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I was recently trying to think of a simple example that demonstrates that the natural inclusion of an abelian group $G$ into $$\widehat{\widehat{G}}=\text{Hom}\_{\mathsf{Ab}}(\text{Hom}_{\mathsf{Ab}}(G,\mathbb{C}^\times),\mathbb{C}^\times)$$ is not necessarily an isomorphism. Note that I'm looking at all homomorphisms; no topology on the group is involved (or, if you prefer, they are all discrete).

Obviously, $G$ has to be infinite. However, the only example I could think of (much less actually prove worked) was $G=\mathbb{Z}$, in which case $\widehat{G}=\text{Hom}\_{\mathsf{Ab}}(\mathbb{Z},\mathbb{C}^\times)\cong\mathbb{C}^\times$, so that $$\widehat{\widehat{G}}\cong\text{Hom}_{\mathsf{Ab}}(\mathbb{C}^\times,\mathbb{C}^\times)$$ which is uncountable due to the existence of uncountably many automorphisms of the field $\mathbb{C}$, and therefore not isomorphic to $\mathbb{Z}$. However, (my understanding is that) the existence of the anything more than the two obvious automorphisms of $\mathbb{C}$ requires the axiom of choice. So my questions are,

Does the claim that $\mathbb{Z}\not\cong\text{Hom}_{\mathsf{Ab}}(\mathbb{C}^\times,\mathbb{C}^\times)$ require AC?

and, if the answer to the above is yes,

Does the claim that there exists some abelian group $G$ such that $G\not\cong\widehat{\widehat{G}}$ require AC? Is it equivalent?

I initially asked this on math.SE but I suspect it's harder than I initially thought.

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If $A$ is a torsion-free, divisible group, then $Hom(G,A)$ is also torsion-free and divisible. I don't think this requires the axiom of choice. –  Steve D Mar 1 '11 at 2:21
    
But don't roots of unity mean that $\mathbb{C}^\times$ isn't torsion-free? –  Zev Chonoles Mar 1 '11 at 2:26
    
$\mathbb{C}^\times\cong \mathbb{R}/\mathbb{Z}\times \mathbb{R}$ –  Steve D Mar 1 '11 at 2:27
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No, I didn't confuse them, it's just that the only way I could think of how to prove the automorphisms of $\mathbb{C}^\times$ as an abelian group were uncountable was to use that the automorphisms of the field $\mathbb{C}$ are uncountable. I think that if I hadn't known the fact about $\text{Aut}(\mathbb{C})$ being uncountable, I would have approached this question in with my group theory hat on and (hopefully) gotten the answer quite quickly - I've learned my lesson about not letting unnecessarily powerful results obscure the easier approaches to a problem. –  Zev Chonoles Mar 1 '11 at 2:47
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The use of widehat suggests that you are thinking of this as a duality. For abstract abelian groups, duality is usually given by homomorphisms to $\mathbb{Q}/\mathbb{Z}$. For commutative topological groups, duality is usually given by continuous homomorphisms into the circle group $U(1)$. For commutative algebraic groups over the complex numbers, duality is usually given by algebraic homomorphisms to $\mathbb{C}^\times$. –  S. Carnahan Mar 1 '11 at 3:29
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up vote 10 down vote accepted

Since $\mathbb{C}^\times \cong \mathbb{R}\_{+}\times S^1$ by polar coordinates, it suffices to show that $\text{Hom}({\mathbb R}\_+,{\mathbb R}\_+)$ is uncountable. But for any real number $a$, $x\mapsto x^a$ gives an endomorphism of ${\mathbb R}_+$. Explicitly, there are uncountably many group homomorphisms from ${\mathbb C}^\times$ to itself given by $re^{i\theta}\mapsto r^ae^{i\theta}$, for any real $a$.

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Let me give a context where I think a modified version of your question will sound more natural.

For an abelian Hausdorff topological group $G$, let's write $\widehat{G}$ for the continuous homomorphisms from $G$ to $S^1$ and $X(G)$ for the continuous homomorphisms from $G$ to ${\mathbf C}^\times$. Both $\widehat{G}$ and $X(G)$ are groups. There is a natural containment of sets $\widehat{G} \subset X(G) \subset C(G,{\mathbf C})$, where the last space is all continuous functions from $G$ to the complex numbers. When $G$ is also locally compact and we give $C(G,{\mathbf C})$ the compact-open topology then it is Hausdorff and $\widehat{G}$ and $X(G)$ are closed subsets. The topology induced on $\widehat{G}$ and $X(G)$ makes them both Hausdorff topological groups.

Fourier analysis on locally compact Hausdorff abelian groups $G$ is concerned with the interplay between $G$ and $\widehat{G}$, not $G$ and $X(G)$. It is not the case that $X(G)$ for locally compact Hausdorff abelian $G$ is uninteresting, e.g., in Tate's thesis $X(G)$ plays a prominent role for various concrete choices of $G$. But one of the primary features of the construction of $\widehat{G}$, namely that for locally compact Hausdorff abelian $G$ the topological groups $G$ and $\widehat{\widehat{G}}$ are naturally isomorphic, is false for the construction of $X(G)$. That, I think, is the interesting question: what is an example of a locally compact Hausdorff abelian group $G$ such that $X(G)$ is locally compact (it's automatically Hausdorff and abelian) and the topological groups $G$ and $X(X(G))$ are not isomorphic by the natural map $G \rightarrow X(X(G))$? Three counterexamples arise quickly if we compute several examples of $X(G)$.

