HQ is a -subalgebra over R by [Vid59]. Let h HQ. Then s = h + h HQ and let ∗ ∈ ∈ ρ Q be such that ρ2 s2 Q. Then ∈ >0 − ∈ 1 ρ s = (ρ s)2 + (ρ2 s2) Q ± 2ρ ± − ∈ and consequently nρ tr(s) Q. Therefore ± ∈ 1 (nρ)2 tr(s)2 = (ρ s)(ρ + s)(ρ s) + (ρ + s)(ρ s)(ρ + s) Q, − 2nρ − − − ∈ 1 so tr(h)= tr(s) HQ. 2 ∈ Corollary 5.2. A cyclic quadratic module Q is archimedean if and only if there exists ρ Q ∈ >0 such that ρ Ξ Ξt Q. − j j j ∈ t Proof. ( ) isP trivial. Conversely, ρ Ξ Ξ Q implies Ξ HQ for 1 j n, so HQ = T ⇒ − j j j ∈ j ∈ ≤ ≤ n since T is generated by Ξ as a trace -subalgebra over R by Proposition 5.1. n j P∗ It is easy to see that KQ is compact if Q is archimedean. The converse fails already with n = 1 ([Mar08, Section 7.3] or [PD01, Example 6.3.1]). If KQ is compact, say X N for all 2 t k k≤ X KQ, then we can add N Ξ Ξ to Q to make it archimedean without changing KQ. ∈ − j j j P POSITIVE TRACE POLYNOMIALS 21 5.2. Schm¨udgen’s Positivstellensatz for trace polynomials. In this subsection we prove a version of Schm¨udgen’s Positivstellensatz for trace polynomials a that are positive on a compact semialgebraic set KS. 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’s Positivstellensatz is used in the commutative ring Tn[a]. Theorem 5.3 (Schm¨udgen’s Positivstellensatz for trace polynomials). Let S a Sym T ∪ { } ⊂ n be finite. If K is compact and a 0, then a Ttr. S |KS ≻ ∈ S Proof. First we apply Corollary 4.4 to reduce to the case S T . Let T be the preordering in ⊂ n T [a] generated by S Ω . Note that K = K and T Ttr. n ∪ n S T ⊂ S Let b Tn[a] be arbitrary. Since KS is compact, there exists β R 0 such that β b KS 0. ∈ ∈ ≥ ± | Then β b P for every ordering P in T [a] containing T S Ω by Proposition 4.12. In the ± ∈ n ⊃ ∪ n terminology of [Sce03], T is weakly archimedean. Since Tn[a] is a finitely generated R-algebra, T is an archimedean preordering in Tn[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 [a] containing T S Ω , so a T by [Sce03, Proposition 3.3] or [Mon98, Theorem 4.3]. n ⊃ ∪ n ∈ Corollary 5.4. Let S Sym T be finite. Then Ttr is archimedean if and only if K is compact. ⊂ n S S 5.3. Putinar’s Positivstellensatz for generic matrices. Our next theorem is a Putinar- type Positivstellensatz for generic matrices on compact semialgebraic sets, which requires a functional analytic proof. While the proof generally follows a standard outline (using a sepa- ration argument followed by a Gelfand-Naimark-Segal construction), several modifications are needed. For instance, the separation is taken to be extreme in a convex sense, and polynomial identities techniques are applied to produce n n matrices. × A set Q Sym GM is a quadratic module if ⊆ n t 1 Q, Q + Q Q, hQh Q h GM . ∈ ⊆ ⊆ ∀ ∈ n t We say that Q is archimedean if for every h GM there exists ρ Q such that ρ aa Q. ∈ n ∈ >0 − ∈ As in Corollary 5.2 we see that a quadratic module is archimedean if and only if it contains t ρ Ξ Ξ for some ρ Q . − j j j ∈ >0 TheoremP 5.5 (Putinar’s Positivstellensatz for generic matrices). Let Q Sym GMn be an ⊂ archimedean quadratic module and a Sym GM . If a 0, then a Q. ∈ n |KQ ≻ ∈ Proof. Assume a Sym GM Q. We proceed in several steps. ∈ n \ Step 1: Separation. Consider Q as a convex cone in the vector space Sym GMn