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Theorem 1.67 On page 19 of Ken Ono's book The Web of Modularity says:

Every modular form on $SL_2(\mathbb{Z})$ may be expressed as a rational function in $\eta(z)$, $\eta(2z)$ and $\eta(4z)$.

The proof uses the following two facts,

1, As graded algebra, $M(SL_2(\mathbb{Z}))$ is generated by $E_4(z)$ and $E_6(z)$.

2, $E_4(z)=\frac{n(z)^{16}}{\eta(2z)^8}+2^8\frac{\eta(2z)^{16}}{\eta(z)^8}$,

$E_6(z)=\frac{\eta(z)^{24}}{\eta(2z)^{12}}-2^5\cdot3\cdot5\cdot\eta(2z)^{12}-2^9\cdot3\cdot11\cdot\frac{\eta(2z)^{12}\eta(4z)^8}{\eta(z)^8}+2^{13}\frac{\eta(4z)^{24}}{\eta(2z)^{12}}$.

My question is how did people discover the two identities?

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This does not answer your question, but I want to elaborate on my comment to Jeremy's answer that such identities can also be understood from the viewpoint of elliptic functions. I use as a reference Vol. 2 of Erdelyi et al., Higher Transcendental functions. By Eq. (13.20.8), $$g_2=C\left(\theta_2^8+\theta_3^8+\theta_4^8\right), $$ where $\theta_k=\theta_k(0|\tau)$ are Jacobi theta values, $C$ is irrelevant and $g_2$ is essentially your $E_4$. There is a similar formula for $g_3\sim E_6$. The theta values can be expressed in terms of Dedekind's eta function (this has a priori nothing to do with modular forms, it's just a matter of rewriting the classical product formulas in new notation). With $z=\tau/2$ we have $$\theta_2=\frac{2\eta(4z)^2}{\eta(2z)},\quad \theta_3=\frac{\eta(2z)^5}{\eta(z)^2\eta(4z)^2},\quad \theta_4=\frac{\eta(z)^2}{\eta(2z)}. $$ (I took this from Wikipedia, but it can be found in many books.) This already gives a formula expressing $E_4$ in terms of $\eta(z)$, $\eta(2z)$ and $\eta(4z)$, but not quite the one you ask about.

To get closer, we recall Jacobi's quartic identity (13.14.23), $$\theta_2(0|\tau)^4+\theta_4(0|\tau)^4=\theta_3(0|\tau)^4$$ and (13.23.15), which tells us that $$\theta_2(0|2\tau)^2=\frac 12\left(\theta_3(0|\tau)^2-\theta_4(0|\tau)^2\right), $$ $$\theta_3(0|2\tau)^2=\frac 12\left(\theta_3(0|\tau)^2+\theta_4(0|\tau)^2\right). $$ Multiplying together and using the quartic identity gives $$ \theta_2(0|2\tau)^2\theta_3(0|2\tau)^2=\frac 14\left(\theta_3(0|\tau)^4-\theta_4(0|\tau)^4\right)=\frac 14\theta_2(0|\tau)^4.$$ We then obtain from the quartic identity $$\theta_2(0|\tau)^8+\theta_3(0|\tau)^8=\theta_4(0|\tau)^8+2\theta_2(0|\tau)^4\theta_3(0|\tau)^4= \theta_4(0|\tau)^8+\frac 18\,\theta_2(0|\tau/2)^8.$$ Thus, we get the alternative formula $$g_2=C\left(2\theta_4(0|\tau)^8+\frac 18\,\theta_2(0|\tau/2)^8\right). $$ This seems to give $$E_4=\frac{\eta(z)^{16}}{\eta(2z)^8}+2^4\frac{\eta(2z)^{16}}{\eta(z)^8}. $$ Note that Ono has $2^8$ instead of $2^4$. I don't know if the reason is a trivial mistake in my calculations, in the references I used or in Ono's book.

