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I am not at all an expert in p-adic analysis, but I was wondering if there is any sensible (or even generally accepted) way to define the number $\pi$ in $\mathbb Q_p$ or $\mathbb C_p$.

I think that circles, therefore also angles, are problematic in a p-adic context, but $\pi$ appears in many other contexts. Of course there are many known series that sum to $\pi$, some may converge p-adically, but those that converge may have different limits, and I think some more motivation would be needed to designate one as an analog of $\pi$.

Maybe one could find an analog based on $e^{n\pi i} = (-1)^n$, or even $\int_{\mathbb R} e^{-x^2/2}dx = \sqrt\pi$. So my question:

Is there or are there p-adic definitions of $\pi$? If not, could we sensibly define $\pi_p$, and how?

TheSimpliFire
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doetoe
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    Here https://en.wikipedia.org/wiki/P-adic_exponential_function we can read "Note that there is no analogue in $\mathbb C_p$ of Euler's identity, $e^{2\pi i} = 1$. This is a corollary of Strassmann's theorem." – GEdgar Jul 31 '21 at 18:57
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    While there is no analogue of $2\pi i$ (and hence $\pi$) in $\Bbb{C}_p,$ one can move to one of a few larger rings containing $\Bbb{C}_p$ in which there is an element which deserves to be called $2\pi i$ (see the discussion [here](https://sbseminar.wordpress.com/2009/02/18/there-is-no-p-adic-2-pi-i/)). This is [Fontaine's "rings of $p$-adic periods."](https://en.wikipedia.org/wiki/Fontaine%27s_period_rings) Classically, periods are numbers which can be expressed as integrals of algebraic functions over algebraically defined domains. – Stahl Jul 31 '21 at 21:07
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    See also [this](https://www.ihes.fr/~maxim/TEXTS/Periods.pdf) survey by Kontsevich and Zagier for some classical background and [here](http://xavier.caruso.ovh/papers/publis/periods.pdf) for some discussion in the $p$-adic case. A more thorough exposition is [here](https://math.stanford.edu/~conrad/papers/notes.pdf). – Stahl Jul 31 '21 at 21:11
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    @Stahl and GEdgar, thanks, this is interesting. However (@Stahl), I think that not having an analogue of $2\pi i$ in the context of the exponential function doesn't necessarily mean there cannot be an analogue of $\pi$ in some of its other incarnations in ordinary analysis. – doetoe Jul 31 '21 at 22:51
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    @doetoe it is true that you may be able to define an analogue of $\pi$ in another sort of way. If you wanted to make a different definition, you could attempt to do so -- but there remains the question of the utility or relevance of such a definition. Not to say one shouldn't ask such a question! My point was simply that there does exist a $p$-adic analogue of $2\pi i$ which you could then use to define $\pi$. While it does not exist in $\Bbb{C}_p$ itself, this $p$-adic notion of $\pi$ has been studied and is related to much interesting modern research. – Stahl Jul 31 '21 at 23:09
  • @Stahl Yes, your point totally stands and is exactly the kind of answer I was hoping for (except that it is very technical!). It essentially says all there is to say about the existence of a p-adic $\pi$ as in Euler's identity. – doetoe Jul 31 '21 at 23:21
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    The p-adic Gamma function obeys identities of the same form as the classical Gamma function which involves $\pi$, in particular the $\Gamma(z)\Gamma(1-z)=\frac{\pi}{\sin \pi z}$ formula and Gauss multiplication formula, so it's possible to try to define it this way, $\sqrt{\pi}:=\Gamma_p(1/2)$, which for $p=1\mod p$ can be chosen to play the role of a canonical square root of $-1$ in $\mathbb{Q}_p$. I read about this in Alain M Robert's A Course in p-adic Analysis if you're interested to look into that more. – Merosity Aug 01 '21 at 17:29
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    @Merosity Thanks, that is a very interesting one as well! – doetoe Aug 01 '21 at 18:44

1 Answers1

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In a very flattering note @saulspatz has asked me to weigh in on this issue. But many of the commenters to your question have already evinced rather greater familiarity with deep $p$-adic matters than I am in possession of. Nonetheless, here’s my feather’s worth.

My first reaction to your question is, “There can’t be such a thing as $p$-adic $\pi$.” At the outset, however, let me warn you that any time an older scientist says that something is impossible, it’s time for young scientists to prove the old geezer (or old hag) wrong. Just think of Kelvin’s preposterous (in light of the geologic knowledge in his time) estimate of the earth’s (and worse, of the sun’s) age.

What archimedean phenomenon would one want to form an analogy to, for defining a $p$-adic $\pi$? Certainly not the ratio, somehow, of the circumference (or area) of a circle to the radius. For there are no circles in $\Bbb Q_p$, except disks, and they don’t have circumferences. You might try comparing the “interior” $\{z\in\Bbb Q_p:|z|<1\}$ of the unit disk $D_1=\{z\in\Bbb Q_p:|z|\le1\}$, but that’s not its interior, and what would naively be thought to be the “circumference”, the set-difference of the two, has $p-1$ times the measure of the “interior”. No good for anything like the real $\pi$, especially when you see that the same construct delivers different ratios under finite extensions of the field $\Bbb Q_p$.

