equivalence between katz and classical modular forms $\newcommand{\CC}{\mathbb{C}}$
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EDIT: Most of the arguments I've used here are WRONG!
So, I've seen many definitions of modular forms, and I can't seem to find a detailed proof of their equivalence. I'll list the two main definitions I've seen, and try to prove/explain as much of their equivalence as I can, leaving the rest as questions. I hope some of you will find this interesting, and let me know if I'm thinking of something in the wrong way, or if I'm just flat out wrong.
I will highlight my questions or things I'm not clear about in bold.
For simplicity, we'll begin by restricting ourselves to the case of modular forms for $\Gamma(N)$, $N\ge 1$.
Because I'm having trouble writing matrices here, we'll always assume that $\gamma$ is a 2x2 matrix with entries $a,b,c,d$ (left to right, top to bottom).
$\textit{Definition 1 (classical)}$.
A modular form of weight $k$ for $\Gamma(N)$ is a function $f : \hH\rightarrow\CC$ with the following properties:
1(a) - $f$ is holomorphic
1(b) - $f(\gamma(z)) = (cz + d)^k f(z)$ for all $\gamma \in \Gamma(N)$
1(c) - $(cz+d)^kf(\gamma(z))$ has $q^{1/N}$ - expansion $(q = e^{2\pi iz})$ with no negative power terms.
$\qquad$
$\textit{Defintion 2 (Katz)}$
Let $R_0$ be a ring containing $1/N$ and a primitive $N$th root of unity. A modular form of weight $k$ for $\Gamma(N)$ defined over a ring $R_0$ is a function on triples $(E/R, \omega, \alpha_N)$, where $E/R$ is an elliptic curve over an $R_0$-algebra $R$, given by the structure map $p : E\rightarrow R$, $\omega$ is a basis of $\underline{\omega}_{E/R} := p_*\Omega^1_{E/R}$, and $\alpha_N : E[N]\stackrel{\sim}{\longrightarrow} (\ZZ/N\ZZ)_R^2$ is an isomorphism of group schemes, which takes values in $R$, and satisfies the following properties:
2(a) $f(E/R, \omega,\alpha_N)$ depends only on the $R$-isomorphism class of the triple $(E/R, \omega, \alpha_N)$.
2(a') For any $\phi : R\rightarrow R'$, we have $f(E'/R', \omega', \alpha_N') = \phi(f(E/R, \omega, \alpha_N))$, where $E', \omega', \alpha_N'$ denote $E, \omega, \alpha_N$ base changed to $R'$.
2(b) For any $\lambda\in R^*$, $f(E/R, \lambda\omega, \alpha_N) = \lambda^{-k}f(E/R,\omega, \alpha_N)$.
2(c) $f(\text{Tate}(q^n)/$$(\ZZ((q))\otimes_\ZZ R_0)$, $\omega_{\can}, \alpha_N)\in\ZZ[[q]\otimes_\ZZ R_0$ for all level structures $\alpha_N$ on $\Tate(q^n)$.
$\qquad$
This leads me to my first semi-question - $\underline{\omega}_{E/R}$ isn't always free right? So, I'm guessing this $f$ takes as input only triples where $E/R$ is an elliptic curve and $R$ is "small enough" that $\underline{\omega}_{E/R}$ is trivial over $R$?
It ought to be the case that a katz modular form defined over $\CC$, restricted to triples over $\CC$, is equivalent to a classical modular form. First suppose that $N \ge 3$, so that the modular curve $X(N)$ is a fine moduli scheme for a representable moduli problem. Then, we have the universal generalized elliptic curve $\eE := \eE_N\rightarrow X(N)$ with a canonical level $N$ structure $\aA_N$ such that any isomorphism class of generalized elliptic curve with level $N$ structure is a pullback of $(\eE_N,\aA_N)$. Over the affine non-cuspidal piece $Y(N)$, $\eE|_{Y(N)}$ is an elliptic curve, and over the cusps $\eE$ has fibers that are Neron $N$-gons.
In this case, let $U = \Spec R$ be a open set of $Y(N)$ such that $\underline{\omega}_{\eE/R}$ is free, with a basis $\omega_{\eE/R}$. Then, for any Katz modular form $f$ of weight $k$ and level $N$, $f(\eE/U,$ $\omega_{\eE/R}$ $, \aA_N)\in R$, which by definition consists of holomorphic functions on $U$ viewed as a subset of $Y(N)^\an$. Thus, for any map $p : \Spec\CC\rightarrow U$, letting $\tilde{p}$ denote the corresponding map of rings $\tilde{p} : R\rightarrow\CC$, note that this map is just "evaluate $f\in R$ at the point $p$ of $U$". Hence, we see that $f((p^*\eE)/\CC, p^*\omega_{\eE/R}, p^*\aA_N) =$ $\tilde{p}(f(\eE/U,$ $\omega_{\eE/R}$ $,\aA_N))$ is a holomorphic function "in $p$". Now, let $\dD\subset\hH$ be a connected fundamental domain for $\Gamma(N)$ (say, a union of $\text{SL}_2(\ZZ)$-translates of the standard fundamental domain for $\text{SL}_2(\ZZ)$). Then since the points of $\dD$ are in 1-1 correspondence with isomorphism classes of pairs $(E/\CC, \alpha_N)$, we've just shown that $f$ gives a holomorphic function on $U\cap\dD^\circ$ (the $^\circ$ denotes interior). Let $f'$ denote this function. Ie, for any $\tau\in U\cap\dD^\circ$, define $f'(\tau) = f(E_\tau/\CC, \omega_\tau, \alpha_{N,\tau})$, where $\omega_\tau$ is the pullback of the differential $\omega_{\eE/R}$ on $\eE$ via the map $\Spec\CC\mapsto\tau\in U$.
