Ribet's Converse to Herbrand: Part II - Cuspstruction

So around a month back I posted the first post in this 2 (or more who knows?) part series on Ribet’s converse to Herbrand’s theorem. This is the sequel, Cuspstruction, it is basically just my personal notes from my STAGE talk with the same name. We were following Ribet’s paper and this is all about section 3. The goal is to construct a cusp form with some very specific properties, which we can then take the corresponding Galois representation and use that to obtain the converse to Herbrand. In this post though we’ll be focussing on constructing the cusp form, hence, Cuspstruction.

Cuspstruction

We will make use the following building blocks, some specific modular forms of weights 2 and type $\epsilon$ \begin{align*} G_{2,\epsilon} &= L(-1,\epsilon)/2 + \sum_{n=1}^\infty \sum_{d|n} d \epsilon(d) q^n\\ s_{2,\epsilon} &= \sum_{n=1}^\infty \sum_{d|n} d \epsilon(n/d) q^n \end{align*} the latter is not a cusp form (not cuspidal at the other cusp of $\Gamma_1(p)$) we call such forms semi-cusp forms, denote the space of such by $S^\infty$ (not standard notation) we will also use \begin{equation*} G_{1,\epsilon} = L(0,\epsilon) + \sum_{n=1}^\infty \sum_{d|n} \epsilon(d) q^n \end{equation*} the Eisenstein series are all hecke eigenfunctions for $T_n$ $n$ coprime to $p\text{.}$

Fix a prime ideal $\mathfrak p|p$ of $\mathbf{Q}(\mu_{p-1})\text{,}$ can think of $\mu_p\subseteq \mathbf{Q}_p^*$ and take $\omega\colon (\mathbf{Z}/p\mathbf{Z})^* \xrightarrow\sim \mu_{p-1}$ the unique character with $\omega(d)\equiv d \pmod{\mathfrak p}$ for all $d\in \mathbf{Z}\text{.}$

By our choice of $\omega$ we get the desired result for the non-constant terms of the $q$-expansion.

So it remains to prove that \begin{equation*} L(-1,\omega^{k-2})\equiv -\frac{B_k}{k}\pmod{\mathfrak p} \end{equation*} \begin{equation*} L(0,\omega^{k-1})\equiv -\frac{B_k}{k}\pmod{\mathfrak p} \end{equation*} we make use of the following expressions (see probably Washington) \begin{equation*} L(0,\epsilon) = -\frac 1p \sum_{n=1}^p \epsilon(n) (n- \frac p2) \end{equation*} \begin{equation*} L(-1,\epsilon) = -\frac{1}{2p} \sum_{n=1}^p \epsilon(n) (n^2 - pn + \frac{p^2}{6}) \end{equation*} $\omega(n)\equiv n^p \pmod{\mathfrak p^2}$ so \begin{equation*} pL(0,\omega^{k-1}) = -\sum_{n=1}^p \omega^{k-1}(n)(n-p/2) \end{equation*} \begin{equation*} \equiv -\sum_{n=1}^p n^{p(k-1) + 1} \pmod{\mathfrak p^2} \end{equation*} and \begin{equation*} pL(-1,\omega^{k-2}) = -\frac 12\sum_{n=1}^p \omega^{k-2}(n)(n^2 - pn +\frac{p^2}{6}) \end{equation*} \begin{equation*} \equiv -\frac 12\sum_{n=1}^p n^{p(k-2) + 2} \pmod{\mathfrak p^2} \end{equation*} but for all $t\gt 0$ even we have the congruence \begin{equation*} \sum_{n=1}^{p-1} n^t \equiv pB_t \pmod{p^2} \end{equation*} and so \begin{equation*} pL(0,\omega^{k-1}) \equiv -pB_{p(k-1)+1} \pmod{p^2} \end{equation*} cancelling \begin{equation*} L(0,\omega^{k-1}) \equiv -B_{p(k-1)+1} \pmod{p} \end{equation*} \begin{equation*} \equiv -B_{p(k-1)+1} \pmod{p} \end{equation*} \begin{equation*} \equiv -\frac{B_k}{k} \pmod{p} \end{equation*} as $p(k-1) + 1 \equiv k \pmod{p-1}$ using Kummer's congruence. Similarly \begin{equation*} L(-1,\omega^{k-2})\equiv -\frac 12 \frac 2k B_k \pmod{p}\text{.} \end{equation*}

Note: the constant term is a $\mathfrak p$ unit unless $B_n$ or $B_m$ is divisible by $p\text{.}$ We need to remove this condition.

Proof

It suffices to find a form with constant coefficient a $\mathfrak p$-unit. If $p\nmid B_k$ then we can use $G_{2,\omega^{k-2}}$ by lemma 3.1.

