5. Matrix exponentials and Von Neumann s theorem The matrix exponential. For an n n matrix X we define

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1 5. Matrix expoetials ad Vo Neuma s theorem 5.1. The matrix expoetial. For a matrix X we defie e X = exp X = I + X + X2 2! +... = 0 X!. We assume that the etries are complex so that exp is well defied o A, the algebra of matrices. We deote by a orm o A with the property that XY X Y. Such orms are easy to costruct: if is a orm o C we ca take X = sup Xu. u C, u 1 Sice X X, the series for exp X is majorized i orm by the umerical series for e X. This shows that the series for expx is absolutely coverget everywhere ad uiformly o compact =bouded i orm) subsets of A. Hece exp X is a holomorphic matrix valued fuctio o A. Its properties resemble closely those of the ordiary expoetial fuctio. i) exp 0 = I ii) expx + Y ) = exp X expy if X ad Y commute iii) exp X exp X = I. I particular exp takes values i GL, C). iv) d dt exp tx = X exp tx = exp tx)x. v) If X X, the expx = lim I + X ). The proofs of these are very similar to the correspodig proofs i the scalar case except that ii) requires a little more care. I fact, it is oly whe X ad Y commute that we ca write X + Y ) = 0 r 1 ) Y r r

2 so that X + Y )! The, with X ad Y commutig, = 0 r Y r r)!. expx + Y ) = 0,0 r Y r r)! = r r1 s Y s s! = exp X exp Y. For v) we proceed as i the scalar case ad write I + X ) = 0 r ) X r r r = u r ) r 0 where 1)1 1 u r ) = )1 2 r 1 )...1 ) if r 0 if r >. We may assume that X C for some costat C for all ; the we have the estimate u r ) Cr for all r uiformly i, so that lim I + X ) = lim u r ) = lim u r) = r 0 r 0 r 0 = exp X. Besides these we have two less obvious limit formulae. The first oe is a special case of the Trotter product formula valid i vastly greater geerality, i the settig of Hilbert spaces ad expoetials of ubouded self adjoit operators. Propositio 1. We have the followig. i) expx + Y ) = lim exp X ) exp Y )) ii) exp[x, Y ] = lim exp X ) exp Y ) exp X ) exp Y )) 2 2

3 Proof. For i) we use exp X ) = I + X + O 1 2 ) to fid that ) )) X Y exp exp = I + X + Y )) 1 + O 2 ad the limit of the right side is expx + Y ). For ii) we eed to expad the expoetials to the third order. We have exp ) ) X X X2 1 = I O 3. It is the a easy calculatio to fid that ) ) X Y exp exp exp X ) exp Y ) = I + [X, Y ] ) 1 + O 2 3 from which ii) follows at oce. Remark. All the above results are true if we replace C by R Proof of Vo Neuma s theorem. Vo Neuma s theorem is the followig. Theorem Vo Neuma). Let G be a closed subgroup of GL,R). The G is a submaifold whose coected compoets all have the same dimesio. I particular G is a Lie group. Proof. We itroduce g = { X gl,r) exp tx G for all t R }. Select a matrix orm o gl,r). It is immediate from Propositio 1 that if X, Y g, the cx, X + Y, [X, Y ] are all i g for c R. Hece g is a Lie algebra. We select a liear space a gl,r) such that g a = gl,r). Let E be the map g a GL,R) defied by EX, Y ) = expx exp Y. It is immediate that E is aalytic ad its differetial is bijective at 0, 0). I fact de 0,0) X, Y ) = de 0,0) X, 0) + de 0,0) 0, Y ) = X + Y 3

4 so that de 0,0) is surjective, hece bijective. Hece E is a diffeomorphism at 0, 0). So there is a umber a > 0 such that E maps g a a a diffeomorphically oto a ope eighborhood G a of I i GL,R); here, for ay subspace m gl,r) we write m a for the ope ball of ceter 0 ad radius a i m. Thus ay elemet x G a ca be writte uiquely a x = exp A expb where A g a, Ba a ; if x 1, the A, B 0. We claim that for some b > 0 with 0 < b < a, E maps g b oto G b G. If this were ot true, we ca fid x G a, x 1 but x = exp A exp B where A g a, B a a with A, B 0 ad B 0 for all. If y = exp A )x, the y G, y 1, y = exp B where B a, B 0, B 0. The B are very small ad so we wat to blow them up to look more closely at them. Sice B 0 we ca fid a iteger r 1 such that r B 1, r + 1)B > 1. The sequece r B ) must have a coverget subsequece, ad so, replacig it by this subsequece we may assume that X = lim r B exists. Clearly X 1; o the other had, r B r +1)B B 1 B ad so, lettig, we have X 1 also. Hece X = 1, i particular, X 0. We claim that exp tx G for all t R. It is eough to show this for all ratioal t > 0. Writig t = c/k where c, k are itegers 1, it is eough to show that exp1/k)x) G for all itegers k 1. We use the argumet that if y m has a limit where the m are itegers 1, the G for all ad so G. Write r = ks + t where 0 t < k. The this limit must be i G. Certaily expr B ) = y r exp X = lim y r r ) ) ) exp k B t = exps B ) exp B. k Sice t /k)b B 0, it follows that ) lim exps B ) = lim ys 1 = exp k X ad so exp 1/k)X) G as we wated. 4

5 E is thus a diffeomorphism of g b a b with G b ad we have i additio that G b G correspods to g b uder E. It is immediate that G b G is a submaifold of G b. This fiishes the proof of the theorem. Remark. The result is false for complex groups; just cosider the uitary groups U) GL,C). But if we ca prove that g as defied above is closed uder multiplicatio by i = 1, the the proof will go through ad establish that G is a complex submaifold, hece a complex Lie group. We call g the Lie algebra of G ad deote it by LieG). Problems 1. Determie LieG) for the classical groups. 5