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Fractional linear transformations and elliptic curves

The following two lemmas might be encountered in a basic course in complex analysis (the first in a basic course in group theory, even).

Lemma 1: Fix a field $F$. The group of fractional linear transformations $PGL_2(F)$ acts triple transitively on $\mathbb{P}^1(F)$ and the stabilizer of any triplet of distinct points is trivial.

Lemma 2: The group of fractional linear transformations on $\mathbb{P}^1(\mathbb{C})$ preserving the upper half plane $\mathbb{H} = \{ z \in \mathbb{C} | \text{Im}(z) > 0 \}$ is $PSL_2(\mathbb{R})$.

I used to only know extremely boring computational proofs of both of these statements. However, I now know better! Today I’d like to give shorter and conceptual proofs of both of these, and then briefly discuss how they come about in the study of elliptic curves (a subject I’d like to talk about in more detail once this semester is over).

Background

Fix a field $F$. The vector space $F^2$ has automorphism group the general linear group $GL_2(F)$, the group consisting of all $2 \times 2$ invertible matrices with coefficients in $F$. Linear transformations preserve lines through the origin, so $GL_2(F)$ also acts on the set of such lines. (More generally, linear transformations act on Grassmannians.) The space of lines in $F^2$ is denoted $\mathbb{P}^1(F)$, the projective line over $F$, because it is an essentially “one-dimensional” object. In some basis, it consists of the span of the vectors $(a, 1), a \in F$ and the span of the additional vector $(1, 0)$; the former is referred to as an “affine slice” $\mathbb{A}^1(F)$ of $\mathbb{P}^1(F)$, and the latter is referred to as the “point at infinity.” In algebraic geometry this construction is important for providing the extra points that make nice theorems such as Bezout’s theorem true; it can be thought of as a kind of compactification.

$GL_2(F)$ acts on lines as follows. A linear transformation

$\displaystyle \left[ \begin{array}{cc} a & b \\ c & d \end{array} \right]$

sends the line spanned by $(z, 1)$ (hereafter denoted by projective coordinates $(z : 1)$) to $(az + b : cz + d)$, which, if $cz + d \neq 0$, is the same as $\left( \frac{az+b}{cz+d} : 1 \right)$. (If $cz + d = 0$, then we are instead at the point at infinity.) Thus we can identify $SL_2(F)$ with the group of fractional linear transformations acting on $\mathbb{P}^1(F)$, up to the fact that scalar multiples of the identity act trivially on lines. So the group actually acting is $GL_2(F)/F^{\times} = PGL_2(F)$, the projective general linear group.

When $F = \mathbb{C}$ (the case that is particularly important to complex analysis), the projective general linear group coincides with the projective special linear group $PSL_2(F)$, since the coefficients of any invertible matrix can be scaled until the matrix has determinant $1$. The projective line $\mathbb{P}^1(\mathbb{C})$ admits a famous interpretation here: under stereographic projection it can be identified with the Riemann sphere, and $PSL_2(\mathbb{C})$ is often called the group of Mobius transformations in this context. As it so happens, $PSL_2(\mathbb{C})$ is precisely the group of biholomorphic automorphisms of the Riemann sphere, which follows from the fact that the only meromorphic functions on the Riemann sphere are the rational functions.

Lemma 1

The key to a conceptual proof of Lemma 1 is the following observation: in the action of $PGL_2(F)$ on $\mathbb{P}^1(F)$, the stabilizer of the point at infinity is the group $z \mapsto az + b$ of affine linear transformations. It follows that we may understand the action of $PGL_2(F)$ on triplets of points $p, q, r$ by first sending the first point to infinity. The remaining two points can only be acted on by affine linear transformations, but the affine linear transformations clearly act transitively on pairs of distinct points, and do so with trivial stabilizer!

The previous proof I knew of this lemma involved showing that the particular triplet of points $0, 1, \infty$ had trivial stabilizer and then showing that any triplet of points could be sent to this one by writing down the coefficients of a transformation doing the job explicitly.

