Moduli Space of Higgs Bundles Instructor: Marco Gualtieri

Moduli Space of Higgs Bundles Instructor: Lecture Notes for MAT1305 Taught Spring of 2017 Typeset by Travis Ens Last edited January 30, 2017

Contents Lecture Guide Lecture 1 [11.01.2017] 1 History Hyperkähler geometry and instantons Hitchen (87-88) Kapustin, Witten ( 05) 2 Course Outline I. Hyperkähler Geometry II. Hitchin papers III. Recent advances 3 Kähler Geometry The basic structure of tangent bundles in Kähler geometry is T g T I T ω where I is a complex structure (so I 2 = 1), g = g(, ) is a Riemannian metric (so g = g and g is positive definite), and ω is a symplectic form (so ω = ω and ω(x, Y ) = ω(y, X).) Compatability of any two of these structures composite is as required. Example 1. I, ω are compatible I ω = ( ωi) = ωi. Equivalently I ω + ωi = 1 ω(x, IY ) + ω(ix, Y ) = 0 I ωi = ω ω(ix, IY ) = ω(x, Y ) 1

Because of the complex structure I, T C = T 1,0 T 0,1 where T 1,0 is the +i eigenspace and T 0,1 is the i eigenspace. So, and 2 T C = 2 T 1,0 (T 1,0 T 0,1) Ω 2 C = Ω 2,0 Ω 1,1 Ω 0,2. The compatibility condition then becomes ω Ω 1,1. 2 T 0,1 3.1 Integrability of Kähler Structures I: [T 1,0, T 1,0 ] T 1,0. Define the Nijenhuis Tensor N 2 T 1,0 T 0,1 by N(X, Y ) = π 0,1 [X, Y ] where X and Y are of type 1, 0. I is integrable of this tensor vanishes. A result of Newlander- Nirenberg says that I is locally isomorphic to C n when I is integrable. ω: ω is integrable if dω = 0. By a theorem of Darboux, if ω is integrable then ω is locally isomorphic to the standard symplectic structure (R 2n, ω std ). If both integrability conditions are satisfied then this is a Kähler structure. Remark 1. It is immediate from ω Ω 1,1 and dω = 0 that ω is locally i k for k C (U, R) by the lemma. k is called the Kähler potential. The de Rham differential splits as Ω p+1,q 1 N Ω p+1,q d Ω p,q : Ω p,q N Ω p,q+1 Ω p 1,q+2 and I is integrable d = +. Lemma 1. I, ω are each integrable if and only if I = 0 ω = 0. Here is the Levi- Civita connection, ie. the unique connection on T which preserves g (Xg(Y, Z) = g( X Y, Z + g(y, X Z) g = 0) and is torsion free ( X Y Y X = [X, Y ]). Proof. If I is integrable, IX = ix, IY = iy and we want to show I[X, Y ] = i[x, Y ]. But The rest is an exercise. I( X Y Y X) = ( X (IY ) Y (IX)) = i[x, Y ]. 2

Remark 2. Given an n-dim. complex manifold with hol. tangent and ccotangent bundles T 1,0 (with coordinates z i ) and T1,0 (with coordinates dz i ). Let K = n T1,0 be the canonical holomorphic line bundle. Kähler metric = Hermitian Metric on K = (unique) Chern connection. Curvature(Chern) = F Ω 2 R. By Poincaré-Lelong, F = i log s g where s g = det g ij (s is the hol. section dz 1 dz n.) In the Kähler case, F = Ric g (I, ). Thus the Einstein equation Ric = 0 K is flat in the Kähler case. 3.2 Holonomy Groups We have a tangent bundle with structures I, ω, g which satisfy I = 0, ω = 0 and g = 0. Hence parallel transport preserves the entire Kähler structure. In particular for a loop γ at x 0, we get an automorphism O(n) P γ : T x0 T x0. Varying over all paths get {P γ } Lie Subgroup O(n), the holonomy group of (M, g). In the Kähler case, get Special Holonomy: Hol(M, g) U(n) SO(2n, R). ( ) aut of (R 2n, I) GL(2n, R) GL(n, C) Sp(2n, R) ( ) aut of (R 2n, ω std ) max cpt subgp U(n) 3

