LECTURE 8-9: THE BAKER-CAMPBELL-HAUSDORFF FORMULA

LECTURE 8-9: THE BAKER-CAMPBELL-HAUSDORFF FORMULA As we have seen, 1. Taylor s expanson on Le group, Y ] a(y ). So f G s an abelan group, then c(g) : G G s the entty ap for all g G. As a consequence, a() 0. It follows that the Le algebra of an abelan Le group s also abelan,.e., Y ] 0 for all, Y g. Conversely, one can prove (usng proposton 3.1 n lecture 6) that f G s connecte an g s abelan, then G s also abelan. In other wors, the Le bracket operaton on g easures the non-coutatvty of the ultplcaton operaton on G. In what follows we woul lke to characterze ths quanttatvely. In other wors, we woul lke to fn out the fferent between exp() exp(y ) an exp( + Y ) for a general Le group. Let G be a Le group an g a left nvarant vector fel on G. Then (f)(a) a f L a e f e (f L a ) t f(a exp(t)). for any f C (G) an any a G. More generally, for any t R, (f)(a exp(t)) s f(a exp(t) exp(s)) s0 s f(a exp((t+s))) f(a exp(t)). s0 t Usng ths an nucton, one can see that for any k 0, In partcular, ( k f)(a exp(t)) k (f(a exp(t))). tk ( k f)(a) k t k f(a exp(t)). The forulae above can be generalze to ult-varable case. In fact, f 1,, k g, then ( 1 2 f)(a) ( 2 f)(a exp(t 1 1 )) t 1 t 1 t 2 f(a exp(t 1 1 ) exp(t 2 2 )), t2 0 an n general t1 0 ( 1 k f)(a) t1 0 k t 1 t k f(a exp(t 1 1 ) exp(t k k )). t1 t k 0 As a consequence, we get the followng Taylor s expanson forula 1

2 LECTURE 8-9: THE BAKER-CAMPBELL-HAUSDORFF FORMULA Proposton 1.1. If f s sooth on G, then for sall t, f(exp(t 1 ) exp(t n )) f(e)+t { f(e)+ t2 2 f(e) + 2 } j f(e) +O(t 3 ). 2 We reark that the prevous forulae hol for vector-value functons as well. Our an result n ths secton s Theore 1.2. Let n 1 an 1,, n g. Then for t suffcently sall, exp(t 1 ) exp(t n ) exp(t 1 n + t2 2 1 n, j ] + O(t 3 )). Proof. We apply proposton 1.1 to the nverse of the exponental ap near e,.e. the ap f efne by f(exp(t)) t for t sall enough. Then obvously, f(e) 0. For any g, (f)(e) t f(exp(t)) t (t) an for any n > 1, ( n f)(e) n t n f(exp(t)) n t n (t) 0. Notce 2 + 2 j ( 1 + + n ) 2 +, j ], t follows that f(exp(t 1 ) exp(t n )) t + t2, j ] + O(t 3 ). 2 On the other han, by the efnton of f, exp(t 1 ) exp(t n ) exp(f(exp(t 1 ) exp(t n ))). Ths copletes the proof. In partcular, we see exp(t) exp(ty ) exp(t + ty + t2 2, Y ] + O( t 3 ) for t sall. So, Y ] onates the fference between exp() exp(y ) an exp( + Y ), an thus onates the non-coutatvty of the group ultplcaton.

LECTURE 8-9: THE BAKER-CAMPBELL-HAUSDORFF FORMULA 3 2. The Baker-Capbell-Hausorff Forula Now the queston s: What are the hgher orer ters n O(t 3 ) above? For splcty we wll enote by log the nverse of exp near 0 g. Let µ(, Y ) log(exp() exp(y )) for, Y close to 0 g. We have seen above that µ(, Y ) + Y + 1 2, Y ] + O( 3, Y 3 ) for, Y sall. A rearkable fact about the reaner ters s that they nvolves only Le brackets! In other wors, we have Theore 2.1 (The Baker-Capbell-Hausorff forula (exstence)). For an Y sall, we have µ(, Y ) + Y + 2 P (, Y ), where P (A, B) s a Le polynoal of orer,.e. P (, Y ) s a cobnaton of neste coutators n, Y that nvolves 1 Le brackets. Although n applcaton, the above exstence result s suffcent, we wll prove the followng explct forula: Theore 2.2 (Dynkn s forula). For an Y sall, ( 1) k ( 1) (l + ) (ay ) l 1 µ(, Y ) +Y + k+1 l 1 + +l k +1 l 1! k1 (a) 1 1! (ay )l k l k! where the secon suaton s over l 1,, l k, 1,, k 0, l + > 0. As a consequence, we can wrte own P (, Y ) for sall. Of course, P 2 (, Y ) 1, Y ]. 2 The next ter P 3 (, Y ) coes fro the followng ters n Dynkns s forula: (1) k 1, l 1 1, 1 1 1 1Y,, Y ]]; 2 2 (2) k 1, l 1 0, 1 2 1 1 1,, Y ]]; 2 1 2 (3) k 2, l 1 1, 1 0, l 2 0, 2 1 1 1Y,, Y ]]; 3 2 (4) k 2, l 1 0, 1 1, l 2 0, 2 1 1 1,, Y ]]; 3 1 It follows P 3 (, Y ) 1 (,, Y ]] Y,, Y ]]). 12 Slarly one can calculate the next ter an get P 4 (, Y ) 1, Y, Y, ]]]. 24 (a) k (Y ), k!

