Let j be the usual j-invariant $j(z) = 1/q + 744 + 196884q + ...$

where $ q = $exp$(2 \pi i z),$ and let $x_n$ denote the exponent of $1 - q^n$ in the product decomposition

$j(z) = (1/q) (1 - q)^{x_1} (1-q^2)^{x_2} .... .$

What, if anything, is known about p-adic continuity of the function $n \mapsto x_n$, especially for p = 2, 3, and 5?

  • 1
    $\begingroup$ What, if anything, have you found in your own numerical experiments? $\endgroup$ – KConrad Jul 2 '11 at 21:06
  • 1
    $\begingroup$ I've computed the exponents $x_n$ for $1 \leq n \leq 10000$. The data is consistent with the hypotheses that the function $n \mapsto x_n$ is p-adically uniformly continuous for p = 2, 3, and 5. I first computed the corresponding exponents in the same range for the normalized Eisenstein series of weights 4 and 6 for the full modular group. Analogous statements are consistent with that data. Just wondering if they're plausible. Details available. $\endgroup$ – Barry Brent Jul 3 '11 at 5:12

The function $x_n$ will not be continuous.

Note that $j$ is a function of $\tau$, and $$\frac{1}{2 \pi i} \frac{d}{d \tau} = q \frac{d}{d q}.$$ If $f$ is a meromorphic modular function, then $dlog(f)$ is a meromorphic modular form of weight two (easy exercise). Applying this with $f = j$, we find that $d log(j)$ is such a function, which one easily computes to be $-E_6/E_4$. Hence $$\frac{-E_6}{E_4} = \frac{j'}{j} = q \frac{d}{dq} \left(- \log(q) + \sum_{n=1}^{\infty} x_n \log(1 - q^n) \right) = - 1 + \sum_{n=1}^{\infty} \frac{n x_n q^n}{1 - q^n}.$$

Expanding the RHS in the usual way, we find that $$\frac{-E_6}{E_4} = - 1 + \sum_{n=1}^{\infty} q^n \sum_{d|n} n x_n.$$

Suppose the form on the left is overconvergent, which it is whenever $i$ is a supersingular $j$-invariant (so in particular for $p=2$, $p = 3$, and $p = 5$). Then one (roughly) expects a decomposion into (generalized) overconvergent eigenforms of weight two, so

$$\frac{E_6}{E_4} = \lambda E^{*}_2 + \sum \lambda_i f_i,$$

where $E^{*}_2$ is the Eisenstein series of weight $2$ and level $\Gamma_0(p)$, and $f_i$ are (non-classical) cuspidal generalized eigenforms (when $p = 2$, this is actually a theorem in this case of David Loeffler). Write $E_6/E_4 = \sum a_n q^n$. If $x_n$ is $p$-adically continuous, then $a_l \equiv a_{l'}$ for $l \equiv l'$ modulo a high power of $p$. This is easily seen to be true for $E^*_2$. Is it also true for $f_i$? What would it mean if $a_l$ was continuous as a function of the primes $l$? Remember that associated to $f_i$ is a Galois representation $\rho_{i}$ such that the trace of Frobenius at $l$ is $a_l(f_i)$. If this was a continuous function for primes $l$, then by Cebotarev density, it would follow that the corresponding Galois representation $\rho_{i}$ would factor (modulo $p^n$) through an abelian extension. In particular, $\rho_{f_i}$ itself would have to be reducible, contradicting known facts. So the chances that $x_n$ are continuous are zero. (With more effort I could produce a rigorous proof of this fact, but it is not worth it.)

Numerical computation will be misleading in this case, for a reason first noted by Serre and Swinnerton-Dyer. Take Ramanujan's function $\Delta = q \prod_{n=1}^{\infty} (1-q^n)^{24}$. Then for primes $l$ it appears that $\tau(l)$ is $2$-adically continuous. This is related to the fact that $\rho_{\Delta}$ is essentially abelian modulo quite a large power of $2$, something like $2^{11}$. But it cannot be so in chararacteristic zero because $\rho_{\Delta}$ is irreducible, even though showing this by naive computation is surprisingly hard. The forms $f_i$ will have a similar property, which is why your computations falsely suggest that the $x_n$ are continuous.

