Bourbaki

Exercise 10, § 3

General Topology, VI

In a projective space Pn\mathbf{P}_n, let Hn,pH_{n,p} be the "quadric" defined by the equation

x02+x12+⋯+xp−12−xp2−⋯−xn2=0(1⩽p⩽n). x_0^2 + x_1^2 + \cdots + x_{p-1}^2 - x_p^2 - \cdots - x_n^2 = 0 \qquad (1 \leqslant p \leqslant n).

Show that Hn,1H_{n,1} and Hn,nH_{n,n} are homeomorphic to Sn−1S_{n-1}. If 2≤p≤n−12 \leq p \leq n-1, Hn,pH_{n,p} is homeomorphic to the space obtained by identifying every point of the product Sp−1×Sn−pS_{p-1} \times S_{n-p} (considered as a subspace of Rp×Rn−p+1\mathbf{R}^p \times \mathbf{R}^{n-p+1}) with its diametrically opposite point. Every point of Hn,pH_{n,p} has an open neighbourhood homeomorphic to Rn−1\mathbf{R}^{n-1}.

Show that H3,2H_{3,2} is homeomorphic to S1×S1S_1 \times S_1 [identify S1×S1S_1 \times S_1 with T×TT \times T, a pair of diametrically opposite points of S1×S1S_1 \times S_1 being identified with a pair of points (u,v)(u, v) and (1/2+u,1/2+v)(1/2 + u, 1/2 + v) of T×TT \times T; then consider the mapping (u,v)→(u+v,u−v)(u, v) \to (u + v, u - v) of T×TT \times T into itself].

II) In a projective space PnP_n, let Cn,pC_{n,p} be the "quadric cone" defined by the equation

x12+x22+⋯+xp2−xp+12−⋯−xn2=0(1≤p≤n−1). x_1^2 + x_2^2 + \cdots + x_p^2 - x_{p+1}^2 - \cdots - x_n^2 = 0 \quad (1 \leq p \leq n-1).

Show that the complement of {0}\{0\} in Cn,pC_{n,p} is homeomorphic to R×Hn−1,p\mathbf{R} \times H_{n-1,p}, with the notation of Exercise 10.

Solution

No solution has been written yet.

Transcribed from the printed volume. Not translated.