Example 1: $X({\mathbf Z}) = {\mathbf C}^\times$.

Example 2: $X({\mathbf R}) = {\mathbf C}$ by analysis: the continuous homs ${\mathbf R} \rightarrow {\mathbf C}^\times$ are parametrized by complex numbers via the formula $\chi_s(x) = e^{xs}$ for $s \in {\mathbf C}$.

Example 3: $X(G) = \widehat{G}$ for compact $G$, and in particular $X(G) \cong G$ for finite $G$ and $X(S^1) = \widehat{S^1}$, which is isomorphic to ${\mathbf Z}$ by the parametrization $\chi_n(z) = z^n$ for $n \in {\mathbf Z}$.

Example 4: Although ${\mathbf Q}_p$ is not compact, $X({\mathbf Q}_p) = \widehat{{\mathbf Q}_p}$ because of the $p$-adic topology: if $\chi \colon {\mathbf Q}_p \rightarrow {\mathbf C}^\times$ is a continuous homomorphism then $\chi({\mathbf Z}_p) \subset S^1$ by compactness of ${\mathbf Z}_p$ and for any $x \in {\mathbf Q}_p$ we have $\chi(x)^{p^n} = \chi(p^nx)$, which is in $S^1$ for $n$ large since $p^nx$ is in ${\mathbf Z}_p$ for $n$ large. Therefore $\chi(x) \in S^1$ for any $x$ in ${\mathbf Q}_p$. It is part of Fourier analysis that $\widehat{{\mathbf Q}_p} \cong {\mathbf Q}_p$, so we get $X({\mathbf Q}_p) \cong {\mathbf Q}_p$, which is a contrast to the case of ${\mathbf R}$ in Example 2: $X({\mathbf R}) \not\cong {\mathbf R}$ as topological groups since ${\mathbf R}$ and ${\mathbf C}$ are not homeomorphic.

To make more computations, we record a result about products. For two locally compact Hausdorff abelian groups $G_1$ and $G_2$, the natural map $X(G_1 \times G_2) \rightarrow X(G_1) \times X(G_2)$ is a topological group isomorphism.

Example 5: $X({\mathbf R}^n) \cong {\mathbf C}^n$.

Example 6: $X({\mathbf R}^\times) \cong X(\{\pm 1\} \times {\mathbf R}) \cong X(\{\pm 1\}) \times X({\mathbf R}) \cong \{\pm 1\} \times {\mathbf R} \cong {\mathbf R}^\times$.

Example 7: $X({\mathbf C}^\times) \cong X(S^1 \times {\mathbf R}) \cong X(S^1) \times X({\mathbf R}) \cong {\mathbf Z} \times {\mathbf C}$.

Thus we find $X(X({\mathbf Z})) \cong X({\mathbf C}^\times) \cong {\mathbf Z} \times {\mathbf C}$, so ${\mathbf Z}$ and $X(X({\mathbf Z}))$ are not isomorphic as topological groups (or even as groups: one is countable and the other is not). Another counterexample is ${\mathbf R}$: $X(X({\mathbf R})) \cong X({\mathbf C}) \cong {\mathbf C}^2 \cong {\mathbf R}^4$. A third counterexample is $S^1$: $X(X(S^1)) \cong X({\mathbf Z}) \cong {\mathbf C}^\times$, so $S^1$ and $X(X(S^1))$ are not isomorphic as topological groups since one is compact and the other is not.

The natural isomorphism of $G$ with $\widehat{\widehat{G}}$ for all locally compact Hausdorff abelian $G$ depends crucially on the use of $S^1$ as the receiver of character values. By this I don't just mean that if we use ${\mathbf C}^\times$ as the receiver of character values then such a duality isomorphism breaks down in specific examples. There is a theorem somewhere in Pontryagin's Topological Groups (maybe it is also in Weil's book on integration on topological groups) which says something like the following: if a locally compact Hausdorff abelian group $A$ could be used as the receiver of character values for a duality isomorphism on all locally compact Hausdorff abelian groups then $A$ must be isomorphic to $S^1$. Perhaps someone else can track down the reference.

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Very nice. What's a good reference for this (that is, where did you learn this from)?? –  Matthew Daws Mar 1 '11 at 12:44
    
What does "this" refer to, specifically? I wrote a lot of different things in the answer. To a large extent I worked out most of it while teaching a course on Tate's thesis. –  KConrad Mar 1 '11 at 22:25
    
I meant, for example, is there a nice reference for $X(G)$ (in the way that there are lots of book about $\hat G$). But perhaps, to quote, you just worked it out for yourself... –  Matthew Daws Mar 2 '11 at 10:29
    
When I was giving the course I consulted Ramakrishna-Valenza's "Fourier Analysis on Number Fields" and Hewitt & Ross's "Abstract Harmonic Analysis", so maybe there is something in one of those two books about X(G), but I am not checking on that before posting this reply. –  KConrad Mar 2 '11 at 19:49
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