Several days ago, I posted this question but didn't get enough conclusion.
The question was about the space $M^!_k(\Gamma_0(N);K)$ where $K$ is a subfield of $\mathbb{C}$.
This is the space of weight k weakly holomorphic modular forms defined on the modular curve $Y_0(N)(K)$ over the field K.
There were no more explanations so I will state what I thought.
If there is something wrong or comments, then please tell me I will appreciate it.
I will assume weight $k$ is zero. I.e focus on the modular function.
First, $Y_0(N)(K)$:
We can view $Y_0(N)$ as an algebraic curve over $\mathbb{Q}$
Moreover, $Y_0(N)(\mathbb{C})\cong\Gamma_0(N)\backslash\mathbb{H}$ as a riemann surface.
Viewing $Y_0(N)$ as an algebraic curve over $K$, can we regard $Y_0(N)(K)$ as a submanifold of $\Gamma_0(N)\backslash\mathbb{H}$? (I think that this is wrong)
Second, function on $Y_0(N)(K)$:
Consider an element in the function field $K(Y_0(N))$. Regard element in this field as a function $Y_0(N)(K)\rightarrow\mathbb{P}^1(K)$.
If we cannot view $Y_0(N)(K)$ as a riemann surface, then how can we define meromorphicity?
For $K=\mathbb{C}$, we get the space of the modular functions of level N, namely $M^!_0(\Gamma_0(N))$
Finally, element in $M^!_0(\Gamma_0(N))$ can be regarded as a meromorphic function $f$ on $\mathbb{H}$ which is invariant under $\Gamma_0(N)$.
From $f(z+1)=f(z)$, we get a $q$ exapansion of $f$ at infinity.
How do we get a $q$ expansion of element in $M_0(\Gamma_0(N);K)$?(It looks like we should get a element in $K((q))$)
Hecke operator and atkin-Lehner operator act on the space $M^!_0(\Gamma_0(N))$
How do this operators act on $M^!_0(\Gamma_0(N);K)$?
First, $Y_0(N)(K)$ is the set of points corresponding to solutions in $K$ of the (rational) equation defining $Y_0(N)$. Thus, its image in $\Gamma_0(N) \backslash \mathbb{H}$ is the set of $\Gamma_0(N)\tau$ such that $j(\tau), j(N\tau) \in K$.
This isn’t a rigorous argument, but it sounds fairly clear that if $K$ isn’t $\mathbb{R}$ or $\mathbb{C}$, then it’s totally disconnected, and thus the image of $Y_0(N)(K)$ in $\Gamma_0(N) \backslash \mathbb{H}$ is also completely disconnected. So it can only be a submanifold if all the points are isolated (and then, it’s of dimension zero, so it’s not very interesting).
Second, you have to understand that (when $K$ is an arbitrary field, we’re not assuming it’s algebraically closed) $Y_0(N)(K)$ contains much less information that the full curve $Y_0(N)$ defined over $K$. In a way, $Y_0(N)$ is the equation, while $Y_0(N)(K)$ is just the set of solutions in $K$.
As an example, if $K$ is a number field, for all but finitely many $N$ (all $N \leq 21$, $N=24,25,27,32,36,49$) $Y_0(N)(K)$ contains finitely many points.
Functions in $K(Y_0(N))$ are meromorphic by definition. That’s one of the possible definitions for an algebraic curve (up to compactification issues, but they’re mostly irrelevant here): a finitely generated field extension of transcendance degree one over a fixed base field.
Finally, to get the $q$-expansion of a function in $M_0(\Gamma_0(N),K)$, there are two possibilities:
realize that it is thus an element of $M_0(\Gamma_0(N),\mathbb{C})$, and thus has a Fourier expansion there – and it will turn out that the coefficients are in $K$. That’s actually one of the possible definitions for $M(\Gamma_0(N),k)$: the set of modular forms whose $q$-expansion is in $K$.
the more algebraic method. Then, you embed $Y_0(N)$ into a projective curve $X_0(N)$ – just by adding two new points (defined over $K$). We’ll call them $0$ and $\infty$ because this is what they correspond to (as limits, at least) in $\Gamma_0(N) \backslash \mathbb{H}$.
It turns out that the function $F=1/j(\tau)$ (whose $q$-expansion is in $q(1+q\mathbb{Z}[[q]])$) is in $\mathbb{Q}(Y_0(N))$ (hence defined over $K$), regular near $\infty$ and vanishes at $\infty$ with multiplicity one, which means that the completion of $\mathcal{O}_{X_0(N),\infty}$ (the ring of stalks of regular functions on $X_0(N)$ near infinity) is naturally isomorphic to $K[[F]]$.
In this completion, we see that any function in $K(Y_0(N))$ regular near infinity can be written as a formal power series in $F$. But we clearly have an inclusion $K[[F]] \rightarrow K[[q]]$ given by expanding $F$ itself as a formal power series in $q$, giving us the $q$-expansion map from $M_0(\Gamma_0(N),K)$ to $K[[q]]$.
It’s not completely trivial (but it shouldn’t be too hard to see) that these constructions yield the same map.