over R. Since Q is archimedean, for every s Sym T there exists ε Q such that 1 εs Q, which in terms of [Bar02, ∈ n ∈ >0 ± ∈ Definition III.1.6] means that 1 is an algebraic interior point of the cone Q in Sym Tn. By the Eidelheit-Kakutani separation theorem [Bar02, Corollary III.1.7] there exists a nonzero R-linear functional L0 : Sym GMn R satisfying L0(Q) R 0 and L0(a) 0. Moreover, L0(1) > 0 → ⊆ ≥ ≤ because Q is archimedean, so after rescaling we can assume L (1) = 1. Let L : GM R be 0 n → the symmetric extension of L , i.e., L(f)= 1 L (f + f t) for f GM . 0 2 0 ∈ n Step 2: Extreme separation. Now consider the set of all linear functionals L′ : GMn R satisfying L′(Q) R 0 and C → ⊆ ≥ L (1) = 1. This set is nonempty because L . Endow GM with the norm ′ ∈C n p = max p(X) : X M (R)g, X 1 . k k {k k2 ∈ n k k2 ≤ } By the Banach-Alaoglu theorem [Bar02, Theorem III.2.9], the convex set is weak*-compact. C Thus by the Krein-Milman theorem [Bar02, Theorem III.4.1] we may assume that our separating functional L is an extreme point of . C 22 I. KLEP, S.ˇ SPENKO,ˇ AND J. VOLCIˇ Cˇ Step 3: GNS construction. On GM we define a semi-scalar product p,q = L(qtp). By the Cauchy-Schwarz inequality for n h i semi-scalar products, t = q GM L(q q) = 0 N ∈ n | is a linear subspace of GMn. Hence t (5.1) p, q = L(q p) h i is a scalar product on GM / , where p = p + denotes the residue class of p GM in n N N ∈ n GM / . Let H denote the completion of GM / with respect to this scalar product. Since n N n N 1 , H is non-trivial. 6∈ N Next we show that is a left ideal of GM . Let p,q GM . Since Q is archimedean, there N n ∈ n exists ε> 0 such that 1 εptp Q and therefore − ∈ t t t (5.2) 0 L(q (1 εp p)q) L(q q). ≤ − ≤ Hence q implies pq . ∈N ∈N Because is a left ideal, we can define linear maps N M : GM / GM / , q pq p n N → n N 7→ for p GM . By (5.2), M is bounded and thus extends to a bounded operator Mˆ on H. ∈ n p p Step 4: Irreducible representation of GMn. The map π : GM (H), p Mˆ n →B 7→ p is clearly a -representation, where (H) is endowed with the adjoint involution . Observe that ∗ B ∗ η = 1 H is a cyclic vector for π by construction and ∈ (5.3) L(p)= π(p)η, η . h i Write = π(GM ). We claim that the self-adjoint elements in the commutant of in (H) A n A′ A B are precisely real scalar operators. Let P be self-adjoint. By the spectral theorem, P ∈ A′ decomposes into real scalar multiples of projections belonging to P . So it suffices to { }′′ ⊆ A′ assume that P is a projection. By way of contradiction suppose that P / 0, 1 ; since η is cyclic ∈ { } for π, we have P η = 0 and (1 P )η = 0. Hence we can define linear functionals L on GM by 6 − 6 j n π(p)Pη,Pη π(p)(1 P )η, (1 P )η L (p)= h i and L (p)= h − − i 1 P η 2 2 (1 P )η 2 k k k − k for all p GM . One checks that L is a convex combination of L and L . Since also L , ∈ n 1 2 j ∈ C we obtain L = L = L by the extreme property of L. Let λ = P η 2; then (5.3) implies 1 2 k k π(p)η, λη = λ π(p)η, η = π(p)Pη,Pη = Pπ(p)η,Pη = π(p)η,Pη h i h i h i h i h i for all p GM . Therefore P η = λη since η is a cyclic vector for π. Then λ 0, 1 since P is ∈ n ∈ { } a 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 and pt = p, then → ∈ n ± π(p)P = Pπ(p)P = (Pπ(p)P )∗ = (π(p)P )∗ = Pπ(P ). ± ± Consequently P and hence P R. Since P is an orthogonal projection, we have P 0, 1 , ∈A′ ∈ ∈ { } so π is irreducible. Step 5: Transition to n n matrices. × We claim that is a prime algebra. Indeed, suppose a b = 0 for a, b . If b = 0, then there A A ∈A 6 is a u H with bu = 0. Since π is irreducible, the vector space V = bu is dense in H. Now ∈ 6 A aV = 0 implies aH = 0, i.e., a = 0. POSITIVE TRACE POLYNOMIALS 23 Since the -center of equals R, we have π(C )= R, so is generated by π(Ξ ) for 1 j g ∗ A n A j ≤ ≤ as an R-algebra with involution. Let f R Lemma 5.6. Let Q Sym T be an archimedean cyclic quadratic module and c T . If ⊂ n ∈ n c > 0, then c Q. |KQ ∈ Proof. Assume c Sym T Q. Steps 1–4 in the proof of Theorem 5.5 work if we replace GM ∈ n \ n with T . Hence we obtain a Hilbert space H, a -representation π : T (H) with π(T )= R, n ∗ n →B n and a cyclic unit vector η H for π such that the linear functional ∈ L : T R, p π(p)η, η n → 7→ h i Q satisfies L( ) R 0 and L(c) 0. Let φ = π Tn : Tn R. By the proof of Theorem 4.8, φ ⊆ ≥ ≤ | → extends to a trace preserving -homomorphism Ψ : T M (R). Let X = Ψ(Ξ ) M (R) for ∗ n → n j j ∈ n j = 1,...,g. Because π(T ) = R, we have L = π and therefore tr(q(X)) = φ(tr(q)) 0 n |Tn |Tn ≥ for every q Q, so X KQ by Corollary 4.5 and c(X)= φ(c) 0. ∈ ∈ ≤ Theorem 5.7 (Putinar’s theorem for trace polynomials). Let Q Sym T be an archimedean ⊂ n cyclic quadratic module and a Sym T . If a 0, then a Q. ∈ n |KQ ≻ ∈ j Proof. Let σ = tr( a) and assume a 0. Since KQ is compact, there exists ε > 0 such j ∧ |KQ ≻ that (σ ε) > 0 for all 1 j n. By Lemma 5.6 we have σ ε Q for all j. Let c ,...,c j − |KQ ≤ ≤ j − ∈ 1 N be the generators of T as an R-algebra. Since Q is archimedean, there exist ρ , . . . , ρ Q n 1 N ∈ >0 such that ρ c2 Q. Write i − i ∈ S = σ ε, ρ c2 : 1 j n, 1 i N j − i − i ≤ ≤ ≤ ≤ and let Q be the quadratic module in Tn[a] generated by S Ωn. Clearly we have Q Q and ∪ ⊂ a > 0. Let H T [a] be the subring of bounded elements with respect to Q, i.e., b T [a] |KQ Q ⊆ n ∈ n such that ρ b2 Q for some ρ Q . Because c generate T , we have T H . Since H is − ∈ ∈ >0 i n n ⊆ Q Q integrally closed in T [a] by [Bru79, Section 6.3] or [Scw03, Theorem 5.3], we also have a H . n ∈ Q Hence Q is archimedean. If P is an ordering in T [a] containing S Ω , then a P P by n ∪ n ∈ \− 24 I. KLEP, S.ˇ SPENKO,ˇ AND J. VOLCIˇ Cˇ Proposition 4.12. Therefore a Q Q by Jacobi’s representation theorem [Mar08, Theorem ∈ ⊂ 5.4.4]. 6. 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 Tn there exists a finite set S Qtr ∈ ⊂ tr( s ) such that K s = KS. Let us give a concrete example of such a set for n = 3. Let { } { } σ = tr( js); using the Cayley-Hamilton theorem and the relations between σ and tr(si) it is j ∧ j easy to check that tr(s)= σ1, tr (s σ )s(s σ ) = σ σ + 3σ , − 1 − 1 1 2 3 2 2 tr (s σ1s + σ2 )s(s σ1s + σ2) = σ2σ3, − − tr (s σ1 1)s(s σ1 1) = σ1 + 4σ2 + 3σ3 + σ1σ2. − − − − Qtr Denote these elements by c1, c2, c3, c4 tr( s ). We claim that K s = K c1,c2,c3,c4 . It suffices ∈ { } { } to prove the inclusion . { } ⊇ t To simplify the notation let s = s M (R). If s 0, then σ < 0 for some j. If σ < 0, ∈ 3 j 1 then c < 0. Hence assume σ 0. If σ ,σ are of opposite sign, then c < 0. If σ > 0 and 1 1 ≥ 2 3 3 1 σ ,σ 