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We have

$$ j(\tau)=\frac{1}{q}+\sum_{n=0}^{\infty}a_nq^n, a_n\in\mathbb{Z},q=e^{2\pi i\tau} $$

Then it is said that because $j$'s only pole is simple, $j$ has degree 1 as a map $j:X(\text{SL}_2(\mathbb{Z}))\rightarrow\mathbb{C}\cup\{\infty\}$.

My question is, why can we deduce this from that "$j$'s only pole is simple"?

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2 Answers

up vote 2 down vote accepted

The $j$-functions is a holomorphic map from one compact Riemman surface (the completion of $\mathcal H/\mathrm{SL}_2(\mathbb Z)$) to another (the Riemann sphere). The only place it has a pole is at the "cusp" $\infty$ of its domain, and this pole is simple. Thus the preimage of $\infty$ in the Riemann sphere (i.e. the polar part of the divisor of $j$) consists of a single point. But the number of points (counted with correct multiplicties) in the preimage of any point of a degree $d$ map between compact Riemann surfaces is equal to $d$. (If the points lying above a given point have ram. degrees $e_1,\ldots,e_n$, then this is the formula $\sum_i e_i = d$.) Applying this to $j$ we find that $d = 1$.

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Very clear, thanks very much! –  hxhxhx88 Jan 11 '13 at 12:49
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$j$ is periodic of period 1, so $j$ is actually defined on $\mathcal{H}/\mathbb{Z}$ which is biholomorphic to the unit disc under the map $\tau \to e^{2 \pi i \tau}$. Thus when we regard $j$ as a meromophic function on the unit disk, by the series you have written down, you can see that $j$ only has one pole (at 0 of the disk, corresponding to $\infty$ of the upper half plane). The leading term being $1/q$ means that the pole is simple.

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Yeah I know $j$ has a simple pole, my question is why this fact can lead to the conclusion that $j$ has degree 1. Matt's answer makes me clear now. BTW, thank you for some help in my another post. I'm new to this area and learning modular form by myself out of interests. Sometimes maybe some questions are trivial or stupid, really thanks to those kind people like you. –  hxhxhx88 Jan 11 '13 at 12:54
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