## Some thoughts on a provoking discussion

So I recently stumbled upon (via Izabella Laba) the discussion at Scott Aaronson’s blog that arose from the events surrounding the dismissal of Walter Lewin.

Amazingly enough, I actually read through the entire 593 comment responses. This was a surprisingly intelligent a civil discussion (on the internet!) between people who don’t completely see eye-to-eye about everything, and about sexism, no less!

The discussion is a little disheartening for the first (roughly) hundred comments or so. However, starting some time around the linked comment, things get a lot better—as a whole, the major players in the discussion actually listened and seemed to empathize with one another, if not perfectly all the time.

A few thoughts:

1. I think that Scott (and the main people in the discussion) did a great job of ignoring the more troll-ish posts. There are a number scattered throughout—towards the end, in particular, there is a post which calls for the ban of Amy (if only for a few days!), which thankfully is largely ignored. However…

Comments such as these are an interesting instance of Lewis’ Law. I do believe that Scott is a good person who—as much as possible—eschews overly sexist views. But it’s interesting that underscoring a discussion about the role of women in STEM fields that there is—quite literally!—a constant low-level buzz of commentary that at the least borders on anti-feminist. So if you were someone reading this post who held views similar to those of Amy (which are not radical in the least), on one hand I would be welcomed that the major discussion is civil and interesting. On the other hand, it’s also believable that you might also feel like the room in which the discussion is happening is subtly hostile to you and your views. Is it surprising that women might be discouraged from self-advocacy in situations like this?

I really should stress that I think that Scott did a wonderful job in this discussion of staying on point, not engaging the trolls, etc. But the existence of these background comments really does suggest something, I think.

2. On that note, seriously? Amy is by no means a “radical feminist” in her postings. I would describe her as pretty middle-of-the-road (although that may say more about me than anything, I guess). She advocates for communication and being aware of the existence of structural imbalances. CRAZY AND RADICAL INDEED.
3. Reading through this sort of discussion really makes me think again about the difficulty of communication when we don’t define our terms—or in this case, when either the context is difficult to convey, or the terms themselves may not be easily definable. Many of the flare-ups that occured throughout the discussion often seemed to result from a mis-reading of what one of the other posters was trying to convey. Not all, certainly, since not everyone agreed on a variety of issues. But there were still many of them.

Anyhow, it was a surprisingly edifying read, although I can’t really say if I would recommend reading through all 593 comments, which will take quite a long time regardless. Still, I’m glad to see that civil discussion about sexism among people who do not agree can take place in this day and age. Kudos to Scott, Amy, Gil, Vijay, dorothy, and a few others.

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## Projectivity (continued)

So what does it mean for a variety to be projective? Well, that’s easy: A variety is projective if you can embed it in projective space.

That’s easy, but that’s not particularly helpful.

What are the benefits of being projective? Why is it something that we should care about?

The way I see it, the main advantage of projectivity is that any analytic projective variety is in fact algebraic i.e. it can be described in terms of zero sets of polynomials, and not just analytic functions. This is essentially a loose paraphrase of Chow’s theorem.

So this explains why projectivity is a good thing, but it doesn’t tell us how to detect it. To help with this, let’s consider what we do get if a variety (we will only really care about tori, but for now we will be more general) is projective.

On $\mathbb{P}^N$, we have the line bundle $\mathcal{O}(1)$. Thus, given a morphism $f : X \to \mathbb{P}^N$, we can pull this line bundle back to obtain a line bundle $L := f^*\mathcal{O}(1)$. This is an ample bundle; that is, if we take a sufficiently high power $L^{\otimes n}$, then sections of this new bundle will in fact yield an embedding into projective space of some dimension. More specifically, choose a basis $s_0, \ldots, s_N$ for $H^0(X, L^{\otimes n})$. Then as this line bundle is base-point free (it comes from a map into projective space), we can consider the map

$X \to \mathbb{P}H^0(X, L^{\otimes n}) \qquad x \mapsto (s_0(x) : \cdots : s_N(x))$

then this map will be an embedding.

Given such a pair $(X, L)$ consisting of a complex manifold $X$ together with an ample line bundle $L$, then we can see that it must be projective. Such a pair is called a polarized variety*.

