Let $G$ be an abelian group. The character group of $G$ is the group of all homomorphisms from $G$ to $\mathbb{C}^\times$ under pointwise multiplication, denoted $\widehat{G}$. I am mainly interested in the case $G$ is a finitely generated abelian group. As groups, $\widehat{G}$ is isomorphic to $F \times (\Bbb C^\times)^{r}$ with $F$ is a finite abelian group and $r=\mathrm{rank}(G)$. The latter group is a (compact abelian) Lie group/direct product of a finite group with an (abelian connected complex linear) algebraic group/topological group.

Is it enough to say that $\widehat{G}$ has all the properties that a Lie/algebraic/topological group has? To put it another way, how much of extra algebra/topology can we earn from the group isomorphism $\widehat{G}\cong F \times (\Bbb C^\times)^{r}$? For example, the cardinality of $F$ enumerates the number of connected components of $F \times (\Bbb C^\times)^{r}$. Does this make sense if we speak of "connected components" of $\widehat{G}$?


$\mathbb{C}^{\times}$ is a Lie group and the character group inherits its Lie group structure (in the finitely generated case). This is because every f.g. abelian group is a quotient of a f.g. free abelian group and so every character group of such a group is obtained from the character group of $\mathbb{Z}^n$, which is $(\mathbb{C}^{\times})^n$ (with its usual Lie group structure), by considering the subgroup respecting some relations (which inherits a Lie group structure).

Said another way, taking the character group defines a contravariant functor from f.g. abelian groups to Lie groups. If you replace $\mathbb{C}^{\times}$ with $S^1$ you get a nicer functor which is a special case of Pontryagin duality, and which among other things is fully faithful.

  • $\begingroup$ Can I say that $\widehat{G}$ has a structure of a Lie group? The isomorphism mentioned above is just for groups, and probably the isomorphism between Lie groups needs more than that. $\endgroup$ – user Jan 20 '18 at 15:54
  • $\begingroup$ @user: you can transport the Lie group structure across the isomorphism. There's an additional argument you need to make to show why this Lie group structure doesn't depend on a choice of presentation, but it's not hard. $\endgroup$ – Qiaochu Yuan Feb 10 '18 at 2:34
  • $\begingroup$ Can you elaborate more on it? Maybe it relies on the fact that we have the group isomorphism not just bijection of sets. $\endgroup$ – user Feb 10 '18 at 4:16

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