# Topological Modular Forms

This will be my first and last post on this topic, since it will take us too far from the theme for this year which is arithmetic geometry. It took awhile for me to write this because something feels wrong in the last post and I wanted to correct it before doing this one. Unfortunately, I can only make a guess at what is happening. I’ll explain it when it comes up. Thus as a warning everything in the last post and in this post should be taken as approximately true (of course, this is a blog, so this warning should probably always be in place).

Recall briefly that we now have a description of (weight ${2}$) modular forms just as a global ${1}$-form on a certain moduli space of elliptic curves with level ${N}$ structure. To get the weight ${2k}$ modular forms we just take tensor powers, so ${H^0(X_0(N), \Omega^{\otimes k})\simeq M_{2k}(\Gamma_0(N))}$. It is funny to notice that a priori it is completely unclear that the collection of all modular forms of a fixed level should form a graded commutative ring, but with this description it falls right out. We define the graded ring of modular forms of level ${N}$ to be ${\displaystyle M(\Gamma_0(N))=\bigoplus_{k=1}^\infty H^0(X_0(N), \Omega^{\otimes k})=\bigoplus_{k=1}^\infty M_{2k}(\Gamma_0(N))}$.

Now notice that if we take ${N=1}$ in our moduli problem we are just taking an elliptic curve plus a cyclic subgroup of order ${1}$, i.e. we are marking the identity. Thus what ought to be the case is that ${\overline{\mathcal{M}_{1,1}}=X_0(1)}$. This is the part that confuses me. Last time I said that ${X_0(N)}$ was a smooth Riemann surface, but ${\overline{\mathcal{M}_{1,1}}}$ is a DM stack. My guess at what is going on is that since we only defined the moduli functor for ${X_0(N)}$ for elliptic curves over ${\mathbb{C}}$, we are maybe just taking the ${\mathbb{C}}$-valued points. Thus maybe ${\overline{\mathcal{M}_{1,1}}(\mathbb{C})\simeq X_0(1)}$. In any case, there is certainly some relation between the two so it isn’t unreasonable to try to figure out what happens when we replace ${X_0(1)}$ with ${\overline{\mathcal{M}_{1,1}}}$.

Now we’ll start the crazy generalizations. There is something called a derived DM stack. Since it would take a lot to define, we’ll just say that it is one of these things where “${\infty}$-blah” gets thrown around. The important idea here is that we can take ${\pi_0}$ and get back an honest DM stack. The big theorem of Hopkins, Miller, and Lurie is that there exists a derived DM stack ${(\mathcal{M}, \mathcal{O})}$ whose underlying DM stack is ${\overline{\mathcal{M}_{1,1}}}$ such that ${\pi_{2k}\mathcal{O}\simeq \omega^k}$ and ${\pi_{2k+1}\mathcal{O}=0}$.

Now “tmf” is something called a commutative ring spectrum and it is formed by taking the derived global sections of ${\mathcal{O}}$. Generalities give us a descent spectral sequence ${H^s(\mathcal{M}, \omega^t)\Rightarrow \pi_{2t+s}\mathbf{tmf}}$. An open and interesting question is to determine which modular forms give homotopy classes in ${\mathbf{tmf}}$ since the classical modular forms form the ${0}$-row of this spectral sequence. I’ll just end by pointing out how mind boggling this is. Modular forms have had such great success in number theory. Now they are successfully being used to understand homotopy groups of spheres and other extremely topological questions. The reverse has been done as well. Topological methods can transfer information back to modular forms and give number theoretic theorems such as congruence relations between ${p}$-adic modular forms. How amazing!

# Geometric Construction of Modular Forms

I was going to move on to deformations of Galois representations to try to talk about some of the ideas involved in proving Taniyama-Shimura, but after a random conversation I decided it might be more fun to try to talk a little about topological modular forms. I know nothing about these things, but it is a pretty big current research interest right now and they are somehow related to the modular forms I just talked about, so let’s give it a try.