Now, many manifolds come with natural choices of polarizations; for example, all non-elliptic curves have either their canonical or anti-canonical bundle which are ample, and so they are just naturally polarized. Elliptic curves are as well, but you can’t use their canonical bundle, since it is trivial.

The same is of course true with complex tori; their canonical bundles are trivial, and so these do not provide us with a projective embedding. So let’s see what else a polarization gives us.

Let’s consider the first chern class of our line bundle. We have (since we are working with the complex numbers) the exponential sequence of sheaves

$0 \to \mathbb{Z} \to \mathcal{O} \to \mathcal{O}^\times \to 0$

which yields the long exact sequence some of whose low degree terms are

$\cdots \to H^1(X, \mathcal{O}) \to H^1(X, \mathcal{O}^\times) \to H^2(X, \mathbb{Z}) \to \cdots$

where $H^1(X, \mathcal{O}^\times)$ is the Picard group of $X$ (denoted $Pic(X)$); that is, the group of line bundles on $X$. The map to $H^2(X, \mathbb{Z})$ is the map which takes a line bundle to its first chern class $c_1(L)$. It is this that we use to understand what makes a manifold projective.

In the case of tori, we know very well what $H^2(X, \mathbb{Z})$ (henceforth we will omit the coefficient ring if it is the integers) is. In fact, due to the Künneth theorem and the fact that topologically, a complex torus is simply a product of circles, we have the isomorphisms

$H^2(X) \cong \Lambda^2 H^1(X) \cong \Lambda^2 H_1(X)^\vee \cong \big(\Lambda^2 H_1(X)\big)^\vee$

Exercise: Check these!

That is, an element of $H^2(X)$ can be though of as an alternating bilinear form $E$ on the underlying lattice $H_1(X)$. In particular, the first chern class of a polarization on a complex torus $X = \mathbb{C}^k / \Gamma$ is an alternating form on its underlying lattice $\Gamma = H_1(X)$.

Now, it is not too hard to see (and you should check this) that there is a bijective correspondence between alternating bilinear forms $E$ on a lattice $\Gamma \subset \mathbb{C}^k$ which satisfy

$E(iv, iw) = E(v,w)$

and hermitian forms $H$ on $\mathbb{C}^k$ which satisfy $\mathfrak{Im}\, H(\Gamma, \Gamma) \subset \mathbb{Z}$; this is given by the bijection

$E(-,-) \qquad \iff \qquad E(i-,-) + iE(-,-)$

Another way to say this is that alternating bilinear forms on $\Gamma$ which are compatible with the complex structure on $\mathbb{C}^k$ are (essentially) the same as hermitian forms on $\mathbb{C}^k$ whose imaginary parts take integer values on $\Gamma$.

And the magic about this is that an element $E \in H^2(X)$ is the first chern class of an ample line bundle if and only if this latter condition is satisfied.

*Well, that’s not exactly correct. It isn’t the line bundle $L$ that is the polarization, but the class of the line bundle in the Neron-Severi group of $X$. But it’s close enough.

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## Why are elliptic curves projective?

Before we go on to discuss what makes a complex torus algebraic, let us perhaps return to the case of elliptic curves. The claim is that all elliptic curves are algebraic; other than the one case explicitly provided in the first post, this has by no means been shown, so let us dwell on this a little further.

There are a few ways that we can see this fact. We will go over the most obvious one first.

Fix an element $\tau$ in the upper half-plane (and so in effect, fix a lattice $\Lambda_\tau$ and hence an elliptic curve). Consider the complex function

$\displaystyle \wp(z) = \frac{1}{z^2} + \sum_{w \in \Lambda_\tau}\!\!{}'\ \Big(\frac{1}{(z + w)^2} - \frac{1}{w^2}\Big)$

where by convention the primed summation means that we sum over all non-zero elements of the lattice. This function is called the Weirstrass $\wp$-function. Note that this function satisfies

$\wp(z + w) = \wp(z)$

for all $z \in \mathbb{C}$ and $w \in \Lambda_\tau$. While it is not holomorphic (it has a pole of order 2 at every lattice point), it is meromorphic, and it is translation invariant for every element in the lattice. It thus yields a well-defined meromorphic function on the elliptic curve $E_\tau = \mathbb{C}/\Lambda_\tau$.

