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Can you please explain for me how we can calculate the coefficients of $q$-expansion series of a modular form function $f$? I am really confused.

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At the very least you should explain what you do know. Your question provides absolutely no context. –  Mariano Suárez-Alvarez Feb 4 '13 at 3:47
    
@MarianoSuárez-Alvarez While I agree that the OP should have definitely provided more, this is a legitimately confusing subject (or at least it was for me!) the firs time I saw it. –  Alex Youcis Feb 4 '13 at 6:38

1 Answer 1

The key to this is a little theorem. Let $\mathcal{O}(\mathbb{D}-\{0\})$ denote holomorphic functions on the punctured disc, and let $X=\left\{f\in\mathcal{O}(\mathfrak{h}):f(z+1)=f(z)\text{ for all }z\right\}$ be the set of all $1$-periodic holomorphic functions on the upper half-plane. Then, the function $e(z)=\exp(2\pi i z)$ is a map $\mathfrak{h}\to\mathbb{D}-\{0\}$. This induces a map $e^\ast:\mathcal{O}(\mathbb{D}-\{0\})\to X$ defined by $e^\ast(f)(z)=f(e(z))$. Then, we have the following theorem

Theorem: The map $e^\ast:\mathcal{O}(\mathbb{D}-\{0\})\to X$ is a bijection.

The fact that $e^\ast$ spits out $1$-periodic functions is clear since $e(z)$ is $1$-periodic. To see why this is actually a bijection we first note that $e^\ast$ is an injection because $e$ is a surjection. The fact that $e^\ast$ is a surjection is slightly harder. Namely, let $f\in X$ and define $g$ by $g(z)=f(w)$ for any $w$ with $w=e(z)$. This is well-defined precisely because $f$ is $1$-periodic. Moreover, since locally $g$ is just a log function it is holomorphic. Thus, $g\in\mathcal{O}(\mathbb{D}-\{0\})$ and $e^\ast(g)=f$. Thus, the theorem is proven.

Now, let's suppose that $f\in \mathcal{M}_k(\text{SL}_2(\mathbb{Z}))$. Then, applying weak-modularity to $\begin{pmatrix}1 & 1\\ 0 & 1\end{pmatrix}$ gives you that $f$ is $1$-periodic. Thus, by the above theorem there exists some $g\in\mathcal{O}(\mathbb{D}-\{0\})$ such that $f(z)=g(e(z))$. Now, since $g\in\mathcal{O}(\mathbb{D}-\{0\})$ is has a Laurent expansion

$$g(w)=\sum_{n\in\mathbb{Z}}a_n w^n$$

with

$$a_n =\frac{1}{2\pi i}\oint_{\Gamma_R} \frac{g(w)}{w^{n+1}}\, dw$$

where $\Gamma_R$ is any circle of radius $0<r<1$ centered at $0$. Note then that

$$f(z)=g(e(z))=\sum_{n\in\mathbb{Z}}a_n e(z)^n=\sum_{n\in\mathbb{Z}}a_n e^{2\pi i n z}=\sum_{n\in\mathbb{Z}}a_n q^n$$

Of course, if you don't want to calculate the Laurent series and you just want to work with $f$ itself you can make the observation that if you parameterize the circle $\Gamma_R$ as $Re^{2\pi i t}$ for $t\in [0,1]$ you get the equality

$$a_n=\frac{1}{R^n}\int_0^1 \frac{g(Re^{2\pi i t})}{e^{2\pi i nt}}\, dt$$

Now, writing $Re^{2\pi i t}=\exp\left(2\pi i \left(\frac{\log(R)}{2\pi i}+z\right)\right)$ and applying the relation between $g$ and $f$ we see that

$$a_n=\frac{1}{R^n}\int_0^1 f\left(\frac{\log(R)}{2\pi i }+t\right) e^{-2\pi i nt}\, dt$$

I'll leave it to you to show that letting $R\to 1^-$ gives us that

$$a_n=\int_0^1 f(t)e^{-2\pi i n t}\, dt$$

which is the classic formulation of the Fourier coefficients.