0, then c < 0. Finally, if σ = 0 and one of σ ,σ equals 0, then σ < 0 implies c < 0 2 3 ≤ 2 1 2 3 3 2 and σ2 < 0 implies c4 < 0. Example 6.2. The denominator in Lemma 2.5 is unavoidable even if f is a hermitian square t or f R[ξ]. For example, let n = 4 and a =Ξ Ξ . Then det(a) is a square in R[ξ]. Suppose ∈ 1 − 1 det(a) Ω4, i.e., ∈ t t det(a)= tr(h h ) tr(h h ) i1 i1 · · · imi imi i X t for some h T . Because a is independent of Ξ +Ξ and Ξ for j > 1, we can assume that ij ∈ n 1 1 j h are polynomials in a and tr(ak) for k N. Moreover, det(a) is homogeneous of degree 4 with ij ∈ respect to the entries of a, so hij are of degree at most 2 with respect to a. Finally, tr(a) = 0, so we conclude that det(a)= tr (α a2 + β tr(a2))2 , α , β R. i i i i ∈ i X If R is a real closed field containing R(ξ), then there exist λ, µ R that are algebraically ∈ independent over R such that 0 λ 0 0 − λ 0 0 0 a = 0 0 0 µ − 0 0 µ 0 after an orthogonal basis change. Therefore 2 2 2 2 2 2 2 2 2 (λµ) = 2 (αiλ + 2βi(λ + µ )) + (αiµ + 2βi(λ + µ )) , i X which is clearly a contradiction. POSITIVE TRACE POLYNOMIALS 25 Example 6.3. Next we show that a traceless equivalent of Theorem 4.13 fails for GM2. A quadratic module T Sym GM is a preordering if T C is closed under multiplication. ⊆ n ∩ n For S Sym GM let Q and T denote the quadratic module and preordering, respectively, ⊂ n S S generated by S. For example, T = Q is the set of sums of hermitian squares in GMn. ∅ ∅ Fix n = 2 and let t t s =Ξ +Ξ , a =Ξ Ξ , f = [s2, a][s, a]. 1 1 1 − 1 Since f = tr(s)[s, a]2, it is clear that f 0. We will show that there do not exist t ,t T |K{s} 1 2 ∈ S and k N such that ∈ 2k (6.1) ft1 = t1f = f + t2. It clearly suffices to assume g = 1. If R is a real closed field containing R(ξ), then after diagonalizing s we may assume that λ 0 0 µ s = 1 , a = − 0 λ µ 0 2 for some λ , λ ,µ R that are algebraically independent over R. Let h GM be homogeneous 1 2 ∈ ∈ 2 of degree (d, e) with respect to (s, a). Then it is not hard to check that 2 t λ h˜(λ , λ ) 0 (6.2) hsh = µ2e 1 1 2 0 λ h˜(λ , λ )2 2 2 1 for some homogeneous polynomial h˜ R[y ,y ] of degree d. Therefore ∈ 1 2 t h sh C h = 0 i, i i ∈ 2 ⇒ i ∀ Xi so we deduce that (6.3) T s = (T C2) Q s . { } ∅ ∩ · { } Now suppose that (6.1) holds for some t ,t T and k N. Since f is homogeneous of 1 2 ∈ S ∈ degree (5, 2) with respect to (s, a), t1,t2 can be taken homogeneous as well. Then t2 is of degree (10k, 4k) and t1 is of degree (10k 5, 4k 2). In particular, the total degrees of t1 and t2 are − − T t odd and even, respectively, so by (6.3) we conclude that t2 and t1 is of the form i hishi ∈ ∅ for h GM . Hence (6.2) implies i ∈ 2 P ˜ 2 4k 2 λ1 i hi(λ1, λ2) 0 t = µ − 1 0 λ h˜ (λ , λ )2 P 2 i i 2 1 for some homogeneous h˜ R[y ,y ]. Now P i ∈ 1 2 f = (λ + λ )(λ λ )2µ2 1 2 1 − 2 implies λ h˜ (λ , λ )2 0 (6.4) ft = (λ + λ )µ4k(λ λ )2 1 i i 1 2 . 1 1 2 1 − 2 0 λ h˜ (λ , λ )2 P 2 i i 2 1 The nonempty set P X M (R): det(s(X)) < 0 and tr(s(X)) > 0 { ∈ 2 } is open in the Euclidean topology, so by (6.4) there exists X M2(R) such that (st1)(X) is 2k ∈ nonzero and indefinite. However, this contradicts ft1 = f + t2 T . ∈ ∅ 26 I. KLEP, S.