# Schwartz–Bruhat Functions on a Non-Archimedian Local Field

I am in the process of reading through Ramakrishnan and Valenza's book Fourier Analysis on Number Fields, and there's this one statement that seems either trivial and unhelpful or simply not true; so, I imagine I'm misunderstanding something.

The statement occurs in the proof of part (i) of Theorem 7.2 in which we're considering a Schwartz-Bruhat function $f:F\to \mathbb{C}$ in the case where $F$ is a non-Archimedean, local field.

... Since $f$ is locally constant with compact support, it factors through a finite quotient group of the form $$\pi_F^m\mathfrak{o}_F/\pi_F^n\mathfrak{o}_F$$ for some integers $m$ and $n.$ Hence, by linearity and the translation invariance of the Haar measure, it suffices to check the assertion for functions $f$ that are merely characteristic functions of the various ideals $\pi_F^j\mathfrak{o}_F$. ...

Where $\pi_F$ is a uniformizing parameter for the maximal ideal of $\mathfrak{o}_F$.

My confusion lies in why this map is a homomorphism.

That is, $f$ is not a homomorphism, right? So, even though I can construct a composition of maps $F\to \pi_F^m\mathfrak{o}_F/\pi_F^n\mathfrak{o}_F\to \mathbb{C}$ it doesn't seem useful in any way since they're not homomorphisms, right? Am I missing something here?

I feel that you can just say that such a function must be a sum of characteristic functions on sets of the form $a+\pi^k\mathfrak{o}_F$ just from the fact that it's locally constant, and then there are finitely many because it has compact support.

I would appreciate any input on why saying this factors through such a quotient is helpful. Thanks in advance!

## 1 Answer

Of course a Schwartz-Bruhat function won't be a homomorphism in general. But I guess the formulation is okay, if one interprets the word "factorization" here just set-theoretically. Factoring refers to writing something as a composition (product) of two maps (which applies not just to homomorphisms). I guess your confusion comes about because you might have only heard this term in algebraic contexts.

And your observation is correct. I will spell out some details.

Let's consider more generally a topological space $X$ and a function $f : X \rightarrow \mathbb{C}$ which is locally constant and compactly supported. Then the set $S:= \{x \in X \, :\, f(x) \neq 0\}$ is open and compact and equals the support of $f$. We can cover $X$ by the open and compact sets $f^{-1}(z)$ for $z \in \mathbb{C}^{\times}$. By compactness, we find pairwise distinct $z_1, \dots, z_n \in \mathbb{C}^{\times}$ such that $f = \sum_{i=1}^n{z_1 \mathbf{1}_{f^{-1}(z_i)}}$.

Suppose now that $X = F$ is a non-archimedan local field and consider a single compact open set $U = f^{-1}(z_i)$. Write $U$ as a union of basic open sets $a_i + \pi^{n_i}\mathfrak{o}_F$ where $n_i \in \mathbb{Z}$ and $a_i \in F$. Take a finite subcover. After possibly discarding some of the remaining finitely many cosets (two cosets will either be disjoint or one of them will be contained in the other one) you've written $U$ as a finite disjoint union of such cosets. Hence the function $f$ we started with is indeed a finite linear combination of indicatar functions of cosets $a + \pi^k \mathfrak{o}_F$.

• This actually clears up a lot of my confusion! I guess I was confused why it obvious/helpful that it factored through this quotient vs. seeing that it could be written as the sum without noticing this, but you've helped shed some light on this. Thanks again! – Laarz May 2 '18 at 21:21