How to show that $\Delta\left(\frac{az+b}{cz+d}\right)=(cz+d)^{12}\Delta(z)$? Let $(a, b; c, d) \in SL_2(\mathbb{Z})$ and $\Delta(z)=q\prod_{n=1}^{\infty} (1-q^n)^{24}$, $q=e^{2\pi i z}$. How to show that $\Delta(\frac{az+b}{cz+d})=(cz+d)^{12}\Delta(z)$? Thank you very much.
I think that $ad - bc =1$ and hence $a=(1+bc)/d$. We can substitute $a$ by $(1+bc)/d$ in $\Delta(\frac{az+b}{cz+d})$. But I am not able to simplify the result.
 A: There is a proof (somewhat nontrivial) in Don Zagier's notes.
A: Somehow it turns out to not be reasonable to try to prove this by direct substitution and manipulation.
Proofs were known in the 19th century (see also things about "Dedekind's $\eta$-function", which is the $24$-th root of Ramanujan's $\Delta$).
Succinct, but subtle, proofs were given by Siegel and Weil: these are recalled, with citations to the original sources, in various places, for example my notes http://www.math.umn.edu/~garrett/m/mfms/notes_2013-14/08a_product_expansion.pdf 
A: First of all, using the fact that $SL(2, \mathbb Z)$ is generated by $\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$ and $\begin{pmatrix} 0 & -1 \\
1 & 0 \end{pmatrix}$, and the functional equation for the former is obvious,
you are reduced to the case of the functional equation for the substitution
$\tau \mapsto -1/\tau$.
One approach is to take logarithmic derivatives with respect to $\tau$ (or, up
to a $2\pi i$ factor, applying $q \dfrac{d}{dq}$ to $\log \Delta$), so that 
you are instead studying how the series
$$1 - 24 \sum_{n = 1}^{\infty} \bigg(\sum_{d | n} d \bigg) q^n$$
transforms under $\tau \mapsto -1/\tau$.
If you take the logarithmic derivative of the formula
$\Delta(-1/\tau) = \tau^{12} \Delta(\tau),$ you will
find that it should almost transform like a weight $2$ modular form,
but not quite.
The discussion of Eisenstein series in Serre's Course in arithmetic will
show that (up to a scalar) this function is an Eisenstein series of weight $2$,
namely
$$\sum_{(m,n) \neq (0,0)} \frac{1}{(m\tau + n)^2}.$$
If this series converged absolutely, it would actually give a modular form of weight $2$.  But it doesn't converge absolutely, so (once you interpret what this series even means), when you make the substitution $\tau \mapsto -1/\tau$ and then rearrange, something slightly non-trivial happens.  But you can figure out exactly what does happen, and so derive the 
correct behaviour under $\tau \mapsto -1/\tau$, and then argue backwards to get the desired modularity property for $\Delta$.
(I think this is all discussed in Course in arithmetic.)
