I have been reading Iwaniec's spectral theory of Automorphic forms, and in one of its definition, its defined that a function $k: \mathbb{H}× \mathbb{H} \rightarrow C$ is point pair invariant if $k(gz,gw)=k(z,w)$ for all $g \in G=SL_{2}(\mathbb{R})$. But I am not understanding how we can set $k(z,w)=k(u(z,w))$ where k(u) is a function of one variable with $u≥0$ here u is distance function on upper half plane. I can understand that if you give me a function f from $\mathbb{R}$ to $\mathbb{R}$ which is even then then $f(u(z,w))$ will give me point pair invariant but how to say converse that is if k is point pair invariant then $k(z,w)=k(u(z,w))$.
Thanks in advance for any hint.
The point is that if $k$ is point point pair invariant, then it only depends on the distance between the points. Basically you should use that $G$ acts transitively on pairs of points of a given distance $d$. You need to know some basic facts about the action of $G$ on the upper half plane. Here is a rough geometric sketch. Let $(z,w)$ and $(z',w')$ be pairs of points in $\mathbb H$, where $z$ and $w$ are distance $d$ apart, and similarly for $z', w'$.
First, how can you prove transitivity on $\mathbb H$? I don't know what Iwaniec does, but one way is to draw a geodesic from $z$ to $z'$ and then flow along this geodesic. So you can send $z$ to $z'$ and $w$ goes to some point $w''$ on a circle of radius $d$ about $z'$. Now use the fact that $G$ contains SO(2) (and conjugates), to show that you can rotate about $z'$ to move $w''$ to $w'$.