This lemma appears when studying elliptic curves in the following way. One can classify smooth projective algebraic curves over $\mathbb{C}$ by the number of “holes” they have topologically, or more precisely by their genus. An incredibly powerful tool to help us do this is the Riemann-Roch theorem, which I need to learn about at some point. Using Riemann-Roch one can show that the only smooth projective curve of genus zero is the projective line $\mathbb{P}^1(\mathbb{C})$ (and in fact this holds over more general fields and has number-theoretic significance), so the next obvious thing to do is classify curves of genus one.

The Riemann-Roch theorem implies that any smooth projective curve of genus one can be written in the form

$\displaystyle a_1 y^2 + a_2 xy + a_3 y = a_4 x^3 + a_5 x^2 + a_6 x + a_7$

(where I’m using affine coordinates for the sake of familiarity.) After a change of coordinates, it’s possible to write any such curve in the form

$\displaystyle y^2 = (x - a)(x - b)(x - c)$

for some distinct $a, b, c \in \mathbb{C}$ (the curve cannot be smooth at a repeated root of the RHS). In fact, any curve of the form

$\displaystyle y^2 = (x - a)(x - b)(x - c)(x - d)$

where $a, b, c, d$ are distinct points on the Riemann sphere is a smooth projective curve of genus one (where if one of the points is at infinity one should interpret $x - a$ as $1$). As it turns out, two such curves are isomorphic if and only if the roots of the corresponding quartic polynomials can be sent to each other by a fractional linear transformation, so one can identify the space of isomorphism classes of smooth projective curves of genus one with the space of isomorphism classes of (unordered) 4-tuples of distinct points on $\mathbb{P}^1(\mathbb{C})$ up to projective linear transformation.

An elliptic curve is a smooth projective curve of genus one equipped with a distinguished point. This point serves as the identity for a group law defined on any elliptic curve, which comes abstractly from an identification of an elliptic curve with its Jacobian variety. It turns out that in order for the group law to have the “usual” definition in terms of the chord-tangent construction, the distinguished point on $y^2 = (x - a)(x - b)(x - c)(x - d)$ needs to be one of the four points $a, b, c, d$.

Given an elliptic curve presented in the form $y^2 = (x - a)(x - b)(x - c)(x - d)$ in which $d$ is the distinguished point, send $d$ to infinity by some fractional linear transformation. After applying this fractional linear transformation, we now have an elliptic curve of the form $y^2 = (x - a')(x - b')(x - c')$ whose group law is given by the chord-tangent construction and whose identity is at infinity. Now the remaining roots $a', b', c'$ can be transformed further by fractional linear transformations, but in order for two elliptic curves written in this way to be isomorphic the group identity (infinity) needs to be preserved, so instead of fractional linear transformations we can only transform the roots via affine linear transformations. Sending $a' \to 0, b' \to 1$, we finally get an elliptic curve in Legendre normal form

$\displaystyle y^2 = x(x - 1)(x - \lambda)$

where $\lambda \in \mathbb{C} -\{0, 1 \}$ is the remaining root. This can be understood in terms of Lemma 1 as follows: every equivalence class of ordered 4-tuples of points on the Riemann sphere has a unique representative in the form $(\infty, 0, 1, \lambda)$. However, since order is irrelevant, we could have chosen to send any of $a', b', c'$ to $0, 1$ to get $\lambda$, so the Legendre normal form is unique only up to the action of $S_3$ on the three roots of the RHS not at infinity. One can write down the fractional linear transformations giving these permutations explicitly, and this gives a natural identification

$\left\{ \text{isomorphism classes of elliptic curves over } \mathbb{C} \right\} \simeq \left( \mathbb{C} - \{ 0, 1 \} \right)/S_3$.

A generic orbit of $\mathbb{C} - \{0, 1\}$ has size six, but there is a special point with an orbit of size three and another special point with an order of size two. (If you don’t know what these points are, it’s a fun exercise to figure them out.) The corresponding elliptic curves are very special!

