How to give a "quick proof" of properties of the Hecke-operators? In "Analytic Number Theory" by Kowalski, Iwaniec p. 370 the authors prove a formula for the Fourier-coefficients of the Hecke-operators. Namely: If $$T(n) f(z) = \frac{1}{n} \sum_{ad = n} a^k \sum_{0 \leq b < d } \sum_m a_f(m) e\left( m \frac{az+b}{d} \right) = \sum_{m=0}^\infty b(m)e(mz)$$
is the Hecke operator $T(n)$ applied to a weight $k$ modular form $f = \sum_{n\geq 0} c(n) e(nz)$ (let's assume, contrary to the book, that $f$ is a modular form over the full modular group), then the formula derived for $b(m)$ then is $$\sum_{d \mid (m, n)} d^{k-1} c\left( \frac{mn}{d^2}\right)$$
The authors then state that this can be used to "give [a] quick proof" of $$T(n) T(m) = \sum_{d \mid (m, n)} d^{k - 1} T\left( \frac{mn}{d^2} \right)$$
I've tried to do the proof using that the Fourier-coefficients determine a modular form. But I got only long expressions for the Fourier coefficients of the LHS and the RHS, and couldn't really conclude anything. So my question is: How can I prove this quickly using the fomula for the Fourier-coefficients?
Thanks!
Added: I've had the following idea: One can prove that the cusp-forms have a basis of normalized eigenforms. For eigenforms one can derive the desired formula by our expression of the Fourier-coefficients (which happen to be the eigenvalues of the eigenforms), which then holds for the whole space of cusp-forms. However, showing that the space of cusp-forms has a basis of normalized eigenforms is not a quick proof. I still think I'm missing something here...
 A: First of all, the "quick proof" that Iwaniec and Kowalski are referring to is probably the same as in Iwaniec's Topics in Classical Automorphic Forms, Theorem 6.6, pg. 96, which uses upper-triangular representatives of the action of the Hecke operators.
But looking into your initial approach involving expanding the LHS and RHS expressions of Fourier coefficients, here is what I think is an alternative combinatorial proof (which I have not seen in the literature).
Let $f(z) = \sum_{l \geq 0} c(l) e(lz)$ be the Fourier expansion of our modular form.  We know that
$$T(m)f(z) = \sum_{l \geq 0} \sum_{e|(l,m)} e^{k-1} c\left(\frac{lm}{e^2}\right) e(lz).$$
Therefore,
$$T(n)T(m)f(z) = \sum_{l \geq 0} \sum_{d|(l,n)} \sum_{e|(ln/d^2,m)} (de)^{k-1} c\left(\frac{lmn}{(de)^2}\right) e(lz).$$
On the other hand,
$$\sum_{d|(m,n)} d^{k-1} T\left(\frac{mn}{d^2}\right) f(z) = \sum_{l \geq 0} \sum_{d|(m,n)} \sum_{e|(mn/d^2,l)} (de)^{k-1} c\left(\frac{lmn}{(de)^2}\right) e(lz).$$
So to show that both these expressions are equal to each other, we need to prove the following proposition:
Proposition Let $l,m,n$ be positive integers, and let $F$ be a function on the positive integers. Then
$$\sum_{d|(l,n)} \sum_{e|(ln/d^2,m)} F(de) = \sum_{d|(m,n)} \sum_{e|(mn/d^2,l)} F(de).$$
For this we'll need the following lemma:
Lemma Let $a,b,c$ be nonnegative integers, and let $f$ be a function on the nonnegative integers.  Let
$$G(a,b,c) = \sum_{i = 0}^{\min(a,b)} \sum_{j = 0}^{\min(a+b-2i,c)} f(i+j).$$
Then $G(a,b,c)$ is symmetric with respect to $a,b,c$, i.e.
$$G(a,b,c) = G(a,c,b) = G(b,a,c) = G(b,c,a) = G(c,a,b) = G(c,b,a).$$
Proof of Lemma  Let $\operatorname{med}(a,b,c)$ denote the median of $a,b,c$.  Write
$$G(a,b,c) = \sum_{k \geq 0} m(k) f(k).$$
A rather tedious case-by-case calculation shows that (and assuming I didn't make an error somewhere):
$$m(k) = \begin{cases}
k+1 & \text{if } k \leq \min(a,b,c),\\
\min(a,b,c)+1 & \text{if } \min(a,b,c) \leq k \leq \operatorname{med}(a,b,c),\\
\max(0,\min(a,b,c)+1+\operatorname{med}(a,b,c)-k) & \text{if } \operatorname{med}(a,b,c) \leq k \leq \max(a,b,c),\\
\max(0,a+b+c+1-2k) & \text{if } k \geq \max(a,b,c),\\
\end{cases}$$
In particular, we see that the expression for $m(k)$ is symmetric in $a,b,c$.
Proof of Proposition  We can apply the Lemma to prove the Proposition for prime powers:
$$\sum_{d|(p^a,p^c)} \sum_{e|(p^ap^c/d^2,p^b)} F(de) = \sum_{d|(p^b,p^c)} \sum_{e|(p^bp^c/d^2,p^a)} F(de).$$
The extension to the general case is straightforward.
If someone else can come up with a slicker proof of the Lemma or the Proposition (something better than just checking cases), I'd love to see it.
