A 3 3 DILATION COUNTEREXAMPLE

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1 A 3 3 DILATION COUNTEREXAMPLE MAN DUEN CHOI AND KENNETH R DAVIDSON Dedicated to the memory of William B Arveson Abstract We define four 3 3 commuting contractions which do not dilate to commuting isometries However they do satisfy the scalar von Neumann inequality These matrices are all nilpotent of order 2 We also show that any three 3 3 commuting contractions which are scalar plus nilpotent of order 2 do dilate to commuting isometries 1 Introduction Seminal work of SzNagy [8] showed that every contraction A has a coextension to an isometry of the form [ ] A 0 S = This provides a simple proof of von Neumann s inequality: p(a) p := sup p(z) z 1 for all polynomials Indeed, this remains valid for matrices of polynomials [ pij (A) ] [ ] pij := sup [ p ij (z) ] z 1 A decade later, Ando [1] showed that two commuting contractions have simultaneous coextensions to a common Hilbert space which are commuting isometries This yields the 2-variable matrix von Neumann inequality However Varopoulos [10] showed that there exist three commuting contractions which do not satisfy von Neumann s inequality, and therefore do not have a simultaneous coextension to three commuting isometries In the appendix, he and Kaijser provide an example with 5 5 matrices A related example of Parrott [6] provides three commuting contractions which do satisfy the scalar von Neumann inequality but 2010 Mathematics Subject Classification 47A20, 47A30, 15A60 Key words and phrases dilation, von Neumann inequality Partially supported by grants from NSERC 1

2 2 MD CHOI AND KR DAVIDSON fail the matrix version Thus they also do not dilate A dilation theorem of Arveson [2] shows that there is a simultaneous coextension to commuting isometries if and only if the matrix von Neumann inequality holds A nice treatment of this material is contained in Paulsen [7], and the earlier material is contained in the classic text [9] Holbrook [4] found three 4 4 matrices which are commuting contractions yet do not coextend to commuting isometries He also showed [5] that for 2 2 matrices, arbitrary commuting families of commuting contractions do have commuting isometric coextensions See also [3] Holbrook asks what the situation is for 3 3 matrices Sometimes these results are stated instead for unitary (power) dilations of the form U i = 0 0 A i 0 on a Hilbert space K = K H K + If these unitaries commute, then the restriction of U i to H K +, namely the lower 2 2 corner, yields commuting isometric coextensions S i ; (and the compression to K H yields commuting coisometric extensions of the A i ) Conversely, suppose that commuting contractions A i have coextensions to commuting isometries S i An old result of Ito and Brehmer [9, Proposition I62] shows that the S i dilate to commuting unitaries U i so that P H U k 1 1 U ks s H = A k 1 1 A ks s for k i 0 It follows that they simultaneously have a triangular form as given above Therefore these two formulations are equivalent In this note, we provide an example of four commuting contractions in the 3 3 matrices M 3 which cannot be coextended to commuting isometries The question of whether there are three commuting 3 3 contractive matrices which cannot be coextended to commuting isometries remains open Our examples are nilpotents of order 2 We show that any finite family of commuting contractions of this form always satisfy the scalar von Neumann s inequality It would be of interest to know whether the scalar von Neumann inequality holds for commuting 3 3 contractions For our counterexample, we exhibit a specific matrix polynomial which shows that the matrix valued von Neumann inequality fails We also show that any three commuting 3 3 contractions which are of the form scalar plus nilpotent of order 2 always do have commuting isometric coextensions

