Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Given the j-function $j(\tau)$,

$j(\tau) = 1728J(\tau)$,

where $J(\tau)$ is Klein’s absolute invariant, the Dedekind eta function $\eta(\tau)$, and the following Eisenstein series,

$\begin{align} E_4 (q) &= 1+240\sum_{n=1}^{\infty} \frac{n^3q^n}{1-q^n}\\[1.5mm] E_6 (q) &= 1-504\sum_{n=1}^{\infty} \frac{n^5q^n}{1-q^n}\\[1.5mm] E_8 (q) &= 1+480\sum_{n=1}^{\infty} \frac{n^7q^n}{1-q^n}\\ \end{align}$

where,

$q = \exp(2\pi i \tau)$

Are the following relations true?:

$\begin{align} 1.\;\; \eta^{24}(\tau) &= \frac{{E_4}^3(q)}{j(\tau)}\\[1.5mm] 2.\;\; \eta^{24}(\tau) &= \frac{{E_6}^2(q)}{j(\tau)-1728}\\[1.5mm] 3.\;\; \eta^{48}(\tau) &= \frac{{E_8}^3(q)}{j^2(\tau)}\\ \end{align}$

share|improve this question
    
These are not too difficult to prove. See my answer. –  glebovg Sep 6 at 8:33

2 Answers 2

up vote 4 down vote accepted

Yes, these three relations are all true. Since $\eta^{24}$ is the weight 12 level 1 cusp form $\Delta$, you can write them as relations between level 1 modular forms, and these are easy to check because the relevant modular form spaces have small finite dimensions.

share|improve this answer
    
Thank you, David. –  Tito Piezas III Mar 20 '12 at 15:29

The first and third identity are easy to prove. Recall that the modular discriminant $\Delta = \Delta(\omega_1, \omega_2)$ is defined by $$\Delta = g_2^3 - 27g_3^2,$$ where $g_2 = g_2(\omega_1, \omega_2) = 60G_4(\omega_1, \omega_2)$ and $g_3 = g_3(\omega_1, \omega_2) = 140G_6(\omega_1, \omega_2)$ are the Weierstrassian invariants. Moreover, we define the Klein invariant $J = J(\omega_1, \omega_2)$ and the $j$-function by $$J = \frac{g_2^3}{\Delta} \quad \text{and} \quad j = 12^3 J,$$ respectively.

Let $\lambda \ne 0$. Because $g_2$ and $g_3$ are of order $-4$ and $-6$, respectively, we have $$g_2(\lambda \omega_1, \lambda \omega_2) = \lambda^{-4} g_2 \quad \text{and} \quad g_3(\lambda \omega_1, \lambda \omega_2) = \lambda^{-6} g_3,$$ and $$\Delta(\lambda \omega_1, \lambda \omega_2) = \lambda^{-12} \Delta,$$ accordingly. Moreover, since $g_2^3$ and $\Delta$ are of the same order, $J(\lambda \omega_1, \lambda \omega_2) = J(\omega_1, \omega_2)$. In particular, $$g_2(\tau) = \omega_1^4 g_2, \quad g_3(\tau) = \omega_1^6 g_3, \quad \text{and} \quad \Delta(\tau) = \omega_1^{12} \Delta$$ On the other hand, $$J = J(\tau) \quad \text{and} \quad j = j(\tau),$$ that is to say, both $J$ and $j$ are functions of $\tau$ alone.

Using the classical identity $\Delta(\tau) = (2\pi)^{12} \eta^{24}(\tau)$, where $\eta(\tau)$ is the Dedekind eta function defined by $$\eta(\tau) = q^{1/12} \prod_{n = 1}^\infty (1 - q^{2n}), \quad q = e^{\pi i \tau}, \quad \textbf{I}[\tau] > 0,$$ we obtain $$\eta^{24}(\tau) = \frac{\Delta(\tau)}{(2\pi)^{12}} = \frac{\omega_1^{12} g_2^3}{(2\pi)^{12} J(\tau)} = \frac{12^3 \omega_1^{12} g_2^3}{(2\pi)^{12} j(\tau)}.$$ In view of a known result, $$E_{2k}(q) = 1 - \frac{4k}{B_{2k}} \sum_{n = 1}^\infty \frac{n^{2k - 1} q^{2n}}{1 - q^{2n}}, \quad q = e^{\pi i \tau}, \quad \textbf{I}[\tau] > 0,$$ where $B_{2k}$ are the Bernoulli numbers and $E_{2k}(q) = G_{2k}(\tau)/2\zeta(2k)$, we get $$g_2(\tau) = 60G_4(\tau) = \frac{4\pi^4 E_4(q)}{3},$$ or, what is the same thing, $$g_2 = \omega_1^{-4} \frac{4\pi^4 E_4(q)}{3},$$ Therefore, $$\eta^{24}(\tau) = \frac{12^3 \omega_1^{12}}{(2\pi)^{12} j(\tau)} \left(\omega_1^{-4} \frac{4\pi^4 E_4(q)}{3}\right)^3 = \frac{E_4^3(q)}{j(\tau)}.$$ Lastly, using the fact that $E_8(q) = E_4^2(q)$ leads to $$\eta^{48}(\tau) = \frac{E_8^3(q)}{j^2(\tau)}.$$

Now, the second identity follows from the first and the fact that $$J(\tau) = \frac{E_4^3(q)}{E_4^3(q) - E_6^2(q)}.$$ Indeed, we have $$E_6^2(q) = E_4^3(q) - 12^3 \frac{E_4^3(q)}{j(\tau)},$$ but in view of the first identity this becomes $$E_6^2(q) = (j(\tau) - 12^3)\eta(\tau)^{24},$$ and the result follows.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.