ˇ SPENKO,ˇ AND J. VOLCIˇ Cˇ Example 6.4. Here we show that the element t1 in Theorem 4.13 cannot be chosen central in general. Let n = 2, s = 1 (Ξ +Ξt ) and S = tr(s)3, det(s)3 . Suppose t s = s2k + t for some 2 1 1 { } 1 2 k N and t ,t Ttr and t T . Let ∈ 1 2 ∈ S 1 ∈ 2 ζ 0 Ψ : T M (R[ζ]), Ξ . → 2 1 7→ 0 1 3 3 3 2 3 2 Since (ζ + 1) = ζ + ( 2 ζ + 1) + 4 ζ , we conclude that Ψ(t1) belongs to the commutative 3 3 preordering generated by ζ and Ψ(t2) belongs to the matricial preordering generated by ζ in 2k M2(R[ζ]). But then Ψ(t1)Ψ(s)=Ψ(s) + Ψ(t2) contradicts [Cim12, Example 4]. t t Example 6.5. Let f = 5 Tr(Ξ Ξ ) 2 Tr(Ξ )(Ξ +Ξ ) T . We will show that f is totally 1 1 − 1 1 1 ∈ 2 positive and write it as a sum of hermitian squares in USA2. Write Ξ = Ξ1 = (ξı)ı and let u = η1 η2 , v = ξ22η1 ξ21η2 ξ12η2 ξ11η1 ξ22η2 ξ21η1 ξ12η1 ξ11y2 . Then ufut can be viewed as a quadratic form in v and 5 α α 2 0 0 00 − α 5 0 α 20 0 00 − − α 0 5 α 20 0 00 − − t t 2 α 2 α 21 0 0 00 ufu = vGαv , Gα = − − − − − 00 0 0 1 α 2 α 2 2 − − − − − 00 0 0 α 2 5 0 α − − 00 0 0 α 2 0 5 α − − 00 0 0 2 α α 5 − for α R. Observe that G is positive semidefinite if and only if 7 α 5 . Hence f is ∈ α − 2 ≤ ≤ − 2 indeed totally positive and a sum of hermitian squares in M2(R[ξ]). By diagonalizing Gα at α = 5 we obtain − 2 5 1 t 1 t f = (ξ ξ )2 + H˜ H˜ + H˜ H˜ , 2 12 − 21 2 2 2 2 3 3 where ξ + ξ ξ ξ 2(ξ + ξ ) ξ 3ξ H˜ = 12 21 22 − 11 , H˜ = 12 21 11 − 22 . 2 ξ ξ ξ + ξ 3 ξ 3ξ 2(ξ + ξ ) 11 − 22 12 21 22 − 11 12 21 t t Note that while H˜ H˜ , H˜ H˜ Sym T , we can compute (H˜ )= (H˜ ) = 0, so H˜ , H˜ / T . 2 2 3 3 ∈ 2 R2 2 R 3 2 3 ∈ 2 However, if we set t t t t t t H =Ξ Ξ ,H = ΞΞ Ξ Ξ,H =Ξ2 2ΞΞ + 2Ξ Ξ (Ξ )2, 1 − 2 − 3 − − then t t t t t H H = (ξ ξ )2,H H = (ξ ξ )2H˜ H˜ ,H H = (ξ ξ )2H˜ H˜ 1 1 12 − 21 2 2 12 − 21 2 2 3 3 12 − 21 3 3 and so 5 t 1 1 t t 1 1 t t f = H H + H− H H H− + H− H H H− . 2 1 1 2 1 2 2 1 2 1 3 3 1 A. Constructions of the Reynolds operator In this appendix we describe a few more ways of constructing for the action of O (R) on Rn n Mn(R[ξ]) defined in Subsection 2.2. We refer to [Stu08] for algorithms for finite group actions. Let : R[ξ] T be the restriction of : M (R[ξ]) T , i.e., the Reynolds operator for Rn′ → n Rn n → n the action of On(R) on R[ξ] given by u t t f = f(uΞ1u , . . . , uΞgu ) for f R[ξ] and u O (R). ∈ ∈ n POSITIVE TRACE POLYNOMIALS 27 A.1. Computing . We start by describing two ways of obtaining . Rn′ Rn′ A.1.1. First method. We follow [DK02, Section 4.5.2] to present an algorithm for computing . We define a linear map c R[O (R)] : Rn′ ∈ n ∗ n d d c(r)= r (1 + se )(1 + te ) r (1 + se )(1 + te ) , dt ds ij ij − ij ji i,j=1 s=t=0 X where e , 1 i, j n, denote the standard matrix units in M (R). (In fact, c equals the ij ≤ ≤ n Casimir operator of the Lie algebra on of skew symmetric matrices of the group On(R) up to a scalar multiple.) For example, if n = 2, then d d c(u u ) = (1 + se )(1 + te ) (1 + se )(1 + te ) 11 22 dt ds 12 12 11 12 12 22 − (1 + se )(1 + te ) (1 + se )(1 + te ) + − 12 21 11 12 21 22 + (1 + se )(1 + te ) (1 + se )(1 + te ) 21 21 11 21 21 22 − (1 + se )(1 + te ) (1 + se )(1 + te ) − 21 12 11 21 12 22 s=t=0 d d = (1 st + 1 st) dt ds − − s=t=0 = 2. − For f R[ξ], u = (u ) O (R), write f u as f µ , where f are linearly independent ∈ ı ı ∈ n i i i i polynomials in the variables ξ and µ are polynomials in