2

Let $T^n=(S^1)^n$ be the torus. We can consider the dual group $\widehat{T^n}=Hom(T^n, \Bbb{C}^*)=\{\phi:T^n\to \Bbb{C}^* : \phi \text{ a group homomorphism}\}$ then $\widehat{T^n}\cong \Bbb{Z}^n$. How does one prove this? I can see for every integer vector $(a_1, \dots, a_n)$ the map $(t_1, \dots, t_n)\mapsto t_1^{a_1}\cdots t_n^{a_n}$ is a homomorphism. But how to show this is the only type of homomorphism? This seems to be a basic fact but I am unable to figure it out. Thanks for the help.

Chanel Rose
  • 422
  • Note that (1) those homomorphisms has to be assumed to be continuous, otherwise it is not true (because algebraically $S^1$ is isomorphic to $\mathbb{R}\oplus(\mathbb{Q}/\mathbb{Z})$ but not topologically), (2) in such situation the image of $T^n\to\mathbb{C}^$ has to be contained in $S^1$, and (3)* if you know homomorphisms from $S^1$ you should easily find homomorphisms from $T^n$. – freakish Mar 13 '23 at 13:13

0 Answers0