A matrix problem and a geometry problem

Transcription

1 University of Szeged, Bolyai Institute and MTA-DE "Lendület" Functional Analysis Research Group, University of Debrecen Seminar University of Primorska, Koper

2 Theorem (L. Molnár and W. Timmermann, 2011) Let H be a complex separable Hilbert space with dim H 3. Assume φ: B s (H) B s (H) is a bijection such that [φ(a), φ(b)] = [A, B] (A, B B s (H)).

3 Theorem (L. Molnár and W. Timmermann, 2011) Let H be a complex separable Hilbert space with dim H 3. Assume φ: B s (H) B s (H) is a bijection such that [φ(a), φ(b)] = [A, B] (A, B B s (H)). Then there exist either a unitary or an antiunitary operator U on H and a function f : B s (H) R such that φ(a) = ±UAU + f (A)I (A B s (H)).

4 The proof uses the following theorem: Theorem (L. Molnár and P. emrl, 2005) Let H be a complex separable Hilbert space with dim H 3 and let φ: B s (H) B s (H) be a bijective transformation which preserves commutativity in both directions.

5 The proof uses the following theorem: Theorem (L. Molnár and P. emrl, 2005) Let H be a complex separable Hilbert space with dim H 3 and let φ: B s (H) B s (H) be a bijective transformation which preserves commutativity in both directions. Then there exists either a unitary or an antiunitary operator U on H and for every operator A B s (H) there is a real valued bounded Borel function f A on σ(a) such that φ(a) = Uf A (A)U (A B s (H)).

6 The proof uses the following theorem: Theorem (L. Molnár and P. emrl, 2005) Let H be a complex separable Hilbert space with dim H 3 and let φ: B s (H) B s (H) be a bijective transformation which preserves commutativity in both directions. Then there exists either a unitary or an antiunitary operator U on H and for every operator A B s (H) there is a real valued bounded Borel function f A on σ(a) such that φ(a) = Uf A (A)U (A B s (H)). Question What happens in two dimensions?

7 Commutativity preservers on B s (C 2 ) Two linearly independent operators A, B B s (C 2 ) commute α, β R s.t. αa + βb = I.

8 Commutativity preservers on B s (C 2 ) Two linearly independent operators A, B B s (C 2 ) commute α, β R s.t. αa + βb = I. E.g. if φ: B s (C 2 ) B s (C 2 ) is non-singular and linear, then φ preserves commutativity in both directions φ(i ) (R \ {0}) I.

9 Commutativity preservers on B s (C 2 ) Two linearly independent operators A, B B s (C 2 ) commute α, β R s.t. αa + βb = I. E.g. if φ: B s (C 2 ) B s (C 2 ) is non-singular and linear, then φ preserves commutativity in both directions φ(i ) (R \ {0}) I. So the preservation of commutativity provides too few information.

10 The rst step: linear commutativity preservers Theorem (with G. Nagy, 2014) Suppose that d N, is an arbitrary unitarily invariant norm and φ: B s (C d ) B s (C d ) is a (real-)linear transformation such that [φ(a), φ(b)] = [A, B] (A, B B s (C d )).

11 The rst step: linear commutativity preservers Theorem (with G. Nagy, 2014) Suppose that d N, is an arbitrary unitarily invariant norm and φ: B s (C d ) B s (C d ) is a (real-)linear transformation such that [φ(a), φ(b)] = [A, B] (A, B B s (C d )). Then there exist either a unitary or an antiunitary operator U on C d and a linear functional f : B s (C d ) R such that φ(a) = UAU + f (A)I (A B s (C d )) or φ(a) = UAU + f (A)I (A B s (C d )).

12 What can we do in general? Problem (Reformulation) Characterize those maps φ: B s (C 2 ) B s (C 2 ) s.t. det[a, B] = det[φ(a), φ(b)] (A, B B s (C 2 )).

13 What can we do in general? Problem (Reformulation) Characterize those maps φ: B s (C 2 ) B s (C 2 ) s.t. det[a, B] = det[φ(a), φ(b)] (A, B B s (C 2 )). Let Z 2 := {A B s (C 2 ): Tr A = 0} and φ: B s (C 2 ) Z 2 B s (C 2 Tr φ(a) ), φ(a) = φ(a) I. 2

14 What can we do in general? Problem (Reformulation) Characterize those maps φ: B s (C 2 ) B s (C 2 ) s.t. det[a, B] = det[φ(a), φ(b)] (A, B B s (C 2 )). Let Z 2 := {A B s (C 2 ): Tr A = 0} and φ: B s (C 2 ) Z 2 B s (C 2 Tr φ(a) ), φ(a) = φ(a) I. 2 We dene the following mapping: ψ := φ Z 2 : Z 2 Z 2.

