By A. Bak

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Extra resources for Algebraic K-theory, number theory, geometry, and analysis: proceedings of the international conference held at Bielefeld, Federal Republic of Germany, July 26-30, 1982

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Since for all t ,s E IR we have c1(t+s) = d(expoht+1)(E) = Ad(exp(s()-')ocl(t) = exp(-ad + cl(s) s t)ocl(t) + cl(s). Differentiating with respect to s at s=O yields + i1(0) = -ad oc (t) + id. ( 1 cl(t) = -adEocl(t) 9 28 Chapter 1 On the other hand the tangent vector of c2 at t is given by h2(t) = C $ t p=o (-ad )p < = id - ad oc (t). ( 2 Thus both curves satisfy the same differential equation and initial condition and therefore are identical. In particular c l ( l ) = c2(l) which yields the assertion.

This means that GL+(n) and in turn GL-(n): = (1 E GL(n)/det 1 < 0) are connected. Example 4: SO(n) Again consider IRn and a non degenerate symmetric bilinear form b : Rn x R n 4 R . of type (p,n-p). Write SL(n) instead of SL(Rn). The intersection O(p,n-p) fl SL(n) is a closed subgroup of O(p,n-p) and thus a Lie group. It is called SO(p,n-p). 3). Let us determine its connected component.

Then a o i, where denotes the adjoint, is a self-adjoint positive N isomorphism. Then i. E P(n) of 46 Chapter 1 a=gop with g = a 0 p -1 E O(n). The decomposition of a into g o p is unique as easily shown. Since the derivative of II is invertible everywhere, ll is a diffeomorphism. The next goal is to show that GL(n) has two components only. Let GL+(n) denote the set of all automorphisms of Rn with positive determinant and SO(n): = O(n) n GL+(n). Consider 1 E GL+(n) and its polar decomposition into 1 = g o f, where g E O(n) and f E P(n).

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