1 A linear representation of the conformal group We wish to find a linear representation of the conformal group. On flat spacetime, in Cartesian coordinates, we can represent the action of the conformal group as the set of transformations a a b x˜ = L bx x˜a = xa + ba x˜a = lxa a 2 a a x + x c x˜ = a 2 2 1 + 2cax + c x where c d hcdL aL b = hab a so that the L b are Lorentz transformations. The second transformation is a translation, the third a dilatation and the final transformations, called special conformal transformations may be seen to be conformal by considering the inversion, xa x˜a = x2 Inversion is a discrete transformation, and since xax 1 s˜2 = a = s2 (x2)2 (x2)2 it is conformal. A special conformal transformation is an inversion followed by translation, followed by a second inversion: a xa xa x + ca xa + x2ca xa ! ! + ca ! x2 = x2 x2 b 2 1 + 2c xa + c2x2 x + cb a x2 Notice that there are 15 conformal transformations: 6 Lorentz, 4 translation, 1 dilatation and 4 special conformal transformations. We define the conformal transformations as those transformations preserving the light cone. This is equivalent to preserving angles, and also equivalent to preserving ratios of lengths. Since a light cone centered at a point ab may be described by the equation b b 0 = l x − a (xb − ab) 2 b 2 = lx − 2lx ab + la conformal transformations will be those transformations which map l;ab ! s;bb Label these parameters as Bb = lab B4 = l 1 B5 = la2 2 Then BA = Ba;B4;B5 satisfies a 4 5 B Ba − 2B B = 0 a a b where B Ba = habB B . Define a 6-dim metric, 0 1 hab hAB = @ 0 −1 A −1 0 1 so that A B hABB B = 0 Any transformation preserving this null quadratic form induces a change in l;ab that maps to another 4-dim light cone. Although we want hAB in this particular form, we notice that if we were to diagonalize it the lower right corner ±1 0 becomes , so h as four positive and two negative eigenvalues. The linear transformations preserving 0 ∓1 AB b hAB therefore comprise the group O(4;2). Since we may recover coordinates, a , near the origin and coordinatesm ab a2 , near infinity, on Minkowski space by forming the ratios Bb ab = B4 ab Bb = a2 2B5 we see that any scaling, BA ! aBA has no effect on spacetime. We therefore restrict to SO(4;2). Finally, note that 2B5 a2 = B4 2 The special orthogonal group, SO(4,2) Next, we find the generators of SO(4,2). Let g = 1 + e be infinitesimally near the identity and require C D hAB = hCDg Ag B C C D D = hCD dA + e A dB + e B C D 2 = hAB + hCBe A + hADe B + O e and therefore, using the metric to lower the upper index on e, eBA = −eAB as we have found before for pseudo-orthogonal groups. We divide the antisymmetric matrices into four types: e [M ] = ab ab 0 0 0 −1 1 B 0 1 C B C B 0 1 C [Pa] = B C B 0 1 C B C @ 1 −1 −1 −1 0 A 0 0 0 −1 1 B 0 1 C B C B 0 1 C [Ka] = B C B 0 1 C B C @ 0 A 1 −1 −1 −1 0 0 0 1 B 0 C B C B 0 C [D] = B C B 0 C B C @ 0 −1 A 1 0 2 a 1 a b b bc bc The first of these generate Lorentz transformation, L b = exp 2 w be a , where e a = h eca = −h eac, and we will A say no more about them. We raise the first index on the remaining generators to recover e B, 0 0 −1 1 0 1B 0 1 C hab B C A B 0 1 C [Pa] = @ 0 −1 AB C B B 0 1 C −1 0 B C @ 1 −1 −1 −1 