By Bryant R.L.

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**Extra resources for An Introduction to Lie Groups and Symplectic Geometry**

**Example text**

Then there exists a smooth mapping ψ: B d → G (where B d is an open ball about 0 in Rd ) so that ψ(0) = e and so that g is the direct sum of the subspaces gm and V = ψ (0)(Rd ). 4 41 Tm M is injective. Thus, by restricting to a smaller ball in Rd if necessary, I may assume henceforth that φ ◦ ψ: B d → M is a smooth embedding. Consider the mapping Ψ: B d × Gm → G deﬁned by Ψ(x, g) = ψ(x)g. I claim that Ψ is a diﬀeomorphism onto its image (which is an open set), say U = Ψ(B d × Gm ) ⊂ G. ) To see this, ﬁrst I show that Ψ is one-to-one: If Ψ(x1 , g1 ) = Ψ(x2 , g2 ), then (φ ◦ ψ)(x1 ) = ψ(x1 ) · m = (ψ(x1 )g1 ) · m = (ψ(x2 )g2 ) · m = ψ(x2 ) · m = (φ ◦ ψ)(x2 ), so the injectivity of φ ◦ ψ implies x1 = x2 .

For any Lie algebra g, there is a canonical bilinear pairing κ: g × g → R, called the Killing form, deﬁned by the rule: κ(x, y) = tr ad(x)ad(y) . 4 36 (i) Show that κ is symmetric and, if g is the Lie algebra of a Lie group G, then κ is Ad-invariant: κ Ad(g)x, Ad(g)y = κ(x, y) = κ(y, x). Show also that κ [z, x], y = −κ x, [z, y] . A Lie algebra g is said to be semi-simple if κ is a non-degenerate bilinear form on g. (ii) Show that, of all the 2- and 3-dimensional Lie algebras, only so(3) and sl(2, R) are semi-simple.

Thus, the ﬁbers of the map φ are the left Gm -cosets in G. ¯ G/Gm → G · m. In particular, the map φ establishes a bijection φ: First, I specify the topology on G· m to be quotient topology induced by the surjective map φ: G → G · m. Thus, a set U in G · m is open if and only if φ−1 (U) is open in G. Since φ: G → M is continuous, the quotient topology on the image G · m is at least as ﬁne as the subspace topology G · m inherits via inclusion into M. Since the subspace topology is Hausdorﬀ, the quotient topology must be also.