  • $\begingroup$ Yes very nice. My instinct was to decompose into eigenforms -- but I hadn't thought through the consequences. I suspected it would show something but I hadn't realised it would show that continuity was unlikely. $\endgroup$ – Kevin Buzzard Jul 4 '11 at 18:05

I do not have Ken Ono's book "The web of modularity: arithmetic of the coefficients of modular forms and $q$-series" around, but it contains the product expansion for the Eisenstein series $$ E_4=1+240\sum_{n=1}^\infty q^n\sum_{d\mid n}d^3 $$ of the form $\prod_{n=1}^\infty(1-q^n)^{a_n}$ (the Borcherds product). The exponents $a_n$ correspond to the expansion of a certain weak modular form. Furthermore, the weight 12 cusp form $\Delta$ is defined as $q\prod_{n=1}^\infty(1-q^n)^{24}$ and, finally, $$ j=\frac{E_4^3}{\Delta}. $$ The data will hopefully explicify your $p$-adic considerations.

Addition. Example 4.8 of Ono's book discusses the product expansion of $E_4(z)$, but already Example 4.7 gives the product for $j(z)$, so I just copy the details.

Let $\theta(z)=\sum_{n\in\mathbb Z}q^{n^2}$ denote the Jacobi theta function and notation $E_k(z)\in1+q\mathbb Q[[q]]$ stand for the Eisenstein series. Define $$ \begin{aligned} f(z) &=\frac{3E_{10}(4z)\delta\theta(z)}{2\Delta(4z)} -\frac{3\theta(z)V_4(\delta E_{10}(z))}{10\Delta(4z)}-\frac{456}5\theta(z) \cr &=\frac3{q^3}-744q+80256q^4-257985q^5+5121792q^8-12288744q^9+\cdots \cr &=\frac3{q^3}+\sum_{n=1}^\infty A(n)q^n, \end{aligned} $$ where $$ \delta:\sum_{n=0}^\infty a(n)q^n\mapsto\sum_{n=0}^\infty na(n)q^n \quad\text{and}\quad V_4:\sum_{n=0}^\infty a(n)q^n\mapsto\sum_{n=0}^\infty a(n)q^{4n}. $$ Then $f(z)$ is a weight 1/2 meromorphic modular form and $$ j(z)=\frac1q\prod_{n=1}^\infty(1-q^n)^{A(n^2)}. $$

  • 2
    $\begingroup$ For what it's worth, the function $f(z)$ above is a meromorphic $p$-adic weight $1/2$ modular form (for all $p$), whose only pole is at infinity. To prove $p$-adic continuity statements one might want to try writing $f$ as a sum of eigenforms -- but not every $p$-adic form is a sum of eigenforms. For example David Loeffler proved that the $j$-invariant was a sum of 2-adic eigenforms but his proof had some computational (in the sense that it won't work for all $p$) aspects in it. $\endgroup$ – Kevin Buzzard Jul 3 '11 at 11:25
  • $\begingroup$ Kevin, thanks for this $p$-adic addition. I really wonder about how much can be actually done $p$-adically for this particular modular form, as there are more conjectures than theorems in the area... And you are one of the very few who know the business. $\endgroup$ – Wadim Zudilin Jul 3 '11 at 11:35
  • $\begingroup$ Wadim -- I don't know how much one can do. In some sense I'm not a good person to ask -- I have limited experience in weight 1/2 forms. I know that in weight 1/2 the Hecke operators do involve $A(n^2)$ rather than $A(n)$ because, basically, $T_\ell$ doesn't work so well and you have to use $T_{\ell^2}$. But if you want to use Hecke operators to compare the $A(n^2)$s then somehow I am wondering whether one really needs to be able to reduce the situation to one involving eigenforms. Perhaps Nick Ramsey can say something coherent when he next comes around. $\endgroup$ – Kevin Buzzard Jul 3 '11 at 19:39

My question should have been more precise. I'm trying to decide how hard to push my experiments. So, is $p$-adic continuity ($p = 2, 3$, or $5$) of $n \mapsto x_n$, or the lack of it, already a theorem in the literature? I judge from the responses, probably not. Aside from my data, which picks out those primes, one reason I ask is that in his Bull. London Math. Soc., 9 (1977) paper, Koblitz also relates the primes $p = 2, 3, 5$ to $j$, as follows: $p$-adic modular functions for these values of $p$ have "natural expansions" in negative powers of $j$ times a certain differential because there is "one supersingular value $\beta \equiv 0$ (mod $p$)." For larger $p$, the natural expansion Koblitz specifies involves other $\beta$ as well. Just a guess, but this seems more likely to connect to my observation than Borcherds' result, simply because it describes such a relationship.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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