The Banach{Tarski Paradox and Amenability Lecture 4: Free Groups and Orthogonal Groups 9 August 2012 Paradoxical decomposition of free groups Definition Let G be a group. Then G is paradoxical if there are two proper disjoint subsets A and B of G such that A ∼ G and B ∼ G. Theorem A free group of rank 2 is paradoxical. Proof. Let F = F (a; b) be the free group of rank 2. Let A+ be the set of all reduced words beginning with a and A− the set of all reduced words beginning with a−1. Define B+ and B− similarly. Let A = A+ [ A− and B = B+ [ B−. To see that A ∼ F , observe that F is the disjoint union of A+ and aA−. Similarly, B ∼ F . Hence F is a paradoxical group. Paradoxical decomposition of sets As an important special case of the Proposition at the end of Lecture 2: Corollary Suppose F = F (a; b) acts freely on a set X . Then X is F {paradoxical. Proof. Using the Axiom of Choice, let M be a set of representatives of the + F {orbits on X . Let XA+ = fwm j w 2 A ; m 2 Mg and define XA− , XB+ and XB− similarly. Check that: I XA+ , XA− , XB+ and XB− are pairwise disjoint (using the freeness of the F {action on X ) I X = XA+ [ aXA− and X = XB+ [ bXB− . ∗ ∗ ∗ Putting A = XA+ [ XA− and B = XB+ [ XB− , it follows that A and B∗ are proper disjoint subsets of X such that A∗ ∼ X and B∗ ∼ X . Orthogonal and special orthogonal groups n Let h·; ·i be the usual inner product on R hx; yi = x1y1 + x2y2 + ··· + xnyn Let Mn(R) be the set of all n × n matrices with entries in R. Definitions 1. The orthogonal group n O(n; R) := fA 2 Mn(R) j hAx; Ayi = hx; yi for all x; y 2 R g 2. The special orthogonal group SO(n; R) := fA 2 O(n; R) j det(A) = 1g Orthogonal and special orthogonal groups Orthogonal groups and special orthogonal groups are examples of Lie groups and of algebraic groups. A (real) Lie group is a group G which is also a finite-dimensional (real) smooth manifold, such that the group operations are compatible with the smooth structure. That is, the multiplication map G × G ! G is smooth, and inversion G ! G is smooth. The reason O(n; R) and SO(n; R) are Lie groups is that they are n2 groups which are closed subsets of the smooth manifold R , so that the group operations are smooth. An algebraic group is, roughly speaking, a group whose elements are the solutions of a collection of polynomial equations. Orthogonal and special orthogonal groups Lemma n−1 The groups O(n; R) and SO(n; R) act on the sphere S . Proof. It suffices to show that elements of O(n; R) preserve the length of n vectors. Given A 2 O(n; R) and x 2 R we have: jjAxjj2 = hAx; Axi = hx; xi = jjxjj2 as required. The cases n = 1 and n = 2 When n = 1 we have O(1; R) = {±1g, the group of order 2, and SO(1; R) = f1g the trivial group. Clearly these don't have a free subgroup! When n = 2, any element of SO(2; R) has the form cos θ − sin θ ; θ 2 [0; 2π) sin θ cos θ This rotates the plane by angle θ anticlockwise about the origin. Note that SO(2; R) is abelian. So this group can't have a nonabelian free subgroup. The group O(2; R) is generated by the rotations SO(2; R) together with the reflection 1 0 0 −1 and also does not contain a nonabelian free subgroup. Free subgroup of SO(3; R) Theorem (Hausdorff 1914) 3 There are rotations φ and ρ about lines through the origin in R such that φ and ρ generate a free nonabelian subgroup of SO(3; R). Corollary For all n ≥ 3, the group SO(n; R) has a free subgroup of rank 2. Free subgroup of SO(3; R) We will give a proof due to Swierczkowski´ (1958). 1 Let φ be rotation about the z{axis by angle θ where cos θ = 3 . Then 0 p 1 0cos θ − sin θ 01 1 − 2 2 0 p3 3 φ = sin θ cos θ 0 = B 2 2 1 C @ A @ 3 3 0A 0 0 1 0 0 1 1 Let ρ be rotation about the x{axis by angle θ where cos θ = 3 . Then 0 1 01 0 0 1 1 0 0 p ρ = 0 cos θ − sin θ = B0 1 − 2 2 C @ A @ p3 3 A 0 sin θ cos θ 2 2 1 0 3 3 We will show that no nonempty reduced word w in fφ±1; ρ±1g is equal to the identity. Free subgroup of SO(3; R) Let w be a nonempty reduced word in fφ±1; ρ±1g. Then w ≡ I () φwφ−1 ≡ I () φ−1wφ ≡ I so without loss of generality we may assume w = vφ±1 where v is a possibly empty reduced word not ending in φ∓1. Now assume by way of contradiction that w ≡ I . Then w(1; 0; 0) = (1; 0; 0). We will obtain a contradiction by showing: p ! a b 2 c w(1; 0; 0) = ; ; (1) 3k 3k 3k where a; b; c are integers, k is a positive integer and b is not divisible by 3. We prove (1) by induction on the length of w. Free subgroup of SO(3; R) Let w be a nonempty reduced word in fφ±1; ρ±1g, wlog w = vφ±1, and assume w ≡ I . We first show p ! a b 2 c w(1; 0; 0) = ; ; 3k 3k 3k where a; b; c are integers and k a positive integer (the length of w). Base case of induction: w = φ±1. Then p ! 1 ±2 2 w(1; 0; 0) = ; ; 0 3 3 Now suppose w = φ±1w 0 or w = ρ±1w 0 where w 0 has length k − 1 p 0 a0 b0 2 c0 0 0 0 and w (1; 0; 0) = 3k−1 ; 3k−1 ; 3k−1 , a ; b ; c 2 Z. Then if w = φw 0 p ! a0 − 4b0 (2a0 + b0) 2 3c0 w(1; 0; 0) = ; ; 3k 3k 3k and in other cases similarly we obtain integers a; b; c. Free subgroup of SO(3; R) We complete the proof by showing that in the expression p a b 2 c w(1; 0; 0) = 3k ; 3k ; 3k the integer b is not divisible by 3. Consider cases: w = φ±1ρ±1u, w = ρ±1φ±1u, w = φ±1φ±1u and w = ρ±1ρ±1u, where u is possibly empty. p a00 b00 2 c00 For example when w = φφu let u(1; 0; 0) = 3k−2 ; 3k−2 ; 3k−2 then w(1; 0; 0) equals p ! p ! a00 − 4b00 (2a00 + b00) 2 3c00 a0 b0 2 c0 φ ; ; = φ ; ; 3k−1 3k−1 3k−1 3k−1 3k−1 3k−1 p ! p ! a0 − 4b0 (2a0 + b0) 2 3c0 a b 2 c = ; ; = ; ; 3k 3k 3k 3k 3k 3k where by inductive assumption b0 is not divisible by 3. So b = 2a0 +b0 = 2(a00 −4b00)+b0 = (2a00 +b00)−9b00 +b0 = 2b0 −9b00 is not divisible by 3. The special orthogonal group SO(3; R) With respect to a suitable basis, every element of SO(3; R) has matrix 0cos θ − sin θ 01 @sin θ cos θ 0A ; θ 2 [0; 2π) 0 0 1 So every nontrivial element of SO(3; R) is rotation about a line 3 through the origin in R ..