We claim the following relationship holds:

$\displaystyle \big(\wp(z)'\big)^2 = 4\big(\wp(z)\big)^3 - g_2\wp(z) - g_3$

where $g_2, g_3$ are given by the expressions

$\displaystyle g_2 = 60 \sum_{w\in \Lambda_\tau}\!\!{}'\ w^-4$

$\displaystyle g_3 = 140 \sum_{w\in \Lambda_\tau}\!\!{}'\ w^-6$

It is interesting to remark that these expressions $g_2, g_3$ are in fact the Eisenstein series of weights 4 and 6, respectively. This can be checked by verifying how they transform as functions of $\tau$ under the two transformations

$\tau \mapsto \tau + 1 \qquad \tau \mapsto -\frac{1}{\tau}$

and noting that they are thus modular forms of weights 4 and 6, respectively; if you happen to know that these spaces are one-dimensional, then you are done. If not, then the following exercise is somewhat instructive.

Exercise Show that the function $g_2(\tau)$ (where we now mention its explicit $\tau$-dependence) can also be written as

$\displaystyle g_2(\tau) = \frac{4\pi^4}{3}\Big(1 + 240\sum_{k=1}^\infty \sigma_3(k)e^{2\pi i k \tau}\Big)$

$\pi \cot \pi x = \sum_{k \in \mathbb{Z}} \frac{1}{x + k}$

which can be checked by looking at the zeros and poles of these two functions.

Now, the shown identity can be verified by comparing the poles on the left- and right-hand sides of the expression; as they match up, the two expressions must be equal (why?)

The point of all this, of course, is that the expression above provides us an explicit description of our elliptic curve as the affine plane curve

$y^2 = 4x^3 - g_2x - g_3$

and in particular, we see that every elliptic curve can be written as the curve cut out by a cubic polynomial.

Now, while this is all true, it is not particularly enlightening. If, after reading this, someone were to ask you “Why are elliptic curves projective?”, all you could answer would be “Because the Wierstrass $\wp$-function exists and satisfies a certain differential equation”. It would not answer, in any way, the question as to why complex tori are not necessarily projective. So if our goal was to answer that question, then this approach, while interesting, fails.

For us to move on, we need to perhaps consider more what exactly it means for a variety to be projective.

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## Abelian varieties

We would like to generalize one of the nice properties of Elliptic curves, namely that maps between them are determined by linear Algebra: Specifically, sublattices of the lattice $\mathbb{Z} \oplus \mathbb{Z}$. To do so we must first introduce the notion of an Abelian variety.

I would like to first address one bit of (somewhat) confusing nomenclature that necessarily arises. An Abelian variety is, first an foremost, a torus. Now, to which torus am I referring?

• Algebraic Torus: The simplest definition of an Algebraic torus is that it is the product of copies of the multiplicative group of a field; since I only work with the complex numbers, it would be isomorphic to $(\mathbb{C}^\times)^k$ for some integer $k$.
• Topological Torus: When topologists speak of a torus, they usually mean specifically $S^1 \times S^1$; that is, an elliptic curve if we forget the complex structure; the surface of a donut. Slightly more generally, an product $(S^1)^{\times k}$ can be called a torus with very little guilt. As we work over the complex numbers, we really only consider even (real) dimensional manifolds, so we would in fact only consider $(S^1)^{\times 2k}$.

Now, there is a relationship between these two, most obviously that $\mathbb{C}^\times$ is homotopic to the circle $S^1$. A little more strongly than that (which doesn’t quite agree with my definition of an Algebraic torus provided above), it turns out that there are two real forms of the torus: $\mathbb{R}^\times$ and $S^1$! So in some sense they are the same, but for my purposes we will only consider the second version of a torus.

So let’s return to the definition of an abelian variety. First and foremost, it is a complex torus. Topologically, it is simply the product $(S^1)^{\times 2k}$. However, to endow it with a complex structure we in fact consider it as the quotient

$\mathbb{C}^k/\Lambda$

where $\Lambda$ is a discrete co-compact subgroup of $\mathbb{C}^k$. It should be clear of course that a one-dimensional (complex dimension, remember) complex torus is nothing but an elliptic curve.

Now, much the same as we classify elliptic curves by looking at the upper half-plane, we would like to classify complex tori with an analogous structure. We begin to do so as follows.