3 A 3 3 DILATION COUNTEREXAMPLE 3 When the second author discovered these examples, he found out that the first author and Y Zhong had a similar example from many years ago which was never published Zhong has left academia, and could not be contacted but he shares in the credit for this work 2 The Example Let H = C 3 have an orthonormal basis f, e 1, e 2 For i = 1, 2, pick θ i (0, π/2) and set c i = cos θ i and s i = sin θ i Define and A 1 = e 1 f A 2 = e 2 f = = A 3 = (c 1 e 1 + s 1 e 2 )f A 4 = (c 2 e 1 + is 2 e 2 )f = c = c s is Observe that A i A j = 0 for all 1 i, j 4 Theorem 21 The matrices A 1, A 2, A 3 and A 4 do not have simultaneous coextensions to commuting isometries Proof Suppose that the A i coextend to commuting isometries [ ] Ai 0 S i = for i = 1, 2, 3, 4 D i V i on a Hilbert space K We may assume that K is the smallest invariant subspace for the S i containing H That is, if we write S k = S k 1 1 S k 2 2 S k 3 3 S k 4 4 for k = (k 1, k 2, k 3, k 4 ) N 4, then K = S k H k N 4 Observe that since A i f = 1, we have Hence Therefore S i f = A i f for i = 1, 2, 3, 4 (c 1 S 1 + s 1 S 2 S 3 )f = 0 0 = S k (c 1 S 1 + s 1 S 2 S 3 )f = (c 1 S 1 + s 1 S 2 S 3 )S k f

4 4 MD CHOI AND KR DAVIDSON Since S i f = e i for i = 1, 2, ker(c 1 S 1 + s 1 S 2 S 3 ) So Similarly, Since S 3 is an isometry, It follows that I = S 3S 3 k N 4 S k span{f, S 1 f, S 2 f} = K S 3 = c 1 S 1 + s 1 S 2 S 4 = c 2 S 1 + is 2 S 2 = c 2 1S 1S 1 + s 2 1S 2S 2 + c 1 s 1 (S 2S 1 + S 1S 2 ) = I + c 1 s 1 (S 2S 1 + S 1S 2 ) S 2S 1 + S 1S 2 = 0 Likewise, using the fact that S 4 is an isometry, Therefore 0 = (is 2 ) S 1 + S 1(iS 2 ) = i(s 2S 1 S 1S 2 ) S 2S 1 = 0 This implies that S 1 and S 2 have pairwise orthogonal ranges, and therefore do not commute This contradiction establishes the result Remark 22 It is possible to coextend any three of these operators to commuting isometries This will follow from Theorem 43 Indeed, the construction given there will simultaneously coextend A 1, A 2 and all matrices A j = (c j e 1 + s j e 2 )f such that c j s j have a common argument So this example could not be given using real matrices Remark 23 It is not even possible to find commuting isometries of the form [ ] Ai B S i = i C i D i As in our proof above, we may suppose that H is a cyclic subspace Exactly the same argument shows that S 3 = c 1 S 1 + s 1 S 2 and S 4 = c 2 S 1 + is 2 S 2 Hence S 2S 1 = 0, which contradicts commutativity

5 A 3 3 DILATION COUNTEREXAMPLE 5 3 Von Neumann s Inequality It is also of interest to find commuting contractions in M 3 that fail the scalar von Neumann inequality Apparently computer searches for such examples have been unsuccessful We show that our example does satisfy the scalar von Neumann inequality So it does not settle this question We provide an explicit matrix polynomial which fails the matrix von Neumann inequality Lemma 31 If A 1,, A n M 3 are commuting nilpotents of order 2, then there is a vector f and vectors v i (Cf) so that either A i = v i f for 1 i n or A i = fv i for 1 i n Proof Observe that a nilpotent A M 3 of order 2 must have rank one Thus it can be expressed as A = vf where f is a unit vector And since 0 = A 2 = v, f A, we have that v, f = 0 If two such non-zero operators A i = v i fi commute, then v 2, f 1 v 1 f 2 = (v 1 f 1 )(v 2 f 2 ) = (v 2 f 2 )(v 1 f 1 ) = v 1, f 2 v 2 f 1 So either v 2, f 1 = v 1, f 2 = 0 or v 2 f2 is a multiple of v 1 f1, in which case this identity remains true So span{v 1, v 2 } span{f 1, f 2 } Furthermore, if n non-zero operators v i fi commute, then the pairwise relations yield span{v 1,, v n } span{f 1,, f n } Therefore one of these subspaces is 1-dimensional By taking adjoints if necessary, we may suppose that span{f 1,, f n } is one dimensional So after a scalar change, we have A i = v i f for i = 1,, n; and each v i belongs to (Cf) Proposition 32 Any finite number of commuting contractions A 1,, A n in M 3 of the form scalar plus order 2 nilpotent satisfy the scalar von Neumann inequality Proof We first suppose that each A i is a nilpotent of order 2 By Lemma 31, we may suppose that there is a unit vector f and vectors v i (Cf) of norm at most 1 so that A i = v i f Since A i A j = 0 for all 1 i, j n, we need concern ourselves only with the linear part of a polynomial

