Spectral curves and the mass of hyperbolic monopoles. Nuno M. Romão. Massachusetts Institute of Technology Cambridge MA, USA

Transcription

1 Nuno M. Romão Massachusetts Institute of Technology Cambridge MA, USA CTQM, University of Aarhus Aarhus, Denmark (joint work with P. Norbury) Workshop on non-linear integral transforms: Fourier Mukai and Nahm CRM, Montreal, August 2007

2 Outline 1 Magnetic monopoles (crash course) 2 3 4

3 Fix a metric on R 3. e.g., for R > 0, ds 2 = dr 2 + R 4 sinh 2 ( r R 2 ) ( dθ 2 + sin 2 θdφ 2) G = SU(2) or another compact Lie group Set up gauge theory for fields (A, Φ)

4 Fields: Φ Ω 0 (R 3, su(2)), Higgs field A Ω 1 (R 3, su(2)), gauge field and then: connection: d A = d + [A, ] curvature: B A = da [A, A] Gauge symmetry: G = Ω 0 (R 3, SU(2))

5 Monopoles are pairs (A, Φ) satisfying R B 3 A 2 + d A Φ 2 < (finite L 2 -norm) B A = d A Φ (Bogomol nyĭ equation) lim x Φ(x) = m > 0 modulo gauge equivalence. (m = monopole mass)

6 Topology: Φ S 2 : S 2 S 2 su(2) = R 3 ) k := deg (Φ S Z + magnetic charge 2 Moduli space of monopoles of given k, M k = {solutions of B A = d A Φ with charge k} {gauge transformations} is a smooth manifold with dim R M k = 4k

7 If ds 2 has constant curvature, the Bogomol nyĭ equations are quite tractable integrability". Then M k can be constructed using holomorphic data. e.g. spectral curves: M k itself has interesting geometry.

8 e.g.: Euclidean metric Then monopoles of different m are trivially related by rescaling: M k = Mk,m=1 R + Structure: S 1 R 3 S1 M 0 k Z k M k,1 R 3 parametrises centre of mass M 0 k contains internal degrees of freedom; SO(3) acts on M 0 k Ṅ.M. Romão

9 M 0 k has rich geometry: Natural L 2 -metric from R 3 B A 2 + d A Φ 2. This is hyperkähler. L 2 -cohomology has implications for electric-magnetic duality. Geodesic flow has natural physical interpretation (slow scattering).

10 e.g.: k = 3, "cyclic scattering" (a geodesic of M 3 ) P.M. Sutcliffe, Nucl. Phys. B 505 ( 97) 517

11 In this talk, we will consider magnetic monopoles in hyperbolic space H 3 instead. More unified symmetry PSL 2 C, but extra difficulties: L 2 -metric diverges! m-dependent holomorphic data AIM: Obtain this mass dependence explicitly, focusing on spectral curves.

12 cf. previous work on M 0 2 : [MNS, 2003]: R m [NR, 2007]: m M. Murray, P. Norbury, M. Singer, Ann. Global Anal. Geom. 23 ( 03) 101

13 H 3 = Hyperbolic space { } (x 1, x 2, x 3 ) R 3, x 3 > 0 S 2 = H 3 = {(x 1, x 2, 0)} { } = P 1 Oriented geodesics: (w, z) ŵ := 1 w Twistor space: Z := {oriented geodesics} = P 1 P 1 P 1

14 Twistor correspondence C µ ν (x 1, x 2, x 3 ) H 3 Z (w, z) C H 3 Z defined by incidence relation: (x 1, x 2, x 3 ) lies on (w, z) µ, ν projections µ ν 1 (w, z) = {points on(w, z)}, geodesic ν µ 1 (x 1, x 2, x 3 ) = P 1, star at (x 1, x 2, x 3 )

15 Equation for star at (x 1, x 2, x 3 ) expresses incidence relation: (x 1 ix 2 )wz (x x x 2 3 )w + z (x 1 + ix 2 ) = 0 A subset of Z invariant under the reality structure σ : (w, z) (ẑ, ŵ) is said to be real. e.g. stars are real.