the variables u . Define jı i P ı c˜(f)= fic(µi). X Find the monic polynomial p of smallest degree such that p(˜c)(f)=0. If p(0) = 0, set (f)= f. 6 Rn′′ If p(0) = 0, write p(t)= tq(t) and define (f)= q(0) 1q(˜c(f)). By [DK02, Proposition 4.5.17], Rn′′ − defines the Reynolds operator for the action of SO (R) on R[ξ]. Since O (R)/ SO (R) = Rn′′ n n n ∼ Z/2Z, setting (f)= 1 ( (f)+ (f)v), where v is an arbitrary element in O (R) SO (R), Rn′ 2 Rn′′ Rn′′ n \ n we obtain the Reynolds operator for the action of On(R) on R[ξ]. A.1.2. Second method. Here we mention another way of computing the Reynolds operator Rn′ in terms of an integral. This approach is based on the way the invariants of On(R) for the action on R[ξ] were described by Procesi in [Pro76]. Let f R[ξ]. We first multihomogenize f as a ∈ function f : M (R)g R, then multilinearize its homogeneous components f and view f as an n → i i element f (M (R) di ) for d = deg(f ). Since M (R) = V V for a n-dimensional vector i ∈ n ⊗ ∗ i i n ∼ ∗ ⊗ space V on which On(R) acts naturally, and V ∗ is isomorphic as a On(R)-module to V , f i can be seen as an element f˜ V 2di . The monomial ξ ξ corresponds to the element i ∈ ⊗ 1ı11 · · · dıdd e e e e , ı1 ⊗ 1 ⊗···⊗ ıd ⊗ d where e , 1 i n, is an orthonormal basis of V . Then we can compute (f ) by integrating i ≤ ≤ Rn′ i the function u (f˜)u over O (R). To obtain (f ) we need to restitute (f ) and multiply 7→ i n Rn′ i Rn′ i the result by a suitable integer. Finally, (f)= (f ). Rn′ Rn′ i P A.2. From to . Once we have , we can compute as follows. Rn′ Rn Rn′ Rn 28 I. KLEP, S.ˇ SPENKO,ˇ AND J. VOLCIˇ Cˇ A.2.1. First method. Let f M (R[ξ]). We can assume that f is independent of Ξ . Let us ∈ n g compute (tr(fΞ )). Since this is an invariant, linear in Ξ , it has the form Rn′ g g ′ (tr(fΞ )) = tr f Ξ Rn g 0 g t t for some f T . Here we used the fact that tr(hΞ ) = tr(h Ξ ) for h T . We define 0 ∈ n g g ∈ n (f)= f . We claim that : M (R[ξ]) T is the Reynolds operator. We have (f)= f Rn 0 Rn n → n Rn for f T since (tr(fΞ )) = tr(fΞ ) as tr(fΞ ) T . For u O (R), f M (R[ξ]) we have ∈ n Rn′ g g g ∈ n ∈ n ∈ n u u u tr((f )Ξg) = tr((fΞg) ) = tr(fΞg) , where the first equality follows as Ξg is an invariant and the second one since tr is linear. Thus, (f u)= (f), and is the Reynolds operator. Rn Rn Rn A.2.2. Second method. If we have n′ , then the Reynolds operator can also be computed by R t expressing an element f M (R[ξ]) as an R(ξ)-linear combination of Ξi Ξ j, 0 i, j n 1. ∈ n 1 1 ≤ ≤ − Note that these elements are linearly independent in Mn(R(ξ)) as there exists X Mn(R) such i tj i tj ∈ that X X , 0 i, j n 1, are linearly independent. We denote Ξ1Ξ1 , 0 i, j n 1, by ≤ ≤ 2 − 2≤ ≤ − y1,...,yn2 . Let c be a n -normal (i.e., multilinear and alternating in the first n -variables) central polynomial of M (R) in 2n2 1 variables. (See e.g. [Row80, Section 1.4] for the construction of n − such polynomials and for the proofs of their properties mentioned below.) Since y1,...,yn2 are independent we can find yn2+1,...,y2n2 1 GMn such that − ∈ 0 = c(y1,...,yn2 ,yn2+1,...,y2n2 1)= z Cn. 6 − ∈ 2 If g n , we can take yn2+1 =Ξ2,...,y2n2 1 =Ξn2 . Then f can be written as follows ≥ − n2 i 1 1 (A.1) f = ( 1) − z− c(f,y1,...,yi 1,yi+1,...,yd)yi. − − Xi=1 