Berger s List (1955) Of Possible Holonomy Groups 1. Locally symmetric spaces (loc. G/K with Hol = K.) Lie theoretic classification by Cartan. 2. M n R Hyperkähler - Define Group O(n) Structure SO(n) orientable U ( ) n Kähler 2 SU ( ) n Calabi-Yau (noncompact Calabi,compact Yau) 2 Sp ( ) n Hyperkähler 4 Sp ( ) n 4 Sp(1) Quaternionic Kähler G 2 O(7) G 2 structure Spin(7) O(8) Spin7 structure (noncompact Bryant, compact Joyce) - Construct: KH reduction, examples - Twistor space - many moduli spaces (Higgs, monopoles,solns to Nahms eqns) coadjoint orbits for complex reductive groups,, have hyperkähler structures. Lecture 2 [13.01.2017] 4 Hyperkähler Structures Definition 1. A hyperkähler manifold is a Riemann manifold with holonomy SP (m) O(n = 4m). Here Sp(m) is the quaternionic unitary group which we now define. Let H = {q = x 0 + ix 1 + kx 2 + kx 3 : x i R} k ImH where i 2 = j 2 = k 2 = ijk = 1. Every quaternion q has a conjugate q = x j 0 ix 1 jx 2 kx 3 and q q = x 2 0 + x 2 1 + x 2 2 + x 2 3. May view H as C 2 via q = (x i 0 + ix 1 ) + (x 2 + ix 3 )j C Cj so the extra structure on C 2 is multiplication by j. More precisely, we have J : L j : C 2 C 2 such that for λ C, J(λv) = λ(jv) ie. J is a complex linear automrophism. So we have J : C 2 C 2 such that C 2 K C 2 J C 2 Any such J is called a quternionic structure on a complex vector space. 1 4

Focus on a certain structure enjoyed by R 4 = H. There are three complex structures L i = I, L j = J, L k = K with the euclidean metric g euc, g(q, q) = q q. These define 3 Kähler structures V g V I,J,K V ω 1 =ω I,ω 2 =ω J,ω 3 =ω K ω 1 = dx 0 dx 1 + dx 2 dx 3 ω 2 = dx 0 dx 2 dx 1 dx 3 ω 3 = dx 0 dx 3 + dx 1 dx 2 In fact, for all (α, βγ) S 2, αi + βj + γk = I (α,β,γ) is also a complex structure. The corresponding Kähler form is ω (α,β,γ) = gi α,β,γ = αω 1 + βω 2 + γω 3 R 4m = H m = (C 2m, J = L J ) w.r.t complex structure defined by L i = I is a left quaternionic module. Have q l = x l 0 + ix l 1 + jx l 2 + kx l 3, l = 1,, m Definition 2. Sp(m) is the stabilizer in O(4m, R) of the structure (g, I, J, K) (and hence ω 1, ω 2, ω 3 ). Sp(m) is a compact (simply) connected Lie group with dim R = 2m 2 + m. m=1: coincides with SU(2) = S 3 m=2: coincides with Spin(5) This is sometimes defined as a subgroup of GL(m, H) preserving (, ). Here GL(m, H) = { [a ij ] m i,j=1 : A ij H } with quaternionic action A (q 1,, q m ) = (q 1,, q m )ĀT. c.f. U(n) GL(n, C) preserving (, ). Similarly, by privileging I, we view ((H m, I), J) = (C 2m, J) as a complex vector space with quaternionic structure. Have complex coordinates and then z 2p 1 = x p 0 + ix p 1, z 2p = x p 2 + ix p 3 g = l dz l 2 ω I = i dz l d z l 2 l ω J + iω K = dz 1 dz 2 + dz 3 dz 4 + Note that ω j + iω K is a holomorphic (complex) symplectic form on C 2m. Hence Sp(m) = U(2m) Sp(2m, C) (Sp(2m, C) is the complex symplectic group.) In fact, the hyperkähler structure is completely determined by (ω I, ω J, ω K ): 5

