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Here's yet another exercise that stumped me:

Let $G$ be a compact abelian topological group. Then $G$ is connected iff its dual $\hat G$ is torsion-free.

Any hints/solutions will be appreciated.

Chera
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2 Answers2

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Mariano's answer is not quite right. For instance if $G=\mathbf{Z}_p$ then $G$ is compact and disconnected, but $G_0$ is trivial so $G/G_0=\mathbf{Z}_p$ is not finite. Instead one must argue as follows.

Let $U$ be a nontrivial closed and open neighbourhood of $1$. Since $G$ is compact we know that $U$ is compact, so by continuity of multiplication there is an open neighbourhood $V$ of $1$ such that $V\cdot U \subset U$. Let $H$ be the group generated by $V$. Then $H$ is open and $H\cdot U\subset U$, so in particular $H\subset U$. So $H$ is a proper open subgroup, so $G/H$ is finite and nontrivial. Now any character of $G/H$ induces a torsion character of $G$.

Sean Eberhard
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  • Your proof really needs the compactness? By continuity of multiplication, you could get open neighbourhoods $V_1,V_2$ of $1$ such that $V_2\subset U$ and $V_1\cdot V_2\subset U$. – Yang Pei Apr 09 '21 at 19:32
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    @YangPei For such $V_1, V_2$ it does not follow that the group generated by $V_1$ is contained in $U$. The implication $G$ connected $\implies$ $\hat G$ torsion-free clearly holds in general (see Mariano's answer), but compactness is essential for the other direction. For example if $G = \mathbf{Q}_p$ then $G$ is totally disconnected even though its dual ($\cong \mathbf{Q}_p$) is torsion-free. – Sean Eberhard Apr 12 '21 at 10:24
  • Ok, thanks for your explanation! – Yang Pei Apr 12 '21 at 11:25
  • I'm sorry, but I do not understand where compactness of $U$ is used. – skewfield May 19 '21 at 21:05
  • @EliaVincenzi It's needed for the existence of $V$. In detail, for every $x \in U$ there is a neighbourhood $V_x$ of 1 and $U_x$ of $x$ such that $V_x \cdot U_x \subset U$, the sets $(U_x : x \in U)$ cover $U$, so by compactness there is a finite subcover $U_1, \dots, U_n$, and $V = V_1 \cap \cdots \cap V_n$. – Sean Eberhard May 22 '21 at 12:50
  • @SeanEberhard yes, now it is clear to me. A last thing: shouldn't we take $V$ symmetric in order to guarantee that its generated group $H$ is such that $H\cdot U\subset U$? Indeed, without symmetry it is not said that if $x\in V\subset U$ then $x^{-1}\in U$, am I right? – skewfield May 24 '21 at 10:56
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    Yes probably we should. – Sean Eberhard May 24 '21 at 12:26
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Suppose $\phi:G\to S^1$ is an element of $\hat G$ and $n\geq1$ are such that $\phi^n$ is the unit element in $\hat G$, that is, $\phi(g)^n=1$ for all $g\in G$. Then $\phi$ takes values in the subgroup of $S^1$ of elements of order divisible by $n$. This subgroup is finite, so it is discrete, so if $G$ is connected, then $\phi$ must be constant (because it is continuous!). We thus see that $G$ connected implies $\hat G$ is torsion-free.

Conersely, suppose $G$ is not connected, and let $G_0$ be the connected component of $1_G$. Then $G/G_0$ is a finite abelian group and non-trivial. In particular, the dual group $(G/G_0)^\wedge$ is also finite and non-trivial (it is non-canonically isomorphic to $G/G_0$, in fact) so it has torsion. To see that $\hat G$ has torsion, we need only observe that there is an injective morphism of groups $(G/G_0)^\wedge\to\hat G$.

Mariano Suárez-Álvarez
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