16 Use of this? How? Solutions of B A = d A Φ can be interpreted as holomorphic bundles over Z.

17 Given: (A, Φ) on trivial bundle E = H 3 C 2 H 3 Obtain family of operators: H : Z End H 0 (µ ν 1 (w, z), E) (w,z) Z H(w, z) := (d A iφ) µ ν 1 (w,z). N. Hitchin ( 82) understood that B A = d A Φ C 2 ker H holomorphic Z

18 Natural line subbundles L ± ker H: } L ± (w,z) {s = ker H(w, z) : lim s(x) = 0. x ε ± (w,z) Spectral curve: Σ := { } (w, z) Z : L + (w,z) = L (w,z).

19 Properties of Σ Σ compact of genus (k 1) 2 Σ O Z (k, k) i.e. given as vanishing locus of a polynomial ψ(w, z) of bidegree (k, k) Σ real: σ(σ) = Σ k (k )( ) k ψ(w, z) = v j, v l ( w) k j z l, v j C k+1 j l j,l=0 Positivity [MNS 03]: ( 1) k 1 ψ( z, z) > 0 z P 1

20 More properties of Σ Σ centred [MNS 03] iff k (2j k) v j 2 = 0, j=0 k 1 (j + 1)(k j) vj, v j+1 = 0. L 2m+k Σ = OΣ j=0 H 0 (Σ, L s (k 2, 0)) = 0 ( L := OP 1 P1(1, 1)) 0<s<2m+2

21 Construction of L λ (k 1, k 2 ) Z Z = U 1 U 2, trivialising cover: U 1 := {(w, z) Z : w, z }, U 2 := {(w, z) Z : w 0, z 0}. Use transition function g 12 : U 1 U 2 C given by g 12 (w, z) = w k 1+λ z k 2 λ. NB: L Z is topologically trivial but holomorphically nontrivial.

22 PROPOSITION: The PGL 2 C-orbit of any spectral curve of given m and k = 2 contains a unique (centred) curve of the form κ sn 2 ρ (w 2 z 2 + 1) + 2 cn ρ dn ρ wz (w 2 + z 2 ) = 0, with 0 κ < 1 and ρ = K (κ) m + 1.

23 Look at limiting cases: 1) κ 0 (axial symmetry): π ρ 2(m + 1) ( ) w 2 π 2 cos wz + z 2 = 0 2(m + 1) ) ) (w e iπ 2(m+1) z (w e iπ 2(m+1) z = 0 Reduces to two P 1 s, which can be interpreted as stars at complex conjugate points. This is the spectral curve of an axially symmetric monopole at (0, 0, 1) [MNS].

24 2) κ 1 (large separation): sn ρ tanh ρ,... and ρ + ( ) ( ) w 2 1 z 2 1 = 0 Interpretation: (w ± 1)(z 1) = 0 limiting stars at (±1, 0, 0) H 3. So this is the limit of infinite core separation".

25 3) m 0 ( 2-nullarons ): sn ρ snk (κ) = 1 cn ρ cnk (κ) = 0 dn ρ dnk (κ) = 1 κ 2 ( κ 1 + w 2 z 2) ( w 2 + z 2) = 0 This can be obtained more directly: R(w) = R(ẑ), R(z) = 1 κ 2 z 2 κ Rat 2(P 1 ) Curves of this type encode solutions to the Potts model, cf. Atiyah & Murray.

26 4) m (euclidean limit): Naively, nothing more than (w z) 2 = 0. In the next asymptotic order, recover spectral curve of euclidean monopole embedded in TP 1. Rescale around (0, 0, 1) H 3.