Anyone would run screaming from that approach, and hope that the first nonzero root of the sine function, or the first nonzero argument $z$ of the complex function $z\mapsto e^{iz}$ to give the value $1$, might help. Even ignoring the question that there can be no good $p$-adically defined notion of firstness, neither of these functions is defined (or, apparently, definable) far enough from zero to be anything other than one-to-one. In the case of the sine function, the series expansion that you know makes perfectly good sense in a neighborhood of $0$, but on its $p$-adic domain of convergence, it’s one-to-one, so has just the one root $0$.

You might think that the exponential offered more hope, but its domain of convergence is the same, $\{z:|z|<p^{-\frac1{p-1}}\}$. (At this point, the multiplicative absolute value confuses me, and I have to use the additive valuation $v_p:\Bbb Q_p\to\Bbb Z\cup\{\infty\}$. It satisfies $v(zz')=v(z)+v(z')$, $v(z+z')\ge\min(v(z),v(z'))$, and $v(0)=\infty$, $v(p)=1$. ) In my preferred notation, the condition for convergence of the exponential is $v(z)>\frac1{p-1}$. But no matter how you look at it, the exponential is one-to-one on that domain, has no other zeros.

Oh well, you say, how about the logarithm? We know that in the right domain of definition, we have $\log(i)=i\pi/2$. Why can’t we use that? But the glory of complex analysis is that the logarithm is not a homomorphism. It tries to satisfy the rule $\log(zz')=\log z+\log z'$ all right, but the logarithm is not defined on a group.

I suppose you can say that the glory of $p$-adic analysis is that the logarithm is a homomorphism, and it’s defined on the huge group $\{z\in\Bbb C_p: v(z-1)>0\}$, the principal units of $\Bbb C_p$. The codomain is $\Bbb C_p$, and it’s an easy exercise to show that the log is surjective (!). But for every $z\in\Bbb C_p$, the inverse image $\log^{-1}(z)$ is infinite, because the kernel of the homomorphism $\log$ is the set of $p$-power roots of unity, an infinite subgroup of the domain group.

You offer another suggestion: maybe some other archimedean series can be used for defining a $p$-adic $\pi$? I don’t think that the $\arctan$ series will help, $\arctan(x)=x-x^3/3+x^5/5-\cdots$. In the real world you can set $x=1$ there, but the series is $p$-adically divergent. What would $\arctan(p)$ a anyway? But one might look at that…

Philosophically speaking, I think you’re chasing a will-o’-the-wisp. I think that $\pi$ is intrinsically an archimedean object, and to ask for a $p$-adic analog is to ask for a contradiction in terms.

J. W. Tanner
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Lubin
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  • Concerning the $p$-adic logarithm from $\{z \in \mathbf C_p : v(z-1) > 0\} = \{z \in \mathbf C_p : |z-1| < 1\}$ to $\mathbf C_p$ being surjective, it's important that the underlying field there is something like $\mathbf C_p$ for the $p$-adic logarithm to be unbounded (in order to achieve all values). On a locally compact field like $\mathbf Q_p$, the domain of the $p$-adic logarithm is $\{x \in \mathbf Q_p : v(x-1) > 0\}$, which is *compact*, so the image of the $p$-adic logarithm on that set is compact and thus bounded. – KCd Aug 01 '21 at 02:40
  • Absolutely right, Keith. I was negligent in that issue. Too late at night to fiddle with it, maybe later. – Lubin Aug 01 '21 at 02:46
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    It may also be interesting to contemplate Hensel’s failed proof of the transcendence of $\pi$ by (erroneously) believing that archimedean and p-adic $\pi$-ish things in this context were genuinely comparable. (For several hours one day some years ago, I did think so, too! :) – paul garrett Aug 01 '21 at 03:40
  • The last paragraph confuses me: The first non-archimedean field that springs to mind for me is hyperreals, and _they_ have a perfectly good $\pi$. Does "archimedean" have two different meanings? – Troposphere Aug 01 '21 at 09:50
  • Sorry for the confusion, @Troposphere . In this racket, “Archimedean” means with the triangle inequality that you’re familiar with, while “nonarchimedean” uses the triangle inequality $|a+b|\le\max(||a|,|b|)$. So $\Bbb Q$ has one arch. absolute value function, and infinitely many mutually incompatible nonarch. absolute value functions, one for each prime $p$. – Lubin Aug 01 '21 at 15:59
  • @Troposphere: See also https://math.stackexchange.com/q/349934/96384 for the two meanings of "non-archimedean". Unfortunately, the accepted answer to that question is, in my view, pretty bad. – Torsten Schoeneberg Aug 05 '21 at 00:09