By Irving Kaplansky

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**Example text**

Note that dim o(n R) = 12 (n ; 1)n. If A is invertible, we get ker df (A) = fY : Y:At + A:Y t = 0g = fY : Y:At 2 o(n R)g = o(n R):(A;1 )t . The mapping f takes values in Lsym(R n Rn ), the space of all symmetric n n-matrices, and dim ker df (A) + dim Lsym(R n Rn ) = 1 (n ; 1)n + 1 n(n +1) = n2 = dim L(R n R n ), so f : GL(n R ) ! Lsym (R n R n ) is 2 2 a submersion. 12 that O(n R) is a submanifold of GL(n R). It is also a Lie group, since the group operations are smooth as the restrictions of the ones from GL(n R).

17. One parameter subgroups. Let G be a Lie group with Lie algebra g. A one parameter subgroup of G is a Lie group homomorphism : (R +) ! e. a smooth curve in G with (s + t) = (s): (t), and hence (0) = e. Lemma. Let : R ! G be a smooth curve with (0) = e. Let X 2 g. Then the following assertions are equivalent. (1) is a one parameter subgroup with X = @t@ 0 (t). (2) (t) = FlLX (t e) for all t. (3) (t) = FlRX (t e) for all t. (4) x: (t) = FlLX (t x) , or FlLt X = (t) , for all t . (5) (t):x = FlRX (t x) , or FlRt X = (t) , for all t.

2) ad(X )Y = X Y ] for X Y 2 g. Proof. (1). LX (a) = Te ( a ):X = Te ( a ):Te( a;1 a ):X = RAd(a)X (a). X1 : : : Xn be a linear basis of g and x X 2 g. 12. 25. Corollary. 23 we have Ad expG = expGL(g) ad Ad(expG X )Y = 1 X 1 k=0 k ad X Y k! 1 X X X Y ]]] + so that also ad(X ) = @t@ 0 Ad(exp(tX )). 26. The right logarithmic derivative. Let M be a manifold and let f : M ! ;We de ne 1 f ( x ) the mapping f : TM ! g by the formula f ( x) := Tf (x) ( ):Tx f: x. Then f is a g-valued 1-form on M , f 2 1 (M g), as we will write later.