Jacobi by pure thought

JB asks whether there is a conceptual proof of Jacobi’s formula:

\(\Delta = q \prod_{n=1}^{\infty}(1 – q^n)^{24}\)

Here (to me) the best proof is one that requires the least calculation, not necessarily the “easiest.” Here is my attempt. We use the following property of \(\Delta\), which follows from its moduli theoretic definition: the only zero of \(\Delta\) is a simple zero at the cusp, moreover, the evaluation of \(\Delta\) on the Tate curve is normalized so that the leading coefficient is \(q\).

Let \(p\) be prime. I claim that

\(\Delta(\tau)^{p+1} \prod_{i=0}^{p-1} \zeta^i = \Delta(p \tau) \prod_{i=0}^{p-1} \Delta \left(\frac{\tau + i}{p}\right).\)

Observe that both these expressions are modular forms of level one and weight \((p+1)\) times the weight of \(\Delta\). One can prove this “by hand,” but also by noting that the RHS is equal to the norm of \(\Delta(p \tau)\) on \(X_0(p)\) down to \(X_0(1)\). On the other hand, the RHS also has a zero of order \(p+1\) at \(q = 0\), from which the result immediately follows, since the ratio will be holomorphic of weight zero. If one defines Hecke operators on \(q\)-expansions in the usual way, it also immediately follows that the logarithmic derivative \(q d/dq \log(\Delta)\) is (as a \(q\)-expansion) an eigenform for \(T_p\) of weight two with eigenvalue \(p+1\) for all primes \(p\). In fact, the same argument as above (with \(X_0(p)\) replaced by \(X_0(n)\)) implies that this derivative is also an eigenform for \(T_n\) with eigenvalue \(\sigma_1(n) = \displaystyle{\sum_{d|n} d}\). This is almost enough to determine the \(q\)-expansion uniquely: in particular, it implies that

\(q \cdot \frac{d}{dq} \log(\Delta) = 1 + m \cdot \sum_{n=1}^{\infty} \sigma_1(n) q^n\)

for some integer \(m\), from which it follows that

\(\Delta = q \prod_{n=1}^{\infty} (1 – q^n)^m.\)

To finish the argument, it suffices to check that \(m = 24\), or that \(\tau(2) =-24\). One way to do this is to note (by uniqueness) that \(\Delta\) is a Hecke eigenform, and then use the equation \(\tau(2) \tau(3) = \tau(6)\) which implies that \(m \in \{0,1,2,3,24\}\); the cases \(m = 1,2,3\) are then ruled out by the equations \(\tau(2) \tau(7) = \tau(14)\) and \(\tau(2) \tau(13) = \tau(26)\), and \(m = 0\) is ruled out by the fact that \(q\) is not a modular form. Curiously enough, this determines \(\Delta\) without ever using the fact that it has weight \(12\). Another (more traditional way) is to show that \(1728 \Delta = E^3_4 – E^2_6= q – 24 q^2 + \ldots\). Is there a way to do this final step by pure thought?