Kähler configurations of points

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1 Kähler configurations of points Simon Salamon Oxford, 22 May 2017

2 The Hesse configuration 1/24 Let ω = e 2πi/3. Consider the nine points [0, 1, 1] [0, 1, ω] [0, 1, ω 2 ] [1, 0, 1] [1, 0, ω] [1, 0, ω 2 ] [1, 1, 0] [1, ω, 0] [1, ω 2, 0] of CP 2. They are the flexes of the non-singular cubic for any c C \ {1, ω, ω 3 } and belong in triples on 12 lines: x 3 + y 3 + z 3 3c xyz = 0

3 A 1-parameter family of nine points 2/24 The representative vectors all have norm 1/ 2, and any two satisfy w, z = 1. It follows that the 9 points are mutually equidistant. The same is true of the set M θ consisting of [0, 1, 1] [0, 1, ω] [0, 1, ω 2 ] [1, 0, 1] [1, 0, ω] [1, 0, ω 2 ] [e iθ, 1, 0] [e iθ, ω, 0] [e iθ, ω 2, 0] Theorem. Any unordered set of nine mutually equidistant points in (CP 2, g) is isometric to M θ for some angle θ. L. Hughston, S. Salamon, in Advances Math. 2016

4 Fubini-Study distance 3/24 Complex projective space CP n 1 is a compact Kähler manifold. Its Riemannian metric g arises from the standard Hermitian form w, z = w z = n w i z i, i=1 on C n, which is invariant by the unitary group U(n). The associated distance d satisfies cos 2 ( 1 2 d([w], [z])) = w, z 2 w 2 z 2 = w, z z, w w, w z, z. The RHS is really a cross ratio of four points [w], [z], [w ], [z ].

5 Moment mapping 4/24 The 2-form ω 0 = i dz i dz i on C n induces a symplectic form on CP n 1, regarded as a symplectic quotient of C n by U(1). Let H = {A C n,n : A = A, tr A = 1} su(n). The mapping S 2n 1 H for which [z] z z = z 1 2 z 1 z 2 z 1 z 3 z 2 z 1 z 2 2 z 2 z 3 z 3 z 1 z 3 z 2 z 3 2 is U(n)-equivariant and defines an isometric embedding CP n 1 H R n2 1.

6 SIC-POVM 5/24 is shorthand for Symmetric Informationally Complete Positive Operator Valued Measure. We shall consider such objects in finite-dimensional Hilbert space C n. Definition 1. A SIC-POVM is a set of n 2 points [z α ] in CP n 1 such that (with the normalization z α = 1) z α, z β 2 = λ, α β, for some fixed λ (0, 1). This defines a subset {P α = z αz α } of H and n 2 { 1 α = β P α = n I, tr(p α P β ) = λ α β. α=1 E. B. Davies: The Quantum Theory of Open Systems, 1976

7 SIC sets 6/24 Definition 2. A SIC set consists of a regular simplex in su(n) whose n 2 vertices lie in the adjoint orbit CP n 1. In fact, n 2 is the maximum number of mutually equidistant points possible in CP n 1, and in this case λ must equal 1 n+1. Proof. Given an equidistant set {P α }, set Q β = P β λi. Then tr(p α Q β ) = { 1 λ α = β 0 α β So {P α } is linearly independent in i u(n). Applying tr(q β ) to I = c α P α, gives 1 λn = c α (1 λ) α. Then n = n 2 c α, and so n(1 λn) = 1 λ.

8 Example: the 2-sphere 7/24 The embedding CP 1 = S 2 R 3 is given by setting ( ) ( ) z1 2 z 1 z a b + ic z 1 z 2 z 2 2 = 1 2. b ic 1 a One SIC set consists of the points [1 + 3, 1 + i] [1 + i, 1 + 3] [1 + 3, 1 i] [ 1 i, 1 + 3] This is an orbit of A, B where A = ( ) ( ) , B = Any SIC set determines an inscribed tetrahedron and any two are congruent by SO(3) = SU(2)/Z 2. If n 3, two SIC sets are not in general congruent by SU(n).

