Matricial real algebraic geometry

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1 Magdeburg, February 23th, 2012

2 Notation R - a commutative ring M n (R) - the ring of all n n matrices with entries in R S n (R) - the set of all symmetric matrices in M n (R) Mn (R) 2 - the set of all finite sums i AT i A i, A i M n (R). Z n (R) - the center of M n (R), i.e. the set R I n.

3 The real spectrum PART 1 The real spectrum of M n (R)

4 Definition of the real spectrum For a given ordering P of R, write psd n (P) n := {A S(R) v T Av P for all v R n } for the set of all P-positive semi-definite matrices in S n (R). The real spectrum of M n (R) is defined by Sper(M n (R)) := {psd n (P) P Sper(R)}. Clearly, P psd n (P) is a one-to-one correspondence from Sper(R) to Sper(M n (R)).

5 Alternative descriptions of the real spectrum (1) For every P Sper(R) there exists a homomorphism φ from R into some real closed field K such that P = {a R φ(a) 0}. Then psd n (P) = {[a ij ] S n (R) [φ(a ij )] is positive semi-definite}. (2) From the usual principal minors test for positive semi-definite matrices over real closed fields, we deduce that psd n (P) = {A S n (R) all principal minors of A belong to P}. (3) Q: Is there an intrinsic description of elements of Sper(M n (R)) (i.e. without reference to Sper(R))?

6 An intrinsic description of the spectrum A subset Q of S n (R) belongs to Sper(M n (R)) (i.e. it is of the form psd n (P) for some P Sper(R)) iff it satisfies (1)-(3) below (1) Q is a proper quadratic module in M n (R), i.e. I n Q, I n Q, Q + Q Q and A T QA Q for every A M n (R). (2) Q is prime in the sense that: for every A S n (R) and B Z n (R) := R I n such that AB 2 Q we have that either A Q or B Q Q. (3) Q Z n (R) is closed for multiplication and Z n (R) Q Q (i.e. Q Z n (R) is an ordering of Z n (R)). Idea of the proof: By reducing to the field case it can be shown that there is a one-to-one correspondence between prime quadratic modules in M n (R) and prime quadratic modules in R.

7 Spectral topologies For every Q = psd n (P) Sper(M n (R)) we define its positive part Q + := {A S n (R) v T Av P + for every v R n \ (supp P) n }. Clearly, Q + is the set of all P-positive definite matrices. For every finite subset G of S n (R) define U G = {Q Sper(M n (R)) G Q + } and V G = {Q Sper(M n (R)) G Q}. The sets U G and V G generate the constructible topology on Sper(M n (R)).

8 Spectral topologies To show that P psd n (P) is a homeomorphism it suffices to prove the following: Theorem 1: (a) For every finite subset G of S n (R) there exists a finite subset G of M G Z n (R) such that U G = U G. (b) For every finite subset G of S n (R) there exists a finite subset G of M G Z n (R) such that V G = V G. Idea of the proof: This is done by induction on n using Schur complements. (Similarly as in Schmüdgen s survey paper.) As a corollary one obtains Artin-Lang theorem for Sper(M n (R)).

9 Part 2: Nichtnegativstellensatz PART 2 The Krivine-Stengle Nichtegativstellensatz for M n (R)

10 Earlier results Gondard & Ribenboim Artin s theorem for matrices: Suppose that a matrix polynomial F S(R[x]) is positive semi-definite in every point of R m. Then there exists a nonzero c R[x] such that c 2 F M n (R[x]) 2.

11 Earlier results Schmüdgen Stengle s theorem for matrices: Given F, G 1,..., G k S n (R[x]), pick f j, g ij R[x] such that {x R m F (x) 0} = {x R m f j (x) 0 for all j} {x R m G i (x) 0 for all i} = {x R m g ij (x) 0 for all i, j}. (They exist by Theorem 1(b)). Let T be the preordering in R[x] generated by all g ij. Then the following are equivalent: 1. F (x) 0 for every x R m such that G i (x) 0 for all i. 2. For every j there exist n j N and s j, t j T such that f j s j = f 2n j j + t j.

