Introduction to Conformal Geometry C. Robin Graham Lecture 1 • Flat model for conformal geometry • Conformal group: O(n + 1; 1) Lecture 2 • Curved conformal geometry • Weyl and Cotton tensors • Characterization of conformal flatness Lecture 3 • Ambient metric • Tractor bundle and connection 2 Riemannian geometry: M: smooth manifold g: metric on M Can measure lengths of tangent vectors: 2 jvj = g(v; v); v 2 TpM Isometry: preserves lengths Conformal geometry: Can only measure angles: g(v; w) θ(v; w); cos θ = jvj jwj Same as knowing g up to scale at each point Conformal mapping: preserves angles Same as preserving g up to scale 3 Flat Model Riemannian geometry: Rn, Euclidean metric: (dyi)2 P Isometries: E(n) = group of Euclidean motions y ! Ay + b; A 2 O(n); b 2 Rn Can view E(n) ⊂ GL(n + 1; R) by: A b (A; b) $ 0 1! y Embed Rn ,! Rn+1 by: y $ 1! A b y Ay + b Then = 0 1! 1! 1 ! 4 Conformal Transformations of Rn Euclidean motions Dilations y ! sy; s 2 R+ Inversions in spheres For unit sphere, center at origin, get y ! y jyj2 Definition: M¨ob(Rn) = Group generated by Euclidean motions, dilations, inversions M¨ob(Rn) provides one approach to the study of the conformal group. But inversions don't map Rn to Rn: center of sphere goes to 1 Suggests compactifying Rn by appending 1 Unnecessary and inappropriate for Euclidean motions 5 Compactify Rn to Sn = Rn [ 1. Metric on Rn+1 induces metric on Sn; hence a conformal structure on Sn. Sn n 1 is conformally equivalent to Rn via stereographic projection. Fruitful point of view: describe conformal geome- try of Sn in terms of Minkowski geometry of Rn+2: quadratic form Q of signature (n + 1; 1) x = (x0; x1; · · · ; xn+1) 2 Rn+2 n Q(x) = (xα)2 − (xn+1)2 αX=0 N = fx : Q(x) = 0g ⊂ Rn+2 n f0g Null cone Pn+1 = ` = [x] : x 2 Rn+2 n f0g lines in Rn+2 n o Q = f` = [x] : x 2 N g ⊂ Pn+1 Quadric Q =∼ Sn : Let y 2 Sn, so y 2 Rn+1; jyj = 1. y The map Sn 3 y ! 2 Q is a bijection. " 1!# π : N ! Q projection 6 n g = (dxα)2 − (dxn+1)2 Minkowski metric αX=0 e I J I 0 = gIJdx dx ; where gIJ = 0 −1! e e n+1 I Rn+2 X = x @xI position vector field on IX=0 If x 2 N , then X(x) 2 TxN I J Have g(X; X) = gIJx x = Q(x) = 0 on N . e e Claim: g(X; V ) = 0 for all V 2 T N . e Proof: Q = 0 on N . So dQ(V ) = 0 if V 2 T N . I J I J Now Q = gIJx x , so dQ = 2gIJx dx . e I J e Gives 0 = dQ(V ) = 2gIJx V = 2g(X; V ). e e 7 Let x 2 N . Then g is degenerate: X ? TxN TxN e (x) But g induces an inner product g on T`Q, TxN e where ` = π(x) = [x]. g(x) is defined as follows: Have π : N ! Q. Gives π∗ : TxN ! T`Q Now π∗(X) = 0, and π∗ : TxN =spanX ! T`Q is an isomorphism (x) Given v, w 2 T`Q, want to define g (v; w) Choose V , W 2 TxN such that π∗V = v, π∗W = w. Unique up to V ! V + cX, W ! W + c0X. Define g(x)(v; w) = g(V; W ). Independent of choices since X ? TxN . e (x) g is a positive definite inner product on T`Q. If 0 6= s 2 R, how are g(sx) and g(x) related? 8 n+2 n+2 Have δs : R ! R ; δs(x) = sx Dilations ∗ 2 g satisfies (δs) g = s g. e e e Have π(sx) = π(x) = `. Suppose v, w 2 T`Q. Choose V , W 2 TxN so that π∗V = v, π∗W = w. Then Vs = (δs)∗V , Ws = (δs)∗W satisfy π∗Vs = v, π∗Ws = w; since π ◦ δs = π. So (sx) g (v; w) = g(Vs; Ws) = g((δs)∗V; (δs)∗W ) ∗ e 2 e 2 (x) = (δs) g(V; W ) = s g(V; W ) = s g (v; w) e e Thus g(sx) = s2g(x). Conclusion: The Minkowski metric on Rn+2 invariantly determines a conformal structure on Q, but not a Riemannian structure. 