4 LECTURE 8-9: THE BAKER-CAMPBELL-HAUSDORFF FORMULA To prove the Dynkn s forula, we wll nee the followng forula that coputes the fferental of the exponental ap at an arbtrary pont. Lea 2.3. For each g, ( exp) (L exp ) e φ(a), where φ s the functon φ(z) 1 e z z 0 ( 1) ( + 1)! z. Proof of Dynkn s forula. Wrte Applyng lea 2.3, we get Z(t) log(exp() exp(ty )). t (exp Z(t)) L exp t (exp ty ) L exp L exp ty φ(a(ty ))(Y ) L exp Z(t) (Y ), where we use the fact φ(a(ty ))(Y ) Y. On the other han, by usng lea 2.3 rectly, t (exp Z(t)) L exp Z(t)φ(aZ(t)) Z t. It follows Z t az(t) I exp( az(t)) (Y ) k 0 1 k + 1 (I exp( az(t)))k (Y ). Notce that by the naturalty of exp an by the efnton of a an A, Thus Z t k 0 k 0 exp( az(t)) A exp( Z(t)) A(exp( ty ) exp( )) A(exp( ty )) A(exp( )) exp( ta(y )) exp( a()). (I exp( tay ) exp( a)) k (Y ) k + 1 ( 1) k k + 1 l 1,,l k, 1,, k 0,l + >0 t l ( 1) l + (ay )l 1 (a) 1 l 1! 1! (ay )l k l k! (a) k Y. k! where n the last step we use the fact that a En(g) s an eleent n a lnear Le group, an thus the exponental ap s exactly the atrx exponental. Now the Dynkn s forula follows fro terwse ntegraton over t fro 0 to 1.

LECTURE 8-9: THE BAKER-CAMPBELL-HAUSDORFF FORMULA 5 3. The Dervatve of the Exponental Map Fnally we prove lea 2.3. We frst show Lea 3.1. Let γ 1 (t), γ 2 (t) be sooth curves on G, an let γ(t) γ 1 (t)γ 2 (t), then γ(t) L γ1 (t)( γ 2 (t)) + R γ2 (t)( γ 1 (t)). Proof. Notce the fact γ(t) µ(γ 1 (t), γ 2 (t)), where µ s the ultplcaton operaton on G. So the forula above follows fro the followng forula we have proven, µ a,b ( a, Y b ) (L a ) b (Y a ) + (R b ) a ( a ). More generally, by usng nucton one can easly see that f γ 1 (t),, γ (t) are sooth curves on G, an let γ(t) γ 1 (t) γ (t), then γ(t) L γ1 (t) L γk 1 R γk+1 (t) R γ(t)( γ k (t)). k1 Now we are reay to prove lea 2.3. For splcty we wll enote ν(, Y ) : t exp( + ty ) ( exp) (Y ). Obvously ν(, Y ) s lnear n Y for each fxe, an lea 2.3 follows fro Lea 3.2. For any, Y g, t exp( + ty ) (L exp ) e φ(a)(y ). Proof. We notce that for any postve nteger, ν(, Y ) t exp( + t Y ] ) (L exp 1 ) k (R exp ) k (L exp 1 (L exp ) k 1 (R exp ) k ν(, Y ) 1 ) 1 (L exp ) k (R exp ) k ν(, Y ). Recall that the fferental of the conjugaton ap c(a) L a R a 1 c(exp( ] k )) A(exp ] k ) s A, so we get exp( a ] k ).

6 LECTURE 8-9: THE BAKER-CAMPBELL-HAUSDORFF FORMULA So we get, for every postve nteger, ν(, Y ) (L exp ) 1 1 Now the result follows snce as, (L exp ) 1 L ( 1) exp 1 exp( a ] k ) ν(, Y ). L exp, ν(, Y ) ν(0, Y ) ( exp) 0(Y ) Y, an, snce a En(g) s a atrx, 1 1 exp( a ] k ) 1 1 exp ( k ) a 1 1 1 ( k ) n n! a n0 1 1 ( ) ] n k ( 1) n (a) n n! n0 1 ] ( 1) x n n x (a) n n0 0 n! ( 1) n (n + 1)! (a)n. n0 Fnally we gve several applcatons. Gven the ervatve of the exponental ap exp at an arbtrary pont, we are reay to answer the followng queston: at whch ponts the ap exp s sngular,.e. ( exp) s not nvertble? Snce ( exp) (L exp ) e φ(a) an L exp ) e s always nvertble, we see that ( exp) s not nvertble f an only f the atrx φ(a) En(g) s not nvertble,.e. 0 s not an egenvalue of φ(a). Snce all egenvalues of φ(a) are of the for φ(λ) 1 e λ, where λ s an egenvalue λ of a En(g), we conclue Corollary 3.3. The sngular ponts of the exponental ap exp : g G are precsely those g such that a En(g) has an egenvalue of the for 2πk, wth k Z \ {0}. As an exaple, we see that f G s an abelan Le group, then exp s non-sngular everywhere. More generally, f g s nlpotent, then exp : g G s non-sngular everywhere.