12 Nichtnegativstellensatz A subset T of M n (R) is a preordering if (a) T is a quadratic module and (b) T Z n (R) is closed for multiplication. The smallest preordering which contains G will be denoted by T G. Theorem 2: For every finite subset G of S n (R) and every element F S n (R), the following are equivalent: (1) F Q V G Q (2) There exist B T G and C T G and s N such that FB = BF = F 2s + C. Note: B can be non-central, so it does not imply Artin s Theorem.

13 Proof (2) (1) is easy. (1) (2) First reduce to the case G = {g 1 I n,..., g m I n } by Theorem 1(b) and write S = {g 1,..., g m }. Let det(f λ I n ) = ( λ) n + c 1 ( λ) n c n be the characteristic polynomial of F. It belongs to R[λ]. Since c 1,..., c n are sums of principal minors of F, they belong (by assumption (1)) to every ordering P of R which contains S.

14 Proof It follows that p(λ) := det(f λ I n ) ( λ) n belongs to every ordering of R[λ] which contains S { λ}. By the usual Positivstellensatz, there exist s(λ), t(λ) T R[λ] S { λ} and k N such that p(λ)s(λ) = p(λ) 2k + t(λ). We can write s(λ) = σ 1 (λ) λσ 2 (λ) and t(λ) = τ 1 (λ) λτ 2 (λ) where σ 1 (λ), σ 2 (λ), τ 1 (λ), τ 2 (λ) T R[λ] S.

15 Proof If follows that σ 1 (F ), σ 2 (F ), τ 1 (F ), τ 2 (F ) T R[F ] S T G. By the Cayley-Hamilton Theorem, p(f ) = ( F ) n. It follows that ( F ) n (σ 1 (F ) F σ 2 (F )) = F 2nk + τ 1 (F ) F τ 2 (F ). If n is even, then we can rewrite this as F (F n σ 2 (F ) + τ 2 (F )) = F 2nk + (τ 1 (F ) + F n σ 1 (F )). where both brackets belong to T G. If n is odd, then we can rewrite this as F (F n 1 σ 1 (F ) + τ 2 (F )) = F 2nk + (τ 1 (F ) + F n+1 σ 2 (F )). where both brackets belong to T G.

16 A counterexample Example: Take G = [ x x 3 ] and F = [ x ]. Clearly, F Q for every Q Sper(M n (R[x])) such that G Q. We claim that there is no b I 2 T G Z n (R[x]) such that Fb = F 2k + C for some k N and C = [c ij ] T G (=sos+x 3 sos). Namely, if such a b exists, then xb = x 2k + c 11 and b = 1 + c 22 for some c 11 = u 1 + x 3 v 1 and c 22 = u 2 + x 3 v 2 with u i, v i R[x] 2, then x(1 + u 2 (x) + x 3 v 2 (x)) = x 2k + u 2 (x) + x 3 v 2 (x). Since x divides x 2k + u 2 (x) which belongs to R[x] 2, it follows that also x 2 divides x 2k + u 2 (x). After canceling x on both sides, we get that the right-hand side is divisible by x while the left-hand side is not.

17 Part 3: Positivstellensatz PART 3 The Krivine-Stengle Positivstellensatz for M n (R)

18 Positivstellensatz Theorem 3: For every finite subset G of S n (R) and every element F S n (R), the following are equivalent: (1) F Q V G Q + (2) There exist b T G Z n (R) and V T G such that F (1 + b) = I n + V. Note: the denominator 1 + b in (2) is central.

19 Proof Suppose that (1) implies (2) for all symmetric matrix polynomials of size n 1 and pick a symmetric F of size n which satisfies (1). We write [ ] f11 g F = g T H and observe that [ ] T [ ] [ ] f11 g f11 g f11 g 0 f 11 I n 1 g T = H 0 f 11 I n 1 [ f f 11 H where H = f 11 H g T [ g. Since F 0] on K S, it follows that f11 g f 11 > 0 on K S, hence is invertible on K 0 f 11 I S. It n 1 follows that H 0 on K S. ] (1)

20 Proof By the induction hypothesis there exist s T S and U T n 1 S such that (1 + s) H = I n 1 + U. (2) On the other hand, there exists by n = 1 elements s 1, u 1 T such that (1 + s 1 )f 11 = 1 + u 1. (3) From (1) we get (with I = I n 1 ) [ f11 4 f11 g g T H ] = [ f11 g 0 f 11 I ] T [ f f 11 H ] [ f11 g 0 f 11 I Now we cancel f 11, multiply by (1 + s)(1 + s 1 ) 4 and use (2), (3)): ] (4)

