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When $N > 1$, the modular forms $\Delta(z)$ and $\Delta(Nz)$ are algebraically independent over the complexes, and the same then is true of their expansions at infinity. But using the fact that the mod $\ell$ reduction of $E_{\ell-1}$ is 1 for $\ell > 3$ one finds that, on passing to any finite characteristic $\ell$, one gets a polynomial relation between these two power series.

The situation is nicest when $\ell$ is 2, 3, 5, 7 or 13. In these cases, $(\mathbb{Z}/\ell)(\Delta)=(\mathbb{Z}/\ell)(j)$, where $j$ is the $j$-invariant (and the abuse of language is obvious). Indeed in these 5 cases, $\Delta$ is respectively $1/j$, $1/j$, $1/j$, $1/(j+1)$ and $1/(j-5)$. Suppose now we fix such an $\ell$; let $N$ be a prime other than $\ell$. Using the modular equation of level $N$ for $j$, we then get an irreducible symmetric $H_N$ in $(\mathbb{Z}/\ell)[x,y]$ of degree at most $N+1$ in $x$, such that $H_N\big(\Delta(z), \Delta(Nz)\big)=0$, (where I continue my abuse).

In an earlier question I asked what could be said about this $H_N$ when $\ell=2$. It turns out that $H_N$ has total degree $N+1$, and highest degree part equal to $(x+y)^{N+1}$.

QUESTION--To what extent does the above result generalize when $\ell$ is 3, 5, 7 or 13?

In all cases I've looked at, the total degree is $N+1$. When $\ell=3$ the highest degree part appears to be $(x-y)^{N+1}$ or $(x+y)^{N+1}$ according as $N$ is 1 or 2 mod 3; I suspect this isn't hard to prove. Here are some results when $\ell$ is 5, 7 or 13.

$\ell= 5$: $H_2=(x+y)^3-xy$ $\qquad$ $H_3=(x+y)^4+2xy(x+y)-xy$ $\qquad$ $H_7=(x+y)^8$ + lower degree terms.

$\ell= 7$: $H_2=(x+y)(x-y)^2-xy$ $\qquad$ $H_3=(x^2+y^2)^2+2xy(x+y)-xy$ $\qquad$ $H_5=(x^2+y^2)^3$ + lower degree terms.

$\ell=13$: $H_2=(x+y)(x^2+8xy+y^2)-xy$ $\qquad$ $H_3=(x^2+5xy+y^2)(x^2+6xy+y^2)+7xy(x+y)-xy$

EDIT: I thank Anna M. not only for her edit, but for her comment which points the way to the solution of the first part of my question. For although the coefficients of her degree $p+1$ polynomial in $X$ do not lie in $\mathbb{Z}[\Delta]$, they do lie in $\mathbb{Z}[\Delta, E_4^3, E_6^2]$, and in the characteristics I'm looking at this is good enough. For in these characteristics $E_4^3$ and $E_6^2$ are $\mathbb{Z}/\ell$-linear combinations of 1 and $\Delta$, and since the coefficient of $X^{p+1-r}$ in her polynomial is a polynomial of total degree $r$ in $\Delta$, $E_4^3$ and $E_6^2$, its reduction is a polynomial of degree at most $r$ in the reduction of $\Delta$.

The result for $E_4^3$ and $E_6^2$ is clear when $\ell$ is 2, 3, or 5 when $E_4$ is 1. It's also clear when $\ell=7$ in which case $E_6$ is 1. To handle $\ell=13$, use the identities $$E_4^3-E_6^2 = 1728\Delta$$ and $$441\, E_4^3+250\, E_6^2=691\, E_{12}.$$ Since the Bernoulli number $B_{12}$ is $\frac{-691}{2730}$,

$691\, E_{12}= 691+(24)(2730)$ (an element of $\mathbb{Z}[[x]]$),

and has mod 13 reduction equal to 691. Reducing the two identities and solving one finds that $E_4^3$ and $E_6^2$ are equal to $1+5\Delta$ and $1+6 \Delta$ in characteristic 13.

EDIT (1/10/14)

  1. Here are remarks that may help those following the link in this question to an earlier question of mine (in characteristic 2). Mod 2, $\Delta$ is $x+x^9+x^{25}+x^{49}+\cdots$; this ties the questions together. I've changed notation completely, and what is $N$ here was called $\ell$ in the last question. My proof in the last edit here that there's a polynomial relation of total degree $N+1$ between $\Delta(z)$ and $\Delta(Nz)$ is much simpler than the proof sketched in the earlier question. Finally I did NOT prove earlier that the highest degree part of this relation is $(x+y)^{N+1}$. But a proof of this can be teased out of other questions and answers of mine--see below.

