Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

For any compact abelian group $K$.$$K\cong H_0\times \mathrm{U}(1)^k,$$where $H_0$ is a finite group.

share|improve this question
1  
I'm guessing that this question won't last long in its current form, but let me point out that you at very least need to add some hypotheses. If the scope of $K$ is "compact topological abelian group" then your statement is false. –  Ramsey Oct 9 '12 at 14:34
    
This is clearly not true, consider for example $\mathbb Z_2^{\mathbb N}$. I am trying to think of some context where this question makes sense. Is $k$ finite? –  Stefan Geschke Oct 9 '12 at 15:05
1  
Presumably he means compact Hausdorff abelian Lie groups, where $k$ is finite. –  Todd Trimble Oct 9 '12 at 15:49
    
Try looking here for instance: hopf.math.purdue.edu/Dwyer-Wilkerson/lie/liegroups.pdf –  Todd Trimble Oct 9 '12 at 15:51
    
Oh sorry, two comments ago, I should have said "he or she". –  Todd Trimble Oct 9 '12 at 15:52
add comment

closed as off topic by Henry Cohn, Emil Jeřábek, Stefan Geschke, Felipe Voloch, Todd Trimble Oct 9 '12 at 15:53

Questions on MathOverflow are expected to relate to research level mathematics within the scope defined by the community. Consider editing the question or leaving comments for improvement if you believe the question can be reworded to fit within the scope. Read more about reopening questions here.If this question can be reworded to fit the rules in the help center, please edit the question.

1 Answer

up vote 2 down vote accepted

Any topological group of the form $K\cong H_0\times U(1)^k$ (with $H_0$ finite and $k$ a positive integer) is a closed subgroup of $U(1)^h$ (for some positive integer $h\geq k$). Furthermore, all the closed subgroups of $U(1)^h$ are of this form.

The proof is an easy application of the Pontryagin-Van Kampen duality. In fact, such groups are the duals of the finitely generated Abelian groups (which are quotients of $\mathbb Z^h$). It is well known that such groups are of the form $F\times \mathbb Z^k$ (with $F$ a finite group).

To find a general form for a compact abelian group is as difficult as giving a structure theorem for discrete abelian group (which is known to be quite a difficult, and fairly open, problem in general, even if there are nice results for countable torsion groups).

EDIT: just to answer also to the comment of Stefan Geschke. Finitely generated groups in $\mathrm{Mod}(\mathbb Z)$ can be characterized as the Noetherian objects of the category. So I guess that (by duality) the objects of the form $K\cong H_0\times U(1)^k$ should be the Artinian objects in the category of compact abelian topological ($T_2$) groups. (N.B.= here by Artinian I mean the category-theoretical notion)

share|improve this answer
add comment

Not the answer you're looking for? Browse other questions tagged or ask your own question.