Lemma 2. Proof. Stab(ω I, ω J, ω K ) = Sp(m) V g V I ω ω Warning 1. Not so for I,J,K. ω 1 = gi ω 2 = gj ω 1 ω 2 = I 1 g 1gJ = K ω K ( K) = g Example 2. Consider dimension 4 on R 4 = V. Lemma 3. (ω 1, ω 2, ω 3 ) has stabilizer Sp(1) { ω 1 ω 1 = ω 2 ω 2 = ω 3 ω 3 0 ω 1 ω 2 = ω 1 ω 3 = ω 2 ω 3 = 0 Proof. Consider Ω 1 = ω 2 + iω 3 (I holomorphic symplectic form dz 1 dz 2.) Then Ω 1 Ω 1 = 0 so Ω 1 decomposes as Ω 1 = θ 1 θ 2 ω 3 Ω 2 ω 2 ω 1 Ω 1 Ω 2 = 2ω1 2 0 θ 1 θ 2 θ 1 θ 2 0 so we have a basis (θ 1, θ 2, θ 1, θ 2 ) for V C. But this is equivalent to a decomposition V C = θ 1, θ 2 θ 1, θ 2. This yields a complex structure by regarding θ 1, θ 2 as the 1, 0 eigenspace and θ 1, θ 2 as the 0, 1 eigenspace. So far, we have a complex structure I on Stab GL(2, C) but Ω 1 is also preserved which imples Stab SL(2, C) and since ω 1 is also preserved, ω 1 Ω 1,1 I. Expand ω 1 = α,β=1,2 h α βθ α θ β where h α, β is a 2 2 Hermitian matrix and so h α, β is conjugate to ( ) λ 0 0 µ We have λµθ 1 θ 2 θ1 θ2 = (det h α, β)θ 1 θ 2 θ1 θ2 = ω 2 1 = 1 2 Ω Ω = 1 2 θ 1 θ 2 θ 1 θ 2 which implies both λ, µ have the same sign. Note 1. M 4 a compact oriented manifold, H 2 H 2 H 4 = R a symmetric, nondegenerate inner product on H 2. b + (M 4 ) = # of positive directions signature (b +, b 2 b + ). In order to have a Kähler structure need what? k3 has signature (3, 19), dim H 2 = 22 supports a hyperkähler structure. 6

The Manifold Case If (M 4m, g) has hol( LC ) Sp(m) then all the Sp(m)-invariant structure on H m passes to corresponding structure on T M, which is flat. So we get a Riemannian manifold (M, g, I, J, K) with three complex structures, 3 Kähler structures and is flat: I = J = K = 3 Kähler structures. If we privilege I we get = complex manifold (M, I) of complex dimension 2m = Ω 1 = ω 2 + iω 3 holomorphic symplectic (2, 0) form 2m = Ω m 1 T 1,0 = K the canonical bundle = K = O trivial CY manifold. Recall LC, K flat so curv(chen) = RicI = 0 Lecture 3 [18.01.2017] 4.1 Examples of Hyperkähler Structures in dimension 4 Recall that we showed a hyperkähler structure in dimension 4 is a triple (ω 1, ω 2, ω 3 ) such that ω 2 i = 0, ω 1 ω 2 = ω 1 ω 3 = ω 2 ω 3 = 0 and dω i = 0 (integrable.) Example 3 (Local). Take an open in R 4 = (C 2, (z 1, z 2 )) with I = i, Ω 1 = dz 1 dz 2 = ω 2 + iω 3 ω 1 = i φ φ C (U, R) = i 2 φ z i z j dz i d z j For example can take φ = z 1 z 1 + z 2 z 2 so i φ = i(dz 1 d z 1 + dz 2 d z 2 ). This simply gives H 4. The only remaining condition is ( ) ω1 2 2 φ = vol Euc det = 1 z i z j ( ) The equation det 2 φ z i z j = 1 is the (complex) Monge-Ampère equation. Calabi: Suppose z i = x i +y i and φ(z 1, z 2, z 1, z 2 ) = φ(x 1, x 2 ). Then we get the real Monge-Ampère equation: ( ) 2 φ det = 1. x i x j This is a nonlinear PDE so assume φ is SO(2) invariant (ie. radial) to get an ODE "cohomgeneity one". Then can solve the equation to get φ(x) = x 0 (c + r 2c ) 1 2 dr. For example, for c = 1 get 1 2 (c + r) 1 2 + 1 2 sinh 1 (r) + K. Diagrams missing 7

