Bourbaki

Exercise 18, § 9

Lie Groups and Lie Algebras, III, 401

Let G be a connected real Lie group, Z its centre, Z_0 the identity component of Z, g the Lie algebra of G and z\mathfrak{z} the centre of g. Then Z and Z_0 are Lie quasi-subgroups of G with Lie algebra z\mathfrak{z} (Exercise 3). An acceptable norm on g is a norm defining the topology of g and making g into a normed Lie algebra.

(a) For all r>0r > 0 and z∈zz \in \mathfrak{z}, there exists an acceptable norm on g whose value at z is <r. (Let q be an acceptable norm on g. Show that, for suitably chosen λ>0\lambda > 0, the function x↦∥x∥=λq(x)+inf⁡y∈zq(x+y)x \mapsto \|x\| = \lambda q(x) + \inf_{y \in \mathfrak{z}} q(x + y) has the required properties.)

(b) Show that the following conditions are equivalent: (i) Z0Z_0 is simply connected; (ii) for every acceptable norm on g, the restriction of exp⁡G\exp_G to the open ball of centre 0 and radius π\pi is injective; (iii) there exists r>0r > 0 such that, for every acceptable norm on g, the restriction of exp⁡G\exp_G to the open ball of centre 0 and radius r is injective.

(To prove (iii) ⇒\Rightarrow (i), use (a). To prove (i) ⇒\Rightarrow (ii), suppose that x, y are in g, ∥x∥<π\|x\| < \pi, ∥y∥<π\|y\| < \pi, x≠yx \neq y, exp⁡Gx=exp⁡Gy\exp_G x = \exp_G y. Then exp⁡ad⁡x=exp⁡ad⁡y\exp \operatorname{ad} x = \exp \operatorname{ad} y and hence ad⁡x=ad⁡y\operatorname{ad} x = \operatorname{ad} y by Proposition 17 of § 6 applied to a complexification of g. Hence there exists a non-zero z in z\mathfrak{z} such that exp⁡Gz=e\exp_G z = e; it follows that (i) is false.)

(c) By considering the group G′G' of Exercise 13 (b), show that the conclusion of (b) is no longer true if π\pi is replaced by a number >π> \pi. (In the notation of Exercise 13, use the norm ae1+be2+ce3↦(a2+b2+c2)12ae_1 + be_2 + ce_3 \mapsto (a^2 + b^2 + c^2)^{\frac{1}{2}} on L(G′)L(G').)

Solution

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Transcribed from the printed volume. Not translated.