9 Arbitrary dimensions 8/24 Conjecture 1. CP n 1 possesses a SIC set for all n. Algebraic solutions are known for n = 2, 3, 4,..., 15, 19, 24, 35, 48. Extensive numerical verification has been carried out for n 151. Conjecture 2. For all n, there is a SIC set that is an orbit of the (Weyl-)Heisenberg group H (Z/nZ) 2. A vector z or point [z] whose orbit H [z] is a SIC set is called fiducial. An example for CP 3 is [ ] [z] = s i(r +s), 1 r + i, s + i(s r), 1+r + i. where r = 2 and s = G. Zauner, PhD thesis, Vienna, 1999

10 Example: a SIC set in CP 7 9/24 Using the identification C 8 = H 4, consider the groups V 1, right multiplication by 1, i, j, k Sp(1) V 2, double sign changes in H 4 V 3, double transpositions of the coordinates. Then G = V 1 V 2 V 3 = (Z2 ) 6 is a subgroup of Sp(4) SU(8). Fix unit quaternions p = 1 2 (1 + i + j k), q = 1 2 (1 + i j k). Proposition. G [0, p, q, j] is a SIC set in CP 7. This orbit cannot project neatly to HP 2 because no S 2 fibre can contain 4 points. S. G. Hoggar in Geometria Dedicata, 1998

11 Analogue: a Gosset polytope 10/24 Its 56 vertices form the orbit under S 8 Z 2 of (1, 1, 1, 3, 3, 1, 1, 1) R 8. They lie in a subspace R 7 and are the weights for the action of E 7 on C 28. All ( ) 56 2 inner products equal 8 or 8, defining 28 mutually equidistant points on RP 6.

12 A 2-torus acting on CP 2 11/24 Fix T = { diag(e iθ 1, e iθ 2, e iθ 3 ) : θ i = 0 } in SU(3). Consider the moment map for the action of T : µ([z]) = 1 z 2 ( z 1 2, z 2 2, z 3 2 ) = (a 2, b 2, c 2 ). The image of µ is a 2-simplex whose inscribed circle C will play a vital role: Let m 1, m 2, m 3 be the midpoints, and set C i = µ 1 (m i ) CP 2. Note that M θ C 1 C 2 C 3.

13 Correct separation 12/24 We set a= z 1 / z, b = z 2 / z, c = z 3 / z. Then C is the intersection of the plane a 2 + b 2 + c 2 = 1 and the sphere a 4 + b 4 + c 4 = 1 2. Lemma. The following are equivalent: µ([z]) C (a+b+c)( a+b+c)(a b+c)(a+b c) = 0 the three points z=[z 1, z 2, z 3 ], [z 1, ωz 2, ω 2 z 3 ], [z 1, ω 2 z 2, ωz 3 ] are the correct distance (d = 2 arccos 1 2 = 2π/3) apart to form part of a SIC set.

14 The generic point in a SIC set 13/24 Let S be a SIC set in CP 2. Up to the action of SU(3), we are free to assume that S contains the two points of C 1 given by z 1 = 1 2 (0, 1, ω), z 2 = 1 2 (0, 1, ω 2 ). Any other point [z] of S satisfies z, z j 2 = 1 4 z 2 for j = 0, 1. Lemma. µ([z]) C and we can take ] [z] = z[σ, φ] = [e iσ cos φ, cos(φ + 2π 3 ), cos(φ + 4π 3 ) for some π < σ π and π 2 < φ π 2. Thus, [z] lies in a 2-torus, pinched to a point where φ = π/2.

15 The pinched torus T 14/24 containing z[0, φ + kπ 3 ] with k = 0, 1, 2, and two points in both C 2, C 3, forming a SIC set L φ when [z 1 ], [z 2 ] are added: C 2 C 3

16 The same thing in a rectangle 15/24 Recall that z[σ, φ] = [ e iσ cos φ, cos(φ + 2π 3 ), cos(φ + 4π 3 )]. Lemma. There exists a SIC set L φ containing [z 1 ], [z 2 ] and z[0, φ] for any φ with π 2 < φ π 2 :

17 Equivalent SIC sets 16/24 The remaining six points are then z[0, φ + π 3 ] z[ 2π 3, π 6 ] z[ 2π 3, π 6 ] C 2 z[ 2π 3, π 6 ] z[ 2π 3, π 6 ] C 3 z[0, φ π 3 ] Lemma. X L φ = M 2φ+π C 1 C 2 C 3, where ( ω 2 ) ω 1 X = 1 1 ω ω 2 3 U(3)

18 The Clifford group 17/24 is the normalizer G of H in U(n), isomorphic (modulo phase) to SL(2, Z n ) (Z n ) 2. For n = 3, G has order 216 and contains X, since Moreover, X 3 = iω 2 I. XAX 1 = ωb, XBX 1 = ω 2 A 1 B 1. Conjecture 3. A fiducial vector z can always be found in an eigenspace of the unitary matrix X where X ij = 1 n τ 2ij+j2, τ = e (n+1)πi/n. For n = 5 see: G. Horrocks, D. Mumford, in Topology 1973

19 Trigonometry 18/24 Let S be a SIC set containing [z 1 ], [z 2 ], and [z 3 ] = z[σ, φ], [z 4 ] = z[τ, ψ], [z 5 ] = z[υ, χ] T. Set x = tan φ, y = tan ψ, z = tan χ. Key lemma. [z 3 ] and [z 4 ] are distance 2π/3 apart iff cos(σ τ) = 1 If A + B + C = 0 then 9(1 + 2 cos 2(φ ψ)) 4(4 cos 2 φ cos 2 ψ + 3 sin 2φ sin 2ψ) = x 2 + 9y 2 27x 2 y 2 24xy, 16(1 + 3xy) cos 2 A + cos 2 B + cos 2 C = cos A cos B cos C.