This post is filling in more information about modular forms. Rather than make you look up this definition again, I’ll copy it back here and give some motivation for it. A modular form of weight ${k}$ and level ${N}$ is an element of the vector space ${M_k(\Gamma_0(N))}$ which consists of holomorphic functions on the upper half plane ${f:\mathcal{H}\rightarrow \mathbf{C}}$ satisfying the additional transformation property that

$\displaystyle \displaystyle f \left(\frac{az+b}{cz+d}\right)=(cz+d)^kf(z)$

for all matrices ${\left(\begin{matrix} a & b \\ c & d \end{matrix}\right)\in SL_2(\mathbf{Z})}$ such that ${c\equiv 0 \mod N}$ (plus the “cusp” condition). The ones that vanish at the cusp are called cusp forms and they form a vector space denoted ${S_k(\Gamma_0(N))}$.

Now all of this looks extremely strange (especially this extending to be holomorphic across these bizarrely defined cusps). But now I’ll show you why it isn’t strange at all. The matrices that become upper triangular mod ${N}$ (the ones we defined the transformation law for) are a finite index subgroup of ${SL_2(\mathbb{Z})}$ notated ${\Gamma_0(N)}$. The group acts naturally on the upper half plane via linear fractional transformations, and we can form the quotient ${Y_0(N):=\Gamma_0(N) \setminus \mathbf{H}}$.

This is a nice space, but we’d really like it to be a nice compact Riemann surface, so we compactify. Equivalently we have to add in finitely many points. There we have it! These are the cusps. So if we take the extended upper half plane I defined before and then take the quotient we get a smooth Riemann surface ${X_0(N):=\Gamma_0(N) \setminus \mathbf{H}^*}$. Suppose I have some global ${1}$-form on ${X_0(N)}$, i.e. a global section of ${\Omega^1}$. First note that this will be holomorphic across the cusps automatically. The condition that seemed superficial before happens for free when we think in these terms.

Second, this weird transformation law is saying exactly that it is something that can descend to be a ${1}$-form on the surface. Thus a modular form is actually a ${1}$-form on the modular curve ${X_0(N)}$. Or more precisely if we take the quotient map ${\pi: \mathbf{H}^*\rightarrow X_0(N)}$, then a ${1}$-form ${\omega}$ can be pulled back ${\pi^*\omega=f(z)dz}$, and ${f}$ is the “form” satisfying these properties. All of our conditions come out naturally. The cusp forms we were considering were just the ones that vanished at the cusps (the points we threw in to compactify). In algebraic geometry symbols we have ${H^0(X_0(N), \Omega^1)\simeq S_2(\Gamma_0(N))}$. We have to do something a little more subtle to get all our weights, but we’ll come back to that.

Probably the hardest part of breaking into the modular forms literature is that there are so many equivalent ways to think about all these things. We actually need one more interpretation of ${X_0(N)}$ in order to motivate the types of generalizations that happen in the definition of topological modular forms. It turns out that ${X_0(N)}$ represents a moduli functor, i.e. it is a “moduli space”. If you’ve read Mumford’s early stuff on ${\mathcal{M}_{1,1}}$, then you may be able to guess what types of things it parametrizes. It turns out that ${X_0(N)}$ is the moduli space of elliptic curves with certain “level ${N}$” structure. It parametrizes pairs ${(E, C)}$ where ${E}$ is a an elliptic curve and ${C\subset E}$ is a cyclic subgroup of order ${N}$. Two pairs are isomorphic if the isomorphism preserves the subgroup. Surprisingly this moduli space is as nice as can be ${X_0(N)}$.

We could spend months talking about all these modular curves and their properties, but for now we’ll have to leave it at this. Next time we’ll start working towards the generalization.

# Taniyama-Shimura 4: The Conjecture

We’ve done a lot of work so far just to try to define the terms in the Taniyama-Shimura conjecture, but today we should finally make it. Our last piece of information is to write down what the L-function of a modular form is. Since I don’t want to build a whole bunch of theory needed to define the special class of modular forms we’ll be considering, I’ll just say that we actually need to restrict our definition of “modular form” to “normalized cuspidal Hecke eigenform”. I’ll point out exactly why we need this, but it doesn’t change anything in the conjecture except that every elliptic curve actually corresponds to an even nicer type of modular form.