A (discrete) rank $2k$ sublattice of $\mathbb{C}^k$ can be described by choosing $2k$ column vectors in $\mathbb{C}^k$ which span the whole space (as an $\mathbb{R}$-vector space). If we write these as a matrix $\Pi$ (which we call the period matrix), then this latter condition is equivalent to the matrix

$\displaystyle \begin{pmatrix} \Pi \\ \overline{\Pi}\end{pmatrix}$

being non-degenerate.

Exercise: Check this at least for the case $k = 1$, i.e. for elliptic curves. It will be easier to see this if you make a simplifying assumption that is to follow.

Now, much as the case for elliptic curves, multiplication of this basis by any matrix in $GL_k\mathbb{C}$ will yield an isomorphic complex torus (if you don’t believe me, you should check this). In particular, if we write $\Pi$ as

$\Pi = \begin{pmatrix} Z_1 & Z_2 \end{pmatrix}$

with $Z_i$ being $k \times k$ complex matrices (which are necessarily invertible), then multiplying by $Z_1^{-1}$ yeilds the simpler form

$\Pi \equiv \begin{pmatrix} Id_k & Z_1^{-1}Z_2 \end{pmatrix} \pmod {GL_k\mathbb{C}}$

more simply, we have that up to isomorphism, every complex torus has a period matrix of the form

$\Pi = \begin{pmatrix} Id_k & Z \end{pmatrix}$

where $Z$ is an invertible $k \times k$ complex matrix.

Now, as in the case of elliptic curves, maps between complex tori are particularly simple. In particular, they always factor into a linear map on the underlying vector space (which fixes the lattice; equivalently, a map of the lattices) followed by a translation. That is, if we denote by $t_y : X \to X$ the map given by $x \mapsto x + y$, then a given morphism $f : Y \to X$ can be written as $f = t_{f(0)} \circ g$ where $g$ lifts to a linear map preserving the lattices on the respective universal covers.

Now, this would be all well and good but for one important fact. As stated before, elliptic curves can always be embedded into projective space as a plane cubic. In particular, as a closed subvariety of projective space, they are in fact algebraic.

However, from the definition above, it turns out that “most” (in a suitably precise sense) complex tori are not algebraic. This is less than ideal, and so we would like to restrict ourselves to those tori which are algebraic. We call those tori Abelian Varieties.

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## On to higher genus curves

Let us summarize what we have discussed about Elliptic curves. First of all, we can parameterize Elliptic curves by either studying the moduli of lattices in the complex plane, or by looking at the Hodge decomposition of their first cohomology group, $H^1(E, \mathbb{C})$.

We also discussed how maps between elliptic curves are in fact determined by maps on the underlying lattices, in effect reducing the problem of studying maps to one of linear algebra. Let us discuss this somewhat further.

The first thing to note is that any map between two elliptic curves is unramified. That is, a map $E_1 \to E_2$ is in fact a covering map. This can be seen quickly with the Riemann-Hurwitz formula: given a map $f : C \to C'$ of curves, we have the relationship

$\chi(C) = d \cdot \chi(C') - \deg R$

where $d$ is the degree of the map, and $R$ is the ramification divisor. Since $\chi(E) = 0$, it follows that $\deg R = 0$, and so the map is unramified.

Consequently, through the theory of covering spaces, there is a Galois correspondence between covers of an Elliptic curve $E$ and subgroups of the fundamental group $\pi_1(E) \cong \mathbb{Z} \oplus \mathbb{Z}$. It then follows that degree $d$ covers of an elliptic curve are in bijective correspondence with sublattices of $\mathbb{Z} \oplus \mathbb{Z}$ of index $d$. As an aside, it is easy enough to show that there are exactly

$\displaystyle \sigma_1(d) = \sum_{k \mid d} k$

such covers.

So how can we carry this on to higher genus curves? Since maps between higher genus curves may be ramified (for example, any map of a higher genus curve to an elliptic curve necessarily has ramification), we can no longer look directly to the theory of covering spaces and the resulting analysis of the fundamental group. Instead what we will do, in some sense, is to linearize our maps of curves. To do so, we will introduce the Jacobian of a curve. It will naturally follow that Elliptic curves are those curves which are isomorphic to their Jacobians, and so all of the above discussion can be seen in some sense as a toy model of what is to follow.