Now, if you are working not with $\text{SL}_2(\mathbb{Z})$ but a congruence subgroup $\Gamma$ of level $N$ the only difference in the above is that you'll have to work relative to a different power of the exponential. Namely, you know that $\begin{pmatrix}1 & N\\ 0 & 1\end{pmatrix}\in\Gamma(N)\subseteq\Gamma$ and thus there is a well-defined $n_0$ which is defined to be the minimal polsitive integer $n$ such that $\begin{pmatrix}1 & n\\ 0 & 1\end{pmatrix}\in\Gamma$. Then, using the same trick as before you can show that any $f\in\mathcal{M}_k(\Gamma)$ is $n_0$-periodic and that if $e_{n_0}(z)=\exp\left(\frac{2\pi i z}{n_0}\right)$ that $e_{n_0}$ is a surjection $\mathfrak{h}\to \mathbb{D}-\{0\}$ and that $e_{n_0}^\ast$ defines a bijection between holomorphic functions on the punctured disc and the $n_0$-periodic functions on the upper half-plane. The rest of the procedure is a straightforward analogy to the case $n_0=1$.


Note not only does this derivation give you how to calculate the Fourier series for a modular form it should also give you an intuition as to why the classic definition of modular form makes sense! Indeed, let's stick with modular forms $f$ in $\mathcal{M}_k(\Gamma(1))$ (where, of course, $\Gamma(1)=\text{SL}_2(\mathbb{Z})$). There are usually three conditions that we want $f$ to satisfy:

$\displaystyle \begin{aligned}&\mathbf{(1)}\quad f:\mathfrak{h}\to\mathbb{C}\text{ is holomorphic}\\ &\mathbf{(2)}\quad f(\gamma(z))=j(\gamma,z)^k f(z)\text{ for all }\gamma\in\Gamma(1)\\ &\mathbf{(3)}\quad f(z)=\sum_{n=0}^{\infty}a_n q^n\end{aligned}$

where $j(\gamma,z)$ is the function $cz+d$ if $\displaystyle \gamma=\begin{pmatrix}a & b\\ c & d\end{pmatrix}$.

While there are many, many ways one could motivate this definitions let me take one that lends itself to things that interest me, albeit with the possibility of exceeding my pay grade. Namely, let $Y(1)$ denote the coset space $\Gamma(1)/\mathfrak{h}$ (i.e. the space of orbits of the action of $\Gamma(1)$ on $\mathfrak{h}$). Then, it is easy to show that $Y(1)$ has the natural structure of a Riemann surface. Moreover this space is a moduli space for elliptic curves. Now, while there is a formal definition of a moduli space a) I am not the right person (at this time in my career!) to ask about such things and b) it's not entirely important to understand the general definition. The point is that $Y(1)$ is a geometric space whose points are indexed by isomorphism classes of elliptic curves in such a way that the geometry of $Y(1)$ tells us things about the isomorphism classes of elliptic curves.

As is standard though, it's hard to work with an open Riemann surface, and so instead of working with $Y(1)$ directly we work with its compactification (or projective closure) $X(1)=\overline{Y(1)}$ which is also a Riemann surface. Now that we have our space of interest and we know its geometry tells us interesting things we scramble to find anything which can shed geometric light on $X(1)$. Well, in a way that is made precise by things like Yoneda's lemma if we want to understand an object, we should look at maps out (and in!) of that object. This is made extremely clear in the theory of Riemann surfaces where a substantial amount of geometry of a surface $X$ can be gleaned from it's field of meromorphic functions $\text{Mer}(X)$. In fact, this is, in a sense, the whole game for compact Riemann surfaces. So, wanting to learn about the geometry of $X(1)$ we attempt to find some meromorphic functions $X(1)\to\mathbb{P}^1$.