Let ${f\in S_k(\Gamma_0(N))}$ be a weight ${k}$ cusp form with ${q}$-expansion ${\displaystyle f=\sum_{n=1}^\infty a_n q^n}$. Since this is an analytic function on the disk, we have the tools and theorems of complex analysis at our disposal. We can perform something called the Mellin transform. It is just a standard integral transform given by the formula $\displaystyle {\Lambda (s) = \int_0^\infty f(it)t^s\frac{dt}{t}}$.

After some computation you find that this transformed function is a product of really nice functions. We get $\displaystyle {\Lambda (s)=\frac{N^{s/2}}{(2\pi)^s}\Gamma(s)L(f,s)}$, where ${\Gamma(s)}$ is the usual Gamma function. Now if you actually went through and worked this out you would find out that ${L(f,s)}$ has a really nice form in terms of the Fourier coefficients. The so-called L-series associated to the Mellin transform is given by

$\displaystyle \displaystyle L(f,s)=\sum_{n=1}^\infty \frac{a_n}{n^s}$.

If your eyes glazed over for the Mellin transform talk, then just think of the L-function of the modular form as taking all of its Fourier coefficients and throwing them in the numerator of this series to make a new function. A quick remark is that if all the ${a_n}$ are ${1}$ (this won’t happen) we recover the Riemann zeta function. Thus you could think of the L-function we get as some sort of generalization of the zeta function. If you’ve been through some elementary number theory you have probably even seen a proof that $\displaystyle {\sum_{n=1}^\infty \frac{1}{n^s}=\prod \frac{1}{1-p^{-s}}}$ where the product is over all primes called an Euler product. Now in general if I hand you a sequence of integers ${a_n}$ that has some reasonable growth condition, then ${\sum_{n=1}^\infty \frac{a_n}{n^s}}$ will be a nice convergent series, probably with an analytic continuation to the plane. The tricky part is to figure out what types of sequences allow this Euler product decomposition.

This is where we have to use that ${f}$ was of this special form. In the theory of modular forms there is something called Atkin-Lehner theory which tells us that the ${a_n}$ for a cusp form of this special type actually satisfy some nice relations such as ${a_{nm}=a_na_m}$ when ${(m,n)=1}$. These relations are precisely the ones needed to conclude that there is a nice Euler product expansion and it is given by

$\displaystyle \displaystyle L(f,s)=\prod_{p|N}(*)\prod_{p\nmid N} \frac{1}{1-a_pp^{-s}+p^{k-1-2s}}.$

We say that a variety is modular if ${L(X,s)}$ coincides with ${L(f,s)}$ up to finitely many primes for some ${f\in S_k(\Gamma_0(N))}$. We’ve been ignoring the technicalities of dealing with the primes of bad reduction and the primes that divide the level (a surprisingly hard problem to determine when these are the same set!), but now we see that for the definition of a variety being modular this doesn’t even matter. There are other subtleties in defining all of this for when the variety does not have ${2}$-dimensional middle cohomology, but again for our immediate purposes you can trust that people have made the suitable adjustments.

Now we see the truly shocking results of Taniyama-Shimura. We take this incredibly symmetric analytic object (so symmetric it is surprising any exist at all) and we take this completely algebraic variety defined over ${\mathbb{Q}}$ and the conjecture claims that we can always find one of these symmetric things that match up with this action on the cohomology. Wiles and Taylor are often credited with proving it in 1994, but it wasn’t actually proved until 2003 by Breuil, Conrad, Diamond, and Taylor. This was the elliptic curve case.

Just last year Gouvea and Yui proved that all rigid Calabi-Yau threefolds are modular. It is a conjecture that all Calabi-Yau varieties over ${\mathbb{Q}}$ should be modular, so this includes K3 surfaces. It might seem weird that K3 surfaces haven’t been proven but the threefold case has been. This just has to do with those technicalities of what to do if the middle cohomology is bigger than 2-dimensional, which it always is. There you have it. The famous Taniyama-Shimura conjecture which led to a proof of Fermat’s Last Theorem.

# Taniyama-Shimura 3: L-Series

For today, we assume our ${d}$-dimensional variety ${X/\mathbb{Q}}$ has the property that its middle etale cohomology is 2-dimensional. It won’t hurt if you want to just think that ${X}$ is an elliptic curve. We will first define the L-series via the Galois representation that we constructed last time. Fix ${p}$ a prime not equal to ${\ell}$ and of good reduction for ${X}$. Let ${M=\overline{\mathbb{Q}}^{\ker \rho_X}}$. By definition the representation factors through ${{Gal} (M/\mathbb{Q})}$. For ${\frak{p}}$ a prime lying over ${p}$ the decomposition group ${D_{\frak{p}}}$ surjects onto ${{Gal} (\overline{\mathbf{F}}_p/\mathbf{F}_p)}$ with kernel ${I_{\frak{p}}}$. One of the subtleties we’ll jump over to save time is that ${\rho_X}$ acts trivially on ${I_{\frak{p}}}$ (it follows from the good reduction assumption), so we can lift the generator of ${{Gal} (\overline{\mathbf{F}}_p/\mathbf{F}_p)}$ to get a conjugacy class ${{Frob}_p}$ whose image under ${\rho_X}$ has well-defined trace and determinant.

We define

$\displaystyle L(X,s) =(*)\prod_{p \ good} \frac{1}{1-{tr}(\rho_X({Frob}_p))p^{-s}+\det(\rho_X({Frob}_p))p^{-2s}}$

where ${(*)}$ is a product of terms at the bad primes. Note that since this is a two-dimensional representation basic linear algebra tells us that the product is over the simpler expression ${(\det(I-\rho_X({Frob}_p)p^{-s}))^{-1}}$.

If you don’t like all this Galois representation stuff, we can describe this L-series without reference to the Galois representation at all. In order to ease notation we will denote the reduction of ${X}$ at a fixed good prime ${p}$ by ${Y:=X_{\mathbf{F}_p}}$ and base changing to the algebraic closure ${\overline{Y}:=X_{\overline{\mathbf{F}_p}}}$. To simplify notation let ${k=\mathbf{F}_p}$.

We have several natural Frobenius actions on ${\overline{Y}}$. The first we will call the absolute Frobenius which we will denote ${F_{ab}:\overline{Y}\rightarrow \overline{Y}}$. This is the identity on the topological space and the ${p}$-th power map on the structure sheaf. On affine patches ${{Spec} A\rightarrow {Spec} \overline{k}}$ the map is the one induced by ${a\mapsto a^p}$ on ${A}$. We can check directly that the map on topological spaces is the identity. For any prime ideal ${\frak{q}\in{Spec} A}$ the contraction ${\frak{q}^c=\{a\in A: a^p\in \frak{q}\}=\{a\in A: a\in\frak{q}\}=\frak{q}}$ by the property of ${\frak{q}}$ being prime. This map translates in the language of schemes to ${(id, F): (\overline{Y}, \mathcal{O}_{\overline{Y}})\rightarrow (\overline{Y}, \mathcal{O}_{\overline{Y}})}$ where ${F}$ is raising sections of the sheaf to the ${p}$-th power.

Note that the absolute Frobenius is not a map of ${\overline{Y}}$ over ${\overline{k}}$. The map is also not the pullback despite making the same commutative diagram.

Let the standard structure map be ${\phi: \overline{Y}\rightarrow {Spec}\overline{k}}$ and define ${\overline{Y}^{(p)}=\overline{Y}\otimes\overline{k}}$ to be the pullback of Frobenius acting on the base field. Since we have a commutative diagram we get by the universal property of a pullback diagram some map ${F_r: \overline{Y}\rightarrow \overline{Y}^{(p)}}$ called the relative Frobenius. We define the arithmetic Frobenius to be the projection on the first factor ${F_{ar}:\overline{Y}^{(p)}\rightarrow \overline{Y}}$. A nice exercise to see if you understand these would be to write down a big commuting diagram that relates all these. Due to wordpress constraints, I won’t actually do this.