21 Proof [ ] [ ] T (1 + s)(1 + s 1 )(1 + u 1 ) 3 f11 g 1 + u1 (1 + s g T = 1 )g H 0 (1 + u 1 )I [ (1 + s)(1 + u1 ) 2 ] [ ] u1 (1 + s 1 )g 0 (1 + s 1 ) 2 (5) (I + U) 0 (1 + u 1 )I Since [ (1 + s)(1 + u1 ) (1 + s 1 ) 2 (I + U) for some W TS n, it follows that = [ 1 + u1 (1 + s 1 )g 0 (1 + u 1 )I for some W T n S. ] = I n + W [ ] (1 + s)(1 + s 1 )(1 + u 1 ) 3 f11 g g T = H ] T [ 1 + u1 (1 + s 1 )g 0 (1 + u 1 )I ] + W (6)

22 Proof Write g = (1 + s 1 )g and c = g g T R 2 and note that σ := ci n 1 g T g M n 1 (R) 2. Write v = 1 + c and multiply (6) by v(1 + v) to get [ ] v(1 + v)(1 + s)(1 + s 1 )(1 + u 1 ) 3 f11 g g T = H [ (1 + u1 ) = v(1 + v) 2 (1 + u 1 ) g (1 + u 1 ) g T g T g + (1 + u 1 ) 2 I + ] + v(1 + v)w = [ v(1 + u1 ) = (v(1 + u 1 ) 2 + v 2 (2u 1 + u1 2 ) + 1)I + (v + 1)σ [ ] T [ v(1 + u1 ) (1 + v) g v(1 + u1 ) (1 + v) g which clearly belongs to I n + TS n. It is also clear that v(1 + v)(1 + s)(1 + s 1 )(1 + u 1 ) 3 belongs to 1 + T S. ] + ] + v(1 + v)w

23 Part 4: Schmüdgen s Positivstellensatz PART 4 Schmüdgen s Positivstellensatz for M n (R[x])

24 Schmüdgen s Positivstellensatz Theorem 4: Suppose that G = {G 1,..., G k } S n (R[x]) are such that the set K G := {x R m G 1 (x) 0,..., G k (x) 0} is compact. Then the preordering T G is an archimedean q.m. Proof: Write R = R[x]. By Theorem 1(b), we can pick a finite set G = {g 1 I n,..., g r I n } M G Z n (R) such that K G = K G. Since K G is compact, so is K {g1,...,g r }. By the usual Schmüdgen s Positivstellensatz, T {g1,...,g r } is an archimedean preordering in R. The following two observations then imply that T G is archimedean. T G = T {g1,...,g r } M n (R) 2. For every B S n (R) there exists σ R 2 such that σ I n B M n (R) 2. Since T G T G, it follows that T G is archimedean.

25 Schmüdgen s Positivstellensatz Combining Theorem 4 with Hol-Scherer Theorem, we get a matrix version of Schmüdgen s Positivstellensatz: Corollary. Suppose that G = {G 1,..., G k } S n (R[x]) are such that the set K G := {x R m G 1 (x) 0,..., G k (x) 0} is compact. Then every F S n (R[x]) which satisfies F (x) 0 on K G belongs to T G. Recall that Hol-Scherer Theorem (2006) says that for every archimedean q.m. M in M n (R[x]) and every F S n (R[x]) which satisfies F (x) 0 on K M, we have that F M.

26 Multivariate matrix moment problem Similarly, we can obtain a solution of the multivariate matrix moment problem by combining Theorem 4 with the following result of Ambrozie & Vasilescu (2003): Proposition: Let M be an archimedean quadratic module on R[x] and L a linear functional on M n (R[x]) such that L(m A T A) 0 for every m M and A M n (R[x]), then there exists a measure µ on K M with values in positive semi-definite real n n matrices such that for every F M n (R[x]) L(F ) = tr(f dµ). K M

A brief history of noncommutative Positivstellensätze. Jaka Cimprič, University of Ljubljana, Slovenia

A brief history of noncommutative Positivstellensätze Jaka Cimprič, University of Ljubljana, Slovenia Hilbert s Nullstellensatz: If f, g 1,..., g m R := C[X 1,..., X n ] then the following are equivalent: