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Pontryagin Duality for locally compact abliean groups gives plenty of continuous (unitary) characters $\chi : A \to \mathbb{R} / \mathbb{Z}$, but if we do not assume local compactness, can anything be said? In particular, is the following true?

Every abelian Hausdorff topological group has a nontrivial continuous (unitary) character

I believe this to be false, but cannot find a concrete example.

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The trivial group is a concrete example. –  YCor Oct 28 '13 at 9:48

3 Answers 3

up vote 13 down vote accepted

The infinite-dimensional sphere $S^\infty$ (the evident colimit of finite-dimensional spheres) admits a topological group structure whose underlying abelian group is a torsion group (in fact a structure of $\mathbb{F}_2$-vector space). See this answer here: http://mathoverflow.net/a/43047/2926.

If $\phi: S^\infty \to S^1$ is a continuous character, then $\phi(S^\infty)$ is evidently a connected torsion subgroup of $S^1$, which can only be trivial since the torsion subgroup $\mathbb{Q}/\mathbb{Z}$ of $S^1$ is totally disconnected.

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If a topological vector space $X$ is not locally convex, then it usually has not non-zero linear continuous functionals, and this means that there are no non-trivial continuous characters on $X$. For example, you can take the space of all measurable functions on $[0,1]$ with the metrics $$ d(x,y)=\int_0^1\frac{|x(t)-y(t)|}{1+|x(t)-y(t)|}d t. $$

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This is a very elementary example. Nice :-) –  Mariano Suárez-Alvarez Oct 28 '13 at 4:38
    
I guess we could also use a Banach space such that the existence of nonzero linear functionals on it is independent of ZF? –  Qiaochu Yuan Oct 28 '13 at 4:41
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I thought that each Banach space has enough linear continuous functionals (by the Hahn-Banach theorem). :) –  Sergei Akbarov Oct 28 '13 at 4:44
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@SergeiAkbarov, that depends on the Axiom of Choice, which ZF does no include. (The two things are not equivalent, though) –  Mariano Suárez-Alvarez Oct 28 '13 at 4:47

I'd think that a linearly ordered group extension of   $\mathbb R$,   especially any non-standard real field wouldn't have any non-trivial (Pontryagin) character.

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Which topology you have in mind? For the topologies I can think of, there are proper open subgroups, and hence the discrete quotient admits Pontryagin characters. –  YCor Oct 28 '13 at 12:05
    
I thought of the linearly ordered topology (induced by the standard inequality relation). –  Wlodzimierz Holsztynski Oct 29 '13 at 1:33
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Then the convex hull of the standard reals is an open subgroup and is not the whole group. So the quotient is a nontrivial (indeed infinite) discrete abelian group. Thus this topological group admits plenty of characters. –  YCor Oct 29 '13 at 22:47
    
Thank you, @Yves. –  Wlodzimierz Holsztynski Oct 31 '13 at 7:54

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