Why are Poincare series defined as they are?

Question

We know the Poincare series are defined as the following:

The $m^{th}$ Poincare series of weight $k$ for $\Gamma$ is: $$ P_{m}^{k} (z) = \sum_{(c,d)=1} (cz+d)^{-k} e^{2 \pi in(\tau z)}. $$ The definition yields that, via the Petersson inner product, we obtain: $$\langle f, P_{m}^{k}\rangle = c_{k,m}a_{m},$$ where $c_{k,m} = \frac{(k-2)!}{(4\pi m)^{k-1}}$, if $f = \sum_{n \in \mathbb{Z}} a_{n}e^{2 \pi inz}$ is a cusp form, and we can further show that the Poincare series generate $\mathcal{S}_{k}$, the space of cusp forms of weight k.

I can't see that, we already know the $a_{n}$'s, why can't we define the $m^{th}$ Poincare series of weight $k$ as: $$ P_{m}^{k} (z) = \frac{1}{c_{k,m}}\sum_{(c,d)=1}(cz+d)^{-k}e^{2 \pi in(\tau z)}\;? $$ It seems to me that this definition also works.

Answer 1

Ultimately this is bound to be a matter of taste -- you can't "prove" that one normalisation is better than another -- but let me try to justify why it is conventional to normalise this way.

The point of Poincare series isn't just to prove there exists a modular form satisfying $\langle f, P^k_m \rangle = (const) \times a_m(f)$ for all $f \in S_k$. It's obvious that some such form exists (because the Petersson product is nondegenerate). The point is that the series having this property has a nice explicit form, and that allows you to prove various fun things you couldn't otherwise prove; renormalising as you suggest to make the constant be 1 makes the explicit form slightly less nice.