Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $k$ be the completion of an algebraic number field at a prime divisor $\mathfrak{p}$. We note that $k$ is locally compact. Let $k^{+}$ be the additive group of $k$ which is a locally compact commutative group.

Tate's Thesis Lemma 2.2.1 states that

If $\xi \rightarrow \chi(\xi)$ is one non-trivial character of $k^{+}$, then for each $\eta \in k^{+}$, $\xi \rightarrow \chi(\eta\xi)$ is also a character. The correspondence $\eta \leftrightarrow \chi(\eta\xi)$ is an isomorphism, both topological and algebraic, between $k^{+}$ and its character group.

The proof of this lemma is divided up into 6 steps, one step is to show that the characters $\chi(\eta\xi)$ are everywhere dense in the character group. Tate writes

$\chi(\eta\xi) = 1$, all $\eta \implies k^{+}\xi \neq k^{+} \implies \xi = 0$. Therefore the characters of the form $\chi(\eta\xi)$ are everywhere dense in the character group.

My question is: How does he get from showing that the $\xi = 0$ to the the result that the $\chi(\eta\xi)$ are everywhere dense?

share|improve this question
    
This feels like a weak-$*$ topology thing, but this isn't linear and the topology on the character group is probably the compact-open topology, so I'm getting confused. Nice question! –  Dylan Moreland Feb 18 '12 at 2:13
    
Following up: I think what I said above is on the right track. If you look at Section 4.1 of Folland's book, he shows that the topology on $\widehat G$ coincides with the weak$*$ topology it inherits as a subset of $L^\infty(G)$. I need to sort this out for my own purposes, so I'll try to summarize the argument in the morning, if no one else has done so by then. –  Dylan Moreland Feb 18 '12 at 7:26
add comment

1 Answer 1

up vote 2 down vote accepted

Denote the character $\xi\rightarrow\chi(\eta\xi)$ by $\chi_\eta$. We want to show that image of the map $f_\chi:\eta\rightarrow\chi_\eta$ is dense in $\hat k$. Take a closed subgroup $H$ of $k$ and set $N_H=\lbrace \xi\in k:\chi_\eta(\xi)=1\ {\rm for\ all}\ \eta\in H\rbrace$. This is also a closed subgroup. We have the short exact sequence $$0\rightarrow N_H\rightarrow k\rightarrow k/N_H\rightarrow 0$$ and the functoriality of Pontryagin duality turns this into $$0\rightarrow \widehat{k/N_H}\rightarrow \widehat{k}\rightarrow \widehat{N_H}\rightarrow 0$$ We have an isomorphism $\widehat{k/N_H}\simeq f_\chi(H)$ (this is basically Theorem 4.39 in Folland's "A Course in Abstract Harmonic Analysis"). Now, setting $H=k$, we see that $N_k=\lbrace 0\rbrace$, so the short exact sequence becomes $$0\rightarrow f_\chi(k)\rightarrow \widehat{k}\rightarrow 0\rightarrow 0$$ Hence $f_\chi(k)\simeq \widehat k$. This is mildly stronger than what Tate has done at this point, but I'm not worried, since we're incorporating the topology directly in the argument.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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