This is pretty much straightforward curiosity. Is there anything known about Newton polygons of classical modular polynomials (polynomial relations between $j(\tau)$ and $j(n\tau)$)? I understand that modular polynomial is not really a practical way of approaching modular curves, still it would be interesting to see if properties of the Newton polygon had some relevance to the geometry and arithmetic of the curve.



Newton polygons of modular polynomials
The polynomial $\Phi\_n(x,y)$ has a degree of $n \prod\_{p|n } \frac{p+1}{p}$ in each variable, so the Newton polygon is contained in a square of that side length. Call this $n'$. For each fixed value of $j\_1$, there are $n'$ possible values of $j\_2$ with multiplicity, so plugging in any value of $x$ or $y$ gives a degree $n'$ polynomial. This means $(n',0)$ and $(0,n')$ are in the Newton polygon.


If there is no cyclic isogeny of degree $n$ from a curve of $j$ invariant $0$ to itself, then $(0,0)$ is in the polygon, which determines the bottom half. If there is a unique such isogeny that is a point of multiplicity $1$, so $(1,0)$ and $(0,1)$ are in the polygon. If there are multiple isogenies, I am not sure exactly how to describe the bottom half. I think the different isogenies always form distinct branches of the modular curve so if $k$ is the number of isomorphism classes of isogenies then $(k,0)$ and $(0,k)$ are the only vertices in the bottom half.


Edit: At each point $x$ of the abstract modular curve $X\_0(N)$ such that $j\_1(x)=0$, $j\_1$ has a zero of order $1$ or $3$. Similarly if $j\_2=0$ then $j\_2$ has a zero of order $1$ or $3$. We will show that these two orders are the same. The order is $1$ if and only if the point is an elliptic point - that is, there is an automorphism of order three of the curve that preserves the cyclic subgroup. Modding out by this subgroup, we get an automorphism of order three of the isogenous curve that fixes the dual subgroup. So $j\_1$ has degree $1$ if and only if $j\_2$ does. What this means is that locally the equation relating $j\_1$ and $j\_2$ looks like $aj\_1+bj\_2+$ higher-order terms with $a\neq 0$, $b\neq 0$. Multiplying together one of these equations for each point with $j\_1=j\_2=0$, we a polynomial of degree $k$ with the coefficients of $x^k$ and $y^k$ nonzero as the lowest-order term, so the Newton polygon is as I described.


The divisor corresponding to this side of the polygon should just be the sum of all the points with $CM$ by $Z[\omega]/(\omega^2+\omega+1)$ where the isogenous curve also has CM by the same ring, with multiplicity $1$ for the elliptic points and $3$ for the CM points. I'm not sure exactly how this relates to the Heegner point construction.


Understanding the top half is basically understanding the behavior as the $j$ invariants of both isogenous curves aprroach $\infty$. How this works out is that one $j$ invariant looks like a power of the other: $j\_1 = j\_2^{a/b}$, where $ab=n$. This means that the slopes of the top half of the Newton polygon are exactly these rational numbers $a/b$. To make it add up I'm guessing the part of length $a$ travels a distance $(a,b) \* \phi ( gcd(a,b))/gcd(a,b)$. I believe there is an explicit combinatorial way of computing this with the modular group. I figured something like this out once. There's probably also a nice reference somewhere.


Basically the idea is that one can view cusps as double cosets $\Gamma\_0(n) \backslash \Gamma / \langle T \rangle$. Then the order of the pole of $j$ on that cusp is the size of the double coset (viewed as a $T$-orbit in $\Gamma/\Gamma\_0(n)$). The isogeny acts by viewing $\Gamma/\Gamma\_0(n)$ as cyclic subgroups of order $n$ of $(\mathbb Z/n )^2$ and sending them to a dual subgroup. You can use this to count the orbits of each degree that are sent to each other degree, which gives the top of the Newton polygon.


The divisors on the top are sums of cusps - basically you're summing all the cusps of a certain type. The divisors on the sides are summing all the points with $j$ invariant $0$, so they're basically the pullback of $O(1)$ from the classical modular curve. The divisors on the bottom corner near $0$ are sums of CM points, so they may be related to Heegner points.