Gibbons Hawking Examples 1. U R 3 U R 3, M = S 1 U with coordinates (θ, x 1, x 2, x 3 ), V (x 1, x 2, x 3 ) > 0. where g m = V 1 (dθ+a) 2 + V (dx 2 1 + dx 2 2 + dx 2 3) A = A 1 (x)dx 1 + A 2 (x)dx 2 + A 3 (x)dx 3 with A i C (U, R) We have a trivial S 1 bundle with principal connection dθa on the principal S 1 bundle M π U. A is called the connection one form. g M diagonal = on basis for T e 0 = V 1 2 (dθ + A) e 1 = V 1 2 dx1 e 1 = V 1 2 dx2 e 1 = V 1 2 dx3, g M = e 2 0 + e 2 1 + e 2 2 + e 2 3. ω 1 = (dθ + A) dx 1 + V (dx 2 dx 3 ) ω 2 = (dθ + A) dx 2 V (dx 1 dx 3 ) ω 3 = (dθ + A) dx 3 + V (dx 1 dx 2 ) which has stabiliser Sp(1). To get a hyperkähler structure we need dω i = 0: dω 1 = da dx 1 + dv dx 2 dx 3 = 0 ( A2 A 3 + V ) dx 1 dx 2 dx 3 = 0 x 3 x 2 x 1 This is a scalar equation in R 3. Including the other equations dω i = 0, get da = R 3 dv, or equivalently, V = 0, so V is harmonic on R 3 = dv closed = exact if U is simply connected = A with da = V. On all of R 3, V = 1 and so just get R 3 Euc S1 (A is unique up to gauge.) Observe that if we use V = 1 r in R3 \{0} (warning: H 2 0), dv = ( r 2dr) = r 2 (r 2 dω) = dω S 2 dv 0 = dv da 8

Generalization: M 4 π R 3 \{p 1,, p k } = B a principal S 1 bundle which is Hopf about each point p i, ie. c 1 (M) = 1 on each Sp 2 i or M S 2 pi is the Hopf fibre S 3 S 2. Diagrams Missing Now, choose a principal connection η Ω 1 (M) which is S-invariant (L θ η = 0) and ι θ = 1 where θ is the vector field generated by the S 1 action. - ker η is the horizontal distribution. - dη = π F, F is the curvature of η Ω 2 (B) (since ι θ dη = L θ η = 0) Given M B R 3, η Ω 1 (M) and V C (B, R) can build the metric g = V 1 η 2 + V π (dx 2 1 + dx 2 2 + dx 2 3) As before, define e 0 = V 1 2 η, e i = V 1 2 dx i, w 0 =. Then we get a hyperkähler structure F = dv (so V is harmonic) and by assumption S 2 p i F = 1 = S 2 p i dv = 1 so V 1 r at each p i. V = c + k i=0 1 x p i R 3 For c 0 these are called (multi) Taub-NUT Hyperkähler metrics (ALF (asymptotically locally flat): R 3 S 1 ) For c = 0 A k ALE (asymptotically locally euclidean: R 4 ) hyperkähler structures. These are also called graviational instantons in the physics literature. These two cases are very different in terms of asymptotic geometry but both have two amazing properties: 1. Extend smoothly to M {p 1,, p k } to give a smooth 4-dimensional manifold similar to the way R 4 Hopf R 3. 2. Resulting (M, g ) is complete. Example 4. c = 0, k = 0 (one point), V = 1 r givese (R4, g Euc ). Have R 4 = C 2 with S 1 action Map e iθ (z 1, z 2 ) = ( e iθ z 1, e iθ z 2 ) θ = i(z 1 z 2 ) g( θ, θ ) = V 1 (η( θ )) 2 = V 1 θ = V 1 = z 1 2 + x 2 2 π :R 4 R 3 (z 1, z 2 ) { x 1 = z 1 2 z 2 2 x 2 + ix 3 = 2z 1 z 2 (map is quadratic and restricted to S 3 is just the Hopf map.) Get a principal S 1 bundle on R 3 \0 with critical value 0. ) 2 r 2 = x 2 1 + x 2 2 + x 2 3 = ( z 1 2 z 2 2 ) 2 + 4 z 1 2 z 2 2 = ( z 1 + z 2 2 V 1 = r 2 = r V = 1 r 9

