2 Some Property of Exponential Map of Matrix

Transcription

1 Soluion Se for Exercise Session No8 Course: Mahemaical Aspecs of Symmeries in Physics, ICFP Maser Program for M 22nd, January 205, a Room 235A Lecure by Amir-Kian Kashani-Poor kashani@lpensfr Exercise Session by Tasuo Azeyanagi asuoazeyanagi@physensfr Inegral Curve Recall ha he inegral curve of a vecor field X on a manifold M is defined as a curve c : a, b M saisfying dc d/ X c We denoe as c x, y Then from he definiion, by using C funcion f, we have X c f y f +x f, x y c c dc d/ f df c dx f + dy f, x y c c and hus by comparing hese we obain dx y, dy x By solving his, we obain since d 2 x/ 2 x x A cos + B sin, y A sin B cos, where A and B are consans The iniial condiion, x0, y0 x 0, y 0 deermines A and B as A x 0 and B y 0 Therefore he inegral curve is x x 0 cos y 0 sin, y x 0 sin + y 0 cos 2 Some Propery of Exponenial Map of Marix By aking he derivaive direcly, we obain d ea d + A + 2! 2 A 2 + 3! 3 A 3 + A + A 2 + 2! 2 A 3 + A + A + 2! 2 A 2 + Ae A

2 + A + 2! 2 A 2 + A 2 Since [A, B] 0, we obain e A A e A e B m0 l0 k0 l0 l0 m! Am l! l n0 n! Bn k! Ak l k! Bl k l k0 A + Bl l! expa + B l! k!l k! Ak B l k In he middle we have defined l m + n and k m and used [A, B] 0 3 We firs noice ha d n n e A Be A dn n [A, e A Be A ] dn 2 n 2 [A, [A, e A Be A ]] [A, [A, [A, e A Be A ]] ] Here in he final expression here are n [A, ] s Then by Talyor expanding e A Be A wih respec o around 0, we obain e A Be A B + [A, B] + 2! 2 [A, [A, B]] + 3! 3 [A, [A, [A, B]]] + Now le us consider he case wih [A, B] B In his case we have [A, [A, B]] [A, B] B More generally, we obain Therefore we finally obain [A, [A, [A, B]] ] B e A Be A B + B + 2! 2 B + 3! 3 B + e B 4 When [A, B] C and [A, C] B are saisfied, we have [A, [A, B]] [A, C] B and [A, [A, [A, B]]] [A, [A, C]] [A, B] C Thus when here are even number 2

3 of he commuaors we have [A, [A, [A, B]] ] B, while for odd number of hem we have [A, [A, [A, B]] ] C Therefore, we finally obain e A Be A B + C + 2! 2 B + 3! 3 C + [ + ] [ 2! 2 + B + + ] 3! 3 + C cosh B + sinh C 5 When A is diagonalizable, we can wrie A as A MDM where M is an n n square marix and D diagd, d 2, is a diagonal marix Then we have deexpa demm expa dem expam Since M expam M k! Ak k0 M k! M AM k k0 d k 0 0 d k 2 k! k0 expd 0 0 expd 2 In he second equaliy, we have insered MM n beween A and A here n is hen n uni marix Thus we have deexpa i expd i exp i d i On he oher hand, we have expra exprmm A exprm AM exp d i Therefore we have proved he desired relaion 6 In general, one can ake an appropriae n n marix M o wrie A as A MJM, i where J is he Jordan canonical form J J 2 J, wih J i Jk λ i λ i λi, 3

4 where λ i is a number and J i is a n i n i marix i, 2, k We noe ha n k i n i We noe ha J i is wrien as J i λ i ni + N i where ni is he n i n i uni marix and 0 N i 0 0 This N i saisfies N i n i 0 and obviously [ ni, N i ] 0 We also noe ha rn i 0 Now we can compue deexpa as deexpa deexpm AM deexpλ i ni + N i i deexpλ i ni expn i i deexpλ i ni deexpn i i expn i λ i deexpn i i Now we evaluae deexpn i Since expn i m! N i m m0 n i m! N i m m Because of his upper riangle form, we can easily obain deexpn i Therefore, we have deexpa k i expn iλ i On he oher hand, we can compue expra as expra exprm AM 4

5 k exp rλ i ni + N i j k exp rλ i ni i exp rλ i ni i exp n i λ i i Therefore, we have proved he desired relaion 3 Lie Group and Lie Algebra We firs denoe he lef-invarian vecor field corresponding o X as X, and X a g GLn, R is denoed as X g As we have seen in Problem Se No7, we have X g gx where gx is defined as i,j,k c f ika kj x ij Now we derive he inegral curve c : c a, b GLn, R by definiion saisfying for a C funcion f on GLn, R d dc f X c f f c ik A kj x ij i,j,k c We also noe ha d dc f df c i,j dc ij f c x ij Thus we obain dc ca The soluion of his equaion saisfying c I n is 2 c expa n0 n n! An We noice ha c T c I n from which we obain by aking he derivaive wih respec o and denoing d/c evaluaed a as c c T c + cc 0 5

6 By by using c expa and c T expa T and evaluaing his a 0, we obain A T + A 0 Thus A is n n real marix saisfying A ij 0 for i j and A ij A ji for i j NOTE: I hink I happened o skip he following explanaion on de c condiion in he exercise session Sorry We also noe ha de c Since deexpm exprm for a general square marix M, we have from de c expra Since ra i A ii 0, his equaliy is saisfied auomaically for X saisfying A T + A 0 2 From he previous problem, we can see ha A is an n n real marix saisfying A ij 0 for i j and A ij A ji for i j Thus, here are nn /2 independen componens in A We hus conclude ha dim son, R dim T In SOn, R nn /2 3 From he above problem, i is obvious ha one can wrie A as given in he problem One can also evaluae he commuaors sraighforwardly 3 In a similar way, we use he resul from We noice ha c T Jc J where c expa Then by aking he derivaive wih respec o and seing 0, we obain A T J + JA 0 Now we denoe A as p q r s, where p, q, r, s are real n n marices Then he above relaion for A is equivalen o p T q T 0 In 0 In p q r T s T + 0 I n 0 I n 0 r s r T p T r s s T q T p q We firs noe ha q saisfies q T q and hus q has nn /2+n nn+/2 independen componens The r also saisfies r T r and hus has nn /2 + n nn + /2 independen componens On he oher hand, p, s saisfy p T s Thus s is deermined compleed once p is deermined Thus in p and s, here are n 2 independen componens in oal To summarize, dim sp2n, R dimt I2n Sp2n, R nn + /2 2 + n 2 2n 2 + n 6