Dual thing

Shadow · **Posted:** Tue, 20 Mar 2012 03:43:09 UTC

Correct me if I'm wrong, but the Pontrjagin dual of $\mathbb{Z}_p$ should be itself, right? The proof for a completion, k^+

, of $\mathbb{Q}$ doesn't seem to use anything other than the fact that k^+

is an integral domain which is locally compact, and $\mathbb{Z}_p$ is actually globally compact, so the proof should carry through without alteration, right?

Also, on that note, for the proof for k^+

(the + is just to emphasize that we are only dealing with

as an abelian group) part of it is $\mathbf{X}(\eta \xi)=1$ for every $\eta\in k^+$ implies $\xi k^+\ne k^+$ hence $\xi=0$ (this is a place where we use the fact that k^+

is an integral domain), and from what I understand, this is supposed to prove that the characters, $\mathbf{X}(\eta -)$ are everywhere dense, and I don't see how that's supposed to follow. (Here $\mathbf{X}$ is any nontrivial character of k^+

). I just don't seem to get the topology on $\widehat{k^+}$ , as I know this must be obvious for some reason.

outermeasure · **Posted:** Tue, 20 Mar 2012 07:26:38 UTC

Shadow wrote:

Correct me if I'm wrong, but the Pontrjagin dual of $\mathbb{Z}_p$ should be itself, right? The proof for a completion, k^+

, of $\mathbb{Q}$ doesn't seem to use anything other than the fact that k^+

is an integral domain which is locally compact, and $\mathbb{Z}_p$ is actually globally compact, so the proof should carry through without alteration, right?

Also, on that note, for the proof for k^+

(the + is just to emphasize that we are only dealing with

as an abelian group) part of it is $\mathbf{X}(\eta \xi)=1$ for every $\eta\in k^+$ implies $\xi k^+\ne k^+$ hence $\xi=0$ (this is a place where we use the fact that k^+

is an integral domain), and from what I understand, this is supposed to prove that the characters, $\mathbf{X}(\eta -)$ are everywhere dense, and I don't see how that's supposed to follow. (Here $\mathbf{X}$ is any nontrivial character of k^+

). I just don't seem to get the topology on $\widehat{k^+}$ , as I know this must be obvious for some reason.

Are you sure? IIRC taking Pontryagin dual swaps compact with discrete, and so the dual of $\mathbb{Z}_p$ is
$\widehat{\mathbb{Z}_p}=\widehat{\varprojlim\mathbb{Z}/p^n\mathbb{Z}}=\varinjlim\widehat{\mathbb{Z}/p^n\mathbb{Z}}=\varinjlim\mathbb{Z}/p^n\mathbb{Z}=\mathbb{Z}(p^\infty),$
the Pr\"{u}fer group (with discrete topology).

You probably meant $\mathbb{Q}_p$ instead? In which case yes it is self-dual.

Shadow · **Posted:** Tue, 20 Mar 2012 07:47:37 UTC

outermeasure wrote:

Shadow wrote:

Correct me if I'm wrong, but the Pontrjagin dual of $\mathbb{Z}_p$ should be itself, right? The proof for a completion, k^+

, of $\mathbb{Q}$ doesn't seem to use anything other than the fact that k^+

is an integral domain which is locally compact, and $\mathbb{Z}_p$ is actually globally compact, so the proof should carry through without alteration, right?

Also, on that note, for the proof for k^+

(the + is just to emphasize that we are only dealing with

as an abelian group) part of it is $\mathbf{X}(\eta \xi)=1$ for every $\eta\in k^+$ implies $\xi k^+\ne k^+$ hence $\xi=0$ (this is a place where we use the fact that k^+

is an integral domain), and from what I understand, this is supposed to prove that the characters, $\mathbf{X}(\eta -)$ are everywhere dense, and I don't see how that's supposed to follow. (Here $\mathbf{X}$ is any nontrivial character of k^+

). I just don't seem to get the topology on $\widehat{k^+}$ , as I know this must be obvious for some reason.