Let zi(f)= n′ (c(f,y1,...,yi 1,yi+1,...,yd)). We define R − n2 i+1 1 (f)= ( 1) z− z (f)y . Rn − i i Xi=1 If f T , then the coefficients in the expression (A.1) are already in T , so in this case (f)= ∈ n n Rn f. Note that (z (f u)) = (z (f)) for u O (R) and f M (R[ξ]). Therefore (f u) = Rn′ i Rn′ i ∈ n ∈ n Rn (f) and : M (R[ξ]) T is the Reynolds operator. Rn Rn n → n 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 = RN . The invariant ring R[W ]G is a finitely generated R-algebra; let p ,...,p be its generators. Let ( , ) denote a 1 m · · G-invariant inner product on W and its dual on W . Since the differentials dp : W W are ∗ i → ∗ G-equivariant, we have (dp , dp ) R[W ]G. Finally let i j ∈ H = (dp , dp ) M (R[W ]G). i j i,j ∈ m 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 a → real closed field R R. Then the following are equivalent: ⊇ (i) φ extends to an R-algebra homomorphism ϕ : R[W ] R; → POSITIVE TRACE POLYNOMIALS 29 G 2 (ii) φ(R[W ] R[W ] ) R 0; ∩ ⊆ ≥ (iii) φ(H) M (R) is positive semidefinite. ∈ mP 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). Without ⇒ loss of generality we can assume that φ(R[W ]G) generates R as a field. First we observe that if : R[W ] R[W ]G is the Reynolds operator for the action of G, R → then ( R[W ]2) R[W ]2. Note that since G acts linearly on W , the action of G on R[W ] R ⊆ does not increase the degree of polynomials. Since G is compact, is given by the integration P P R formula (f) = f g dµ(g), where µ is the normalized left Haar measure. If f R[W ] is of R G ∈ degree d, then (f 2) is a limit of sums of squares of degree 2d. Using Carath´eodory’s theorem RR [Bar02, Theorem I.2.3] it is easy to see that the cone of sums of squares in R[W ] of degree at most 2d is closed in the space of polynomials of degree at most 2d (cf. [Mar08, Section 4.1]). Hence we conclude that (f 2) is indeed a sum of squares in R[W ]. R 1 Now let T R[W ] be the preordering generated by φ− (R 0). We claim that 1 / T . ⊆ 2 ≥ 1 − ∈ Otherwise 1 = s0 + i 1 siti for some si R[W ] and ti φ− (R 0). By applying we − ∈ ∈ ≥ R get ≥ P P (B.1) 1= (s )+ (s )t . − R 0 R i i i 1 X≥ By the above observation we have (s ) R[W ]G R[W ]2, so (B.1) implies R i ∈ ∩ 1= φ( 1) = φ ( (s )) + Pφ ( (s )) φ(t ) 0, − − R 0 R i i ≥ i 1 X≥ a contradiction. Therefore we can extend T to an ordering P R[W ], which gives rise to a ⊂ homomorphism ϕ : R[W ] R , where R is the real closure of the ordered field of fractions of 0 → 0 0 R[W ]/(P P ). Since ∩− ker φ T T P P = ker ϕ , ⊆ ∩− ⊆ ∩− 0 we see that ϕ extends φ and hence R R . By the Artin-Lang homomorphism theorem 0 ⊆ 0 [BCR98, Theorem 4.1.2] there exists a homomorphism ϕ : R[W ] R satisfying ker ϕ = ker ϕ . → 0 Thus ϕ extends φ. 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 R extends to an R-algebra homomor- n → phism ϕ : R[ξ] R if and only if φ(Ωn) R 0. → ⊆ ≥ On(R) Let us outline the proof. Let p1,...,pm be generators of the R-algebra Tn = R[ξ] . Their g g differentials dpi : Mn(R) (Mn(R) )∗ are On(R)-equivariant maps. Since we can identify g g → g (Mn(R) )∗ with (Mn(R)∗) , and the On(R)-equivariant polynomial maps Mn(R) Mn(R) are g → g precisely trace polynomials [Pro76, Theorems 7.1 and 7.2], we have dp Tn. OnM (R) there i ∈ n is an On(R)-invariant inner product t (X,Y )=tr X Y . j j Xj Finally, let H = ((dp , dp )) M (T ). i j i,j ∈ m n Let φ : T R be an R-algebra homomorphism. Then Theorem B.1 implies that φ extends to n → an R-algebra homomorphism ϕ : R[ξ] R if and only if φ(H) M (R) is positive semidefinite. → ∈ m To prove the non-trivial direction in Corollary B.2 it therefore suffices to show the following. t m Lemma B.3. If φ(Ωn) R 0, then α φ(H)α 0 for all α R . ⊆ ≥ ≥ ∈ 30 I. KLEP, S.