Given the Dedekind eta function $\eta(\tau)$. Define $m = (p-1)/2$ and a $24$th root of unity $\zeta = e^{2\pi i/24}$.

  1. Let p be a prime of form $p = 12v+5$. Then for $n = 2,4,8,14$: $$\sum_{k=0}^{p-1} \Big(\zeta^{m k}\, \eta\big(\tfrac{\tau+m k}{p}\big)\Big)^n = -\big(\sqrt{p}\;\eta(p\tau)\big)^n\tag1$$

  2. Let p be a prime of form $p = 12v+11$. Then for $n = 2,6,10,14, and\; 26$: $$\sum_{k=0}^{p-1} \Big(\zeta^{m k}\, \eta\big(\tfrac{\tau+m k}{p}\big)\Big)^n = \big(\sqrt{p}\;\eta(p\tau)\big)^n\tag2$$

Note: One can also use the identity $\zeta^N \eta(\tau) = \eta(\tau+N)$ for a further simplification.

It is easily checked with Mathematica that $(1)$ and $(2)$ hold for hundreds of decimal digits, but are they really true? (Kindly also see this related post. I have already emailed W. Hart, and he replied he hasn't seen such identities yet.)


The $n = 26$ for proposed identity $(2)$ was added courtesy of W.Hart's answer below. (I'm face-palming myself for not checking $n = 26$.)

  • $\begingroup$ When you say that these hold to hundreds of decimal digits, are you plugging in some particular $\tau$? Or are you speaking of comparing coefficients in the $q$-expansions somehow? $\endgroup$
    – Ramsey
    Commented Sep 16, 2012 at 21:33
  • $\begingroup$ Yes, I was plugging in various $\tau$. For example, say $\tau = \sqrt{-d}$, it is easy to let Mathematica check d = 1 to 1000. $\endgroup$ Commented Sep 17, 2012 at 15:31

2 Answers 2


This question asks in effect to show that $\eta^n$ is a $\pm p^{n/2}$ eigenfunction for the Hecke operator $T_p$. The claim holds because each of these $\eta^n$ happens to be a CM form of weight $n/2$, and $p$ is inert in the CM field ${\bf Q}(i)$ or ${\bf Q}(\sqrt{-3})$. In plainer language, the sum over $k$ takes the $q$-expandion $$ \eta(\tau)^n = \sum_{m \equiv n/24 \phantom.\bmod 1} a_m q^m $$ and picks out the terms with $p|m$, multiplying each of them by $p$; and the result is predictable because the only $m$ that occur are of the form $(a^2+b^2)/d$ or $(a^2+ab+b^2)/d$ where $d = 24 / \gcd(n,24)$, and the congruence on $p$ implies that $p|m$ if and only if $p|a$ and $p|b$.

For $n=2$ this is immediate from the pentagonal number identity, which states in effect that $\eta(\tau)$ is the sum of $\pm q^{c^2/24}$ over integers $c \equiv 1 \bmod 6$, with the sign depending on $c \bmod 12$ (and $q = e^{2\pi i \tau}$ as usual). Thus $$ \eta^2 = \sum_{c_1^{\phantom0},c_2^{\phantom0}} \pm q^{(c_1^2+c_2^2)/24} = \sum_{a,b} \pm q^{(a^2+b^2)/12} $$ where $c,c' = a \pm b$.

Once $n>2$ there's a new wrinkle: the coefficient of each term $q^{(a^2+b^2)/d}$ or $q^{(a^2+ab+b^2)/d}$ is not just $\pm 1$ but a certain homogeneous polynomial of degree $(n-2)/2$ in $a$ and $b$ (a harmonic polynomial with respect to the quadratic form $a^2+b^2$ or $a^2+ab+b^2$). Explicitly, we may obtain the CM forms $\eta^n$ as follows:

@ For $n=4$, sum $\frac12 (a+2b) q^{(a^2+ab+b^2)/6}$ over all $a,b$ such that $a$ is odd and $a-b \equiv1 \bmod 3$. [This is closely related with the fact that $\eta(6\tau)^4$ is the modular form of level $36$ associated to the CM elliptic curve $Y^2 = X^3 + 1$, which happens to be isomorphic with the modular curve $X_0(36)$.]

@ For $n=6$, sum $(a^2-b^2) q^{(a^2+b^2)/4}$ over all $a \equiv 1 \bmod 4$ and $b \equiv 0 \bmod 2$.

@ For $n=8$, sum $\frac12 (a-b)(2a+b)(a+2b) q^{(a^2+ab+b^2)/3}$ over all $(a,b)$ congruent to $(1,0) \bmod 3$.

@ For $n=10$, sum $ab(a^2-b^2) q^{(a^2+b^2)/12}$ over all $(a,b)$ congruent to $(2,1) \bmod 6$.

@ Finally, for $n=14$, sum $\frac1{120} ab(a+b)(a-b)(a+2b)(2a+b)q^{(a^2+ab+b^2)/12}$ over all $a,b$ such that $a \equiv 1 \bmod 4$ and $a-b \equiv 4 \bmod 12$.

I can't give a reference for these identities, but once such a formula has been surmised it can be verified by showing that the sum over $a,b$ gives a modular form of weight $n/2$ and checking that it agrees with $\eta^n$ to enough terms that it must be the same.

  • $\begingroup$ Thanks, N. Elkies, for the detailed response! Based on W. Hart's answer below, it turns out proposed Identity 2 can be extended to $n = 26$ (which I should have checked!) thus completing the family. $\endgroup$ Commented Sep 17, 2012 at 15:14

(Intended as a comment to Noam Elkies' response.)

My colleague Marco Streng was kind enough to point out that according to "Eta Products and Theta Series Identities", a book of Guenther Koehler MR2766155: "Serre [128] proved that the Fourier series of a modular form f is lacunary if and only if f is of CM-type, i.e., if f is a linear combination of Hecke theta series. In [129] he showed that eta^r for r=2,4,6,8,10,14,26 are the only even powers of eta which are lacunary."

Here, [129] is J.-P.Serre, Sur la lacunarité des puissances de $\eta$, Glasg. Math. J, (27) 1985, 203--221.

Note that this is only for even powers of eta, not for odd powers or other kinds of eta product, of which there are numerous lacunary expressions.

Anyway, this suggests that there ought to be an identity for n = 26, (and possibly others for various eta products?).

  • 2
    $\begingroup$ Turns out proposed Identity 2 can be extended to $n = 26$. $\endgroup$ Commented Sep 17, 2012 at 15:16
  • 4
    $\begingroup$ I just stumbled on a paper, "The 26th power of Dedekind's eta-function" by Chan, Cooper, and Toh. See unimodular.net/archive/26thpowerofeta.pdf $\endgroup$ Commented Sep 18, 2012 at 11:00

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