Instead, we’ll do an example. Let ${Y={Spec} k[t]}$ (recall that ${k=\mathbf{F}_p}$). This means that ${\overline{Y}={Spec} \overline{k} [t]={Spec} (k[t]\otimes_k \overline{k})}$. The descriptions in terms of the ring homomorphism that induces the map on the spectra are as follows. The absolute Frobenius is still just ${f\mapsto f^p}$. The relative Frobenius is ${F_{ab}\otimes id}$. Since the absolute raises elements of ${k[t]}$ to the ${p}$, everything in ${k}$ is fixed by this map, and on ${\overline{k}}$ it is defined to be fixed. This means that the relative Frobenius only alters the ${t}$ by ${t\mapsto t^p}$. This is sometimes referred to as “raising coordinates to the ${p}$-th power”. The arithemtic Frobenius does nothing to the ${k[t]}$ part, but raises the ${\overline{k}}$ coefficients to the ${p}$, so ${\sum a_nt^n\mapsto \sum a_n^p t^n}$. Likewise, the geometric Frobenius takes the ${p}$-th root of the coefficients.

Straightforward (but non-trivial) computations also give that the map that the absolute Frobenius induces on the \'{e}tale site is trivial. If we look at our diagram we see that ${F_{ab}=F_{ar}\circ F_r}$. Since the induced map on cohomology is contravariant this gives ${F^*_r\circ F^*_{ar}=id}$. This means that on cohomology ${F^*_r=(F^*_{ar})^{-1}=F^*_{ge}}$ by definition of the geometric Frobenius.

Now the smooth, proper base change theorem for \'{e}tale cohomology tells us that ${H^d_{\textit{\'{e}t}}(\overline{Y}, \mathbb{Q}_\ell)\simeq H^d_{\textit{\'{e}t}}(\overline{X}, \mathbb{Q}_\ell)}$ which is two-dimensional. Since the ${F_r}$ action here is a linear operator on a vector space it makes sense to take the trace and determinant. We can define the L-series without use of the Galois representation as:

$\displaystyle L(X,s)=(*)\prod_{p \ good} \frac{1}{1-{tr}(F_r^*)p^{-s}+\det(F_r^*)p^{-2s}}$

where again the ${(*)}$ is a product of terms involving primes of bad reduction. Since there are only finitely many this is irrelevant for the definition of modularity. Of course we could have defined this without all the different Frobenius actions (we only used the relative one), but now we can get to the punchline. These two L-series are actually the same.
We just sketched above that the action of ${F_r}$ and ${F_{ge}}$ were the same on the \'{e}tale site. But ${F_{ge}=1\times {Frob}_p^{-1}}$ where ${{Frob}_p}$ is the canonical generator of ${{Gal} (\overline{k}/k)}$. We have a surjection ${{Gal} (\overline{\mathbb{Q}}/\mathbb{Q})\rightarrow {Gal} (\overline{k}/k)}$ and if we consider ${{Frob}_p}$ a lift of this element, then by the functoriality and equivariant isomorphisms above we get that ${{tr}(\rho_X({Frob}_p))={tr}(\rho_X'({Frob}_p^{-1}))={tr}(F_r^*)}$. The determinant term turns out to always be ${p^3}$ since it can be checked to be the third power of the ${\ell}$-adic cyclotomic character in both cases. Thus the two L-series are the same. This also tells us the representation is odd.

Note that they appear to be off by an inverse, but we actually took the contragredient representation of the one that acts on ${H^d_{\textit{\'{e}t}}(\overline{X}, \mathbb{Q}_\ell)}$, so the inverse corrects for this and they are actually the same.

This got a little technical at parts, so the one thing to take away from this post is that to any ${X/\mathbb{Q}}$ with two-dimensional middle cohomology we can produce some function which is just defined in terms of the trace and determinant of certain operators on the cohomology. This is called the L-series and will be crucial in the definition of modularity.

# Taniyama-Shimura 2: Galois Representations

Fix some proper variety ${X/\mathbb{Q}}$. Our goal today will seem very strange, but it is to explain how to get a continuous representation of the absolute Galois group of ${\mathbb{Q}}$ from this data. I’m going to assume familiarity with etale cohomology, since describing Taniyama-Shimura is already going to take a bit of work. To avoid excessive notation, all cohomology in this post (including the higher direct image functors) are done on the etale site.