6 6 MD CHOI AND KR DAVIDSON Consider a polynomial p(z) = c + n a i z i + q(z) where q(z) consists of higher order terms Without loss of generality, we may suppose that some a i 0 Let λ i be scalars of modulus 1 so that a i = λ i a i Set Clearly B 1 Then ( n ) 1 n B = a i a i A i p(a 1,, A n ) = ci + n a i A i = ci + n a i B = p(λ 1 B,, λ n B) = q(b), where q(x) = p(λ 1 x,, λ n x) Hence by von Neumann s inequality, p(a 1,, A n ) = q(b) q p Now suppose that A i = λ i I + N i where Ni 2 = 0 If N i 0, the fact that A i 1 implies that λ i < 1 If some A i = λ i I with λ i = 1, replace it with A i = (1 ε)λ i I for ε > 0 small If we establish the inequality for this new n-tuple, we recover the case we desire by letting ε tend to 0 Define Möbius maps Then b i (z) = z λ i 1 λ i z for 1 i n B i := b i (A i ) = (1 λ i 2 ) 1 N i are commuting nilpotents of order 2, and A i = b 1 i (B i ) Moreover, B i are contractions by the one variable von Neumann inequality (or by direct computation) If p is a polynomial, then q(z) = p(b 1 1 (z 1 ),, b 1 n (z n )) is a function in the polydisc algebra A(D n ) of the same norm as p because each b 1 i takes the circle T onto itself Hence p(a 1,, A n ) = q(b 1,, B n ) q = p Therefore the scalar von Neumann inequality is satisfied

7 A 3 3 DILATION COUNTEREXAMPLE 7 If the matrix von Neumann inequality holds for a 4-tuple of matrices A 1,,A 4, then the canonical map from A(D 4 ) into A = Alg({A i }) taking z i to A i for 1 i 4 is completely contractive Therefore Arveson s Dilation Theorem [2] shows that the 4-tuple does dilate to commuting unitaries Hence the restriction to the common invariant subspace generated by the original space H yields a coextension to commuting isometries Hence there is a matrix polynomial which shows that this inequality fails for our example in Theorem 21 We will exhibit one Recall the notation of Theorem 21 Apply the Gram-Schmidt process to the vectors u 1 and u 2 to get orthogonal unit vectors f 1 and f 2, where 1 0 α 1 β 1 u 1 = 0 c 1, u 2 = 1 s 1, f 1 = α 2 α 3, and f 2 = β 2 β 3 is 2 α 4 β 4 c 2 Proposition 33 Let p(z) = ] [ᾱ1 z 1 + ᾱ 2 z 2 + ᾱ 3 z 3 + ᾱ 4 z 4 β 1 z 1 + β 2 z 2 + β 3 z 3 + β 4 z 4 Then p = sup p(z) < 2 = p(a 1, A 2, A 3, A 4 ) z i =1 So the matrix von Neumann inequality fails for (A 1, A 2, A 3, A 4 ) Proof Let z = (z 1, z 2, z 3, z 4 ) t with z i = 1 Since f 1 and f 2 are orthonormal, p(z) 2 = z, f z, f 2 2 z 2 = 2 This inequality is strict unless z span{f 1, f 2 } = span{u 1, u 2 } By compactness, the norm of p is exactly 2 only if this value is obtained However we claim that no z with z i = 1 lies in this subspace Indeed, suppose that We require z = z 1 z 2 z 3 z a = a 0 c 1 + b 1 s 1 = b ac 1 + bs 1 c 2 is 2 ac 2 + ibs 2 1 = a = b = ac 1 + bs 1 = ac 2 + ibs 2 Arguing as in the proof of Theorem 21, we find that Re(āb) = 0 = Re(iāb) Thus āb = 0, contradicting a = b = 1 So p < 2