The unfortunate thing about Riemann surfaces is what while their rigidity is what allows us to glean so much information from their meromorphic functions it is precisely what it makes the construction of such functions so difficult. From complex analysis we may be tempted to try to construct meromorphic functions on a surface $X$ by taking quotients of holomorphic functions on $X$. But, the sad fact of it all is that compact Riemann surfaces have no non-constants holomorphic functions (apply the maximum principle!)! Thus, taking quotients of holomorphic functions on $X$ won't help us. But, if we're in the case of a space $X(1)$ where we have a natural covering $\mathfrak{h}\to X(1)$ there is a very fruitful way to create meromorphic functions $X(1)\to\mathbb{P}^1$. Namely, take meromorphic functions on $\mathfrak{h}$ which are $\Gamma(1)$-invariant and descend them to $X(1)$!

This is nice because while $X(1)$ has no non-zero holomorphic functions $\mathfrak{h}$ has them in abundance. But, we can't take the quotient of $\Gamma(1)$-invariant holomorphic functions to get a $\Gamma(1)$-invariant meromorphic functions since $\Gamma(1)$-invariant holomorphic functions on $\mathfrak{h}$ are constant (they are basically just holomorphic functions on $X(1)$). But, if instead of $\Gamma(1)$-invariant functions we took ones that transformed by the $\Gamma(1)$ action in a predictable way we could perhaps take ratios giving us a) a meromorphic function and b) if we created the functions correctly (namely how they transform with respect to the $\Gamma(1)$ action) a $\Gamma(1)$-invariant function.

This is precisely where the first two conditions for an element of $\mathcal{M}_k(\Gamma(1))$ come in. If we divide two functions satisfying $\mathbf{(1)}$ and $\mathbf{(2)}$ we will get a meromorphic function which is $\Gamma(1)$-invariant. So, where does $\mathbf{(3)}$ come into play? The issue, if you recall, is that we are not dealing with $Y(1)$ but instead with $X(1)$ which is the closure of $Y(1)$. And, if you think about it we see that conditions $\mathbf{(1)}$ and $\mathbf{(2)}$ give us functions whose ratio is a meromorphic function on $Y(1)$. But, whose to say that when we extend $Y(1)$ to $X(1)$ the functions will staty well-defined? They could pick up singularities.

This is where $\mathbf{(3)}$ comes into play though. It can be shown that for $\Gamma(1)$ we only have to add in one point to compactify $Y(1)$. This point (which should be clear by examining the fundamental domain for $Y(1)$)can be thought of as being $i\infty$--a point infinitely far up the $y$-axis. Now, since $X(1)$ is just $Y(1)\cup\{i\infty\}$ to check that we have a holomorphic function on $X(1)$ we need only check that it is holomorphic at $i\infty$. Thus, we would like our functions that satisfy $\mathbf{(1)}$ and $\mathbf{(2)}$ to be holomorphic at $i\infty$. But, the beauty of the map $e(z)$ as described above allows us to turn this into a simple question. Namely, since $\text{Im}(z)\to \infty$ is true if and only if $e(z)\to 0$ we see that the point $i\infty$ in $\mathfrak{h}$ corresponds to $0$ in $\mathbb{D}-\{0\}$.

So, to see if $f$ is holomorphic at infinity we need only check that $(e^\ast)^{-1}(f)$ is holomorphic at $0$. But, if we have a function meromorphic on a punctured disc, we know that the functions continues holomorphically to the full disc if and only if its Laurent series at $0$ has no non-negative terms. But, by the above the negative terms of the Laurent series for $(e^\ast)^{-1}(f)$ are just the negative terms of the Fourier expansion for $f$. Thus, we see that $f$ will be holomorphic at $i\infty$ if and only if $f$ has no negative terms in its Fourier series. This motivates condition $\mathbf{(3)}$.

I apologize if this was more than you were looking for, this is just something that interests me and I thought I'd share. Please feel free to ask any questions you might have!


EDIT: For anyone that is interested in the above aspects of modular forms I highly recommend the book of Diamond and Shurman.

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Many thanks Alex Youcis for the great answer –  Hakeem Mar 4 '13 at 8:05

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