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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?

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    $\begingroup$ What, if anything, have you found in your own numerical experiments? $\endgroup$ – KConrad Jul 2 '11 at 21:06
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    $\begingroup$ 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
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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.

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  • $\begingroup$ 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
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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 $$ j(z)=\frac1q\prod_{n=1}^\infty(1-q^n)^{A(n^2)}. $$

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    $\begingroup$ 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
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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.

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