15 What can we do in general? Problem (Reformulation) Characterize those maps φ: B s (C 2 ) B s (C 2 ) s.t. det[a, B] = det[φ(a), φ(b)] (A, B B s (C 2 )). Let Z 2 := {A B s (C 2 ): Tr A = 0} and φ: B s (C 2 ) Z 2 B s (C 2 ), We dene the following mapping: ψ := φ Z 2 : Z 2 Z 2. We can prove the following: ( φ(a) = ±ψ A Tr A ) 2 I Tr φ(a) φ(a) = φ(a) I. 2 (A B s (C 2 )).

16 Now, we identify elements of Z 2 with vectors of R 3 using the vector space isomorphism ( ) ι: R 3 a b + ic Z 2, (a, b, c), b ic a

17 Now, we identify elements of Z 2 with vectors of R 3 using the vector space isomorphism ( ) ι: R 3 a b + ic Z 2, (a, b, c), b ic a and we dene the following transformation: ξ : R 3 R 3, ξ = ι 1 ψ ι.

18 Now, we identify elements of Z 2 with vectors of R 3 using the vector space isomorphism ( ) ι: R 3 a b + ic Z 2, (a, b, c), b ic a and we dene the following transformation: ξ : R 3 R 3, ξ = ι 1 ψ ι. A rather simple calculation shows the following two equations: det[ι(a 1, b 1, c 1 ), ι(a 2, b 2, c 2 )] = 4 (a 1, b 1, c 1 ) (a 2, b 2, c 2 ) 2 and

19 Now, we identify elements of Z 2 with vectors of R 3 using the vector space isomorphism ( ) ι: R 3 a b + ic Z 2, (a, b, c), b ic a and we dene the following transformation: ξ : R 3 R 3, ξ = ι 1 ψ ι. A rather simple calculation shows the following two equations: det[ι(a 1, b 1, c 1 ), ι(a 2, b 2, c 2 )] = 4 (a 1, b 1, c 1 ) (a 2, b 2, c 2 ) 2 and ξ(a1, b 1, c 1 ) ξ(a 2, b 2, c 2 ) = (a1, b 1, c 1 ) (a 2, b 2, c 2 ).

20 Problem (Reformulation #2) Characterize those maps φ: R 3 R 3 s.t. ( a, b) = (φ( a), φ( b)) ( a, b R 3 ) where ( a, b) = a b =

21 Problem (Reformulation #2) Characterize those maps φ: R 3 R 3 s.t. ( a, b) = (φ( a), φ( b)) ( a, b R 3 ) where ( a, b) = a b = a 2 b 2 a, b 2 =

22 Problem (Reformulation #2) Characterize those maps φ: R 3 R 3 s.t. ( a, b) = (φ( a), φ( b)) ( a, b R 3 ) where ( a, b) = a b = a 2 b 2 a, b 2 = a 2 b ( a b 2 a 2 b 2 ) 2. }{{} Heron's formula

23 Problem (Reformulation #2) Characterize those maps φ: R 3 R 3 s.t. ( a, b) = (φ( a), φ( b)) ( a, b R 3 ) where ( a, b) = a b = a 2 b 2 a, b 2 = a 2 b ( a b 2 a 2 b 2 ) 2. }{{} Heron's formula Of course, the similar question could be asked for an arbitrary real Hilbert space E. This is the Rassias-Wagner problem.

24 Problem (Reformulation #2) Characterize those maps φ: R 3 R 3 s.t. ( a, b) = (φ( a), φ( b)) ( a, b R 3 ) where ( a, b) = a b = a 2 b 2 a, b 2 = a 2 b ( a b 2 a 2 b 2 ) 2. }{{} Heron's formula Of course, the similar question could be asked for an arbitrary real Hilbert space E. This is the Rassias-Wagner problem. (Beckmann-Quarles; Lester-Martin, linear R-W)

25 Theorem (G.) Let E be a real (not necessarily separable) Hilbert space and φ: E E be an arbitrary transformation such that ( a, b) = (φ( a), φ( b)) ( a, b E). (1)

26 Theorem (G.) Let E be a real (not necessarily separable) Hilbert space and φ: E E be an arbitrary transformation such that ( a, b) = (φ( a), φ( b)) ( a, b E). (1) (i) If dim E = 2, then there exists a linear operator A: E E with det A = 1 such that the following holds: φ( a) = ±A a ( a E). (2)

27 Theorem (G. continued) (ii) If 2 < dim E <, then there exists an orthogonal linear operator R : E E such that φ( a) = ±R a ( a E) is satised.