Are you sure? IIRC taking Pontryagin dual swaps compact with discrete, and so the dual of $\mathbb{Z}_p$ is
$\widehat{\mathbb{Z}_p}=\widehat{\varprojlim\mathbb{Z}/p^n\mathbb{Z}}=\varinjlim\widehat{\mathbb{Z}/p^n\mathbb{Z}}=\varinjlim\mathbb{Z}/p^n\mathbb{Z}=\mathbb{Z}(p^\infty),$
the Pr\"{u}fer group (with discrete topology).

You probably meant $\mathbb{Q}_p$ instead? In which case yes it is self-dual.

Yes, I saw that the proof for $\mathbb{R}$ pulls through to that for $\mathbb{Q}_p$ (aside from the detail I mentioned in my first post), however I don't see why the same proof doesn't pull through for $\mathbb{Z}_p$ , since it seems to have all the necessary properties that the completion of $\mathbb{Q}$ had, though I suppose the assignment of characters must just somehow fail to be dense in the character group, as the compactness gives that the image ought to be closed (albeit if the image is discrete, this isn't saying much).

outermeasure · **Posted:** Tue, 20 Mar 2012 07:56:20 UTC

Shadow wrote:

outermeasure wrote:

Shadow wrote:

Correct me if I'm wrong, but the Pontrjagin dual of $\mathbb{Z}_p$ should be itself, right? The proof for a completion, k^+

, of $\mathbb{Q}$ doesn't seem to use anything other than the fact that k^+

is an integral domain which is locally compact, and $\mathbb{Z}_p$ is actually globally compact, so the proof should carry through without alteration, right?

Also, on that note, for the proof for k^+

(the + is just to emphasize that we are only dealing with

as an abelian group) part of it is $\mathbf{X}(\eta \xi)=1$ for every $\eta\in k^+$ implies $\xi k^+\ne k^+$ hence $\xi=0$ (this is a place where we use the fact that k^+

is an integral domain), and from what I understand, this is supposed to prove that the characters, $\mathbf{X}(\eta -)$ are everywhere dense, and I don't see how that's supposed to follow. (Here $\mathbf{X}$ is any nontrivial character of k^+

). I just don't seem to get the topology on $\widehat{k^+}$ , as I know this must be obvious for some reason.

Are you sure? IIRC taking Pontryagin dual swaps compact with discrete, and so the dual of $\mathbb{Z}_p$ is
$\widehat{\mathbb{Z}_p}=\widehat{\varprojlim\mathbb{Z}/p^n\mathbb{Z}}=\varinjlim\widehat{\mathbb{Z}/p^n\mathbb{Z}}=\varinjlim\mathbb{Z}/p^n\mathbb{Z}=\mathbb{Z}(p^\infty),$
the Pr\"{u}fer group (with discrete topology).

You probably meant $\mathbb{Q}_p$ instead? In which case yes it is self-dual.

Yes, I saw that the proof for $\mathbb{R}$ pulls through to that for $\mathbb{Q}_p$ (aside from the detail I mentioned in my first post), however I don't see why the same proof doesn't pull through for $\mathbb{Z}_p$ , since it seems to have all the necessary properties that the completion of $\mathbb{Q}$ had, though I suppose the assignment of characters must just somehow fail to be dense in the character group, as the compactness gives that the image ought to be closed (albeit if the image is discrete, this isn't saying much).

Yes, it fails to give enough details about what happens in the middle, analogous to how it failed for $\mathbb{Z}\subset\mathbb{Q}$ when $\hat{\mathbb{Q}}=\mathbb{R}$ but $\hat{\mathbb{Z}}$ is only a quotient $\mathbb{T}=\mathbb{R}/\mathbb{Z}$ . Knowing where it sends things like 1/(things prime to p), which is in $\mathbb{Z}_p$ , doesn't tell you where it sends 1/p and the like, which are in $\mathbb{Q}_p$ but not $\mathbb{Z}_p$ . The local uniform continuity just can't extract enough information from $\mathbb{Z}$ or $\mathbb{Z}_p$ which is not dense, contrast with $\mathbb{Q}$ which is dense.

S.O.S. Mathematics CyberBoard

Dual thing

Who is online