4.4 Geometry of the special unitary group 4.4.1 Special unitary group The special unitary group is the generalization of the special orthogonal group to the complex numbers. The group is defined as SUn = fP 2 Un j det(P ) = 1g: Example 4.4.2. The case n = 1. This is a toy example. It is clear that SU1 is the set of complex numbers having length one, in other words SU1 is the 2 unit circle in the plane R . We have earlier identified the group SO2 with the circle, and therefore also that SO2 = SU1. Lemma 4.4.3. Elements P 2 SU2 are of the form a −b P = b a with any complex numbers a and b such that aa + bb = 1. We have a map 4 ': SU2 −! R , sending a matrix to the vector determined by its first column. 2 2 3 2 4 The map identifies SU2 with the unit 3-sphere fx1 + x2 + x3 + x4 = 1g ⊂ R . In particular the 3-sphere has a group structure. a c Proof. Let P = [ b d ] be an element of SU2. As det(P ) = 1 we have the two following expressions for the inverse of P , namely d −c a b P −1 = = P ∗ = : −b a c d This gives us that d = a and that c = −b. In other words we have that a −b P = ; b a where det(P ) = aa + bb = 1. If we write out the complex numbers as a = x1 + ix2 and b = x3 + ix4; then the map ' takes P to '(P ) = (x1; x2; x3; x4). Then '(P ) is a point on the unit 3-sphere. It is clear that the map ' is injective, and also surjective. Remark 4.4.4. A remarkable result from topology shows that the 1-sphere and the 3-sphere are the only spheres that allow a group structure. 75 a −b Lemma 4.4.5. Let P = b a be an element of SU2. Write a = x1 + ix2. Then, for any unitary matrix Q we have that characteristic polynomial of QP Q∗ is 2 t − 2x1t + 1: Moreover, the eigenvalues of QP Q∗ are complex conjugates λ and λ, only depending on the trace Trace(QP Q) = 2x1. In particular the matrix P is conjugate to the diagonal matrix λ 0 : 0 λ Proof. As Q is unitary we have that QQ∗ = 1. Therefore the characteristic polynomial cP (t) = det(t − P ) is invariant for conjugate matrices. Indeed, we have that ∗ ∗ ∗ cP (t) = det(Q) det(t − P ) det(Q ) = det(Q(t − P )Q ) = det(t − QP Q ); where the last expression is the characteristic polynomial of QP Q∗. As P is in SU2 is thas determinant one, and a computation gives that 2 2 cP (t) = t − (a + a)t + 1 = t − 2x1 + 1: p 2 The roots of the characteristic polynomial are x1 ± x1 − 1, hence the eigen- values are complex conjugate. Since a unitary matrices are normal we have that P is normal, and by Theorem 4.3.16 we have that P is conjugate to a diagonal matrix. A diagonal matrix must neccessarily have its eigenvalues on the diagonal. Remark 4.4.6. Let P be an element of SU2. With the notation as in the p 2 above lemma we have that the eigenvalues of P are x1 ± x1 − 1. Thus the two eigenvalues λ and λ are distinct as long as x1 6= ±1. Then P is conjugate to two diagonal matrices obtained by changing the order of its eigenvalues. By the lemma above we have that P must be conjugate to at least one of λ 0 0 1 ∗ these, say D = 0 λ . The matrix Q = [ −1 0 ] is unitary, and such that QDQ is a diagonal but with λ interchanged with λ. In the cases when x1 = ± then there is only only diagonal matrix in the orbit, namely I and −I. 4.4.7 Latitudes The points (±1; 0; 0; 0) of the unit 3-sphere are identified as the north and south pole. The latitudes at −1 ≤ c ≤ 1 is then obtained as the intersection of the plane x1 = c and the unit 3-sphere. Thus, the latitude is given by 2 2 2 2 x2 + x3 + x4 = (1 − c ): 76 p The latititudes are 2-spheres with radie ( 1 − c2) when c 6= ±1, and simply the poles when c = ±1. Proposition 4.4.8. The conjugacy classes in SU2 are the latitudes of the unit 3-sphere. For a given −1 ≤ c ≤ 1 the corresponding latitude consists of all matrices P 2 SU2 having trace Trace(P ) = 2c. In particular trace zero matrices form the equator. The two special latitudes c = ±1, i.e. the poles, consits each of one single element fIg and {−Ig, and do together form the center of the group SU2. Proof. The statements follows by Lemma 4.4.5. 