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Usually the question about Iwasawa logarithm is why $\log_p p = 0$, but here I am interested in justifying why $\log_p$ must be null over the group of roots of 1 of order prime to p.

The reason I have seen given is that if $\zeta^n = 1$ then $n\cdot\log_p\zeta = \log_p 1 = 0$, but one could actually argue the same for $e^{i\frac{p}{q}(2\pi)}$ (with $p/q$ any rational), yet no definition of complex logarithm implies it to be null over the unit circle $\mathbb{U}$ (at least as far as I know).

Are there other reasons ?

ClemD
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  • The answer to the question you link to, with comments underneath, kind of addresses the $p$-primary roots as well, doesn't it? Also, since the $p$-adic log is already $0$ on the $p$-power roots of unity, it feels somewhat natural to just do that for all roots of unity. I mean, if the complex logarithm were $0$ on $p$-power roots of unity, it would (if continuous) be zero on the entire unit circle, and hence basically just be $ln$ of the absolute value of $z$, a boring (and non-analytic) function. – Torsten Schoeneberg Jun 23 '21 at 16:29
  • Conversely, in the archimedean world we have, as closure of all roots of unity $\overline{\mu(\mathbb C)}$, that wonderful compact torus $S^1 \simeq \mathbb R/\mathbb Z$ whose additive structure can take all the torsion and hence be a nice part of the codomain of the logarithm (modulo fiddling with branch cuts), keeping it both locally analytical and a group homomorphism. What object should take its place in the $p$-adic setting? As far as I know (happy to learn otherwise), $\mu(\mathbb C_p)$ (esp. its $p$-primary part) is not closely linked to anything with a "nice" structure. – Torsten Schoeneberg Jun 23 '21 at 16:40
  • Thanks :) Not sure I understood you, though. Do we agree that any root of unity is either a $p$-primary xor a power of a $p$-power root ? Mr Lubin stated « the logarithmic series already vanishes at the $p$-primary roots of unity », while you stated « the $p$-adic log is already $0$ on the $p$-power roots of unity », so I guess I'm confused here. Also, how do you then prove these roots are inside the disc of convergence ? Then how do you prove the series converge to zero ? – ClemD Jun 23 '21 at 20:35
  • Ah, my bad: I guess "$p$-primary" should mean the same as "of order a power of $p$". In my above comments, I wrongly used it for the opposite, maybe should have said "of order coprime to $p$". – Torsten Schoeneberg Jun 23 '21 at 20:48
  • So anyway, the roots of unity in $\mathbb C_p$ whose order is a power of $p$ all have the property that $\lvert \zeta - 1 \rvert_p < 1$ i.e. thy lie in the radius of convergence of the $p$-adic logarithm. Since that logarithm series also has the formal property of turning multiplication into addition, and they are torsion elements, they must be sent to $0$. It's not that easy to prove that elementarily, cf. e.g. https://math.stackexchange.com/q/2816184/96384. – Torsten Schoeneberg Jun 23 '21 at 20:58
  • I think that one often misses the real problem: The classical (Archimedean, complex) logarithm is not a homomorphism. For, if you look at the definition of “homomorphism”, you see that it must be defined on a group. And the complex log is not defined on any group. The $p$-adic log, by contrast, is defined on the multiplicative group of all principal units, $1+\mathfrak m$, where $\mathfrak m$ is the maximal ideal in the integers of the complete field under consideration. – Lubin Jun 24 '21 at 00:13
  • @Mr Schoeneberg : so if I'm right, so far we take for granted* that the logarithmic series converge to zero for roots of order a power of $p$. How do we extend this result to other roots ? I guess there is no way to use a continuity argument like we would do in $\mathbb{C}$ (as you said, there is no nice structure in $\mu(\mathbb{C})$).

    *I read the proof you linked (thanks btw), but I do not see how to extend it for every root of order a power of $p$.

