Weight Two

Theorem 4.1   The only prime $p<60000$ such that $d_2(\Gamma_0(p))>0$ is $p=389$. (Except possibly $50923$ and $51437$, which I haven't finished checking yet.)

Proof. This is the result of a large computer computation, and perhaps couldn't be verified any other way, since I know of no general theorems about $d_2(\Gamma_0(p))$. The rest of this proof describes how I did the computation, so you can be convinced that there is valid mathematics behind my computation, and that you could verify the computation given sufficient time. The computation described below took about one week using $12$ Athlon 2000MP processors. In 1999 I had checked the result stated above but only for $p<14000$ using a completely different implementation of the algorithm and a 200Mhz Pentium computer. These computations are nontrivial; we compute spaces of modular symbols, supersingular points, and Hecke operators on spaces of dimensions up to $5000$.

The aim is to determine whether or not $p$ divides the discriminant of the Hecke algegra of level $p$ for each $p<60000$. If $T$ is an operator with integral characteristic polynomial, we write $\disc(T)$ for $\disc(\charpoly(T))$, which also equals $\disc(\mathbb{Z}[T])$. We will often use that

$$\disc(T)\!\!\!\!\mod{p} = \disc(\charpoly(T)\!\!\!\!\mod p).$$

Most levels $p<60000$ were ruled out by computing characteristic polynomials of Hecke operators using an algorithm that David Kohel and I implemented in MAGMA, which is based on the Mestre-Oesterle method of graphs (our implementation is ``The Modular of Supersingular Points'' package that comes with MAGMA). I computed $\disc(T_q)$ modulo $p$ for several primes $q$, and in most cases found a $q$ such that this discriminant is nonzero. The following table summarizes how often we used each prime $q$ (note that there are $6057$ primes up to $60000$):

$q$	number of $p<60000$ where $q$ smallest s.t. $\disc(T_q)\neq 0$ mod $p$
2	5809 times
3	161 (largest: 59471)
5	43 (largest: 57793)
7	15 (largest: 58699)
11	15 (the smallest is 307; the largest 50971)
13	2 (they are 577 and 5417)
17	3 (they are 17209, 24533, and 47387)
19	1 (it is 15661 )

The numbers in the right column sum to 6049, so 8 levels are missing. These are

$$389,487,2341,7057,15641,28279, 50923,$$    and $$51437.$$

(The last two are still being processed. $51437$ has the property that $\disc(T_q)=0$ for $q=2,3,\ldots,17$.) We determined the situation with the remaining 6 levels using Hecke operators $T_n$ with $n$ composite.

$p$	How we rule level $p$ out, if possible
389	$p$ does divide discriminant
487	using charpoly($T_{12}$)
2341	using charpoly($T_6$)
7057	using charpoly($T_{18}$)
15641	using charpoly($T_6$)
28279	using charpoly($T_{34}$)

Computing $T_n$ with $n$ composite is very time consuming when $p$ is large, so it is important to choose the right $T_n$ quickly. For $p=28279$, here is the trick I used to quickly find an $n$ such that $\disc(T_n)$ is not divisible by $p$. This trick might be used to speed up the computation for some other levels. The key idea is to efficiently discover which $T_n$ to compute. Though computing $T_n$ on the full space of modular symbols is quite hard, it turns out that there is an algorithm that quickly computes $T_n$ on subspaces of modular symbols with small dimension (see §3.5.2 of my Ph.D. thesis). Let $M$ be the space of mod $p$ modular symbols of level $p=28279$, and let $f=\gcd(\charpoly(T_2),\deriv(\charpoly(T_2)))$. Let $V$ be the kernel of $f(T_2)$ (this takes 7 minutes to compute). If $V=0$, we would be done, since then $\disc(T_2)\neq 0\in\mathbb{F}_p$. In fact, $V$ has dimension $7$. We find the first few integers $n$ so that the charpoly of $T_n$ on $V_1$ has distinct roots, and they are $n=34$, $47$, $53$, and $89$. I then computed $\charpoly(T_{34})$ directly on the whole space and found that it has distinct roots modulo $p$. $\qedsymbol$