Math 319 Problem Set 4: Bilinear forms and the Lorentz group Lie Groups Review pp. 1{5 from Taylor and Wheeler's Spacetime Physics. For a little more on the invariance of the interval, see pp. 22{26. We de¯ne a new \dot product" called the Lorentz product on R2 by (t1; x1) ¢ (t2; x2) = t1t2 ¡ x1x2 : For this problem set, when we write v ¢ w for vectors v and w in R2, we mean the Lorentz product of the vectors. The Lorentz product is motivated by the de¯nition of the interval on p. 24 of Taylor and Wheeler: If v = (t1; x1) and w = (t2; x2) are two events in space-time = R2, the interval between them satis¯es (interval)2 = (v ¡ w) ¢ (v ¡ w) : (It would be more natural to write an equation for the interval itself, but | since (v ¡ w) ¢ (v ¡ w) can be negative | in order to handle all cases, we'd have to work over the complex numbers. For now, we'll stick to R.) 1. Check that the Lorentz product has the following properties. a. v ¢ w = w ¢ v for all v; w in R2. b. For all u; v; w in R2, (u + v) ¢ w = u ¢ w + v ¢ w u ¢ (v + w) = u ¢ v + u ¢ w c. (av) ¢ w = a(v ¢ w) = v ¢ (aw) for all v; w in R2 and all a in R. d. If v is a particular vector satisfying v ¢ w = 0 for every w in R2 then v must be the zero vector (0; 0). Formal language: Property (a) means the Lorentz product is symmetric; proper- ties (b) and (c) mean it is bilinear; property (d) means it is non-degenerate. The usual dot product on Rn also has these three properties. (The argument that the usual dot product on Rn is non-degenerate is essentially the same as problem (4c) in Problem Set 2.) The more general term, encompassing all such \dot products" is symmetric, non-degenerate bilinear form. A vector space together with a symmetric, non-degenerate bilinear form is called a metric vector space. And linear transforma- tions of the vector space that preserve the bilinear form are called isometries of the metric vector space. For the rest of this problem set, assume the dot product is the Lorentz product. 2. In this problem you'll use the language of transformations to show that L = fT : R2 ! R2 : T linear and T (v) ¢ T (w) = v ¢ w for all v; w 2 R2g 1 is a group. It is called the Lorentz group. a. Show that L contains the identity transformation I. b. Assume S and T are in L. Show that T ± S preserves the Lorentz product. Since we know that the composition of linear transformations is linear, this shows that T ±S is in L, so L is closed. c. Show that if v is a particular vector satisfying T (v) = (0; 0), then v = (0; 0). [Hint: Use the property in (1d).] d. Result (c) implies that if T is in L, then T is invertible; write T ¡1 to denote its inverse. Show that T ¡1 preserves the Lorentz product. Since we know that the inverse of an invertible linear transformation is a linear transformation, this shows that T ¡1 is in L and completes the proof that L is a group. d. Let L+ be the subset of L consisting of those transformations in L having deter- minant +1. Show that L+ is a subgroup of L. [Compare the proof that SO(n; R) is a subgroup of O(n; R) in Problem Set 2.] 3. Show that v ¢ w = vT Cw for · ¸ 1 0 C = : 0 ¡1 4. a. Using (3) above and reasoning as in (5) from Problem Set 2, explain why an equivalent description of L is given by L = fM 2 M(2; R): M T CM = Cg: b. Use the description in (a) and the language of matrices to show L is a group. Taylor and Wheeler's de¯nition of the interval (and our de¯nitions of the Lorentz form and the Lorentz group for R2) are really specializations of the de¯nitions for R4: we think of spacetime as 4-dimensional, with an event at time t and in location (x; y; z) in R3 having coordinates (t; x; y; z). In R4, the Lorentz product corresponds to v ¢ w = vT Cw for 2 3 1 0 0 0 6 0 ¡1 0 0 7 C = 6 7 : 4 0 0 ¡1 0 5 0 0 0 ¡1 In other words, we de¯ne the Lorentz product of v = (t1; x1; y1; z1) and w = (t2; x2; y2; z2) by v ¢ w = t1t2 ¡ x1x2 ¡ y1y2 ¡ z1z2: Then the square of the interval between the spacetime events v and w is again (v¡w)¢(v¡w). It is straightforward to check that this product has all the properties in (1). In this case, the set of matrices (or, equivalently, the set of linear transforma- tions) in L is also called the Lorentz group. Context should make clear whether we mean the version with one or three spatial coordinates, although we can write L(2; R) or L(4; R) for emphasis. 