27 Coordinates: η d dζ TP 1 ζ Sequence {(w m, z m )} m, (w, z ) = (z, z ). ( ) (η, ζ) = lim m(z m w m ), z m η 2 = lim m m2 (w m z m ) 2 (2cnρdnρ 2)w m z m + κsn 2 ( ρ w 2 = K (κ) lim m zm ) ρ 0 ρ 2 Get: ( ) ( ) η 2 K (κ) 2 ζ 2 κ κζ 2 1 = 0 cf. Hurtubise,Commun. Math. Phys. 92 ( 83) 195

28 Proof of the Proposition (sketch): (i) Assume Σ is nonsingular. Use symmetry to bring curve to the form w 2 z 2 + u2 2uv + 4 wz u2 4 2(u v) 4(u v) (w 2 + z 2 ) + 1 = 0, u, v ]2, [, u 2 2uv + 4 > 0.

29 (ii) Map Σ P 1 P 1 P 2 = P 1 P 1 /Z 2. Image is a conic, so rational. Construct Σ as a 2-cover Σ P 1.

30 (iii) Interpret L 2m+k Σ = O Σ using reciprocity. Thus calculate u(m, κ) and v(m, κ) = κ + 1 κ.

31 AIM: Construct spectral curves of higher charge k > 2 and any m. Coefficients of ψ will typically involve integrals on higher genus g > 1. Condition H 0 (Σ, L s (k 2, 0)) = 0 becomes nontrivial. Simplifying idea: Explore Platonic symmetries to obtain curves that are Galois covers of an elliptic curve.

32

33 PROPOSITION: Let Σ P 1 P 1 be a smooth (k, k) curve, invariant under G = A 4, S 4 or A 5 SO(3), with E := Σ/G elliptic. Then: G = A 4 k = 3 or k = 4 G = S 4 k = 4 G = A 5 k = 6 These are the lowest k for which a (k, k) curve with the given symmetry exists. A priori, not guaranteed that these (k, k) curves will be spectral curves as yet.

34 Ansätze for Platonic spectral curves: A 4 symmetry: k = 3, α R (w z) 3 + iα(w + z) ( ) (wz) 2 1 = 0 k = 4, α, β R ( ) (w z) 4 + α w 4 z 4 + 6w 2 z 2 + 4wz(w 2 + z 2 ) ( ) iβ(w z)(w + z) (wz) 2 1 = 0

35 S 4 symmetry: k = 4, α R ( ) (w z) 4 + α w 4 z 4 + 6w 2 z 2 + 4wz(w 2 + z 2 ) + 1 = 0. A 5 symmetry: k = 6 candidate is not a spectral curve.

36 Goals: check H 0 (Σ, L s (k 2, 0)) = 0, compute α(m) =? 0<s<2m+2 Will focus on k = 3, A 4 symmetry.

37 THEOREM: There is a unique PGL 2 C-orbit of tetrahedrally symmetric monopoles with mass m > 0. A representative is the centred monopole with spectral curve as above, where α, m satisfy the relation ( 2ϖ1 2m + 3 ) ϖ 1, ϖ 2 = α 2 involving the -function of the elliptic curve with invariants g 2 = α 2, g 3 = α α 4.

38 L 2m+k Σ Proof: Key tool is again a reciprocity law. = O Σ nowhere-vanishing ξ H 0 (Σ, L 2m+k ) ξ represented by ξ j : U j Σ C related on U 1 U 2 Σ by ( w ) 2m+k ξ 1 (w, z) = ξ2 (w 1, z 1 ) z Then ω j := dlog ξ j satisfy ( dw ω 1 = (2m + k) w dz z ) + ω 2. and are therefore differentials of the 3rd kind.

39 Integrality condition on periods: 1 ω 1 = 1 d ξ1 2πi 2πi ξ 1 Z m l := 1 ω 1, n l := 1 ω 1. 2πi b l 2πi a l

40 Let p j, q j be the poles of dw w dz z, j = 1,... 2(k 1). Res pj (ω 1 ) = (2m + k) and Res qj (ω 1 ) = 2m + k The p j, q j lie on two separate A 4 -orbits. Set p := π(p j ), q := π(q j ) for π : Σ Σ/A 4 =: E.

