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?

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4 Responses to Jacobi by pure thought

  1. Dear GR,

    When you use uniqueness to infer the Hecke eigenform property, aren’t you implicitly using weight 12 to obtain uniqueness? Also, in order to get 24 by pure thought, I think the argument
    recalled by BCnrd and Scott Carnahan here could give it, although it involves a detour through the weight two Eisenstein series (see my answer there, but reverse-engineer it, using the BCnrd/SC argument, as a tool for deducing the constant term of the wt. 2 Eisenstein series), and so it is a little roundabout.



  2. Dear Matt, I’m not sure that’s true – certainly T_p \Delta has a zero at the cusp and has the same weight as \Delta, so T_p \Delta/\Delta is a constant. Perhaps something more needs to be said concerning the fact that the zero of \Delta has order one. I agree with your suggestion about ways to see “24”, although I wanted an argument that avoided using any (pseudo) modularity of E_2.

  3. This reminds me a bit of the proof by Kohnen, which also uses a multiplicative version of the Hecke operator (which is the right-hand side of your formula). But he uses the uniqueness of Delta and he computes the 24 by hand.

    (“A Very Simple Proof of the q-Product Expansion of the ∆-Function”, The Ramanujan Journal, 10 (2005), pp. 71-73.)

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