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
 A: Smooth in this setting by definition means locally constant.
A: 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.
A: 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.
A: 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.