If $p|B_k$ try the possible products \begin{equation*} G_{1,\omega^{m-1}}G_{1,\omega^{n-1}} \end{equation*} with $2\le m,n\le p-3$ even as above with $m+n \equiv k \pmod{p-1}\text{.}$ We want to claim that at least one of these must work (i.e. we have a pair $m,n$ with $p\nmid B_m, p\nmid B_n$). If this isn't the case if we let \begin{equation*} t = \#\{2\le n \text{ even} \le p-3: p|B_n\}\text{,} \end{equation*} we must have $t \ge (p-1)/4\text{,}$ assume otherwise, we will derive a contradiction from this.

Greenberg showed that \begin{equation*} \frac{h_p}{h_{\mathbf{Q}(\mu_p)^+}} = h^*_p = 2^? p \prod_{\substack{k=2\\ \text{even}}}^{p-2} L(0,\omega^{k-1}) \end{equation*} (this is obtained by taking a quotient of the analytic class number formulas for $\mathbf{Q}(\mu_p),\mathbf{Q}(\mu_p)^+$) but by lemma 3.1 we know that $\mathfrak p^t$ will divide the product of $L$-values. And so $p^t|h_p^*\text{,}$ we will get a contradiction if we show \begin{equation*} h_p^*\le p^{(p-1)/4}\text{.} \end{equation*}

Work of Carlitz-Olson '55, Maillet's determinant shows that \begin{equation*} h_p^* = \pm\frac{D}{p^{(p-3)/2}} \end{equation*} where $D$ is the determinant of a $(p-1)/2 \times (p-1)/2$ matrix with entries in $[1,p-1]\text{.}$ So recalling Hadamard's inequality \begin{equation*} |\det(v_1\cdots v_n)| \le \prod_{i=1}^n ||v_i||\text{,} \end{equation*} or the simpler corollary \begin{equation*} |A_{ij}| \le B \implies |\det(A)| \le n^{n/2}B^{n} \end{equation*} and applying with $B = p, n=(p-1)/2$ gives \begin{equation*} |D| \le \left(\frac{p-1}{2}\right)^{(p-1)/4} p^{(p-1)/2} \lt 2^{-(p-1)/2} p^{(3p-3)/4} \end{equation*} so \begin{equation*} h_p^* \lt p^{(p+3)/4} 2^{-(p-1)/4}\text{.} \end{equation*} And we are done as $h_p^* = 1$ for $p\le 19$ and as $p\le 2^{(p-1)/4}$ for $p\gt 19\text{.}$

Now we fix $2\le k\le p-3$ even with $p|B_k$ and let $\epsilon = \omega^{k-2}\text{,}$ $k$ must really be at least 4 (or even 10) so $\omega$ is a non-trivial even character, we will work in weight 2, type $\epsilon$ from now on.

Let \begin{equation*} f= G_{2,\epsilon} -cg \end{equation*} with $c = L(-1,\epsilon)/2\text{.}$ This is a semi-cusp form by construction. We get $f\equiv G_{2,\epsilon}\pmod{\mathfrak p}$ because \begin{equation*} c \equiv -B_k/2k \equiv 0 \pmod{\mathfrak p} \end{equation*} by lemma 3.1 (again!) as we assume $p|B_k$ now. Additionally by the same lemma $G_k \equiv G_{2,\epsilon}\pmod{\mathfrak p}\text{.}$

Let's take stock, we have a semi-cuspidal form which mod $\mathfrak p$ looks like $G_k$ and is hence an eigenform mod $\mathfrak p$ but we want an actual eigenform, bro do you even lift?

We start with $f$ from the proposition above it's a mod $\mathfrak p$ eigenform and so we can use Deligne-Serre lifting lemma (6.11 in Formes modulaires de poids 1) to obtain a semi-cusp form $f'\text{,}$ that is an eigenvalue for the Hecke operators stated.

To promote the semi-cusp form to a full blown cusp form we observe that the space $S_2^\infty(\Gamma_1(p),\epsilon)$ is generated by the cusp forms and $s_{2,\epsilon}$ which is also an eigenform we only have to check that $f'$ isn't $s_{2,\epsilon}$ (or it's scalar multiple). So we check the eigenvalues mod $\mathfrak p\text{.}$ \begin{equation*} \epsilon(\ell) + \ell \equiv 1 + \ell\epsilon(\ell)\pmod{\mathfrak p} \end{equation*} implies $\epsilon(\ell) = 1\text{,}$ but $\epsilon$ is non-trivial!

The final challenge is to ensure that $f'$ is also an eigenform for $T_{p^i}\text{.}$

Use the theory of newforms. There are no oldforms for $\Gamma_1(p)$ as \begin{equation*} M_2(\operatorname{SL}_2(\mathbf{Z})) = 0\text{.} \end{equation*} A newform that is an eigenform for all hecke operators coprime to the level $p$ is also an eigenform for the remaining Hecke operators.

So in conclusion:

Remark

Word on the internet is that Mazur, Mazur-Wiles' proof of the Main conjecture of Iwasawa theory is modelled on this.

That's all for now, in the remainder of Ribet's paper he constructs a Galois representation from this and use it to prove the theorem.