8 8 MD CHOI AND KR DAVIDSON Now we compute p(a 1, A 2, A 3, A 4 ) Clearly it suffices to consider the 2, 1 and 3, 1 entries That is p ( [ ] [ [ ] [ ] ᾱ 1 + ᾱ 3 c 1 + ᾱ 4 c 2 u 1, f c1 c2 ),,, = ᾱ 2 + ᾱ 3 s 1 + ᾱ 4 is 2 0 1] s 1 is 2 β 1 + β 3 c 1 + β 4 c 2 = u 2, f 1 β 2 + β 3 s 1 + β u 1, f 2 4 is 2 u 2, f 2 Therefore p(a 1, A 2, A 3, A 4 ) 2 = 2 2 u i, f j 2 = j=1 2 u i 2 = 4 4 Dilating Three 3 3 Nilpotents of Order 2 The purpose of this section is to prove a positive dilation result for certain triples of 3 3 commuting contractions First we need a couple of lemmas Lemma 41 Suppose that three commuting contractions A 1, A 2 and A 3 have coextensions S 1, S 2 and S 3 which are commuting isometries Then if a i 1 for i = 1, 2, 3, the operators a i A i also have coextensions which are commuting isometries Proof Observe that a i A i coextend to commuting contractions a i S i If a 1 < 1, let d i = (1 a i 2 ) 1/2 and coextend a 1 S 1 to a 1 S d 1 S T 1 = 0 S S 1 0 and simultaneously for i = 2, 3, coextend a i S i to a 1 S i a 1 S i 0 0 a i T i = a i S i I = 0 0 a 1 S i a 1 S i It is easy to see that these contractions commute, and T i are isometries Now dilate a 2 T 2 to an isometry, and dilate T 1 and T 3 to commuting isometries in the same manner Finally, repeat a third time to dilate the third term to an isometry

9 A 3 3 DILATION COUNTEREXAMPLE 9 Lemma 42 Given three unit vectors v 1, v 2, v 3 in C 2, there exist three commuting unitaries U 1, U 2, U 3 in M 2 such that U i v j = U j v i for 1 i, j 3 Proof We may choose an orthonormal basis for C 2 in which v 1 = [ 0 1 ] We set U 1 = I 2 If v 2 = e iθ v 1, let U 2 = e iθ I 2 and choose U 3 to be any unitary matrix such that U 3 v 1 = v 3 This works Otherwise, v 1 and v 2 are linearly independent Write v 3 = av 1 + bv 2 For convenience, we may multiply v 3 and v 2 by scalars of modulus 1 so that a and b are real This does not affect the problem Write v 2 = [ α2 β 2 ], v 3 = [ α3 β 3 ], U 2 = A simple calculation shows that au 1 + bu 2 = [ ] β2 α 2 ᾱ 2 β 2 and U 3 = [ ] [ ] a + b β2 bα 2 β3 α = 3 = U bᾱ 2 a + bβ 2 ᾱ 3 β 3 3 [ ] β3 α 3 ᾱ 3 β 3 This shows that the unitary matrices commute By construction, U i v 1 = v i = U 1 v i for i = 2, 3 Finally observe that U 3 v 2 = (au 1 + bu 2 )v 2 = av 2 + bu 2 v 2 = U 2 (av 1 + bv 2 ) = U 2 v 3 Theorem 43 Suppose that A 1, A 2 and A 3 are three commuting 3 3 matrix contractions which are all of the form scalar plus nilpotent of order 2 Then there exist commuting isometric coextensions S 1, S 2 and S 3 of A 1, A 2 and A 3 Proof First assume that the A i are nilpotent of the form A i = v i f for i = 1, 2, 3, where the v i are unit vectors in (Cf) C 2 Choose a basis for C 2 so that v 1 = [ 0 1 ] Let U i be the commuting 2 2 unitaries given by Lemma 42 Then the construction yields the identities U[ i v 1 = ] v i γi α for i = 1, 2, 3 So the second column of each U i is v i, say U i = i δ i β i