Now, using 2(a) and considering an open cover of $Y(N)$ that trivializes $\underline{\omega}_{\eE/R}$, one finds that $f'$ is a holomorphic function on $\dD^\circ$, Then, one can compute that condition 2(b) (weight $k$ homogeneity w.r.t. $\omega$) shows that $f'$ satisfies condition 1(b) (this is spelled out rather clearly in various sources). Now, some care will be needed to show that $f'$ extends to a holomorphic function on the entirety of $\hH$ (ie, want to prove that $f'$ is continuous on the boundaries of translates of $\dD$), though I expect this can be accomplished by considering different fundamental domains?
This leads to another question: What of the case $N < 3$, ie when no universal elliptic curve exists? For example, in the post:
Is there an elliptic surface over $Y(1)$?
Will Sawin gives a proof that there is no neighborhood of the points $j = 0$ or $j = 1728$ that has an elliptic curve above it with the property that the fiber above $j = a$ is an elliptic curve with $j$ invariant equal to $a$. Is there a nice way to see that a Katz modular form must be "continuous" at these points?
Lastly, we have the notion of holomorphicity at the cusps. My understanding of this is as follows. For a classical modular form $f'$, this is defined by looking at the expansion of $f'$ in terms of $q := q_N := q^{1/N} = e^{2\pi i z/N}$. We do this, because on the upper half plane, the map $e^{2\pi i z/N}$ gives a conformal equivalence from a vertical strip of width $N$ to the punctured unit disk. Now, passing to the quotient space $Y(N) = \Gamma(N)\backslash\hH$, we see that $q$ is invariant under the stabilizer of $\infty$ in $\Gamma(N)$, and hence is a holomorphic function in a neighborhood of $\infty\in X(N)$. Looking at fundamental domains, we see that in fact $q$ is a complex analytic local uniformizer at $\infty$, which does not extend to a global meromorphic function on all of $X(N)$. Now, let $t$ be an algebraic uniformizer at $\infty\in X(N)$ - ie, it's a globally meromorphic function with a simple pole at $\infty$, and is in the stalk $\oO_\infty$ of the algebraic structure sheaf of some algebraic model of $X(N)$. Then even though $q\notin\oO_\infty$, we have that $\widehat{\oO}_\infty\cong \CC[[t]] = \CC[[q]]$, where here we view $q$ as a formal power series in $t$, which can be accomplished since $t$ (thinking of it as a modular function, and hence having a fourier expansion), can be seen as a formal power series in $q$ of order 1.
Now, there's a natural map $\Spec\oO_\infty\rightarrow X(N)$, which extends to a map $i : \Spec\CC((t))\rightarrow\Spec\oO_\infty\rightarrow X(N)$. Intuitively speaking, the pullback of $\eE$ along $i$ should correspond to an infinitesimal family of elliptic curves with $j$-invariant and level $N$ structure parametrized by $t$. Now, identifying $\CC(t) = \CC(q)$ as per the previous paragraph, we can consider the pullback of $\eE$ along $i$ to be the same infinitesimal family, now reparametrized by $q$ (viewed as a power series in $t$). Complex analytically, $q$ actually gives a conformal equivalence of a neighborhood of $\infty$ with a open subset $V$ of the punctured unit disk. In this setting, noting that a modular form of level $N$ can be thought of as a function on lattices, and hence on elliptic curves with differential modulo isomorphism (in both cases with some level structure, see section 1.5 in Diamond/Shurman), it's not hard to see that viewing a modular form as a function on the "Tate family" over $V$, its $q$-expansion is actually just a function on $V$, which agrees with the notion of evaluating a Katz modular form at the Tate curve over $\CC((q))$. Hence, the "holomorphicity at $\infty$" conditions 1(c) and 2(c) are equivalent (may need to use the fact that congruence modular forms have bounded denominators).
The above two paragraphs show that the Tate curve is the pullback of $\eE$ by a rather special morphism $\Spec\CC((q))\rightarrow X(N)$. Namely, the morphism is obtained by mapping an algebraic uniformizer $t$ in $\oO_\infty$ to its power series in $q$ given by the realization of $t$ as a periodic meromorphic function on $\hH$.
 A: On $\omega_{E/R}$: Yes. I don't remember if this is always free. But the definition means exactly what it says $E/R$ is an elliptic curve, and $\omega$ is a basis for $\omega_{E/R}$. From this we can conclude that $\omega_{E/R}$ has a basis. I'd also like to point out that we can always make $\omega_{E/R}$ free by making $R$ larger - it's a line bundle, and its sixth power has a nowhere vanishing section, $\Delta$, so it's always trivialized on a sixfold cover.
On $f'$: Yes. The key point is that the definition of the function does not depend on the fundamental domain used, and every point is in the interior of some fundamental domain. (translate your favorite fundamental domain by an appropriate element of $SL_2(\mathbb R)$.) 
On continuity: Use the universal elliptic curve with full level $3$ or $4$ structure or something like that. This extends to $j=0$ and $j=1728$. You then get a function on this. The base curve will have a lot of automorphisms that change the level structure but fix the elliptic curve. Using $2$, we see that the function is invariant to these automorphisms, so it descends to the quotient, which gives the continuity (in fact analyticity) that you want.