## Hodge decomposition

We discussed in the previous post that we can understand the moduli of Elliptic curves by looking at the moduli of rank two lattices. That is, we consider them as $\mathbb{C}/\Lambda_\tau$ for some element $\tau$ in the upper half-plane. It followed that

$\mathcal{M}_{1,1} \cong \mathbb{H}/SL_2\mathbb{Z}$

is the moduli space of elliptic curves. The notation $\mathcal{M}_{1,1}$ will be discussed further at a later point; in brief, it is the moduli space of 1-marked genus 1 curves: Elliptic curves, that is.

However, there is another way of discussing the moduli of Elliptic curves. If we instead consider an Elliptic curve as an abstract 1-dimensional complex manifold, then its genus can be defined as

$g = \dim H^1(E, \mathcal{O}_E) = \dim H^0(E, K_E)$

i.e. we can define the genus to be the dimension of the space of holomorphic 1-forms. Since this is the same as the topological genus, we know this to be one.

Now, we further know that all elliptic curves are homeomorphic. In particular, for every elliptic curve $E$ we have that their cohomology groups are isomorphic, and in particular they all have $H^1(E, \mathbb{Z})$ isomorphic to the fixed lattice

$\Lambda = \displaystyle \Big(\mathbb{Z}^2, \langle\, , \rangle\Big)$

where the intersection form can be written as the matrix

$\begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}$

We will use this to provide a new look at their moduli.

Let us first note that $\Lambda$ has an almost-complex structure: that is, there is a linear endomorphism $I : \Lambda \to \Lambda$ which satisfies

• $I^2 = -Id_\Lambda$
• $\langle Iu, Iv \rangle = \langle u, v \rangle$ i.e. the intersection form is preserved.

Exercise: What is this endomorphism?

We can then look at the complexification $\Lambda_\mathbb{C} = \Lambda \otimes \mathbb{C}$. This splits into $\pm i$-eigenspaces, and so we write

$\Lambda_\mathbb{C} \cong \Lambda^{1,0}\oplus \Lambda^{0,1}$

where $\Lambda^{1,0}$ is the $i$-eigenspace (and vice versa).

Lastly, since $\Lambda$ has an almost-complex structure, it follows that $\Lambda_\mathbb{R} = \Lambda \otimes \mathbb{R}$ can be given the structure of a complex vector space, and moreover, there is thus a complex-linear isomorphism $\Lambda_\mathbb{R} \cong \Lambda^{1,0}$.

So we now return to using this idea to describe the moduli of elliptic curves. In brief, the decomposition of $\Lambda$ into $\Lambda^{1,0} \oplus \Lambda^{0,1}$ is called the Hodge decomposition. This decomposition will also provide a way to discuss the moduli of Elliptic curves.

The idea is as follows. We define a lattice-polarized Elliptic curve to be an elliptic curve E together with a choice of isomorphism

$\phi : H^1(E, \mathbb{Z}) \to \Lambda$

which we know exists because Elliptic curves are homeomorphic. This of course yields a complexified map $\phi_\mathbb{C} : H^1(E, \mathbb{C}) \to \Lambda_\mathbb{C}$.

We also have an integration map

$\iota : H^0(E, K_E) \to Hom\big(H_1(E, \mathbb{Z}), \mathbb{C}\big) \cong H^1(E, \mathbb{C})$

given by mapping an element $\omega \in H^0(E, K_E)$ to the map

$\gamma \mapsto \int_\gamma \omega$

The composition of these two maps is then a non-zero map $\iota \circ \phi_\mathbb{C} : H^0(E, K_E) \to \Lambda_\mathbb{C}$. Since this map is $\mathbb{C}$-linear, it follows that its image must be $\Lambda^{1,0}$.

There are now a few things that we can say about this map. First of all due to the intersection form on $\Lambda_\mathbb{C}$, it follows that for any $\lambda \in \Lambda^{1,0}$ that

$\mathfrak{Im}\,\langle \lambda, \overline\lambda\rangle > 0$

Exercise: Check this!

It follows of course that for any $\omega \in H^0(E, K_E)$ that $\langle\iota \circ \phi_\mathbb{C} \omega , \overline{\iota \circ \phi_\mathbb{C} \omega }\rangle$ has positive imaginary part. The punchline is now the following.