41 Fix a nonzero ω H 0 (E, K E ). Reciprocity law for π ω and ω 1 on Σ: g a l π ω n 2(k 1) l b l π ω m l = (2m + k) l=1 j=1 qj p j π ω

42 Can show m l, n l Z are independent of α. Transfer the calculation to E: define ( g ) c := π j=1 (m ja j + n j b j ) =: l 1 a + l 2 b; then c q ω = 2(2m + k) ω p

43 Choice of path of integration on RHS requires care. One criterion: path is right if l 1, l 2 do not jump when m is varied. Need to calculate c H 1 (E, Z). Use invariant theory to construct E.

44 Geometry of the canonical embedding Long exact sequence of Q := P 1 P 1 Σ φ P g 1 0 O Q ( 2, 2) ψ O Q (k 2, k 2) O Σ (k 2, k 2) 0 implies H 0 (Q, O(k 2, k 2)) = H 0 (Σ, O(k 2, k 2)) and thus a factorisation φ : Σ P 1 P 1 P k(k 2) (w, z) [1 : w :... : w k 2 : wz :... : w k 2 z k 2 ]

45 LEMMA: For any smooth (3,3) curve Σ P 1 P 1, a nontrivial linear equivalence relation is equivalent to p 1 + p 2 + p 3 q 1 + q 2 + q 3 p 1 + p 2 + p 3 O Σ (1, 0) or O Σ (0, 1).

46 ( ) : clear Proof (sketch): ( ) : φ(p 1 ), φ(p 2 ), φ(p 3 ) P 2 P 3 Geometric Riemann Roch implies stronger condition: φ(p 1 ), φ(p 2 ), φ(p 3 ) P 1 P 3. Now consider two cases:

47 1) p 1, p 2 lie on a (0, 1) curve, say. Then φ(p 1 ), φ(p 2 ) span S c = {[1 : w : c : wc] w P 1 }, c = const. and φ(p 3 ) = [1 : w : z : wz] S c iff z = c. 2) Suppose no two points lie on a (1, 0) curve or (0, 1) curve. Then they must lie on a smooth (1, 1) curve. Use PGL 2 C to relate it to the diagonal P 1 : P 1 [1 : w : w : w 2 ]. But three distinct points of this form must be linearly independent!

48 Equivalent statements (D L s (1, 0) crossing Θ): 1. D p 1 + p 2 + p 3 2. f C(Σ) : (f ) + D 0 3. H 0 (Σ, L s (1, 0)) 0 PROPOSITION: H 0 (Σ, L s (1, 0)) = 0 for 0 < s < 2m + 2.

49 Proof of the Proposition: L = O(1, 1) is A 4 SO(3)-invariant. So L Σ = π ˆL, where π : Σ E. Hence L s Σ = π ˆLs, s C, with deg ˆL = 0. E elliptic ˆL s p p dz = p p L s Σ Orb p Orb p

50 Orb p Orb p + O Σ (1, 0) q 1 + q 2 + q 3 ( ) iff Orb p Orb p 0 or Orb p Orb p L 1 Σ = O Σ( 1, 1). LHS of ( ) A 4 -invariant q 1 + q 2 + q 3 gq 1 + gq 2 + gq 3, g A4. But {q 1, q 2, q 3 } is not A 4 -invariant, as A 4 -orbits have size 4,6 or 12.

51 Previous Lemma implies two cases: 1) q 1 + q 2 + q 3 O Σ (1, 0) Orb p Orb p 0 i.e. s = 0 (mod 2m + 3) 2) q 1 + q 2 + q 3 O Σ (0, 1) 0 Orb p Orb p +O Σ (1, 0) (q 1 +q 2 +q 3 ) L s+1 Σ i.e. s = 1 (mod 2m + 3) So 0 < s < 2m + 2 L s (1, 0) Σ does not meet Θ.

52 An argument to compute α(m) for m 1 2 N. Suppose ξ H 0 (Σ, O( 2, k)) = H 0 (Σ, L 2m (k 2, 0)). In C[z, z 1, w, w 1 ], ξ 1 (w, z) = w 2 z k ξ 2 (w 1, z 1 ) + ψ(w, z)χ(w 1, w, z 1, z). Čech-cohomological interpretation: 0 H 0 (Σ, O( 2, k)) H 1 (Q, O( 2 k, 0)) ψ H 1 (Q, O( 2, k)) ξ χ ψχ det Ψ 0 since H 0 (Σ, O( 2, k)) = 0.