For those that are intimately familiar with etale cohomology, we’ll do the quick way first. I’ll describe a more hands on approach afterwards. Let ${\pi: X\rightarrow \mathrm{Spec} \mathbb{Q}}$ be the structure morphism. Fix an algebraic closure ${v: \mathrm{Spec} \overline{\mathbb{Q}}\rightarrow \mathrm{Spec}\mathbb{Q}}$ (i.e. a geometric point of the base). We’ll denote the base change of ${X}$ with respect to this morphism ${\overline{X}}$. Suppose the dimension of ${X}$ is ${n}$.

Let ${\ell}$ be a prime. We consider the constructible sheaf ${R^n\pi_*(\mathbb{Z}/\ell^m)}$. Now we have an equivalence of categories between these sheaves and continuous ${G=Gal(\overline{\mathbb{Q}}/\mathbb{Q})}$-modules by taking the stalk at our geometric point. Thus ${R^n\pi_*(\mathbb{Z}/\ell^m)_v\simeq H^n(\overline{X}, \mathbb{Z}/\ell^m)}$ has a continuous action of ${G}$ on it, and hence we get a continuous representation ${\rho_{X,m}: G\rightarrow Aut(H^n(\overline{X}, \mathbb{Z}/\ell^m)\simeq GL_d(\mathbb{Z}/\ell^m)}$. These all form a compatible family and hence we can take the inverse limit and tensor with ${\mathbb{Q}_\ell}$ to get what is known as an ${\ell}$-adic Galois representation ${\rho_X: G\rightarrow GL_d(\mathbb{Q}_\ell)}$. For a technicality that will come up later, we will abuse notation and now relabel ${\rho_X}$ to be the dual (or contragredient) representation.

If you aren’t comfortable with etale cohomology, then you can just use it as a black box cohomology theory to get the same thing as follows. First take the base change ${\overline{X}\rightarrow \mathrm{Spec} \overline{\mathbb{Q}}}$. Given any element of the Galois group ${\sigma \in G}$ we get an automorphism of ${\overline{\mathbb{Q}}}$. Thus we can fill in the diagram:

${\begin{matrix} \overline{X} & \stackrel{\sigma}{\rightarrow} & \overline{X} \\ \downarrow & & \downarrow \\ \mathrm{Spec} \overline{\mathbb{Q}} & \stackrel{\sigma}{\rightarrow} & \mathrm{Spec} \overline{\mathbb{Q}} \end{matrix}}$

Since ${\sigma}$ was an automorphism, then only thing you have to believe about cohomology is that you then get an isomorphism via pullback ${H^n(\overline{X}, \mathbb{Q}_\ell)\stackrel{\sigma^*}{\rightarrow} H^n(\overline{X}, \mathbb{Q}_\ell)}$. Thus we get a continuous group homomorphism ${G\rightarrow Aut(H^n(\overline{X}, \mathbb{Q}_\ell))}$ as before. Again, we’ll actually use the dual of this in the future.

To return to an elliptic curve ${E}$ over ${\mathbb{Q}}$, we know that these are just tori, and hence the first Betti number is ${2}$. In this case we get that our Galois representation ${\rho_E: G\rightarrow GL_2(\mathbb{Q}_\ell)}$. If you’ve seen Taniyama-Shimura explained before, this should look familiar. This turns out to be exactly the same representation as the one you get from the Galois action on the Tate module. But the definition of the Tate module requires a group law, and hence the ability to get such a representation doesn’t generalize to all varieties in the way that using middle $\ell$-adic cohomology does. This is the standard modern approach to defining modularity for other types of varieties.

# Taniyama-Shimura 1

It’s time to return to plan A. I started this year by saying I’d post on some fundamental ideas in arithmetic geometry. The local system thing is hard to get motivated about, since the way I was going to use it in my research seems irrelevant at the moment. My other option was to blog some stuff about class field theory, since there is a reading group on the topic that I belong to this quarter.

The first goal of this new series is to understand the statement of the famous Taniyama-Shimura conjecture that led to the proof of Fermat’s Last Theorem. A lot of people can probably mumble something about the conjecture if they have any experience in algebraic/arithmetic geoemtry or any of the number theory type fields, but most people probably can’t say anything precise about what the conjecture says (I’ll continue to call it a “conjecture” even though it has been proved).