10 10 MD CHOI AND KR DAVIDSON Observe that if U is the bilateral shift on l 2, then W i = U U i are commuting unitaries of the form: O O O O O O U i O O O O O W i = O U i O O O O O O U i O O O O O O U i O O O O O O U i O on K = K C 4 K + Here O is a 2 2 zero matrix Decompose the central 4 4 block of W i as C H: [ ] [ ] O O = U i O γ i α i 0 0 = 3 u i A i δ i β i 0 0 The spaces K + and H K + are invariant for each W i Therefore W i are commuting unitary (power) dilations of the A i Note that Wi then form commuting unitary dilations of the adjoints, A i ; and H is the difference of invariant subspaces K C 4 and K C By Lemma 31, if A i are commuting 3 3 nilpotents of order 2, then either they have the form A i = v i f or their adjoints do By Lemma 41, it suffices to dilate the normalized matrices Ãi := A i / A i Hence we can assume that each v i = 1 The argument above produces commuting unitary dilations The restriction of these unitaries to the smallest common invariant subspace containing the original 3-dimensional space yields commuting isometric coextensions of the Ãi We reduce the general case to the nilpotent case as in the previous section Suppose that A i = λ i I + N i where Ni 2 = 0 Then λ i A i 1 Moreover, if λ i = 1, then A i = λ i I is already an isometry, and we coextend it to λ i I on the larger space When λ i < 1, define the Möbius map b i (z) = z λ i 1 λ i z

11 A 3 3 DILATION COUNTEREXAMPLE 11 Then b i (A i ) are commuting nilpotents of order 2 Dilate them to commuting unitaries W i as above Then define U i = b 1 i (W i ) These are commuting unitaries dilating A i References [1] T Ando, On a pair of commuting contractions, Acta Sci Math (Szeged) 24 (1963), [2] W Arveson, Subalgebras of C*-algebras, Acta Math 123 (1969), [3] S Drury, Remarks on von Neumann s inequality, Banach spaces, harmonic analysis, and probability theory (Storrs, Conn, 1980/1981), 14 32, Lecture Notes in Math, 995, Springer, Berlin, 1983 [4] J Holbrook, Schur norms and the multivariate von Neumann inequality, Recent advances in operator theory and related topics (Szeged, 1999), Oper Theory Adv Appl 127, , Birkhäuser, Basel, 2001 [5] J Holbrook, Inequalities of von Neumann type for small matrices, Function spaces (Edwardsville, IL, 1990), Lecture Notes in Pure and Appl Math 136, , Dekker, New York, 1992 [6] S Parrott, Unitary dilations for commuting contractions, Pacific J Math 34 (1970), [7] V Paulsen, Completely bounded maps and operator algebras, Cambridge Studies in Advanced Mathematics 78, Cambridge University Press, Cambridge, (2002), xii+300 pp [8] Sz Nagy, B, Sur les contractions de l espace de Hilbert, Acta Sci Math (Szeged) 15 (1953), [9] B Sz Nagy and C Foiaş, Harmonic analysis of operators on Hilbert space, North Holland Pub Co, London, 1970 [10] N Varopoulos, On an inequality of von Neumann and an application of the metric theory of tensor products to operators theory, J Funct Anal 16 (1974), Mathematics Department, University of Toronto, Toronto, ON M5S 2E4, CANADA address: choi@mathtorontoedu Pure Math Department, University of Waterloo, Waterloo, ON N2L 3G1, CANADA address: krdavids@uwaterlooca