20 Equilateral triangles in T 19/24 Specify their vertices by vertical coordinates x, y, z and set p = x + y + z, q = yz + zx + xy, r = xyz. Corollary. If z[σ, φ], z[τ, ψ], z[υ, χ] are all 2π/3 apart then where f (x, y, z) = F (p, q, r) = 0, F = 9 22p 2 + 9p q 126p 2 q + 27p 4 q + 298q 2 226p 2 q 2 +24p 4 q q 3 138p 2 q q q 5 3pr 50p 3 r 15p 5 r +88pqr 48p 3 qr + 234pq 2 r + 18p 3 q 2 r 144pq 3 r + 81pq 4 r + 189r 2 480p 2 r 2 153p 4 r qr 2 306p 2 qr q 2 r 2 486p 2 q 2 r q 3 r q 4 r 2 558pr 3 486p 3 r pqr 3 810pq 2 r r 4 162p 2 r qr q 2 r pr r 6. cf. p 2 3q = 3 4 in the Euclidean case.

21 Hexagons in CP 2 20/24 Let S be a SIC set containing [z 1 ], [z 2 ] and [z i ] = z[σ i, φ i ] for i = 3, 4, 5, 6. Set t = tan φ 1, x = tan φ 2, y = tan φ 3, z = tan φ 4, giving 4 equations in 4 unknowns: f (x, y, z) = f (t, y, z) = f (t, x, z) = f (t, x, y) = 0. If a, b, c, d are the elementary symmetric polynomials in t, x, y, z, we can convert this into a system F 1 (a, b, c, d) = F 2 (a, b, c, d) = F 3 (a, b, c, d) = F 4 (a, b, c, d) = 0. For example, F 1 = f (x, y, z) + f (t, y, z) + f (t, x, z) + f (t, x, y).

22 A quotient ideal 21/24 Solutions for which φ i =±π/6 give rise to L φ, and we may assume G = 1 3a 2 + 6b + 9b 2 18ac 27c d + 54bd + 81d 2 is non-zero. Consider I = F 1, F 2, F 3, F 4, and compute I : G = {r R[a, b, c, d] : r G I }, by finding a Gröbner basis for uf 1, uf 2, uf 3, uf 4, (1 u)g. With approriate orderings, its first element (/G ) is (d 1) 3 (3d 1) 3 (3+b+3d)(9d 1) 3 (1+3b+9d) 3 (19+9b+27d). Corollary. If S is not isometric to L φ, one of the following holds: 1 d = 1, 3, 1 9, b = (3d + 3), 1 3 (9d + 1), 1 9 (27d + 19).

23 Cross field passes 22/24 The case 9b + 27d + 19 = 0 gives rise to a 1-parameter set of solutions to the system F 1 = F 2 = F 3 = F 4 = 0, such as (a, b, c, d) = (0, 22 9, c, 1 9 ). However, this gives rise to 9 points of which only 27 of the ( 9 2) distances equal 2π/3:

24 Conclusion 23/24 Theorem. Any SIC set in CP 2 is isometric to L φ for some φ and therefore to the set we started with. [0, 1, 1] [0, 1, ω] [0, 1, ω 2 ] [1, 0, 1] [1, 0, ω] [1, 0, ω 2 ] [e 2iφ, 1, 0] [e 2iφ, ω, 0] [e 2iφ, ω 2, 0] Classifying SIC-POVM s in higher dimensions appears beyond the scope of present methods. For n = 4, the analogue of C consists of two parallel circles, isolated points of which generate 64 SIC sets in CP 3.

25 More references 24/24 M. Appleby: J. Math. Phys M. Appleby, S. Flammia, G. McConnell, J. Yard: arxiv: S, T. Flammia: J. Phys. A, 2006 J. M. Renes et al: J. Math. Phys A. J. Scott, M. Grassl: J. Math. Phys A. J. Scott: arxiv: G. Zauner: J. Quant. Inf H. Zhu: J. Phys. A 2010

26 Translation-invariant metrics 25/24 Here we see two different fibres µ 1 (p i ) = R 2 /Z 2, for p 1, p 2 C. The red curves are points of distance d from the origin:

27 The pinched torus T 26/24 We may assume that a third (black) point of S is given by z 3 =z[0, φ], so [z 1 ], [z 2 ], [z 3 ] form an equilateral triangle in CP 2. Most points the correct distance from [z 3 ] cannot belong to S :

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