  2. When $\ell$ is 2, 3, 5, 7 or 13, there's just one supersingular $j$-value; it's respectively 0, 0, 0, -1 and 5. Now there's a branched covering $X_0(N) \to X_0(1)$ of moduli spaces in characteristic $\ell$, of degree $N+1$. $X_0 (1)$ is the projective $j$-line and $\Delta(Nz)/\Delta(z)$ is a rational function on $X_0 (N)$. Finding the degree- $(N+1)$ part of $H_N$ amounts to determining the values of this rational function at the $N+1$ points (with multiplicity) lying over the supersingular $j$-value (since $\Delta$ is respectively $1/j$, $1/j$, $1/j$, $1/(j+1)$, and $1/(j-5)$). This looks like the right approach to determining this highest degree part, and I hope someone can follow through on it.

  3. Suppose $\ell=2$. Then there is a branched Galois covering $X(N) \to X(1)=X_0 (1)$ with Galois group $PSL_2(\mathbb{Z}/N)$. This map is ramified over $j=0$ with ramification degree 12, and one wants to show that $j(z)/j(Nz)$ is 1 at each of the $(N^3-N)/24$ points lying over $j=0$. Perhaps this is easy--at any rate my theory of characteristic 2 theta series gives a cumbrous proof. I construct an affine curve, the generators of whose co-ordinate ring are theta series, and identify the function field of this curve with the function field of Igusa's $X(N)$. I find elements $F$, $G$ and $H$ of the co-ordinate ring identifying with the reciprocals of $j(z/N)$, $j(z)$ and $j(Nz)$. The curve has $(N^3-N)/24$ points at infinity which identify with the points lying over $j=0$. Finally I show that at each point at infinity, $G$ and $H$ have poles of order 12 while $G+H$ has a pole of lower order. It follows from this that $H/G$ takes the value 1 at each of these points, giving the desired result.

EDIT (2/11/14)--The remaining part of my question is this. Suppose that ell is 3,5,7 or 13, that N is a prime other than ell, and that H* is the highest degree part (i.e the degree N+1 homogeneous part) of H_N. What does H* look like? I now have a conjectural answer to this, which is precise when ell is 3,5 or 7, and supported by computer evidence. Let (N/ell) be the usual Legendre symbol.

CONJECTURE

1.---When ell is 3 or 5, H* is (x-y)^(N+1) or (x+y)^(N+1) according as (N/ell) is 1 or -1.

2.---When ell is 7 and (N/7) is -1, H* is (xx+yy)^((N+1)/2).

3.---When ell is 7 and (N/7) is 1, N>2, write N as rr+7ss with r =1,2 or 4 mod 7. Then H* is ((x-y)^2a)((x+y)^2b) where a+b=(N+1)/2 and a-b=r.

4.---When ell is 13, H* is a product of powers of linear factors x-cy with each c^7 equal to (N/13). Another way of saying this, when N>2, is the following. Let g(s) in Z/13[x,y] be xx+sxy+yy. Then when (N/13) is 1, H* is a product of powers of g(3), g(5), g(6) and g(-2), while when (N/13) is -1, H* is a product of powers of g(-3), g(-5), g(-6) and g(2). But it's unclear to me what the exponents, a,b,c and d, in these conjectural product decompositions should be.

Remark---I can show that 3a+5b+6c-2d has the same mod 13 reduction as does -N(N/13)tau(N). This and 3. above suggest to me that a,b,c and d may be related to the coefficient of x^N in certain weight 2 modular forms (perhaps of level 169?).

EDIT(1/12/15)

In my edit of 1 year ago I indicate a "cumbrous" proof that when l=2, and N is an odd prime, then H_N is (x+y)^(N+1), and I ask whether a simpler proof might be well-known. I think now that the answer to this last question is "no", for the following reason. Let F and G be the mod 2 reductions of the expansions at infinity of delta(z) and delta(Nz). Then Z/2(F,G) is a degree N+1 extension of Z/2(G) and one can show that T_N (F^k) is the image of Tr(F^k) under the map taking f(x) to f(x^(1/N)) where Tr is the trace in the field extension. Now let P_k be Tr(F^k) and S_k be the kth elementary symmetric function in the conjugates of F. What we know about H_N tells us that S_k is a polynomial in G of degree at most k.

Now if H_N is (x+y)^(N+1), or more generally a square, then when k is odd, S_k would have degree that is strictly less than k. Newton's formula relating power sums and elementary symmetric functions would then give an immediate proof that Tr(F^k) has degree less than k in G, so that T_N (F^k) has degree less than k in F, or in other words that the Hecke algebra acts locally nilpotently on the space of modular forms in characteristic 2 and level 1. This is a famous result whose known proof uses the connection between modular forms and Galois representations as well as a theorem of Tate--if my result about H_N were well-known, I doubt that this simple proof of local nilpotence would have been missed.

