## Basis for $M_k(\Gamma(N))$ with Fourier coeffs in $\mathbb{Z}[\Zeta_N]$?

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Hi all.

Recently i read that the space of completely holomorphic (also at the cusps) modular forms $M_k(\Gamma(N))$ possesses a basis having Fourier coefficients in $\mathbb{Z}[\zeta_N]$ where $\zeta_N = e^{2 \pi i / N}$.

Can somebody point out a reference for this?

I already know the following things: At least for $k \geq 2$, $S_k(\Gamma(N))$ -- the subspace of cusp forms -- possesses a basis having Fourier coefficients in $\mathbb{Z}$ (see Shimura, Thm 3.52). What is missing is the Eisenstein series $G^{v}$ (see Diamond/Shurman, Thm 4.2.3). All the Fourier coiefficients except the first one do indeed lie inside $\mathbb{Z}[\zeta_N]$ (up to a constant in $\mathbb{Q}$) but the constant term of the Eienstein series is (in the case that $v_1 \equiv 0 \mod N$) the term

$\sum_{n \in \mathbb{Z} \setminus \{0\}, n \equiv v_1 \mod N} \frac{1}{n^k}$

This is the Hurwitz Zeta Function up to the term $N^{-k}$. The question here is: is this value in $\mathbb{Z}[\zeta_N]$ (up to some denominator) or is there a completely different way to see that such a basis with Fourier coeffs in $\mathbb{Z}[\zeta_N]$ exists?

Note that i am aware of this post: http://mathoverflow.net/questions/78043/is-there-a-miller-basis-for-m-kn but i was not able to locate the result in these books.

best and thanks!

Fabian Werner

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I guess you mean the value divided by $\pi^k$ ? – François Brunault Feb 6 2012 at 16:12

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Fabian, don't forget that the higher Fourier coefficients are all multiplied by $\pi^k$ (look at the definition of $C_k$ in Diamond/Shurman Theorem 4.2.3). – BR Feb 6 2012 at 20:52

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(So, in fact, the Eisenstein series can be normalized.) – BR Feb 6 2012 at 21:00

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Fabian, I actually don't know any better methods than those Francois mentions in his answer. But, there is a general principle that the zero-th term in the Fourier expansion of a modular form lies in the field generated by the higher Fourier coefficients (see, e.g., math.umn.edu/~garrett/m/v/…), so I knew it had to work even if I didn't know how! – BR Feb 9 2012 at 3:21

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Here is the reference to the rationality principle used by Klingen (in the more general context of Hilbert modular forms) : ams.u-strasbg.fr/mathscinet/search/…*&s6=&s7=&s8=All&vfpref=html&yearRangeFirst=&yearRangeSecond=&yrop=eq&r=2&mx-pid=133304 (page 266). Note that in order to apply it here, you will need to know that $M_k(\Gamma(N))$ admits a basis having all Fourier coefficients in $\mathbf{Q}(\zeta_N)$, so this is a little bit circular... – François Brunault Feb 9 2012 at 10:49
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## 1 Answer

The constant term of the Eisenstein series $G_k^{0,v}$ in Diamond-Shurman is, up to a factor $N^k$, given by

$$\zeta(k,\frac{v}{N}) + (-1)^k \zeta(k,-\frac{v}{N})$$

where $\zeta(s,x) = \sum_{\substack{n \in \mathbf{Q}_{>0}, \ n \equiv x \mod{1}}} \frac{1}{n^s}$ is the Hurwitz zeta function.

You can prove by hand that this constant term indeed lies in $\pi^k \cdot \mathbf{Q}(\zeta_N)$. This is a tedious exercise (which I admit I haven't done) using the functional equation of the Hurwitz zeta function linking $\zeta(s,\cdot)$ and $\zeta(1-s,\cdot)$ and the fact that $\zeta(1-k,x) \in \mathbf{Q}[x]$ for any $k \geq 1$ (it is given by a Bernoulli polynomial). For these two facts see for example Wikipedia.

The more conceptual explanation is that $\Gamma(N) \backslash (\mathcal{H} \cup \mathbf{P}^1(\mathbf{Q}))$ admits a canonical model $X(N)$ defined over $\mathbf{Q}(\zeta_N)$ (see Shimura, Introduction to the arithmetic theory of automorphic functions, Chapter 6). Moreover, there is a more conceptual definition of Eisenstein series of weight $k$ as sections of $\mathcal{L}^{\otimes k}$, where $\mathcal{L}$ is a certain line bundle on $X(N)$ (defined using the universal elliptic curve over $Y(N)$). Since the cusps of $X(N)$ are rational over $\mathbf{Q}(\zeta_N)$, the Fourier coefficients of these Eisenstein series belong automatically to $\mathbf{Q}(\zeta_N)$. It then suffices to check that these Eisenstein series coincide with $G_k^{0,v}$ (suitably divided by $(2\pi i)^k$). One reference I know for this point of view is Kato, $p$-adic Hodge theory and values of zeta functions of modular forms, Astérisque 295, section 3.

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@Francois: I tried the first version but unfortunately i have no idea on how to filter out one particular evaluation of the Hurwitz Zeta function in the functional relation. Was this proposed as an excercise in some book? Were there any hints or so? – Fabian Werner Feb 8 2012 at 9:10
(i.e. the functional equation only seems to say: if one of the values $\zeta(k, \frac{d}{N})$ is nice, so are the other ones) – Fabian Werner Feb 8 2012 at 9:12
@Fabian : I won't write the exercise for you, but here are some hints : take the formula given by Wikipedia and evaluate at $s=k$. On the left hand side you have the value $\zeta(1-k,m/N)$, which you know is rational (by the second fact I mentioned). And on the right hand side you have the values $\zeta(k,\cdot)$, which you're interested in. You can package all the formulas for $1 \leq m \leq N$ by saying the vector of $\zeta(1-k,m/N)$'s is some matrix times the vector of $\zeta(k,m'/N)$'s. Then it suffices to check the matrix has coefficients in $\mathbf{Q}(\zeta_N)$ and is invertible. – François Brunault Feb 8 2012 at 9:39
Maybe i am completely blind but... Lets take $N=3$ and $k$ divisible by 4 then (for $c$ some constant) $\zeta(1-k, \frac{d}{N}) = c * \sum_{v=1}^{N} cos(\frac{k\pi}{2} - \frac{2\pi d v}{N}) \zeta(k, \frac{v}{N})$ now $cos(\frac{k\pi}{2} - \frac{2\pi d v}{N}) = cos(\frac{2\pi d v}{N}) = \frac{1}{2}(e^{2\pi i dv/N} + e^{-2\pi i dv/N}) = 1/2(x^{vd} + \overline{x}^{vd})$ ($x = e^{2\pi i / 3}$) so the matrix is given by $\begin{pmatrix} x+\overline{x} & x^2 + \overline{x}^2 & 2 \\ x^2 + \overline{x}^2 & x + \overline{x} & 2 \\ 2 & 2 & 2\end{pmatrix}$ but this matrix is clearly not invertible... – Fabian Werner Feb 8 2012 at 15:57
I was being somewhat sloppy. What you should consider is the vector of values $\zeta(k,v/N)+(-1)^k \zeta(k,-v/N)$, instead of $\zeta(k,v/N)$. Otherwise, you cannot get algebraic values (this is related to the following recent MO question : mathoverflow.net/questions/87348/…). – François Brunault Feb 8 2012 at 17:01
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