Lecture 4 [20.01.2017] Recall that we constructed the A k ALE hyperkähler manifolds: Given get a metric R 3 \{p 1,, p k } Mprincple S 1 bundle with c 1 = 1 about each p i η Ω 1 ( M) S 1 -invariant connection dη = π dv V a harmonic function on R 3 with 1 r singularities at p i g = V π g R 3 3 + V 1 η 2. Taking V = c + k 1 i=0 x p i with c = 0 gave A k ALE and c 0 gives the multi Taub-NUT metrics. In either case, M = M {p 1,, p k } is a smooth 4-manifold with complete hyperkähler metric. Description of Hyperkähler Structure Finish Diagram Example 5. 1. V = 1 r (R 4, g R 4) flat hyperkähler 2. A 1 diagram missing associated to straightline p p get minimal surface S 2 M = H 2 = S 2 l. If x 2 l = x 3 l = 0, then ω 1 = (dθ + A) dx 1 + V dx 2 dx 3 ω 2 = (dθ + A) dx 2 V dx 1 dx 3 ω 3 = (dθ + A) dx 3 + V dx 1 dx 2 ω 1 is I (Kähler) and Ω 1 = ω 2 + iω 3. So Ω 1 Sl 2 = 0 implies S l 2 is Lagrangian for ω 2, ω 3 and a complex curve with respect to I. In the physics terminology, this is a (B, A, A) brane (the B refers to complex geometry, the A to symplectic.) Fact: For a generic x S cx 2, no line p i p j is parallel to x. Hence (M, I x ) has no rational curves and in fact is an affine algebraic variety but for x parallel to p i p j get rational curves in corresponding complex structure. 10

5 Penrose Twistor Space Start with a hyperkähler manifold (M 4n, g, I, J, K) and let S 2 be the sphere of complex structures with coordinates (x 1, x 2, x 3 ) x 1 I + x 2 J + x 3 K = I x. Then the twistor space is Z = M S 2. Given (m, x) Z, T (m,x) = T m M T x S 2. Since I x T m M and I std T x S 2, I = I x I std T (m,x) Z. Theorem 1. (Z, I) is a complex (2n + 1) C dimensional manifold. Proof. Strategy: Identify generators for Ω (2n+1,0) Z and show these are closed. The holomorphic symplectic structure for I x is what? I x = x 1 I + x 2 J + x 3 K (x 1, x 2, x 3 ) S 2 Ω x = aω 1 + bω 2 + cω 3 [a : b : c] CP 2 a 2 + b 2 + c 2 = 0 smooth conic inp 2 P 1 S 2 conicp 1 P 2 ζ P 1 S 2 conic P 1 P 2 η η η η = 1 1 + ζ ζ (1 ζ ζ, i ( ζ ) ζ), (ζ + ζ) So the holomorphic symplectic form for the complex structure I ζ is Ω ζ = 2iζω 1 + (1 + ζ 2 )ω 2 + i(1 ζ 2 )ω 3 = Ω 1 + 2iζω 1 + ζω 1 + ζ 2 Ω1. Finally for the holomorphic volume form for Z 2n+1. Θ = Ω ζ } {{ } Ω ζ dζ η defines a complex structure and dθ = (d M + d S 2)Θ = 0 Example 6. If H = C 2 then Z 3 is a complex 3-fold with complex structure determined by ω 1 = i 2 (dz 1 d z 1 + dz 2 d z 2 ) ω 2 + iω 3 = dz 1 dz 2 Ω ζ = Ω 1 + 2iζω 1 + ζ 2 Ω1 = dz 1 dz 2 + 2iζ i 2 (dz dz 1 1 + dz 2 d z 2 ) + ζ 2 (d z 1 d z 2 ) ζ(dz 1 d z 1 + dz 2 d z 2 ) = (dz 1 + ζd z 2 ) (dz 2 ζd z 1 ). 11

Hence there are complex coordinates { (ζ, z 1 + ζ z 2, z 2 ζ z 1 ) ζ (ζ 1, ζ 1 z 1 + z 2, ζ 1 z 2 z 1 ) ζ 0 on C C 2. If we glue these, get ) Z = tot (O(1) O(1) P 1 since the transition functions are (ζ, ζ) for the 2 factors. Diagram missing Warning:Each p M defines p S 2 Z and this is obviously a complex curve, hence defines S p H 0 ( }{{} P 1, O(1) O(1)). There exists a real structure on Z, σ : Z C Z preserving =H 0 (O(1)) H 0 (O(1)) all the structure: antiholomorphic involution compatible with Ω ζ sends sections to sections The sections fices by σ are R 4 C 4, R 4 being the real twistor lines and C 4 being all twistor lines. 12