4.4.9 Longitudes A longitude of the 3-sphere S we define as the intersection of S with a 2-dimensional plane passing through the poles. A longitude is then a 1- dimensional circle. There is a unique line passing through the two poles of the sphere S. Thus, if P is any point on the sphere, different from the poles, there will be a unique plane containing P and the poles, hence there is a unique longitude passing through P . 4 Let W ⊆ R be the plane x3 = x4 = 0. For any w = (w1; w2; 0; 0) 2 W we let P = w1+iw2 0 . Then the longitude W \ S are the matrices P w 0 w1−iw2 w such that det(Pw) = 1. Elements in this particular longitude are then the matrices λ 0 W \ S = f j λλ = 1g: 0 λ 4 Lemma 4.4.10. Let Q be an element of SU2. The map W −! R sending 1 w 7! QP Q∗ w 0 is an R-linear, injective map. a −b Proof. Write Q = b a . We then compute that a −b λ 0 a b aaλ + bbλ ab(λ − λ) QP Q∗ = = : b a 0 λ −b a ab(λ − λ) aaλ + bbλ The map in question takes w 2 W to the first column of the matrix QP Q∗. Write now a = x1 + ix2 and b = x3 + ix4. Then we obtain that 2 2 2 2 2 2 2 2 aaλ + bbλ = (x1 + x2 + x3 + x4)w1 + i(x1 + x2 − x3 − x4)w2: 77 And, for ab(λ − λ) we get the expression 2(x2x3 − x1x4)w2 + i2(x1x3 + x2x4)w2: Hence we have that the map is real linear. Infact, the matrix representation of this map in the standard basis is 21 0 3 60 x2 + x2 − x2 − x27 A = 6 1 2 3 47 : 40 2(x2x3 − x1x4) 5 0 2(x1x3 + x2x4) We need to see that the matrix A has rank two. Assume that coefficients 2 2 2 2 of the second column of A vanishes. Using that x1 + x2 + x3 + x4 = 1 2 − 2 4 2 2 1 2 2 together with x1 + x2 x3 − x2 = 0 gives that x1 + x2 = 2 = x3 + x4. Thus q 1 2 q 1 2 x2 = ± 2 − x1 and x3 = ± 2 − x4. Then the two remaining coeffcients 2 2 2 2 2 2 2 2 1 2 1 1 2 give x2x3 + x1x3 = x1x4 + x2x4. Which gives 2 x3 = 2 ( 2 − x3). The solutions 1 are then that all coordinates x1; x2; x3 and x4 are ± 2 . Then, by checking, one sees that one is not able to choose parities such that x2x3 = x1x4 and x1x3 = −x2x4 are satisfied simultaneously. Hence, we have that the matrix A has rank two. Proposition 4.4.11. The longitudes of SU2 are the conjugate subgroups of T , that is subgroups of the form ∗ ∗ Q(W \ S)Q = fQDλQ j Dλ 2 T g; with Q 2 SU2. Proof. We need to show that QT Q∗ is a longitude for any given Q, and that every longitude is of this particular form. By Lemma 4.4.10 we have that ∗ 4 4 QT Q determines a real linear map W −! R . The image VQ ⊆ R is a plane that contains the poles (±1; 0; 0; 0), and consquently VQ \ S is a longitude. ∗ By construction the longitude VQ \ S is the conjugate subgroup QT Q .A point P on the sphere corresponds to a matrix in SU2. By Lemma 4.4.5 the P is conjugate to a diagonal D matrix, hence of the form P = QDQ∗. Thus, all longitudes are conjugate subgroups QT Q∗. Exercises. Special unitary group. 1. Let x and y be two points on the unit 3-sphere S, which is a group. Compute the coordinates for the product x · y. 2. Show that the subgroup SO2 of SU2 is a longitude on the unit 3-sphere S. 78.