2 5. Recall the speci¯c descriptions of matrices in O(2; R) from Problem Set 1; those matrices had entries involving the (circular) trigonometric functions sine and cosine. To obtain speci¯c descriptions of the matrices in the Lorentz group, we need to know about the hyperbolic trigonometric functions. The hyperbolic sine and hyperbolic cosine are de¯ned by e® ¡ e¡® e® + e¡® sinh(®) = cosh(®) = 2 2 (and tanh(®) = sinh(®)= cosh(®), etc.). Assuming the usual properties of the expo- nential function, verify the following hyperbolic trigonometric identities: a. cosh2(®) ¡ sinh2(®) = 1 b. cosh(¡®) = cosh(®) and sinh(¡®) = ¡ sinh(®) c. cosh(® + ¯) = cosh(®) cosh(¯) + sinh(®) sinh(¯) sinh(® + ¯) = sinh(®) cosh(¯) + cosh(®) sinh(¯) d. cosh(®) > 0 for all ® e. (cosh(x))0 = sinh(x) and (sinh(x))0 = cosh(x). The usual trigonometric functions are called circular because every point on the circle with equation x2 +y2 = 1 can be written in the form (cos(®); sin(®)) for some ® in R. These functions are called hyperbolic trigonometric functions because of the properties (a) and (d): every point on the right-hand branch of the hyperbola x2 ¡y2 = 1 can be written in the form (cosh(®); sinh(®)) for some ® in R; every point on the left-hand branch can be written in the form (¡ cosh(®); sinh(®)). To see why these are called hyperbolic trigonometric functions, do part (f) below. See Taylor and Wheeler, page 59, for more on the comparison of the circular and hyperbolic trigonometric functions. f. Use the Taylor series for ex from Problem Set 3 to write the Taylor series for sinh(x) and for cosh(x). Compare these to the series for sin(x) and cos(x). 6. Suppose T (1; 0) = (a; b) and T (0; 1) = (c; d) for T in L(2; R). Assume a > 0. a. Explain why T (1; 0) = (cosh(®); sinh(®)) for some ® in R. b. Use (1; 0) ¢ (0; 1) = 0 to express c in terms of ® and d. c. Use (0; 1) ¢ (0; 1) = ¡1 to ¯nd two solutions for (c; d) in terms of ®. d. Write the matrix of T corresponding to each solution in (c). Find the determinant of each. Following the language in the Euclidean case, call the transformation of determinant +1 a rotation and the one of determinant ¡1 a reflection. 7. Suppose T and S are rotations as in (6), with the matrix of T expressed in terms of ® and that of S expressed in terms of ¯. Find the matrices for T ± S and S ± T . Does the order of composition matter? 8. Same as (7) for T and S reflections. 3 9. Repeat the analysis of (6) under the assumption that T (1; 0) = (a; b) with a < 0. You should get two new matrices for T , again one of determinant +1 and one of determinant¡1. Call these a rotation of the second kind and a reflection of the second kind respectively, to distinguish them from the two cases of (6), which we will call of the ¯rst kind. 10. Show that the composition of two rotations of the second kind is a rotation of the ¯rst kind. Does the order of composition matter? 11. a. Let L++ be the subset of L consisting of rotations of the ¯rst kind. Show that L++ is a subgroup of L. (Some people call L++ the Lorentz group, rather than L.) b. A group is said to be abelian if its group operation is commutative. (The term is in honor of Niels Abel, another famous Norwegian mathematician.) For example, SO(2; R) is abelian, although O(2; R) is not. Is L++ abelian? c. Do the rotations of the second kind form a subgroup of L? Why or why not? 12. This problem relates the form of our matrices in L++ to the form of those given in Taylor and Wheeler on page 42. a. Show that 1 cosh(®) = q 1 ¡ tanh2(®) b. Use (a) and sinh(®) = cosh(®) tanh(®) to rewrite the matrix of a rotation of the ¯rst kind using ¯ = tanh(®). 13. a. Show that 2 3 cosh ® sinh ® 0 0 6 sinh ® cosh ® 0 0 7 M = 6 7 4 0 0 1 0 5 0 0 0 1 is in L(4; R). b. Make up some more matrices in L(4; R), including matrices with determinant ¡1.