The statement of the conjecture is that every elliptic curve over ${\mathbb{Q}}$ is modular. Simple enough, but to unravel what it means to be modular we are going to have to take many posts just for the definition. If you’ve seen this explained before, it might still be interesting to read this series because I’m going to set up the machinery in a slightly different (but equivalent) way so that it will generalize to varieties other than elliptic curves in the future.

We’ll first define modular forms. A modular form of weight ${k}$ and level ${N}$ is an element of the vector space ${M_k(\Gamma_0(N))}$ which consists of holomorphic functions on the upper half plane ${f:\mathcal{H}\rightarrow \mathbf{C}}$ satisfying the additional transformation property that

$\displaystyle \displaystyle f \left(\frac{az+b}{cz+d}\right)=(cz+d)^kf(z)$

for all matrices ${\left(\begin{matrix} a & b \\ c & d \end{matrix}\right)\in SL_2(\mathbf{Z})}$ such that ${c\equiv 0 \mod N}$ (plus something else that we’ll get to shortly).

This is an analytic object if there ever was one. If this is the first time you’ve seen this, then the thing to pay attention to is that these depend on a choice of weight, ${k}$, and level, ${N}$. To get a feel for the level, note that it becomes “easier” to satisfy this transformation law as the level increases, because the amount of matrices we have to check is less. For example, when ${N=1}$ this says our ${f}$ has to behave nicely under every single linear fractional transformation that sends the upper half plane to the upper half plane. One might reasonably guess that ${0}$ is the only holomorphic function with this property. More on this later. The weight is a little harder to get a feel for.

The map ${z\mapsto e^{2\pi i z}}$ is a holomorphic map from the upper half plane onto the punctured unit disk. Note that ${e^{2\pi i z}\rightarrow 0}$ as ${z}$ tends to infinity along the imaginary axis. We can compose with this map and consider our modular form to be a holomorphic function on the punctured disk. This is well-defined because if ${e^{2\pi i z}=e^{2\pi i w}}$, then ${z}$ and ${w}$ differ by an integer and ${f(z+n)=f\left(\left(\begin{matrix}1 & n \\ 0 & 1 \end{matrix}\right)\cdot z\right)=(1)^kf(z)=f(z)}$.

We say ${f}$ extends to be holomorphic at infinity if there is a holomorphic extension to the whole disk. We require modular forms to have this property. Thus a modular form has a Fourier expansion called a ${q}$-expansion denoted

$\displaystyle \displaystyle f=\sum_{n=0}^\infty a_nq^n \ \text{where} \ q=e^{2\pi i z}$

(note that a Fourier series in general involves negative powers, but these would give a pole at infinity). The cusp forms are the subspace denoted $S_k(\Gamma_0(N))$ of the modular forms that vanish at all cusps. To define cusp, just think of the extended upper half plane as ${\mathbf{H}\cup \mathbb{P}^1_{\mathbb{Q}}}$. We stick all the rational numbers along the real line in and also throw in a point at infinity. In practice, we only have to check holomorphic extension across finitely many of these cusps because due to the transformation law we only need to pick on cusp in each equivalence class under the action of the matrix group. When for instance $N=1$ again, all we have to check is that $f$ vanishes at infinity, or upon composing to the disk we get that $a_0=0$.

Do any of these things exist? Well, as we’ve already noted, for small N it seems very hard to satisfy these properties. In fact, our guess was right, $\dim_\mathbb{C} S_k(\Gamma_0(1))=0$ for $1. So until we bump the weight up to 12, we actually only have the 0 function satisfying our properties. For weight 12, there is only one up to scalar multiple. This doesn't look good, but actually when we allow the level to grow we get a lot (even of low weight). But before next time, just ponder how severe the symmetry condition we are imposing is. Somehow every elliptic curve is closely related to one of these which is why the result is so surprising.

Now we have our basic analytic object of the conjecture. The next several posts will go back to the algebraic side of things. Depending on how much detail I decide to give to define the terms in the Taniyama-Shimura conjecture could take anywhere from 4 to 8 or so posts, just to give you an idea of how long you have to hold out for the statement.