Let $G$ be a quasi-split connected reductive group over a $p$-adic field $F$. Let $B$ be a Borel subgroup which is defined over $F$, with $B = TU$, $T$ defined over $F$. The choice of $T$ and $B$ gives a set of nonrestricted roots $\tilde{\Delta}$, which together with an $F$-splitting and a nontrivial unitary character $F \rightarrow S^1$ yields a unitary character $\chi$ of $U(F)$.


Let $V$ be a smooth, irreducible, admissible representation of $G(F)$. A linear functional $\lambda: V \rightarrow \mathbb{C}$ is called a **Whittaker functional** for $\chi$ if for all $u \in U(F)$ and $v \in V$, we have


$$\lambda(u \cdot v) = \chi(u) \lambda(v)$$


It is a theorem that the space of Whittaker functionals for $\chi$ is at most one dimensional. If this dimension is one, then $V$ is called **$\chi$-generic**. 


Fix a nonzero Whittaker functional $\lambda$, and for $v \in V$, define a function $W\_v :G(F) \rightarrow \mathbb{C}$ by $W\_v(g) = \lambda(g \cdot v)$. The set $W = W\_{\lambda}$ of such functions is closed under addition and scalar multiplication, and becomes a representation of $G(F)$ if we set $g \cdot W\_v = W\_{g\cdot v}$. Then the representation $W$ is called a **Whittaker model**, and up to $G(F)$-isomorphism it does not depend on the choice of $\lambda$, since these things are all scalar multiples of each other.


From what I can see, $V \rightarrow W\_{\lambda}, v \mapsto W\_v$, is an isomorphism of $G(F)$-modules. It is clearly surjective, and for injectivity, if we suppose that $0 \neq v \in V$, and $W\_v(g) = 0$ for all $g \in G(F)$, then $\lambda(g \cdot v) = 0$ for all $g \in G(F)$. This is impossible, because $g \cdot v : g \in G(F)$ spans $V$, because $V$ is irreducible. 


So my question is, what is the point of defining Whittaker models if the thing we defined is just isomorphic to our original representation? Why would it be important to regard $V$ as a set of functions $G(F) \rightarrow \mathbb{C}$?



What's the point of a Whittaker model?
This question is a bit like saying "what's the point of the theory of bases for vector spaces -- this just gives you an isomorphism of your space with $\mathbb{R}^n$. What is the point of defining this isomorphism if the thing we defined is just isomorphic to our original representation? Why would it be important to regard $V$ as a set of vectors in $\mathbb{R}^n$?


As you probably know, some questions involving vector spaces do not involve choosing a basis. But sometimes when you want to do some explicit calculation, you have to resort to coordinates.


Here is an example with vector spaces. How are you going to prove that the natural map from a finite-dimensional real vector space to its double-dual is an isomorphism? A natural way to do this would be to check it's an injection and then do a dimension count to check it's an isomorphism. This reduces the question to checking that the dual of an $n$-dimensional vector space is $n$-dimensional, and *if you pick a basis* this reduces the question to checking it for the "model" $\mathbb{R}^n$ of an $n$-dimensional vector space, and this case is easy.


Now consider the following theorem of Casselman. Let $G=GL\_2(\mathbb{Q}\_p)$ and consider a smooth irreducible admissible representation of $G$ on a complex vector space $V$. Casselman observed that if we look at the invariants for the congruence subgroups $\Gamma\_n:=\begin{bmatrix} \* & \* \\ 0 & 1 \end{bmatrix}$ mod $p^n$ as $n$ increases, then there were two possibilities. Either $V$ is 1-dimensional and and the dimensions of the invariants are $0,0,0,\ldots,0,1,1,1,1,\ldots$ (the jump being where the basis vector becomes stable) or $0,0,0,\ldots,0,1,2,3,4,\ldots$ and in particular the first point when the dimension becomes positive (which it has to do at some point by smoothness) it becomes 1.


However are we going to prove this in the infinite-dimensional case? We just have some completely abstract vector space with an action of $G$. How do we even *begin* to get a feeling for any action of any element of $G$ on any vector in $V$ at all? I invite you to go away and think about trying to solve this problem.


Now imagine that someone tells you that there's this completely natural model of $V$ -- of *any* infinite-dimensional smooth irreducible representation of $GL\_2(\mathbb{Q}\_p)$ -- as a certain space of functions, and we know *completely explicitly* how certain elements of $G$ act on this space. All of a sudden we have a completely concrete "model" for the situation and can begin to do calculations! And this gives you a foothold into beginning the proof.


