By Andrew Ranicki

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**Example text**

Hence, it has finitely many connected components. Clearly, for a matrix T ∈ O(n), T ∗ is its transpose. , det(T ) = ±1. It follows that the homomorphism det maps O(n) onto the subgroup {±1} of R∗ . Therefore, SO(n) is a normal Lie subgroup of O(n) of index 2. In particular, SO(n) is open in O(n). Moreover, it has finitely many connected components. The natural action of the group O(n) on Rn preserves the euclidean distance. Therefore, it acts differentiably on the unit sphere S n−1 = {x ∈ Rn | x21 + x22 + · · · + x2n = 1}.

A Lie algebra a is abelian if [x, y] = 0 for all x, y ∈ a. Let a be a Lie algebra. We denote by aopp the opposite Lie algebra of a. It is the same linear space with the commutator (x, y) −→ [x, y]◦ = −[x, y]. Clearly, aopp is a Lie algebra. Moreover, x −→ −x is an isomorphism of a with aopp . Evidently, we have (aopp )opp = a. If dim a = 1, a has to be abelian. If dim a = 2, we can pick a basis (v1 , v2 ) of a and see that [x, y] is proportional to [v1 , v2 ] for any x, y ∈ a. Therefore, we can assume that [x, y] is proportional to e1 for any x, y ∈ a.

7. Lemma. Let ϕ : G −→ H be a Lie group homomorphism of a simply ˜ be the connected, connected Lie group G into a connected Lie group H. Let H ˜ universal covering Lie group of H and p : H −→ H the covering projection. Then ˜ such that p ◦ ϕ˜ = ϕ. there exists a unique Lie group homomorphism ϕ˜ : G −→ H In addition, if ϕ : G −→ H is a Lie group morphism of connected Lie groups, ˜ −→ H ˜ such that the diagram there exists a unique Lie group homomorphism ϕ˜ : G ϕ ˜ ˜ −−− G −→ pG ϕ ˜ H pH , G −−−−→ H where the vertical arrows are covering projections, is commutative.