28 Theorem (G. continued) (ii) If 2 < dim E <, then there exists an orthogonal linear operator R : E E such that φ( a) = ±R a ( a E) is satised. (iii) If dim E = and in addition φ is assumed to be bijective, then there exists a linear, surjective isometry R : E E such that we have φ( a) = ±R a ( a E).

29 Outline of the proof for n = 3 Let us consider the projectivised R 3 which will be denoted by P(R 3 ). The subspace generated by v will be denoted by [ v]. Let g φ : P(R 3 ) P(R 3 ), g φ ([ v]) = [φ( v)] ( v 0). Note that φ( v) = 0 v = 0.

30 Outline of the proof for n = 3 Let us consider the projectivised R 3 which will be denoted by P(R 3 ). The subspace generated by v will be denoted by [ v]. Let g φ : P(R 3 ) P(R 3 ), g φ ([ v]) = [φ( v)] ( v 0). Note that φ( v) = 0 v = 0. STEP 1: We prove that g φ is a homeomorphism.

31 STEP 2: We consider two linearly independent vectors a, b E and let C a, b := { v E \ {0}: ( v, a) = ( v, } b) E (note that C a, b is a plane i a = b )

32 STEP 2: We consider two linearly independent vectors a, b E and let C a, b := { v E \ {0}: ( v, a) = ( v, } b) E (note that C a, b is a plane i a = b ) and PC a, b := {[ v] P(R 3 ): v C a, b }.

33 STEP 2: We consider two linearly independent vectors a, b E and let C a, b := { v E \ {0}: ( v, a) = ( v, } b) E (note that C a, b is a plane i a = b ) and PC a, b := {[ v] P(R 3 ): v C a, b }. We can show that PC a, b contains a loop γ : [0, 1] PC a, b not homotopic to the trivial loop δ : [0, 1] PC a, b, δ γ(0) if and only if a = b holds.

34 STEP 3: Using that g φ is a homeomorphism, we obtain that holds with some λ φ > 0. φ( a) = λ φ a ( a R 3 )

35 STEP 3: Using that g φ is a homeomorphism, we obtain that φ( a) = λ φ a ( a R 3 ) holds with some λ φ > 0. We prove that λ φ = 1, using that g φ is a homeomorphism.

36 STEP 3: Using that g φ is a homeomorphism, we obtain that φ( a) = λ φ a ( a R 3 ) holds with some λ φ > 0. We prove that λ φ = 1, using that g φ is a homeomorphism. STEP 4: Since = (φ( a), φ( b)) = a 2 b 2 a, b 2 = ( a, b) φ( a) 2 φ( b) 2 φ( a), φ( b) 2 ( a, b R 3 ),

37 STEP 3: Using that g φ is a homeomorphism, we obtain that φ( a) = λ φ a ( a R 3 ) holds with some λ φ > 0. We prove that λ φ = 1, using that g φ is a homeomorphism. STEP 4: Since = (φ( a), φ( b)) = we obtain a 2 b 2 a, b 2 = ( a, b) φ( a) 2 φ( b) 2 φ( a), φ( b) 2 ( a, b R 3 ), a, b = φ( a), φ( b) ( a, b R 3 ).

38 STEP 3: Using that g φ is a homeomorphism, we obtain that φ( a) = λ φ a ( a R 3 ) holds with some λ φ > 0. We prove that λ φ = 1, using that g φ is a homeomorphism. STEP 4: Since = (φ( a), φ( b)) = we obtain a 2 b 2 a, b 2 = ( a, b) φ( a) 2 φ( b) 2 φ( a), φ( b) 2 ( a, b R 3 ), a, b = φ( a), φ( b) ( a, b R 3 ). Finally, we apply Wigner's theorem.