9 Conformal Group Idea: Any map of Rn+2 which preserves the linear structure and the Minkowski metric will induce a map of Q preserving the conformal structure. Definition: O(n + 1; 1) = linear tranformations of Rn+2 preserving Q Let L 2 O(n + 1; 1). Then LjN : N ! N . Preserves lines, so induces LQ : Q ! Q. Claim: LQ is a conformal transformation of Q. Proof: Let x 2 N , so π(x) = [x] = ` 2 Q. Then LQ([x]) = [Lx]. Must show that ∗ (Lx) (x) (LQ) g is a multiple of g . Will actually show they are equal. Let v, w 2 T`Q. Choose V , W 2 TxN so that π∗V = v, π∗W = w. Then L∗V satisfies π∗L∗V = (LQ)∗(v), sim. for W . So ∗ (Lx) (Lx) (LQ) g (v; w) = g (LQ)∗(v); (LQ)∗(w) ∗ = g(L∗V; L∗W ) = (L g)(V; W ) = g(V; W ) = g(x)(v; w) e e e 10 Thus O(n + 1; 1) acts on Q by conformal transformations. −I 2 O(n + 1; 1) acts trivially. Definition: The conformal group of Q = Sn is O(n + 1; 1)={Ig. Metrics in the conformal class on Q correspond to sections of N ! Q (modulo 1): Let s : Q ! N be a section of π. If ` 2 Q, then x = s(`) is a point on `. So g(x) is a metric in the conformal class at `. Thus ` ! g(s(`)) is a metric in the conformal class on Q. g(s(`)) determines s(`) up to 1. Explicit realization on Sn Realize Sn =∼ Q by xn+1 = 1. This section s deter- mines the usual metric on Sn, since we can choose V , W so that dxn+1(V ) = dxn+1(W ) = 0, in which n case g becomes (dxα)2. αX=0 e Recall LQ is the map induced on Q by L. n n Write LSn : S ! S for the corresponding con- formal transformation of Sn, when Q is identified with Sn this way. 11 How does L 2 O(n + 1; 1) act on Sn? A b Let L 2 O(n + 1; 1). Write L = t . c d! Let y 2 Sn. So y 2 Rn+1 and jyj = 1. y Set x = 2 N . Then 1! A b y Ay + b Lx = t = t 2 N . c d! 1! c y + d! + Ay+b Ay b t cty+d Now t = (c y + d) c y + d! 0 1 1 @ A and Ay+b = 1. cty+d So obtain Ay+b L n(y) = S cty+d Conclusion: The conformal tranformations of Sn are fractional linear transformations induced by L 2 O(n + 1; 1). 12 This argument also gives the conformal factor associated to LSn. ∗ (Lx) (x) Recall LQ g = g n+1 (x) n If x = 1, then g = gSn is usual metric on S . y Take y 2 Sn and apply above with x = . 1! L n(y) Then Lx = (cty + d) S = sx0 1 ! L n(y) with s = (cty+d); x0 = S ; so (x0)n+1 = 1. 1 ! 0 0 Use g(sx ) = s2g(x ); get ∗ (Lx) ∗ (sx0) 2 ∗ (x0) LQ g = LQ g = s LQ g ∗ t −2 So (LSn) gSn = (c y + d) gSn. Conclusion: The conformal factor associated to the fractional linear transformation Ay+b t −2 L n(y) = is (c y + d) . S cty+d 13 Can conjugate LSn by stereographic projection to obtain formulae for the conformal transformations on Rn. Why are all conformal transformations of this form? Liouville's Theorem: Suppose n ≥ 3. If U ⊂ Sn is open and connected and ' : U ! Sn is a C1 conformal transformation, then ' is the restriction to U of LSn for some L 2 O(n + 1; 1). This is false for n = 2: holomorphic and conjugate- holomorphic maps are conformal. But the global result is true for n = 2: every conformal transfor- 2 mation of S is LS2 for some L 2 O(3; 1). This follows by complex analysis: conformal transfor- mations are holomorphic or anti-holomorphic; the holomorphic ones are in P SL(2; C), isomorphic to the identity component of O(3; 1). Liouville's Theorem is a result about solutions of overdetermined systems of pde's; cf. talks of M. Eastwood. The condition that ' is conformal is '∗g = Ω2g, where Ω > 0 is an unknown function. This is a first order system of pde's for the compo- nents of '. There are n + 1 unknowns: Ω and the n components of '. There are n(n + 1)=2 equa- tions: the components of g. These are equal for n = 2, so problem is determined in that case{lots of solutions. Overdetermined for n > 2. 14 The proof proceeds by first analyzing Ω. We know that Ω for any LSn is of the form Ω(y) = (cty + d)−1. So try to show that Ω−1 is a linear function of y. Actually, the Theorem is proved in its Rn realiza- tion. Corresponding statement is that the anal- ogous Ω−1 on Rn is a quadratic function of the 2 −1 form µjx − x0j + κ.