144169

The space of classical modular cuspforms of level one and weight 24 has dimension two — the smallest weight for which the dimension is not zero or one. What can we say about the Hecke algebra acting on this space without computing it?

Formally, the Hecke algebra $\mathbf{T}$ is a rank two $\mathbf{Z}$-algebra, which is either an order in the ring of integers of a real quadratic field, or a subring of $\mathbf{Z} \oplus \mathbf{Z}.$ Let’s investigate the completion of this algebra at various primes $p.$

Let’s first consider the prime $p =23.$ The curve $X_0(23)$ has genus two, and the corresponding Hecke algebra in weight two is $\mathbf{Z}[\phi],$ where $\phi$ is the Golden Ratio. The prime $p =23$ does not split in this field, and hence modulo $p$ there is a pair of conjugate eigenforms with coefficients in $\mathbf{F}_{p^2}.$ Multiplying by the Hasse invariant, we see that this eigenform also occurs at level one and weight 24 over $\mathbf{F}_{p}.$ It follows that:

$\mathbf{T} \otimes \mathbf{Z}_{23} = W(\mathbf{F}_{23^2}).$

In particular, $\mathbf{T} = \mathbf{Q}(\sqrt{D})$ for some square-free integer $D > 0.$

Now let us consider primes $p < 23.$ Any Galois representation modulo such a prime will occur — possibly up to twist — in lower weight. Yet all the spaces in lower weight have dimension at most one, and hence it follows that the residue fields of $\mathbf{T}$ are all of the form $\mathbf{F}_p.$ Suppose further that $5 \le p < 23.$ Then, using theta operators, we may find two distinct eigenforms in weight 24, from which it follows that $\mathbf{T}$ has two distinct residue fields of characterstic $p,$ and so, for $5 \le p < 23,$ we have:

$\mathbf{T} \otimes \mathbf{Z}_p = \mathbf{Z}_p \oplus \mathbf{Z}_p.$

One expects at level one that $a_2(f)$ always generates the Hecke field. This is still a conjecture, but we may deduce this unconditionally in weight 24 because the dimension of the cuspforms is two, and so this follows automatically from the Sturm bound! Hence we may write:

$\mathbf{T} = \mathbf{Z}[a_2(f)], \quad a_2(f) = \displaystyle{\frac{a + b \sqrt{D}}{2} \in \mathbf{Z} \left[ \frac{1+\sqrt{D}}{2} \right]}$

where $b \ne 0.$ Even better, using Hatada's Theorem — giving congruences for $a_2$ and $a_3$ for eigenforms of level one modulo $8$ and $3$ respectively — we may write

$a_2(f) = 12(a + b \sqrt{D}), \quad a,b \in \mathbf{Z}$

where $b \ne 0.$ This gives an upper bound on $D$ in light of the Deligne bound $|a_2| \le 2 \cdot 2^{23/2}.$ More precisely, we obtain the bound $b^2 D < 2^{27}/24^2,$ and hence that $D < 233017.$

Let's now think more carefully about $p = 2$ and $3.$ For these primes, there will be a unique Coleman family of slope $v(-24) = 3$ for $p =2$ and $v(252) = 2$ for $p = 3.$ I can't quite see a pure thought way of proving this, but at least this would be a consequence of the strong form of the GM-conjecture as predicted by Buzzard. So we should expect that, in these cases

$\mathbf{T} \otimes \mathbf{Z}_p \hookrightarrow \mathbf{Z}_p \oplus \mathbf{Z}_p.$

In addition to congruences for small primes, there will also be congruences between the unique cusp form with an Eisenstein series modulo the numerator of $B_{24},$ which is

$\displaystyle{B_{24} = \frac{-1}{2 \cdot 3 \cdot 5 \cdot 7 \cdot 13} \times 103 \times 2294797.}$

I claim that these primes will also have to split in $\mathbf{T}.$ For example, it is impossible for $b$ to be divisible by $2294797,$ because that would violate the inequality on $b^2 D$ above, and hence it follows that $p = 2294797$ must also split in $\mathbf{T} \otimes \mathbf{Q}.$ The same argument works for $p = 103$ having ruled out some very small $D.$ To summarize, we have the following:

The primes $5 \le p < 23,$ $p = 103, 2294797$ split in $K = \mathbf{Q} (\sqrt{D}),$ but $p = 23$ does not split, and $D < 233017.$ Moreover, we expect that $2$ and $3$ also split.

