By Jacques Faraut
This self-contained textual content concentrates at the standpoint of study, assuming merely ordinary wisdom of linear algebra and uncomplicated differential calculus. the writer describes, intimately, many attention-grabbing examples, together with formulation that have no longer formerly seemed in e-book shape. issues lined comprise the Haar degree and invariant integration, round harmonics, Fourier research and the warmth equation, Poisson kernel, the Laplace equation and harmonic features. excellent for complicated undergraduates and graduates in geometric research, harmonic research and illustration thought, the instruments constructed can be worthwhile for experts in stochastic calculation and the statisticians. With a number of routines and labored examples, the textual content is perfect for a graduate path on research on Lie teams.
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Extra info for Analysis on Lie Groups: An Introduction
2 log F(t) = t(X + Y ) + This proves (i). To prove (ii) put G(t) = exp(t X ) exp(tY ) exp(−t X ) exp(−tY ) t2 2 (X + 2X Y + Y 2 ) + O(t 3 ) 2 t2 · I − t(X + Y ) + (X 2 + 2X Y + Y 2 ) + O(t 3 ) 2 = I + t(X + Y ) + = I + t 2 [X, Y ] + O(t 3 ) , and (ii) follows by considering log G(t). 4 For X, Y ∈ M(n, R), lim (i) (ii) k→∞ lim k→∞ exp exp X Y exp k k k = exp(X + Y ), Y X Y X exp exp − exp − k k k k k2 = exp([X, Y ]). 3 Exercises 29 Proof. From the preceding proposition exp X Y 1 exp = exp (X + Y ) + O k k k exp X Y exp k k k = exp (X + Y ) + O 1 k2 , 1 k , and this proves (i).
Since (z) = ∞ k=0 (−1)k (z − 1)k z, k+1 we have Exp(ad X ) Exp(t ad Y ) Y = ∞ k=0 (−1)k Exp(ad X ) Exp(t ad Y ) − I k+1 k Exp(ad X ) Exp(t ad Y )Y. 5 Exercises 47 Observing that Exp(t ad Y )Y = Y, we obtain Exp(ad X ) Exp(t ad Y ) Y = ∞ k=0 (−1)k · k+1 t q1 +···+qk E(k) (ad X ) p1 (ad Y )q1 . . (ad X ) pk (ad Y )qk (ad X )m Y. q1 ! . m! The convergence of the series is uniform for t in [0, 1]. The statement is obtained by termwise integration since 1 t q1 +···+qk dt = 0 1 . 5 1 1 1 log(exp X exp Y ) = X + Y + [X, Y ] + X, [X, Y ] + Y, [Y, X ] 2 12 12 + terms of degree ≥ 4.
Then put g = exp s X and take the derivatives at s = 0, dπ ([X, Y ]) = dπ (X )dπ(Y ) − dπ(Y )dπ (X ). The adjoint representation π = Ad of G on g is a special case for which the derived representation is the adjoint representation ad of g on g. If π1 and π2 are two equivalent representations, then the derived representations dπ1 and dπ2 are also equivalent. The converse holds if G is connected.
Analysis on Lie Groups: An Introduction by Jacques Faraut