     – ClemD Jun 24 '21 at 10:21
  • @Mr Lubin : is it really that impossible to define the complex log on a group ? I mean, I would naively define $\log : \mathbb{C}^* \to \mathbb{R} \times S^1$ with $\mathbb{R} \times S^1$ being an additive group shaped like a cylinder (I use $S^1$ with the exact same meaning as $\mathbb{U}$ before ; hope it's not a problem). I'm not saying this is useful in any way (though at least this presentation feels much more natural to me than reasoning with principal values) , but then I do not get how not being a homomorphism is the core problem here. – ClemD Jun 24 '21 at 10:36
  • (My bad : $S^1$ and $\mathbb{U}$ can not have the exact same meaning since the latter is a multiplicative group.) – ClemD Jun 24 '21 at 11:34
  • You're right, we are not "forced" to make the $p$-adic log $0$ on the other roots of unity, but I find the fact that it is $0$ on the $p$-power roots a good motivation to do it. And if we want it to be a group homomorphism but not $0$ there, then we have to invent a torsion group as new factor in the codomain. (By the way I agree with your view of the complex log as having codomain $\mathbb R \times S^1$.) -- Ultimately, to answer your question, you first have to say, well, what do you want to do with the $p$-adic logarithm. – Torsten Schoeneberg Jun 24 '21 at 14:58
  • Thanks a lot ! That seems a nice conclusion to me. I don't know much about what mathematicians want to do with $p$-adic logarithm. Should I open a new question ? – ClemD Jun 25 '21 at 08:36
  • Here's an idea for putting a log on the p-power roots of unity. We can fix a choice of roots of unity, ${1, \zeta_p, \zeta_{p^2}, \zeta_{p^3}, ...}$ in a compatible way such that $\zeta_{p^n}^{p^{-k}} = \zeta_{p^{n+k}}$. Now for $x=\sum_{k \ge -a}x_kp^k \in \mathbb{Q}p$ we can define (pardon my silly but suggestive choice of notation) $1^x := \prod{-a \le k \le -1}\zeta_{p^{-k}}^{x_k}$. A corresponding logarithm for this would get you an element of $\mathbb{Q}_p / \mathbb{Z}_p$. I believe this gives you an isomorphism between two representations of the Prufer p-group. – Merosity Jun 25 '21 at 18:23
  • For $|x-1|<1$ we can factor $x=\zeta u$ for $\zeta$ finite order and $u$ infinite order. Then add the usual $\log_p(u)$ defined by its power series with its codomain lying in $\mathbb{Z}_p$ and $\ell(\zeta)$ defined as above with codomain $\mathbb{Q}_p/\mathbb{Z}_p$, so we pick a representative with integer part 0 and unambiguously add them: $L(x)=\ell(\zeta)+\log_p(u) \in \mathbb{Q}_p$. Using identities can cause problems since $\ell(\zeta\omega)=\ell(\zeta)+\ell(\omega)$ is only guaranteed to hold modulo a p-adic integer. This is analogous to the complex logarithm varying modulo $i2\pi$. – Merosity Jun 25 '21 at 19:03
  • @Merosity : are you somehow refering to Fontaine's period rings ? Your interesting comments reminded me of maxmoo's comment here, though yours seem more straightforward. Maybe choosing a family of compatible roots of unity is a standard way of proceeding. – ClemD Jun 28 '21 at 12:12
  • @ClemD No idea it doesn't seem quite like what I'm talking about but I'm not sure since I just came up with it while thinking about your question. It kind of looks like Jay P's comment, but I don't know enough to know if it's equivalent or not just because both depend on a choice of compatible system of primitive $p$-power roots of unity. I imagine if it is, it's an alternate perspective on things, like how automorphisms of the roots changes the logarithm to decouple the choice of roots. I have a lot to learn lol. – Merosity Jun 28 '21 at 14:54
  • Thinking about what Mr Lubin meant, I believe that the "complex log" and "p-adic log" were to be understood as the "log series over the complex or the p-adic". Mr Lubin was perhaps refering to Mr Schoeneberg's assertion that "Since the log series is a morphism, torsion elements must be sent to 0". This argument works in p-adic world since the log series is defined over a multiplicative group, but not in the complex world. Hence e.g. $\log e^{i(2\pi)/17} \neq 0$, though the log series is defined for $e^{i(2\pi)/17}$ and it is a root of unity. I wish Mr Lubin could confirm this. – ClemD Jul 07 '21 at 14:48

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