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## Locally constant functions with compact support = smooth ?

Hello,

I have a trivial question, but I hope that you don't mind helping. I often get confused with basic definitions.

Let F be a p-adic field. Then (from what I understand) $C_c^{\infty}(F)$ is the set of functions $f: F \to \mathbb{C}$ such that $f$ is locally constant and $f$ has compact support. Is the locally constant part the same as smooth or continuous (the $\infty$)? What exactly does smooth mean in this case?

Thanks, Tom

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In the $p$-adic setting, smooth = locally constant. This is not the same as continuous. – Kevin Ventullo May 7 2011 at 23:27
If $F$ is a local field, then there is a continuous surjective map from $F$ to the Cantor set, but such a map can't be smooth. – S. Carnahan May 8 2011 at 1:51

A function $f : F \to \mathbb{C}$ is said to be smooth provided that $f$ is invariant under translation by some open subgroup of $F$ (hence by some compact open subgroup). Notice that when $f$ has compact support, this is equivalent to requiring that $f$ be locally constant. I don't have time to cook up a counterexample at the moment (nor do I have one readily available, which I suppose is a problem), but without the compact support assumption this is definitely a stronger condition than locally constant.

More generally, if $\pi : F \to GL(V)$ is a complex representation of the additive (topological) group of $F$, a vector $v \in V$ is called smooth provided that there exists an open subgroup of $F$ which fixes $v$. Taking $V$ to be the space of all functions $f : F \to \mathbb{C}$ with the action of $F$ given by translations, one arrives at the definition above.

This is relevant when, for example, one studies induced representations of $p$-adic groups.

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 Yes, you're right about the distinction between invariant by an open subset and locally constant. Sorry for my mistake. – Joël Cohen May 8 2011 at 4:23 Justin, for a locally constant but not uniformly locally constant function, just consider a sum of characteristic functions of balls with radii tending to 0, and centres wandering around (i.e., no cluster points). – L Spice May 8 2011 at 18:06 @L Spice: Yes, I was picturing something like this but didn't want to trust my intuition in this non-Archimedean setting... – Justin Campbell May 8 2011 at 21:22 Actually I was rather sloppy: “wandering around” doesn't suggest the need to go very far, but the local compactness of $\mathbb Q_p$ means that I should have said that we were considering balls with centres tending to $\infty$. (I have at least avoided referring to the centre of a ball, that being an eminently Archimedean concept.) – L Spice May 9 2011 at 2:21

To acquire some sense of differential calculus over fields other than the standard real or complex (topological) fields, one might start by getting acquainted with this paper or other similar contributions of the same authors. Every topological field can be considered as a "topologized" module over the discrete topological ring $\mathbb Z$ . So it is meaningfull to speak of differentiability and smoothness of maps between any topological fields in the BGN−sense. However, then the important determination axiom (see iii, p. 8 in the referred paper) is not satisfied, and so there need not be any uniquely determined derivatives.

The basic definition goes as follows. Consider a map $f:E\supseteq U\to F$ where $E,F$ are topologized modules over the topological ring $\boldsymbol R$ and $U$ is open in $E$ . Writing $g^{[0]}=f$ and $U^{[0]}=U$ and $E^{[0]}=E$ , recursively define $U^{[i+1]}=(U^{[i]})^{1}$ where $V^{1}={(w,z,t):w,w+tz\in V}$ , and $E^{[i+1]}=(E^{[i]})^{1}$ where $G^{1}=G\times G\times\boldsymbol R$ and the operation "$\times$" refers to the particular topologized module product inherent in the BGN−axioms. Then $f$ is smooth if there exists a sequence $g^{[i]}:E^{[i]}\supseteq U^{[i]}\to F$ of maps satisfying the property that $g^{[i]}(w+tz)=g^{[i]}(w)+tg^{[i+1]}(w,z,t)$ holds for $(w,z,t)\in U^{[i+1]}$ . The maps $g^{[i]}$ need not be unique when the determination axiom is not satisfied.

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But the point here is that $\mathbb{C}$ is not an $F$-algebra. So there's no calculus involved when studying smooth complex (or $l$-adic) representations of a $p$-adic group. Now when you're interested in $p$-adic representations of $p$-adic groups, that a whole different story (and your paper is indeed useful for that case). – Joël Cohen May 8 2011 at 12:46
"So there's no calculus involved when studying smooth complex (or $l$-adic) representations of a $p$-adic group." Why then use such misleading phrases as $C^\infty$, "smooth"? You already referred to this in you answer. $C^\infty$ and "smooth" in the sense of some kind of differentiability are meaningfull only when basically speaking of a map $f:X\supseteq U\to Y$ where $X,Y$ are (topological or otherwise suitably structured) modules over the same (structured) ring, or more specifically, a field. – TaQ May 8 2011 at 17:56
TaQ: The word "smooth" does not mean anything about differentiability in the context of functions from p-adic fields to the complex numbers. Why use such a misleading term? The functions one cares about in the classical setting (complex domain, complex codomain) are genuinely smooth, so we simply adopt the word smooth to refer to the relevant class of functions in the new setting (p-adic domain, complex codomain) even though it really is not as calculus-like as you might wish. As von Neumann would say, you just get used to it. – KConrad May 8 2011 at 23:31
I edited my answer to explain that it is meaningfull to speak of differentiability of maps between $p$-adic and the complex field. (Unfortunately, the latex server did not manage to do its job properly although I checked that the code is correct!) So the language "one is used to" really is confusing. – TaQ May 9 2011 at 7:55
TaQ: where has the notion of differentiability of maps from Q_p to C which you are advertising been used in applications (i.e., not in papers just about definitions)? Continuous homomorphisms from the additive group Q_p to C* are locally constant, and this illustrates why continuous locally constant functions are the basic functions of interest from p-adic domains to archimedean domains. Look at Tate's thesis for a nontrivial application of these concepts. You're not going to get people to change terminology at this point. – KConrad May 9 2011 at 12:47

Yes :

smooth + compactly supported = locally constant + compactly supported

In this context, smooth means continuous with regards to the discrete topology on $\mathbb{C}$, meaning we just look at the algebraic structure of $\mathbb{C}$ (and we could just as well use $\overline{\mathbb{Q}}$ instead). This is why smooth representations are sometimes also called algebraic in old litterature.

And as to why this is denoted $\mathcal{C}^{\infty}$, the only plausible explanation I could think of is an analogy with the case of real groups. I'd be glad if someone could provide an explanation.

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Smooth in this setting by definition means locally constant.

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So if differentiable means that $f'(x)$ exists, does it imply that infinitely differentiable = locally constant? – Ho Chung Siu May 8 2011 at 3:26
Actually, f′(x) does not make much sense since $f(x) \in \mathbb{C}$ and $x \in \mathbb{Q}_p$. – Joël Cohen May 8 2011 at 4:14