Bourbaki

Exercise 8, § 3

Integration, VII, p. 93

a) Show that there exists on Sn−1S_{n-1} a measure ωn−1\omega_{n-1} invariant under O(n,R)O(n, \mathbf{R}) and, up to a constant factor, only one (regard Sn−1S_{n-1} as a homogeneous space for O(n,R)O(n, \mathbf{R})).

b) The mapping (t,z)↦tz(t, z) \mapsto tz permits identifying the topological space Rn−{0}\mathbf{R}^n - \{0\} with the topological space R+∗×Sn−1\mathbf{R}_+^* \times S_{n-1}. Show that the measure induced on Rn−{0}\mathbf{R}^n - \{0\} by the Lebesgue measure on Rn\mathbf{R}^n may then be identified, up to a constant factor, with tn−1dt⊗dωn−1(z)t^{n-1} dt \otimes d\omega_{n-1}(z). (Make use of the fact that the Lebesgue measure on Rn\mathbf{R}^n is invariant under O(n,R)O(n, \mathbf{R}) and the fact that the Lebesgue measure of the closed ball with center 0 and radius rr is proportional to rnr^n.)

c) Let LhL_h be the intersection of Sn−1S_{n-1} and the hyperplane xn=hx_n = h. The non-empty LhL_h are the intransitivity classes of O(n−1,R)O(n-1, \mathbf{R}) in Sn−1S_{n-1}. Show that ωn−1\omega_{n-1} may be put in the form ∫λhdν(h)\int \lambda_h d\nu(h), where λh\lambda_h is a measure invariant under O(n−1,R)O(n-1, \mathbf{R}) and carried by LhL_h, and ν\nu is a measure on [−1,1][-1, 1]. Identifying LhL_h with Sn−2S_{n-2} by z↦hen+1−h2zz \mapsto he_n + \sqrt{1-h^2} z, one can assume that λh=ωn−2\lambda_h = \omega_{n-2}. Let KθK_\theta be the subset of Sn−1S_{n-1} defined by sin⁡θ≤xn≤1\sin \theta \leq x_n \leq 1. By calculating the Lebesgue measure of the set of tztz (0≤t≤10 \leq t \leq 1, z∈Kθz \in K_\theta), show that ωn−1(Kθ)\omega_{n-1}(K_\theta) is proportional to ∫θπ/2cos⁡n−2φ dφ\int_\theta^{\pi/2} \cos^{n-2} \varphi \, d\varphi. From this, deduce that if one sets h=sin⁡θh = \sin \theta (−π/2≤θ≤π/2-\pi/2 \leq \theta \leq \pi/2), then the measure ν\nu may be identified with cos⁡n−2θ dθ\cos^{n-2} \theta \, d\theta.

d) Conclude, by induction on nn, that

dωn−1=cos⁡n−2θ1cos⁡n−3θ2⋯cos⁡θn−2 dθ1 dθ2⋯dθn−1, d\omega_{n-1} = \cos^{n-2} \theta_1 \cos^{n-3} \theta_2 \cdots \cos \theta_{n-2} \, d\theta_1 \, d\theta_2 \cdots d\theta_{n-1},

where

x1=sin⁡θ1x2=cos⁡θ1sin⁡θ2…xn−1=cos⁡θ1cos⁡θ2⋯cos⁡θn−2sin⁡θn−1xn=cos⁡θ1cos⁡θ2⋯cos⁡θn−2cos⁡θn−1 x_1 = \sin \theta_1 \\ x_2 = \cos \theta_1 \sin \theta_2 \\ \ldots \\ x_{n-1} = \cos \theta_1 \cos \theta_2 \cdots \cos \theta_{n-2} \sin \theta_{n-1} \\ x_n = \cos \theta_1 \cos \theta_2 \cdots \cos \theta_{n-2} \cos \theta_{n-1}
with −π/2≤θi≤π/2-\pi/2 \leq \theta_i \leq \pi/2 for 1≤i≤n−21 \leq i \leq n-2, 0≤θn−1<2π0 \leq \theta_{n-1} < 2\pi.

Solution

No solution has been written yet.

Transcribed from the printed volume. Not translated.