By Bibhutibhushan Datta

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**Example text**

Xj (Fl (t, x)). Xi (Fl (t, x)) satisfy the linear orn d dinary differential equation dt gi (t) = j=1 fij (t)gj (t) and have initial values in the linear subspace E(V)x , so they have values in it for all small t. Therefore T (FlX −t )E(V)FlX (t,x) ⊂ E(V)x for small t. 2) is satisfied and E(V) is integrable. 29. Examples. (1) The singular distribution spanned by W ⊂ Xloc (R2 ) is involutive, but not integrable, where W consists of all global vector fields with ∂ support in R2 \ {0} and the field ∂x 1 ; the leaf through 0 should have dimension 1 at 0 and dimension 2 elsewhere.

An integral manifold of E is called maximal, if it is not contained in any strictly larger integral manifold of E. 22. Lemma. Let E be a smooth distribution on M . Then we have: (1) If (N, i) is an integral manifold of E and X ∈ XE , then i∗ X makes sense and is an element of Xloc (N ), which is i|i−1 (UX )-related to X, where UX ⊂ M is the open domain of X. (2) If (Nj , ij ) are integral manifolds of E for j = 1, 2, then i−1 1 (i1 (N1 ) ∩ i2 (N2 )) −1 and i2 (i1 (N1 ) ∩ i2 (N2 )) are open subsets in N1 and N2 , respectively; furthermore i−1 2 ◦ i1 is a diffeomorphism between them.

Let C(X, Y ) := exp−1 (exp X. exp Y ) for X, Y near 0 in g, and let C(t) := C(tX, Y ). C(t), = k≥0 (k+1)! Draft from April 18, 2007 Peter W. 29 4. Lie Groups I z 49 k z where g(z) := e z−1 = k≥0 (k+1)! C(t). 11). 25) = Ad(exp(tX) exp Y ) = Ad(exp(tX)). ead Y = et. ead Y . If X, Y , and t are small enough we get ad C(t) = log(et. C(t) = g(log(et. C(t). For z near 1 we put f (z) := 1. f (z) = ˙ ˙ X = g(log(et. C(t) = f (et. C(t), ˙ C(t) = f (et. X, C(0) = Y Passing to the definite integral we get the desired formula 1 C(X, Y ) = C(1) = C(0) + ˙ dt C(t) 0 1 f (et.