Lemma 2

The key to a conceptual proof of Lemma 2 is the following observation: any Mobius transformation which fixes the upper half plane $\mathbb{H}$ (considered as a subset of $\mathbb{C}$ after identifying $\mathbb{P}^1(\mathbb{C})$ with the one-point compactification $\mathbb{C} \cup \{ \infty \}$) also fixes the lower half plane, and therefore it also fixes $\mathbb{R} \cup \{ \infty \}$. In particular, its zeroes and poles must all lie on $\mathbb{R} \cup \{ \infty \}$, so it must have the form $e^{i \theta} \frac{az+b}{cz+d}, a, b, c, d \in \mathbb{R}, ad - bc = \pm 1$. But from here it’s clear that $\theta = 0$, so it only remains to decide the sign of the determinant. The sign of the determinant determines, in a straightforward manner, whether $i$ is sent to itself or to $-i$, hence whether the corresponding transformation fixes $\mathbb{H}$ or swaps it with the lower half plane.

The previous proof I knew of this lemma took over two pages; I don’t even want to describe how awful it was!

This lemma appears when studying elliptic curves in the following way. The “analytic” way to define an elliptic curve is as a quotient $\mathbb{C}/\Lambda$ where $\Lambda$ is a free abelian subgroup of $\mathbb{C}$ of rank $2$, i.e. a lattice. (The correspondence between this definition and the one I gave above is not at all obvious, but it is the analytic definition which figures prominently in the classical history of the subject and its relation to the arc-length of ellipses. In the modern treatment the connection between the two definitions goes through the Weierstrass elliptic functions. If you want to learn more about the relationship between the two, I recommend the beautiful exposition by Stevenhagen, which can be found on this page of algebra notes.) Topologically, $\mathbb{C}/\Lambda$ is a torus (a parallelogram with opposite sides identified) regardless of the particular choice of $\Lambda$ is, but it also carries the structure of a Riemann surface, and as it turns out, two such quotients are isomorphic as Riemann surfaces if and only if $\Lambda_1 = c \Lambda_2$ for some $c \in \mathbb{C}^{*}$.

Given an oriented basis $(\omega_1, \omega_2)$ for a lattice, i.e. one such that $\text{Im}\left(\frac{\omega_1}{\omega_2}\right) > 0$, we may quotient by the above equivalence and set $\omega_2 = 1$. This identifies the space of isomorphism classes of elliptic curves $\mathbb{C}/\Lambda$ equipped with an oriented basis for the lattice $\Lambda$ with the upper half plane $\mathbb{H}$. Of course, we don’t want to be content with a choice of basis. Two oriented bases span the same lattice if and only if they can be sent to each other by an element of $SL_2(\mathbb{Z})$ (the positive determinant is necessary to preserve the orientation), so it follows that two points in the upper half plane correspond to the same elliptic curve if and only if one can be taken to the other under the action of $PSL_2(\mathbb{Z})$, which sits naturally inside $PSL_2(\mathbb{R})$ as a discrete subgroup. This gives a natural identification

$\left\{ \text{isomorphism classes of elliptic curves over } \mathbb{C} \right\} \simeq \mathbb{H}/PSL_2(\mathbb{Z})$.

What’s the relationship between this picture and the picture coming from Lemma 1? Well, as it turns out, the points $(a', 0), (b', 0), (c', 0)$ on an elliptic curve in the form $y^2 = (x - a')(x - b')(x - c')$ are precisely the non-identity points of order two, which for the corresponding quotient $\mathbb{C}/\Lambda$, where $\Lambda$ is spanned by $1, \tau \in \mathbb{H}$, are precisely the points $\frac{1}{2}, \frac{\tau}{2}, \frac{1 + \tau}{2}$. Now, since as an abstract group $\mathbb{C}/\Lambda$ is isomorphic to a product of two copies of the circle group, the subgroup consisting of the points of order dividing $N$ (“$N$-torsion”), for any positive integer $N$, is isomorphic to $(\mathbb{Z}/N\mathbb{Z})^2$. When we pick two of the points of order two to send to $0, 1$ to write an elliptic curve in Legendre form, we are in essence picking an ordered basis for their $2$-torsion.

