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Context: I've just started reading Tate's thesis. In it, we start with a local field $k.$ The aim of the section is to describe the structure of the character groups of $k^+$ (the additive group) and $k^*$(the multiplicative group). But for some reason when looking at the character group for $k^+$, we are looking only for the characters $\chi: k^{+} \to S^1$, where $S^1$ is the circle group but in $k^*$, we are looking at quasi characters $\chi^\prime:k^* \to \mathbb{C}^*$. Why are we doing this? @anon answered a related question, Characters of a Group: two definitions, on Math StackExchange regarding this but it really doesn't help much.

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    $\begingroup$ Your link went to the question, not to an answer, so I changed it to point to the sole answer to that question. I hope that was correct. $\endgroup$
    – LSpice
    Jul 17 at 22:01
  • $\begingroup$ @LSpice Yes, thank you. Since there was only one answer, I didn't really think I'd need to link exactly that (and I honestly don't know how to link answers). I'll keep that in mind from next time. $\endgroup$
    – Rits
    Jul 18 at 2:40
  • $\begingroup$ Re, every answer has a row of 'buttons' below it, the first of which is the word 'Share'. Clicking that will get a link to the answer. You can also get a link to a comment, entirely non-obviously, by clicking on the timestamp of the comment. $\endgroup$
    – LSpice
    Jul 18 at 16:05
  • $\begingroup$ @LSpice Ohh, thanks! $\endgroup$
    – Rits
    Jul 19 at 4:29

2 Answers 2

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I think the basic reason for the apparently different definitions boils down to the different topologies of $k^{*}$ versus $k^{+}$.

For simplicity of discussion, let's consider the case that $k= \mathbb{Q}_p$. If $\psi: \mathbb{Q}_p^{+}$ is a continuous additive character, then it must have image in the unit circle. A simple way to see this is to use that $\mathbb{Z}_p$ is compact, so its image under the continuous map $\psi$ is a compact subgroup of $\mathbb{C}^{*}$. It therefore lies in $S^1$. One can apply the same argument to the compact subgroups $p^{-1} \mathbb{Z}_p$, $p^{-2} \mathbb{Z}_p$, and so on.

On the other hand, one can easily define continuous quasi-characters of $\mathbb{Q}_p^*$ that do not have image in the unit circle. For instance, take $\chi(p) = p$ and $\chi(x) = 1$ for $x \in \mathbb{Z}_p^{*}$. It is still true that $\mathbb{Z}_p^*$ is a compact subgroup of $\mathbb{Q}_p^*$, but now we have $\mathbb{Q}_p^* = \bigcup_{j \in \mathbb{Z}} p^j \mathbb{Z}_p^{*}$, and the character is free to send $p$ anywhere in $\mathbb{C}^*$.

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    $\begingroup$ Great explanation, and right to the point. This same idea is, in some very broad sense, at the root of why diagonalizable and unipotent groups behave so differently. $\endgroup$
    – LSpice
    Jul 17 at 21:59
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    $\begingroup$ This is true for $k$ any locally compact non-archimedean field. Any $x \in k$ is contained in a compact subgroup, such as the set of $y \in k$ with $|y| \leq |x|$. The image of any compact subgroup must be contained in $S^1$. $\endgroup$
    – Matt Young
    Jul 17 at 23:00
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    $\begingroup$ @KevinCasto But it is not true for the fields $\mathbb R$ and $\mathbb C$, which also appear in his work, as the identity function gives a quasi-character that's not a character. So maybe taking circle-valued multiplicative characters is a way of forcing the fields to behave like each other, in that $\mathbb C^*$-valued multiplicative characters have different natures for the archimedean and non-archimedean fields. $\endgroup$
    – Will Sawin
    Jul 18 at 0:59
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    $\begingroup$ @KevinCasto Oh, sorry, my comment was completely garbled. It should be characters of the additive group, where the exponential map provides the example to show the archimedean fields behave differently. $\endgroup$
    – Will Sawin
    Jul 18 at 17:20
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    $\begingroup$ More generally, over non-archimedean local fields, unipotent (linear) groups are ascending unions of compact subgroups. (This is what makes Jacquet modules work so well.) Not so in the archimedean case. And not for $GL(1)$ over non-archimedean fields, either. $\endgroup$ Jul 18 at 18:23
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There's a lot to absorb in Tate's thesis, but it is worth the effort. A quick, concrete answer to your question is that is allows, for $s\in\mathbb C$, functions like $n\to n^{-s}$ to be a quasicharacter. This is useful for building $L$-functions

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    $\begingroup$ Thanks for the response. So why not even look at such quasi characters for $k^{+}$? Is it because they are just not interesting? Or maybe there is no proper structure to them, as we have for the $S^{1}$ case? Also, these quasi characters dont have a proper structure to them it seems. If we look at only the characters from $k^{*} \to $S^{1}$, can we expect them to have a nice structure? Have people tried doing it? $\endgroup$
    – Rits
    Jul 17 at 19:29
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    $\begingroup$ @Rits, re, as @‍MattYoung's later answer indicates, we don't look at non-unitary, continuous characters of the additive group of $k$ because there aren't any! $\endgroup$
    – LSpice
    Jul 17 at 22:00
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    $\begingroup$ @LSpice haha yes, I completely failed to realize that. Even to answer my second question, there aren't any characters from $k^{*} \to \mathbb{S}^{1}$, except the trivial character, which is clear from Matt's response. $\endgroup$
    – Rits
    Jul 18 at 2:39
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    $\begingroup$ @Rits, well, it's not quite right to say that non-archimedean $k$ has no unitary characters of $k^\times$: for real $t$, $x\to |x|_p^{it}$ on $\mathbb Q_p^\times$ is such, and these play a large role in treating Hecke characters... $\endgroup$ Jul 18 at 18:26
  • $\begingroup$ @paulgarrett Ohh, right. I missed that. Thanks a lot! $\endgroup$
    – Rits
    Jul 19 at 4:50

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