0 A 0 0 0 1 1 B 0 1 C B C B 0 1 C = B C B 0 1 C B C @ 0 0 A −1 1 1 1 0 0 0 0 1 1 B 0 1 C B C A B 0 1 C [Ka] = B C B B 0 1 C B C @ −1 1 1 1 0 0 A 0 0 0 0 1 B 0 C B C A B 0 C [D] = B C B B 0 C B C @ −1 A 1 We exponentiate each in turn. For Pa, a a g(b ) = exp(b Pa) 0 0 b0 1 B 0 b1 C B C B 0 b2 C = expB C B 0 b3 C B C @ 0 0 A −b0 b1 b2 b3 0 0 0 0 b0 1 0 0 1 B 0 b1 C B 0 C B C B C B 0 b2 C 1 B 0 C = 1 + B C + B C B 0 b3 C 2 B 0 C B C B C @ 0 0 A @ 0 0 A 0 1 2 3 a −b b b b 0 0 b ba 0 with the remaining powers vanishing. Therefore, the four transformations give 0 1 b0 10 B0 1 B 1 b1 CB B1 C B 2 CB 2 C A B 1 b CB B C B˜ = B CB C B 1 b3 CB B3 C B CB C @ 1 0 A@ B4 A 0 1 2 3 1 a 5 −b b b b 2 b ba 1 B 3 0 B0 + b0B4 1 B B1 + b1B4 C B C B B2 + b2B4 C = B C B B3 + b3B4 C B C @ B4 A 5 a 1 4 a B + B ba + 2 B b ba On spacetime, this transformation of BA gives B˜a x˜a = B˜4 Ba + baB4 = B4 = xa + ba and is therefore a translation. On a coordinate centered on infinity, this same translation becomes x˜a y˜a = x˜2 B˜a = 2B˜5 Ba + baB4 = 5 a 1 4 a 2 B + B ba + 2 B b ba a 4 B + ba B 2B5 2B5 = a 4 a 1 + B ba + B b ba B5 2B5 4 ya + B ba = 2B5 1 + 2yab + B4 bab a 2B5 a 5 x2 = 2B = 1 and since B4 y2 , 4 ya + B ba y˜a = 2B5 1 + 2yab + B4 bab a 2B5 a ya + y2ba = a 2 2 1 + 2y ba + y b This means that in a neighborhood of the point at infinity, a translation takes a somewhat complicated form – the same form that a special conformal transformation takes in a neighborhood of the origin. Now consider a special conformal transformation. We have a a g(c ) = exp(c Ka) 0 0 ca 1 = exp@ ca 0 0 A 0 0 0 0 ca 1 0 0 0 0 1 1 a = 1 + @ ca 0 0 A + @ 0 0 cac A 2 0 0 0 0 0 0 1 ca 1 1 a = @ ca 1 2 cac A 0 1 4 Therefore, 0 1 ca 10 Ba 1 ˜A 1 a 4 B = @ ca 1 2 cac A@ B A 0 1 B5 0 Ba + B5ca 1 4 a 1 2 5 = @ B + B ca + 2 c B A B5 The coordinates near the origin and near infinity change according to: B˜a x˜a = B˜4 Ba + B5ca = 4 a 1 2 5 B + B ca + 2 c B a 5 B + B ca = B4 B4 + Ba c + 1 c2 B5 1 B4 a 2 B4 xa + 1 x2ca = 2 a 1 2 2 1 + x ca + 4 c x Now redefine the parameter, replacing ca ! 2ca and we have a 2 a a x + x c x˜ = a 2 2 1 + 2x ca + c x For the inverse coordinate, x˜a y˜a = x˜2 B˜a = 2B˜5 Ba + B5ca = 2B5 1 = ya + ca 2 and making the same change of parameter, we have a simple translation at infinity, y˜a = ya + ca Finally, we find the dilatations, g(l) = exp(lD) 0 0 0 0 1 = exp@ 0 −l 0 A 0 0 l 0 1 0 0 1 = @ 0 e−l 0 A 0 0 el The 6-vector changes to 0 1 0 0 10 Ba 1 A B˜ = @ 0 e−l 0 A@ B4 A 0 0 el B5 0 Ba 1 = @ e−l B4 A el B5 5 so the spacetime coordinates transform as B˜a x˜a = B˜4 Ba = el B4 = el xa and B˜a y˜a = 2B˜5 Ba = e−l 2B5 = e−l ya Notice that the coordinates at the origin and at infinity transform oppositely under dilatations. This will have important consequences. 3 Maurer-Cartan structure equations for SO(4,2) We already know that the Lie algebra for pseudo-orthogonal groups takes the form 1 [M ;M ] = − (h M − h M − h M + h M ) AB CD 2 BC AD AC BD BD AC AD BC where I’ve written the labels down so that the antisymmetrizations, [A;B] and [C;D] are easier to see. Defining a dual basis of 1-forms by A C AC M B;w D = 2DDB the Maurer-Cartan structure equations are A C A dw B = w B ^ w C where the the labels have been raised with hAB.