53 For our k = 3, A 4 Ansatz, Ψ = 1 iα 3 iα iα 3 iα 1 det Ψ = (3 α 2 ) < α < 3.

54 Now consider: 0 H 0 (Σ, O( 3, k+1)) H 1 (Q, O( 3 k, 1)) ψ 1 H 1 (Q, O( 3, k+1)) Again, det Ψ 1 0 whenever m > 1 2, since H 0 (Σ, L 2m 1 (k 2, 0)) = H 0 (Σ, O( 3, k + 1)) = 0 provided 0 < 2m 1 < 2m + 2.

55 Calculate: det Ψ 1 = 4(1 3α 2 ) 2 (3 α 2 ) 2 0. α 1 3 for m > 1 2. In fact, α = 1 3 m = 1 2 since then H 0 (Σ, L 2m 1 (k 2, 0)) = H 0 (Σ, O(k 2, 0)) has nontrivial sections!

56 Repeat the argument stepwise to obtain: m α

57 Calculate rational invariants ˇv, ˇx, ˇy C(Σ) A 4 to map Σ to P 2 with A 4 -orbits as fibres: ˇv = P 3 P 1 3 := (w+z)((wz)2 1) (w z) 3, ˇx = P 4 P 1 4 := w 4 z 4 +w 4 +z 4 +12w 2 z 2 +1 (w z) 4, ˇy = P 6 P 1 6 := (wz)6 ((wz) 2 +1)((w+z) 4 +4wz(w+z) 2 +(wz) 2 )+1 (w z) 6. Relation: 4ˇx 3 11ˇx 2 4ˇy 2 14ˇv 2 27ˇv 4 + 2ˇx(5 + 9ˇv 2 ) 3 = 0.

58 Eliminate ˇv = i α using equation for Σ. Introducing x := ˇx 11 12, y := 2ˇy, get: y 2 = 4x 3 g 2 x g 3 =: F(x) with g 2 = α 2, g 3 = α α 4.

59 Picture is:

60 Complex analysis on elliptic curve E: ω = ± dx. F(x) Poles: p 1 = (e πi/4 α, 0), p 2 = ( e πi/4 α, 0), q 1 = (0, e πi/4 α), [ p := π(p 1 ) = 1 : [ q := π(q 1 ) = 1 : q 2 = (0, e πi/4 α), (1 α 2 : 2i + 1α )] α ( α )] α 2 : +2i α = π(p 2 ), = π(q 2 )

61 Make use of reciprocity relation q ω = 2(2m + 3) ω c p

62 Periods and integral between poles: 2ϖ := 2ϖ := q Reciprocity relation: p b a ω = 2i + dx ω = 2 F(x) ω = 2i α 2 e 2 e1 e 2 dx F(x) dx F (x) l 1 ϖ + l 2 ϖ = 2i(2m + 3) α 2 Can show Re ϖ = ϖ l 2 = 2l 1. dx F (x).

63 Deal with elliptic integrals using uniformisation: (ϖ 2ϖ ( )l (2m + 3) 12 1 ) α 2 mod Λ. Apply to both sides to get ( ) ϖ1 l 1 = 1 2(2m + 3) 12 1 α 2. Finally, calculate l 1 = 4 from half-mass data.

64 Checks: Can compute α(m) for m Q. m gives known euclidean spectral curve: η 3 + i Γ ( ) π 3 ζ(ζ 4 1) = 0.

65 Where is the Nahm transform? The Nahm transform relates solutions of the Bogomol nyĭ equations to solutions of a Lax system. For hyperbolic monopoles of half-integer m, this system is a discretisation of the euclidean Nahm equations. Open question: Describe the Nahm transform of hyperbolic monopoles of arbitrary m. We have obtained explicit spectral curves for any m from the twistor side. What is the underlying integrable system?

66 Reference: P. NORBURY, N.M.R.: Commun. Math. Phys. 270 (2007)