By Khan R.A.

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6. There exist e > 0 and an analytic matrix-valued function R(X,Y ) on Be (0) ⇥ Be (0) such that ⇣ ⌘ 1 exp X expY = exp X +Y + [X,Y ] + R(X,Y ) 2 24 1 Lie Groups and Algebraic Groups when X,Y 2 Be (0). Furthermore, kR(X,Y )k C(kXk + kY k)3 for some constant C and all X,Y 2 Be (0). 6 we now obtain the fundamental identities relating the Lie algebra structure of gl(n, R) to the group structure of GL(n, R). 7. 14) ⇣ 1 ⌘ X exp k 1 ⌘⌘k2 Y . 15) k Proof. 6 implies that exp ⇣1 ⌘ ⇣1 ⌘ ⇣1 ⌘ X exp Y = exp (X +Y ) + O(1/k2 ) , k k k where O(r) denotes a matrix function of r whose norm is bounded by Cr for some constant C (depending only on kXk + kY k) and all small r.

Identify Hn with C2n as a vector space over C by the map a + jb 7! ab , where a, b 2 Cn . Let T = A + jB 2 Mn (H) with A, B 2 Mn (C). (a) Show left multiplication by T on Hn corresponds to multiplication by the ⇥ Athat ¯⇤ B matrix B A¯ 2 M2n (C) on C2n . (b) Show that multiplication by i on Mn (H) becomes the transformation A B¯ iA iB¯ 7! B A¯ iB iA¯ 11. 6) as an H-valued function on C2n ⇥ C2n . 2 The Classical Lie Algebras 13 (a) Show that B(x, y) = B0 (x, y) + jB1 (x, y), where B0 is a C-Hermitian form on C2n of signature (2p, 2q) and B1 is a nondegenerate skew-symmetric C-bilinear form on C2n .

Then H is a closed subgroup of GL(n, R), and Lie(H) = {X 2 Lie(G) : exp(tX) 2 H for all t 2 R} . Proof. It is obvious that H is a closed subgroup of GL(n, R). If X 2 Lie(H) then exp(tX) 2 H ⇢ G for all t 2 R. Thus X 2 Lie(G). t u We consider Lie(Sp(n, C)). Since Sp(n, C) ⇢ GL(2n, C) ⇢ GL(2n, R), we can look upon Lie(Sp(n, C)) as the set of X 2 M2n (C) such that exptX 2 Sp(n, C) for all t 2 R. This condition can be expressed as exp(tX t )J exp(tX) = J for all t 2 R . 19) Differentiating this equation at t = 0, we find that X t J + JX = 0 for all X 2 Lie(Sp(n, C)).