ˇ SPENKO,ˇ AND J. VOLCIˇ Cˇ Proof. Denote h = (dp ) T . If α = (α ) Rm, then ij i j ∈ n i i ∈ t α Hα = αi1 αi2 (dpi1 , dpi2 ) iX1,i2 t = α α tr h h i1 i2 i1j i2j iX1,i2 Xj t = tr α h α h i1 i1j i2 i2j Xj iX1,i2 t = tr αihij αihij . ! ! ! Xj Xi Xi t Therefore φ(tr(hh )) 0 for all h T implies that φ(H) is positive semidefinite. ≥ ∈ n From the proof of Lemma B.3 we see that to prove Theorem 4.8 using the Procesi-Schwarz theorem, one would need to extend the chain of equivalences in Theorem B.1 with the condition t (iii’) α φ(H)α 0 for all α Rm. ≥ ∈ However, (iii’) (iii) fails in our context. ⇒ Example B.4. Let O2(R) act on M2(R) by conjugation, i.e., n = 2 and g = 1 in the setting of this paper. If Ξ = Ξ(2) is a generic 2 2 matrix and × 2 t y1 = tr(Ξ), y2 = tr(Ξ ), y3 = tr(ΞΞ ), O2(R) then R[ξ] = R [y1,y2,y3] (see e.g. [ADS06]; algebraic independence follows from the Jaco- bian criterion). For this choice of generators we have 2 2y1 2y1 H = 2y 4y 4y . 1 3 2 2y1 4y2 4y3 Let R be the real closure of the rational function field R(ε) endowed with the ordering 0 <ε<α for every α R . 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Igor Klep, Department of Mathematics, University of Auckland E-mail address: [email protected] Spelaˇ Spenko,ˇ Departement Wiskunde, Vrije Universiteit Brussel E-mail address: [email protected] Jurij Volciˇ c,ˇ Department of Mathematics, Ben-Gurion University of the Negev E-mail address: [email protected] POSITIVE TRACE POLYNOMIALS 33 NOT FOR PUBLICATION Contents 1. Introduction...... 1 1.1. Main results and reader’s guide...... 4 Acknowledgments ...... 5 2. Preliminaries ...... 5 2.1. Polynomial and trace -identities ...... 6 2.1.1. A particular trace∗ -identity ...... 6 2.2. Generic matrices and the∗ trace ring ...... 8 2.2.1. Reynolds operator ...... 9 2.3. Positive involutions and totally positive elements ...... 9 3. Counterexample to the 3 3 universal Procesi-Schacher conjecture...... 10 4. The Krivine-Stengle Positivstellensatz× for trace polynomials ...... 13 4.1. Cyclic quadratic modules and preorderings...... 13 4.2. An extension theorem ...... 15 4.3. Stellens¨atze...... 18 5. Positivstellens¨atze for compact semialgebraic sets ...... 20 5.1. Archimedean (cyclic) quadratic modules ...... 20 5.2. Schm¨udgen’s Positivstellensatz for trace polynomials...... 21 5.3. Putinar’s Positivstellensatz for generic matrices ...... 21 5.4. Putinar’s theorem for trace polynomials ...... 23 6. Examples...... 24 Appendix A. Constructions of the Reynolds operator ...... 26 A.1. Computing n′ ...... 27 A.1.1. First method...... R ...... 27 A.1.2. Second method...... 27 A.2. From n′ to n ...... 27 A.2.1. FirstR method...... R ...... 28 A.2.2. Second method...... 28 Appendix B. How not to prove the extension Theorem 4.8...... 28 References...... 31