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What happens when λ = 1? 18. Show that a connected Lie group G is abelian if and only if its Lie algebra satisfies [x, y] = 0 for all x, y ∈ g. Conclude that a connected abelian Lie group of dimension n is isomorphic to Rn /Zd where Zd is some discrete subgroup of rank d ≤ n. (Hint: To show “G abelian” implies “g abelian”, look at how [, ] was defined. ) ˜ → G be 19. ) Let G be a connected Lie group and let π: G ˜ can be regarded as the space the universal covering space of G. ) Show ˜×G ˜→G ˜ for which the homotopy class of that there is a unique Lie group structure µ ˜: G ˜ is the identity and so that π is a homomorphism.

It follows that Aut(P ) is the space of sections of the bundle C(P ) = P ×C G where C: G × G → G is the conjugation action C(a, b) = aba−1 . Moreover, it easily follows that the set of vector fields on P whose flows generate 1-parameter subgroups of Aut(P ) is identifiable with the space of sections of the vector bundle Ad(P ) = P ×Ad g. Connections. Let A(P ) denote the space of connections on P . Thus, an element A ∈ A(P ) is, by definition, a g-valued 1-form A on P with the following two properties: (1) For any p ∈ P , we have ι∗p(A) = ωG where ιp : G → P is given by ιp (g) = p · g.

For now, though, I want to stress that there is a sort of analogue of actions for these “infinite dimensional Lie groups”. For example, if M is a manifold and Diff(M) is the group of (global) diffeomorphisms, then we can regard the natural (evaluation) map λ: Diff(M)×M → M given by λ(φ, m) = φ(m) as a faithful Lie group action. If M is compact, then every vector field is complete, so, at least formally, the induced map λ∗ : Tid Diff(M) → X(M) ought to be an isomorphism of vector spaces. If our analogy with the finite dimensional case is to hold up, λ∗ must reverse the Lie bracket.

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An Introduction to Lie Groups and Symplectic Geometry by Bryant R.L.


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