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I am reading the proof of the following theorem in Bump's Automorphic Forms and Representations. However, I don't understand the first part of the proof.

It is clear that $(c,d)\mapsto |cz+d|^2$ is a positive definite quadratic form. Then equation 2.2 gives $\operatorname{im}(g\cdot z)=\frac{\operatorname{im} z}{|cz+d|^2}$, so we want $|cz+d|^2$ to be small. But why does it follow that there exists such a constant $R(z)$? Moreover, in the last line, he seems to use that $z\mapsto R(z)$ is continuous. Why is this true?

Any help is appreciated :)

As an aside, why is such an action called "discontinuous"? I have seen some books call this "properly discontinuously" or in French "propre". Isn't this terminology contradictory ? The action of $SL_2 \mathbf{Z}$ on $\mathbf{H}$ is "continuous" in the sense that $SL_2\mathbf{Z}\times \mathbf{H}\to \mathbf{H}$ is continuous, right?

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Dr. Heinz Doofenshmirtz
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    Small comment about feeling "properly discontinuous action" is silly: you're not alone! See this and its answer. – Jakob Streipel Nov 08 '21 at 13:25
  • @prets thanks for the comment – Dr. Heinz Doofenshmirtz Nov 09 '21 at 13:49
  • This is not explained at all in Bump. But it is true that one can choose $R(z)$, continuous in $z$, such that if $c$ or $d$ are larger than $R(z)$ in magnitude, then $\mathrm{Im}(gz) < \epsilon$. You can show this with a bit of tedious computation. Note that $\lvert x + iy \rvert = x^2 + y^2$, so consider the real and imaginary parts separately. For each $x, y$, you can explicitly determine ranges for $c$ and $d$, and the ranges can be chosen to vary continuously in $z$. But I emphasize that this is not clear from Bump. – davidlowryduda Nov 14 '21 at 00:44

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