# Model Theory and Algebra by D.H. Saracino, V.B. Weispfennig

By D.H. Saracino, V.B. Weispfennig

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In particular, for any x ∈ gl(n, K), exp(x) exp(−x) = 1, so exp x ∈ GL(n, K). (4) For ﬁxed x ∈ gl(n, K), consider the map K → GL(n, K) : t → exp(tx). Then exp((t + s)x) = exp(tx) exp(sx). In other words, this map is a morphism of Lie groups. (5) The exponential map agrees with change of basis and transposition: exp(AxA−1 ) = A exp(x)A−1 , exp(xt ) = (exp(x))t . A full proof of this theorem will not be given here; instead, we just give a sketch. The ﬁrst two statements are just equalities of formal power series in one variable; thus, it sufﬁces to check that they hold for x ∈ R.

11. Exercises −1 1 0 −1 is not in the image of the exponential map. ). 1. Consider the group SL(2, R). 2.

Thus, the derivative ∂ξ f = − dtd tξ f . For example, if ξ = ∂x is the constant vector ﬁeld on R, then the ﬂow on points is given by t : x → x + t, and on functions it is given by ( t f )(x) = f (x − t), so ∂x f = − dtd t f . 25. Let G be a ﬁnite-dimensional Lie group acting on a manifold M , so we have a map ρ : G → Diff (M ). Then (1) This action deﬁnes a linear map ρ∗ : g → Vect(M ). (2) The map ρ∗ is a morphism of Lie algebras: ρ∗ [x, y] = [ρ∗ (x), ρ∗ (y)], where the commutator in the right-hand side is the commutator of vector ﬁelds.