This is enough to determine $D$ completely up to 72 possibilities, and 9 with the unproven assumption at $2$ and $3.$ On the other hand, all of these $D$ are quite large (the smallest are $3251$ and $15791$ respectively), which forces $b$ to be very small. But we also have the congruence

$12(a + b \sqrt{D}) \equiv 1 + 2^{23} \mod 2294797.$

For the remaining $D,$ we can determine, with $|b|$ satisfying the required inequality, whether there exists such a congruence with $|a| \le 2^{27/2}/24 \sim 483.$ A simple check shows that is a unique solution (with the assumption on two or three or not), and hence, by (something close to) pure thought, we have shown that $D = 144169,$ and moreover (using Deligne's bound again) that

$a_2(f) = 12(45 \pm \sqrt{144169}), \qquad \mathbf{T} = \mathbf{Z}[12 \sqrt{144169}].$

One can indeed check this is the case directly, if you like. Curiously enough, this Hecke eigenvalue is quite close to the Deligne bound — the probability it is (in absolute value) this big is, assuming a Sato-Tate distribution, slightly under 5%.

Extra Credit Problem: Hack Ken Ribet’s Yelp password by using the fact that 144169 is his favorite prime number.

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4 Responses to 144169

1. JB says:

Regarding lowest slope families for $p=2,3$… The result for $p=2$ is proven in M. Emerton’s thesis, no (presuming that by Coleman families you mean things passing through integer weights only)? For $p=3$ I can give a short-ish argument (which has an obvious analog for $p=2$). I’m not sure if it qualifies as pure thought or not, but it does not quite use the Buzzard slope hammer.

Let $a_1(w)$ be the function on the disc $v_3(w) > 0$ that is recording $-\tr(U_3)$ where the trace is the trace of $U_3$ acting on overconvergent $3$-adic cuspforms, i.e. $a_1(w)$ is the linear coefficient of the characteristic power series of $U_3$ at weight $w$ (if $k$ is an integer then the corresponding point in the disc is $w(k) := 4^k – 1$; it lies in the region $v_3(w) \geq 1$).

The function $a_1(w)$ is not just analytic in $w$ but given by a power series in $w$ over $\Z_3$. In particular, it has finitely many roots. Using Koike’s $3$-adic computation in the 70s of $\tr(U_3)$ one can check that $3 \ndvd a_1(w)$ and that $a_1(w)$ has exactly two roots. Mucking around a little bit will show that the roots are highly congruent to $w(10)$ and $w(14)$ (side remark: 10 and 14 are the only two weights which have no level one forms and at least two $3$-new forms at level $\Gamma_0(3)$). My computer tells me that one root of $a_1(w)$ is congruent to $w(10)$ modulo $3^8$ and the other is congruent to $w(14)$ modulo $3^5$. In any case, if $v_3(w) > 1$ (e.g. if $w = w(k)$ with $k \congruent 0 \bmod 3$) then $v_3(a_1(w)) = 1+1 = 2$. This looks good to explain the existence of lowest slope 2 family over the whole open disc $v_3(w) > 1$. I don’t need this but it is good to note that if $k \not\congruent 0 \bmod 3$ then $k$ is either congruent to $10$ or $14$ modulo $3$ and so $v_3(a_1(w(k))) > 2$.

To continue look at the whole characteristic power series of $U_3$ acting on tame level one overconvergent $3$-adic cuspforms. This is a two variable power series power series $P(w,t) = 1 + \sum a_i(w)t^i$ and each $a_i(w)$ is a power series over $\Z_3$. A theorem of Roe (generalizing the famous result of Buzzard and Kilford for $p=2$) tells you the following information: $3 \ndvd a_i(w)$ and $a_i(w)$ has exactly $2{i+1\choose 2}$ roots on $v_3(w) > 0$ and every single one of those roots lie in the disc $v_3(a_i(w)) \geq 1$ (this is a concrete interpretation of the behavior of the tame level one $3$-adic eigencurve over the boundary of weight space).

To get the slopes in weight $w$ you can just compute the Newton polygon of $P(w,t)$. But now it follows that for any $w$ with $v_3(w)\geq 1$ the lowest slope in weight $w$ is at least $2{i+1\choose 2}/i = i+1 \geq 2$. This shows that there are no families of slope strictly lower than $2$ over the closed disc $v_3(w)\geq 1$. Taking a $w$ with $v_3(w) > 1$ we just saw that $v_3(a_1(w)) = 2$. Since this is lower than the lower bound the first sentence provide for $i>1$, such $w$ must have a slope $2$ form. But re-using the argument from the first sentence of this paragraph, the next slope in weight $w$ has to be at least ${2{i+1\choose 2} – 2 \over i-1} = i+2 > 2$. And so our slope $2$ form in weight $w$ has multiplicity one.

• Sure, these facts are known and not so hard to prove/compute, but the goal was to avoid computations with modular forms as much as possible. Certainly if you compute the trace of $T_2$ and $T_4$ in weight 24 you can determine $a_2(f)$ immediately.

2. maillard duck says:

Sorry for ignorant question but why is T imaginary quadratic excluded?

• This turns out to be a consequence of the fact that the Hecke operators commute with the Petersson inner product, which is self-adjoint. Roughly, if $T F = a F,$ then

$a \langle F,F \rangle = \langle a F, F\rangle = \langle TF,F \rangle = \langle F, TF \rangle = \langle F, a F \rangle = \overline{a} \langle F, F \rangle.$