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  • $\begingroup$ For $\ell = 3$, we have $H_2 = x^3 - xy + y^3 = (x + y)^3 - xy$ and $H_7 = x^8 + x^7 y + x^6 y^2 + x^5 y^3 + x^4(y^4 - y) + x^3(y^2 + y^5) + x^2 (y^3 + y^6) + x (y^7 - y^4 - y) + y^8$, which is $(x - y)^8 - xy(x^3 - x^2y - xy^2 + y^3) - xy$. $\endgroup$
    – Anna M.
    Jan 8, 2014 at 23:51
  • $\begingroup$ For $N$ prime and $\ell = 2, 3, 5$, since $\Delta = 1/j$, the highest-degree homogeneous part of the modular equation for $\Delta$ is the same as that for $j$. Since the latter has degree $N + 1$, I guess you know the same for the modular equation for $\Delta$. $\endgroup$
    – Anna M.
    Jan 8, 2014 at 23:58
  • $\begingroup$ And I think more generally you know that for $N=p$ prime (and any $\ell$; I guess it works in characteristic 0), the degree of this modular equation for $\Delta$ is always $p + 1$: you write it down as a product $\big(X - \Delta(pz)\big) \prod_i \bigg(X - \Delta\left( \frac{z + i}{p}\right)\bigg)$, just like you do for $j$. $\endgroup$
    – Anna M.
    Jan 9, 2014 at 0:00
  • $\begingroup$ @Anna--It's trickier than you think. What you say in your last comment is wrong; delta(Nz) is transcendental over Q(delta(z)) and the coefficients in your product are not polynomials in delta(z). When ell=11 for example the irreducible relation between delta(z) and delta (2z) has degree 150 and the highest degree part is (xy)^75, But when Z/ell(delta) =Z/ell(j), as happens when ell is 2,3,5,7 or 13, then what you say is right. As for your earlier comment, the level p modular equation for j has total degree 2p and highest degree part (xy)^p and (to be continued)... $\endgroup$ Jan 9, 2014 at 1:07
  • $\begingroup$ ... when delta=1/j, the highest degree part of the modular equation for delta corresponds to the lowest degree part of the modular equation for j. So one needs to determine various low order terms in this last modular equation, which is a delicate business. $\endgroup$ Jan 9, 2014 at 1:13

1 Answer 1

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Recall that H* is the highest degree part (i.e. the degree N+1 homogeneous part) of the "level N modular equation, H, for delta" , assuming the characteristic ell is 3,5,7 or 13, and that N is an odd prime other than ell. I am now certain as to how H* factors. In this answer I'll state what I'm certain of, together with supporting computer evidence provided by Andrew Sutherland. (My claims are easily translated into claims about the classical modular polynomial mod ell, and Sutherland is expert at computing these polynomials). But as I can't prove anything, this answer is highly incomplete.

In the last edit to my question I made precise conjectures as to how H* factors when ell is 3,5 or 7. Sutherland has verified these for N<1000.

Here is my conjecture when ell is 13. In the last edit to my question I suggested that H* should be a product of powers of quadratic polynomials g(3), g(5), g(6) and g(-2) when (N/13) is 1, and a product of powers of g(-3), g(-5), g(-6) and g(2) when (N/13)=-1. I'm now sure of what the exponents a,b,c and d appearing in these products are. Namely, the space of weight 2 cusp forms for Gamma_0 (169) contains unique elements A,B and C whose q-expansions are:

A: q+3*q^2+4*q^3+0*q^4+6*q^5-2*q^6+q^7-6*q^8-8*q^9+...

B: q-4*q^2-3*q^3-7*q^4-q^5+5*q^6+8*q^7+q^8+6*q^9+...

C: q+3*q^2-3*q^3+7*q^4-q^5+5*q^6+q^7+8*q^8-q^9+...

I believe that 7a (resp. 7b, 7c) is (N+1)- the coefficient of q^N in A (resp. B,C). And of course a+b+c+d=(N+1)/2. For example when N=499, my conjecture turns out to be that H* is the product of (xx-3xy+yy)^66, (xx-5xy+yy)^72, (xx-6xy+yy)^81 and (xx+2xy+yy)^31.

Andrew Sutherland has sent me a printout of H* for N<1000. Comparing the printout with my conjectured factorization is a nuisance, but I've checked that my conjecture holds for N<100, and for N=499 as above.

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  • $\begingroup$ The edit to my answer was entirely unnecessary, (and I'll get rid of it); the original argument given in the edit to the question was fine. The point that I should have gathered from Anna M. is that the kth elementary symmetric function in the delta((z+i)/p) and (p^12)*delta(pz) is a weight 12k modular form of level 1. And then of course, for our particular ell, the reduction of its expansion is a polynomial of degree at most k in F. $\endgroup$ Jun 3, 2014 at 0:29

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