Bourbaki

习题 19, § 1

标签 08AI,永久不变。引用为 [Bourbaki, Tag 08AI]。

实变函数, III, 119

Bourbaki marks this one as harder than the rest.

Google 翻译生成的粗略草稿,质量较低。术语依照术语表,但句子可能有误;之后会由仔细的译文取代。

One knows (A, IX, §10, exerc. 2) that in the complex plane C2\mathbf{C}^2 the group AA of angles of directed lines is isomorphic to the orthogonal group O2(C)\mathbf{O}_2(\mathbf{C}), the canonical isomorphism associating the angle θ\theta with rotation through the angle θ\theta; one transports the topology of O2(C)\mathbf{O}_2(\mathbf{C}) (considered as a subspace of the space M2(C)\mathbf{M}_2(\mathbf{C}) of matrices of order 2 over C\mathbf{C}) to the group AA by this isomorphism, so making AA a locally compact topological group;

此外,映射θ↦cos⁡θ+isin⁡θ\theta \mapsto \cos \theta + i \sin \theta是群A在复数≠0\neq 0的乘法群C∗\mathbf{C}^*上的*同构的*,逆同构由公式cos⁡θ=12(z+1/z)\cos \theta = \frac{1}{2}(z + 1/z)定义, sin⁡θ=12i(z−1/z)\sin \theta = \frac{1}{2i}(z - 1/z)。从这些关系推导出加法群 C\mathbf{C} 的每个连续同态 z↦φ(z)z \mapsto \varphi(z) 在 A 上,使得复数函数 cos⁡φ(z)\cos \varphi(z), sin⁡φ(z)\sin \varphi(z)) 在 C\mathbf{C} 上可微,由关系定义cos⁡φ(z)=cos⁡az\cos \varphi(z) = \cos az、sin⁡φ(z)=sin⁡az\sin \varphi(z) = \sin az(aa复数)。

解答

尚未写出解答。

被以下引用

无。

引用了

  • IX, §10 not yet on this site
Google 翻译生成的粗略草稿,质量较低。术语依照术语表,但句子可能有误;之后会由仔细的译文取代。