My conclusion is that even though Ono's formulas may be new, the problem he solves is very classical and was first treated using the theory of elliptic and theta functions rather than the much newer theory of modular forms.

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  • $\begingroup$ You make a convincing case (and I certainly agree) that the tools to prove the identities Ono states have been around for a very long time - definitely predating Dedekind's formal definition of the eta-function. $\endgroup$ – Jeremy Rouse Feb 18 '15 at 13:11
  • $\begingroup$ Just came across this question and wanted to add that expanding a few terms shows that it is indeed 2^4, not 2^8. Definitely a typo in Ono. (gets out red pen) $\endgroup$ – William J. Keith Aug 12 '15 at 0:14
  • $\begingroup$ The correct constant for $E_4$ is $2^8$. The first eta-quotient is $1-16q+\dots$ and the 2nd eta-quotient is $q+\dots$ where $E_4=1+240q+\dots$ thus we need $240=-16+2^8\cdot 1$ for the correct equation. $\endgroup$ – Somos Oct 21 '17 at 22:45
  • $\begingroup$ Both constants are correct for two related values. That is $E_4(q^2) = 1+240q^2+\cdots=t_1+2^4\cdot t_2$ while $E_4(q)=1+240q+\cdots=t_1+2^8\cdot t_2$ where $t_1$ and $t_2$ are the two eta-quotients. $\endgroup$ – Somos Oct 21 '17 at 23:07
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The answer is probably some fairly basic linear algebra. For the first one, each term on the right hand side is a modular form of weight $4$ and level $2$. The space $M_{4}(\Gamma_{0}(2))$ has dimension $2$ and it is a finite computation to enumerate the eta quotients in this space. There are two, and they span the space.

There are no eta-quotients of level $2$ and weight $6$ (because $\Gamma_{0}(2)$ has an elliptic point of order $2$), so it is natural to look for weight $6$ forms of level $4$. (It is a classical result that every level four modular form can be expressed in terms of eta quotients - see Theorem 1.49 in Ono's book.) The space $M_{6}(\Gamma_{0}(4))$ has dimension $4$ and contains $10$ eta quotients. Pick your favorite four that span the space, and express $E_{6}(z)$ in terms of them.

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    $\begingroup$ I am no expert on the history of this topic but I doubt this answer is quite correct. This is how one would prove the identities nowadays, but at least the first one must have been known in some version before modular forms were understood. A closely related problem is to express the Weierstrass invariants of an elliptic curve in terms of theta values. Such questions were studied in the 19th Century before the modern theory of modular forms was developed. I hope someone can give a more complete answer. $\endgroup$ – Hjalmar Rosengren Feb 11 '15 at 5:45
  • $\begingroup$ I confess to not being an expert in the history either, but I think it is plausible for the following reasons. (1) The "modern theory" of modular forms (in reference to dimensions of spaces) can be thought of in fairly elementary terms - this is present even in Mordell's 1917 paper proving Ramanujan's conjecture about the coefficients of $\Delta(z)$. (2) While the Eisenstein series are natural from the perspective of elliptic functions, the Dedekind eta function (first defined in 1877) is less so - it takes a good bit of work to determine the transformation law for it. $\endgroup$ – Jeremy Rouse Feb 16 '15 at 14:06
  • $\begingroup$ It wasn't until Morris Newman's 1957 paper that one had a general criteria about when an eta quotient transforms under $\Gamma_{0}(n)$. (3) Ken Ono told me that (according to his memory), he did not find these identities in an older source, but just computed them himself. $\endgroup$ – Jeremy Rouse Feb 16 '15 at 14:11
  • $\begingroup$ I don't disagree with anything you write, but note that (1) when discussing elliptic functions, 1917 is very modern; (2) you can understand such identities without knowing the modular transformation laws for eta products. The eta products you have are simply theta values. See my more elaborate remark below. $\endgroup$ – Hjalmar Rosengren Feb 18 '15 at 10:34

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