A 14th and 26th-power Dedekind eta function identity? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-18T22:53:05Z http://mathoverflow.net/feeds/question/107327 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/107327/a-14th-and-26th-power-dedekind-eta-function-identity A 14th and 26th-power Dedekind eta function identity? Tito Piezas III 2012-09-16T16:10:47Z 2012-09-17T15:09:20Z <p>Given the <em>Dedekind eta function</em> $\eta(\tau)$. Let <em>p</em> be a prime and define $m = (p-1)/2$.</p> <ol> <li><p>Let <em>p</em> be a prime of form $p = 12v+5$. Then for $n = 2,4,8,14$: $$\sum_{k=0}^{p-1} \Big(e^{\pi i m k/12} \eta\big(\tfrac{\tau+m k}{p}\big)\Big)^n = -\big(\sqrt{p}\;\eta(p\tau)\big)^n$$</p></li> <li><p>Let <em>p</em> be a prime of form $p = 12v+11$. Then for $n = 2,6,10,14, and\; 26$: $$\sum_{k=0}^{p-1} \Big(e^{\pi i m k/12} \eta\big(\tfrac{\tau+m k}{p}\big)\Big)^n = \big(\sqrt{p}\;\eta(p\tau)\big)^n$$</p></li> </ol> <p>It is easily checked with <em>Mathematica</em> that these hold for hundreds of decimal digits, but are they really true? (Kindly also see this related <a href="http://mathoverflow.net/questions/106925/" rel="nofollow">post</a>. I have already emailed W. Hart, and he replied he hasn't seen such identities yet.) </p> <p><strong>----EDIT----</strong></p> <p>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.) </p> http://mathoverflow.net/questions/107327/a-14th-and-26th-power-dedekind-eta-function-identity/107356#107356 Answer by Noam D. Elkies for A 14th and 26th-power Dedekind eta function identity? Noam D. Elkies 2012-09-17T04:11:58Z 2012-09-17T14:42:33Z <p>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 $p|b$.</p> <p>For $n=2$ this is immediate from the <a href="http://en.wikipedia.org/wiki/Pentagonal_number_theorem" rel="nofollow">pentagonal number identity</a>, 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 $c,c' = a \pm b$.</p> <p>Once $n>2$ there's a new wrinkle: the coefficient of each term <code>$q^{(a^2+b^2)/d}$</code> or <code>$q^{(a^2+ab+b^2)/d}$</code> 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:</p> <p>@ 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 $X_0(36)$.]</p> <p>@ For $n=6$, sum $(a^2-b^2) q^{(a^2+b^2)/4}$ over all $a \equiv 1 \bmod 4$ and $b \equiv 0 \bmod 2$.</p> <p>@ For $n=8$, sum $\frac12 (a-b)(2a+b)(a+2b) q^{(a^2+ab+b^2)/3}$ over all $(a,b)$ congruent to $(1,0) \bmod 3$.</p> <p>@ For $n=10$, sum $ab(a^2-b^2) q^{(a^2+b^2)/12}$ over all $(a,b)$ congruent to $(2,1) \bmod 6$.</p> <p>@ 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 $a-b \equiv 4 \bmod 12$.</p> <p>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.</p> http://mathoverflow.net/questions/107327/a-14th-and-26th-power-dedekind-eta-function-identity/107381#107381 Answer by William Hart for A 14th and 26th-power Dedekind eta function identity? William Hart 2012-09-17T13:47:15Z 2012-09-17T14:42:11Z <p>(Intended as a comment to Noam Elkies' response.)</p> <p>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."</p> <p>Here, [129] is J.-P.Serre, Sur la lacunarité des puissances de $\eta$, Glasg. Math. J, (27) 1985, 203--221.</p> <p>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. </p> <p>Anyway, this suggests that there ought to be an identity for n = 26, (and possibly others for various eta products?).</p>