39 Back to the Molnár-Timmermann problem Theorem (G.) Fix a unitarily invariant norm on C d d where d 2. Let φ: B s (C d ) B s (C d ) be an arbitrary transformation for which the following holds: [A, B] = [φ(a), φ(b)] (A, B B s (C d )). (3)

40 Back to the Molnár-Timmermann problem Theorem (G.) Fix a unitarily invariant norm on C d d where d 2. Let φ: B s (C d ) B s (C d ) be an arbitrary transformation for which the following holds: [A, B] = [φ(a), φ(b)] (A, B B s (C d )). (3) Then there exist a function f : B s (C d ) R and a unitary or antiunitary operator U such that φ(a) = ±UAU + f (A)I (A B s (C d )) is satised.

41 Back to the Molnár-Timmermann problem (cont.) Theorem (G.) Let H be a separable Hilbert space and x a unitarily invariant norm on B(H). Let φ: B s (H) B s (H) be a bijection for which the following holds: [A, B] = [φ(a), φ(b)] (A, B H d ). (4)

42 Back to the Molnár-Timmermann problem (cont.) Theorem (G.) Let H be a separable Hilbert space and x a unitarily invariant norm on B(H). Let φ: B s (H) B s (H) be a bijection for which the following holds: [A, B] = [φ(a), φ(b)] (A, B H d ). (4) Then there exist a function f : B s (H) R and a unitary or antiunitary operator U such that φ(a) = ±UAU + f (A)I (A B s (H)) is satised.

43 k-parallelepipeds For any k vectors a 1,... a k E let us denote the k-dimensional volume of the parallelepiped spanned by them by the symbol k ( a 1,... a k ).

44 k-parallelepipeds For any k vectors a 1,... a k E let us denote the k-dimensional volume of the parallelepiped spanned by them by the symbol k ( a 1,... a k ). Theorem (G.) Let E be a real (not necessarily separable) Hilbert space, 2 < k <, k dim E and φ: E E be a transformation such that k ( a 1,... a k ) = k (φ( a 1 ),... φ( a k )) ( a 1,... a k E). (5)

45 Theorem (G., cont.) (i) If dim E = k, then there exists a linear operator A: E E with det A = 1 such that the following holds: φ( a) = ±A a ( a E). (6)

46 Theorem (G., cont.) (i) If dim E = k, then there exists a linear operator A: E E with det A = 1 such that the following holds: φ( a) = ±A a ( a E). (6) (ii) If 2 < k < dim E( ), then there exists a linear (not necessarily surjective) isometry R : E E such that φ( a) = ±R a ( a E) is satised.

47 The proof is a straightforward consequence of the fundamental theorem of projective geometry.

48 Back to the parallelogram case Note that if we knew that [ c] [ a, b], then we could use the fundamental theorem of projective geometry and drop the bijectivity condition.

49 Back to the parallelogram case Note that if we knew that [ c] [ a, b], then we could use the fundamental theorem of projective geometry and drop the bijectivity condition. Question If dim E =, is the condition [ c] [ a, b] = [φ( c)] [φ( a), φ( b)] satised?

50 Back to the parallelogram case Note that if we knew that [ c] [ a, b], then we could use the fundamental theorem of projective geometry and drop the bijectivity condition. Question If dim E =, is the condition [ c] [ a, b] = [φ( c)] [φ( a), φ( b)] satised? Problem Let E be an arbitrary real Hilbert space. Characterize those transformations φ: R 2 E which preservers the area of parallelograms.

51 Problem Describe those transformations φ: E E s. t. ( a, b) = 1 = (φ( a), φ( b)) = 1 holds.

52 Problem Describe those transformations φ: E E s. t. ( a, b) = 1 = (φ( a), φ( b)) = 1 holds. Problem Characterize those transformations φ: E E s. t. ( a, b) = 0 = (φ( a), φ( b)) = 0 and ( a, b) = 1 = (φ( a), φ( b)) = 1 are satised.

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54 L. Molnár and P. emrl, Nonlinear commutativity preserving maps on self-adjoint operators, Quart. J. Math. 56 (2005), L. Molnár and W. Timmermann, Transformations on bounded observables preserving measure of compatibility, Int. J. Theor. Phys. 50 (2011), T. M. Rassias and P. Wagner, Volume preserving mappings in the spirit of the Mazur-Ulam theorem, Aequationes Math. 66 (2003), no. 12, P. emrl, Nonlinear commutativity-preserving maps on Hermitian matrices, Proc. Roy. Soc. Edinburgh, 138A (2008),

55 This research was also supported by the European Union and the State of Hungary, co-nanced by the European Social Fund in the framework of TÁMOP A/2-11/ 'National Excellence Program'

56 Thank You for Your Kind Attention