Consider $\mathbb{P}\Lambda_\mathbb{C}$ (we consider the projectivisation since we only care about the image of the line $\mathbb{C}\omega$). Then the collection of those lines $[\omega]$ satisfying

$\mathfrak{Im}\,\langle \omega , \overline\omega\rangle > 0$

form a moduli space for lattice-polarized Elliptic curves. Moreover, if we quotient out by the natural action of $SL_2\mathbb{Z}$ then we recover the usual moduli space of Elliptic curves.

Another way of saying this is that the complex structure on the Elliptic curve determines (and is determined by) its Hodge decomposition on $H^1(E, \mathbb{C})$.

In further posts, we will consider how to extend these ideas to higher genus curves, and even to higher dimensional analogs of Elliptic curves, namely Abelian Varieties.

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## Moduli

Let $E_1$ and $E_2$ be a pair of elliptic curves. We saw in the previous post that all maps $E_1 \to E_2$ are in fact linear; that is, they arise from linear maps on their universal covers, $\mathbb{C}$. More specifically, they arise from linear maps on $\mathbb{C}$ which preserve the underlying lattices.

The first question which arises is the following.

When are two Elliptic curves isomorphic?

Let’s describe our curves a little more explicitly. We will let $\Lambda_1 = \langle z_1, z_2 \rangle$ and $\Lambda_2 = \langle w_1, w_2\rangle$ denote the underlying lattices of the curves $E_1, E_2$, respectively. There is then the map $z \mapsto z_1^{-1}\cdot z$ which takes the lattice $\Lambda_1$ to the lattice $\Lambda_\tau = \langle 1, \tau\rangle$ where $\tau = z_2/z_1$. In particular, since the map described above is linear, we have an isomorphism

$E_1 \cong E_\tau = \mathbb{C}/\Lambda_\tau$

whose inverse is given by $w \mapsto z_1 \cdot w$. Since isomorphism is an equivalence relation, we see then that $E_1 \cong E_2$ if $z_2 / z_1 = w_2/w_1$. With this being the case, we will often look at the parameter $\tau$ to classify elliptic curves. Due to orientation concerns, we can in fact restrict to the case that $\mathfrak{Im}\tau > 0$. With that in mind, let $\mathbb{H}$ denote the upper half-plane, the collection of those complex numbers with strictly positive imaginary part. We will use this as the starting point to discuss a parameter space, or moduli space for the collection of elliptic curves.

Recall that our question of the day is

When are two Elliptic curves isomorphic?

We have only partially answered this question: We noted that if the ratio of the basis elements are the same, then the curves must be isomorphic. But it seems that the choice of basis elements should also not matter, since an Elliptic curve is defined via a lattice, and not a lattice with a choice of basis. And it turns out that this is indeed true, which has some rather interesting consequences.

Recall that the group $SL_2\mathbb{Z}$ is the group of 2 by 2 invertible integer matrices. This can be thought of as the collection of all possible bases of a rank two free Abelian group. We will use this to identify other isomorphic curves.

Let $E_\tau = \mathbb{C}/\Lambda_\tau$ be an elliptic curve, and let

$\gamma = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$

be an element of $SL_2\mathbb{Z}$. Note that we have chosen as our basis of $\Lambda_\tau$ the elements $1, \tau$, but we could have just as easily have chosen the elements $c\tau + d, a\tau + b$. That is,

$\langle 1, \tau\rangle = \langle c \tau + d, a\tau + b\rangle$

However, but the same discussion as before, we can multiply this right-hand description by the complex number $(c\tau + d)^{-1}$ to see that the lattices $\Lambda_\tau$ and $\Lambda_{\gamma\tau}$ are isomorphic, where

$\displaystyle\gamma\tau = \frac{a\tau + b}{c\tau + d}$

It follows that if we want to describe the correct moduli space for Elliptic curves, that we must consider instead $\mathbb{H}/SL_2\mathbb{Z}$, where $SL_2\mathbb{Z}$ acts on $\mathbb{H}$ via

$\displaystyle \begin{pmatrix} a & b \\ c & d \end{pmatrix} \cdot \tau = \frac{a \tau + b}{c \tau + d}$

This quotient space is a coarse moduli space for classifying Elliptic curves.