Now, there is a natural way to parameterize elliptic curves with a given ordered basis for their $N$-torsion. As before, we start with the identification of $\mathbb{H}$ with oriented bases $(\tau, 1)$. Given such an oriented basis we may choose the ordered basis $\frac{\tau}{N}, \frac{1}{N}$ for the $N$-torsion, and now two elliptic curves with given ordered basis for their $N$-torsion are isomorphic to each other, with the isomorphism respecting the choice of basis, if and only if the corresponding element of $SL_2(\mathbb{Z})$ is an element of a smaller subgroup consisting of those matrices isomorphic to the identity matrix $\bmod N$. This group is denoted $\Gamma(N)$, and the quotient $\Gamma(1)/\Gamma(N)$ is isomorphic to $SL_2(\mathbb{Z}/N\mathbb{Z})$. When $N = 2$, this group is isomorphic to $S_3$! This is because choosing a basis of $\mathbb{F}_2^2$ is equivalent to choosing two of its non-identity points, and it gives the connection between this picture of elliptic curves and the one we described above.

So the story is as follows. Given an elliptic curve $\mathbb{C}/\Lambda$ with $\Lambda$ spanned by $(\tau, 1)$, the ordered basis $\frac{\tau}{2}, \frac{1}{2}$ determines a Legendre form $y^2 = x(x - 1)(x - \lambda)$, and $\lambda$ is invariant under the action of $\Gamma(2)$. This gives a natural map

$\mathbb{H}/\Gamma(2) \xrightarrow{\lambda} \mathbb{C} - \{ 0, 1 \}$

which, if you believe everything I’ve said so far, is a bijection. Quotienting further by the action of $\Gamma$ then gives the action of $S_3$ on $\mathbb{C} - \{ 0, 1 \}$ we saw earlier. In addition, it turns out that $\lambda$ is meromorphic as a function on $\mathbb{H}$ and well-behaved in the technical sense necessary for it to be a modular function of weight $0$ with respect to $\Gamma(2)$, and in fact it generates all such modular functions.

An interesting consequence of everything I’ve said so far: the action of $\Gamma(2)$ on $\mathbb{H}$ satisfies all of the technical conditions required for the above map to be a covering map. (This is not the case with the action of $\Gamma(1)$.) Among Riemann surfaces, $\mathbb{H}$ is very special because it is one of the three simply connected Riemann surfaces, which then implies that it is the universal cover of $\mathbb{C} - \{ 0, 1 \}$ and that $\Gamma(2)$ is isomorphic to the fundamental group of $\mathbb{C} - \{ 0, 1 \}$. But the fundamental group of the plane minus two points is none other than the free group on two generators! (In fact, if I’m not mistaken, these generators can be taken to be $\left[ \begin{array}{cc} 1 & 2 \\ 0 & 1 \end{array} \right]$ and its transpose.)

I’ve been learning all of this stuff through 18.784, the number-theory seminar I’m currently taking with Scott Carnahan on modular forms. The sheer number of different mathematical disciplines related to the subject is mind-boggling; I haven’t even mentioned the hyperbolic metric on $\mathbb{H}$ and how $PSL_2(\mathbb{R})$ also happens to be its group of hyperbolic isometries, or about how all this works over fields smaller than $\mathbb{C}$ and the relationship to number theory, or about the relationship to theta functions of lattices… if you’d like some references, I’ll be adding them to the “Bibliography” page on the left.

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5 Responses

1. […] Posts AboutBibliographyThe definition of a functionFractional linear transformations and elliptic curvesSuggestions and RequestsNationals 2010GILA I: Group actions and equivalence relationsGraded […]

2. Well, regardless, it’s very neat stuff.

3. This is very interesting. I’m currently involved in research relating to hyperbolic geometry, and isometries (formerly) in the upper-half plane (PSL(2,R)). I’ve started working in SU(1,1) since then, but it’s nice to see other people out there doing work here.

• This isn’t “work” so much as “exposition of classical material,” but thanks!

4. It’s amazing how far you’ve related the stuff in 18.784 to. š I didn’t recognize anything at first glance.