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28 votes
1 answer
2k views

Intuitive reason why the $j$-invariant is a cube?

Let $\tau$ be a CM point of discriminant $D$. Assume that $D$ is not divisible by $3$. Then $j(\tau)$ is an algebraic integer of degree equal to the class number $h(D)$. Let $ \gamma_2(\tau)=j(\tau)^{...
Shimrod's user avatar
  • 2,375
23 votes
3 answers
2k views

Why are values of Eisenstein $E_2^*$ algebraic integers?

I'm looking for a proof that the following term is an algebraic integer whenever $\tau_N=\frac{N+\sqrt{-N}}{2}$ is a quadratic irrationality with class number $1$: $$A_N:=\sqrt{-N}\cdot\frac{E_2(\...
L. Milla's user avatar
  • 598
10 votes
1 answer
1k views

Reference for the odd dihedral case of Artin's conjecture

The example that Matt Emerton cited here prompted me to become interested in how one proves that odd two dimensional dihedral Galois representation are modular. This is the first case of the strong ...
Jonah Sinick's user avatar
  • 7,062
8 votes
2 answers
744 views

A question related to Hilbert modular form

This is a question related to Hilbert modular forms. Let $\mathbb{K}=\mathbb{Q}(\sqrt D)$ be an imaginary quadratic field with discriminant $D<0$ and $\zeta (\text{mod } m)$ a Hecke character such ...
user15243's user avatar
  • 424
7 votes
1 answer
175 views

Is every eigenvector sequence for the Hecke operators a eigenform?

Let $f(n) = a_n$ be a sequence taking values in $\mathbb C$ for $n=1,2,...$. Let $T_m$ be the Hecke operators (of a fixed weight $k$) defined as usual in terms of the $a_n$. That is: $$T_m(f)(n) = \...
Asvin's user avatar
  • 7,746
6 votes
0 answers
456 views

Conditions under which an $\eta$-quotient becomes a **weak** modular form (reference request for theorems similar to Ligozat's theorem)

For any $z \in \mathcal{H}$, let $q = e^{2\pi iz}$; and the eta function is defined as ${\displaystyle \eta (q) =q^{\frac {1}{24}}\prod _{n=1}^{\infty }\left(1-q^{n}\right).}$ By an $\eta$-quotient ...
Davood Khajehpour's user avatar
3 votes
1 answer
341 views

Siegel's bad character

Let $K$ be an imaginary quadratic field with discriminant $d_K$. Suppose that $d_K=gt$, where either $g,t$ are discriminants or have the value $g=1,t=d$. Let $f$ be an additional discriminant of a ...
Shimrod's user avatar
  • 2,375
3 votes
0 answers
186 views

Maximum value of newform from Galois representation

One can attach $\ell$-adic Galois representations to holomorphic cuspidal newforms of weight $2$ on the upper half-plane. If a newform is $L^2$-normalized, can one extract its maximum value from the ...
sup's user avatar
  • 39
2 votes
1 answer
159 views

On some claims on cyclic modules over Hecke algebra used in Serre's "Quelques applications du théorème de densité de Chebotarev"

I have been reading section 7 of Serre's "Quelques applications du théorème de densité de Chebotarev" (http://www.numdam.org/item/PMIHES_1981__54__123_0/), and in particular have been trying ...
asrxiiviii's user avatar
2 votes
0 answers
245 views

Ambiguity about the exact definition of coefficients of modular forms

You can see the parts after my questions in the boxes. I received the answer to my first question in the comments. I am confused about the definition of $a_n$ and $b_n$ in Part II below. I know the ...
Tireless and hardworking's user avatar
2 votes
0 answers
270 views

Generalized Siegel Weil formula

I am studying the following Poincare-like series, \begin{equation} F_k(\tau,\bar{\tau})=\sum_{\gamma\in\Gamma_{\infty}\backslash\Gamma}\sqrt{\text{Im}\gamma\tau}(q_{\gamma}\bar{q}_{\gamma})^k, \end{...
Sounak Sinha's user avatar
1 vote
0 answers
132 views

Stabilizers of points in the upper half-plane

Suppose that $\Gamma$ is a group acting discontinuously on $\mathcal{H} = \{ z \in \mathbb{C} : \operatorname{Im}(z) > 1\}$. In order to keep things simple, suppose that $\Gamma \subseteq \...
Oiler's user avatar
  • 163
1 vote
0 answers
129 views

Siegel's formula for generalized theta series with characteristics?

Siegel's formula(Siegel-Weil) directly relates the weighted sum of theta functions to Eisenstein series. (Or equivalently, the weighted sum of the cusp form is zero). I wonder if there is a ...
Y.J.'s user avatar
  • 11