A First Look at SL2Z

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This was originally published as a video on our SeanForScience YouTube channel.

We've been studying the subgroups of $SL_{2}\mathbb{C}$, although these have involved various restrictions from $\mathbb{C}$ to $\mathbb{R}$. Today we’re going to restrict even further, down to $\mathbb{Z}$.

$SL_{2}\mathbb{Z}\subset SL_{2}{C}$ with integral elements. Despite losing access to the rationals, $SL_{2}\mathbb{Z}$ retains its group structure, replete with inverse elements. The actions on the group are curious - if not fascinating - and provide insight in to and connection between fields as disparte as String Theory and Number Theory.

Since this is something of a “getting to know you'“ discussion, we'll start with the basics. Today we'll describe the basics of the group, the group action on the Riemann Sphere and introduce modular forms.

About $SL_{2}\mathbb{Z}$

$SL_{2}\mathbb{Z}$ is generated by two matrices,

$$m_{1} = \left(\begin{array}{cc}1&1\\0&1\end{array}\right),\quad m_{2} = \left(\begin{array}{cc}0&1\\-1&0\end{array}\right)$$

We'll prove that in an appendix to this section, for now let's just work through some exmaples to convince you that it might be true. Even if you've spent a lot of time with the physicists' $\sigma$-matrices, it's worth sinking some explicit calculational practice into these examples. As always with the integers, higher intuition can come slowly.

Example 1:

Powers of $m_{2}$

The matrix $m_2 = i\sigma_{2}$ serves as a representation of the complex number $i$. As we discussed when talking about the quaternions, the complex plane can be modeled by

$$\mathbb{C} \sim \big\{ a\mathbb{1} + b m_{2} | a,b \in \mathbb{R}\big\}.$$

In particular, we have

$$m_{2}^2 = -\mathbb{1},\quad m_{2}^{4} = \mathbb{1}, \quad m_{2}^{-1} = m_{2}^{3}.$$

Example 2:

Powers of $m_{1}$

The upper triangular matrix $m_{1}$ has the familiar property that

$$m_{1}^2 = \left(\begin{array}{cc}1&2\\0&1\end{array}\right).$$

This immediately generalizes to a result valid for all positive $n$.

$$m_{1}^{n} = \left(\begin{array}{cc}1&n\\0&1\end{array}\right).$$

To access negative $n$ - and therefore the inverse of $m_{1}$ - requires a bit more effort.

Example 3:

The inverse of $m_{1}$

Minus signs are generate in part by the matrix $m_{2}$, but it also tends to move individual elements around the array. You might find it instructive to explcitly show that

$$m_{1}^{-1} = m_{2} m_{1} m_{2} m_{1} m_{2} = \left(\begin{array}{cc}1&-1\\0&1\end{array}\right)$$

$SL_{2}\mathbb{Z}$ Actions

$SL_{2}\mathbb{Z}$ can be thought of as the group of automorphisms of the Riemann sphere,

$$\hat{\mathbb{C}} = \mathbb{C}\cup\{\infty\},$$

where complex-infinity is assigned to the north pole, and the origin is assigned to the south. From this perspective, $SL_{2}\mathbb{Z}$ acts on a generic element $\tau \in \hat{\mathbb{C}}$ via Möbius transformation:

$$\gamma = \left(\begin{array}{cc}a&b\\c&d\end{array}\right) : \tau \mapsto \frac{a\tau+b}{c\tau+d},\quad \gamma \in SL_{2}\mathbb{Z}$$

If you're anything like me - your first reaction to that sort of group action is WTF?! But rest assured, this fractional linear transformation leads to a lot of really astonishing properties. To glean insight, it helps to keep the fact that

$$\det\gamma = ad - bc = 1 , \quad \gamma \in SL_{2}\mathbb{Z}.$$

Some Examples

To build an intuition for these actions, it helps to practice with some examples. First off, our generates act by

$$m_{1}: \tau \mapsto \tau + 1,$$

and

$$m_{2}: \tau \mapsto -\frac{1}{\tau}.$$

For $c\neq0$, the map sends $\tau = -d/c$ to $\infty$ - aka the north pole. If $c$ vanishes, then $\gamma$ leaves the north pole invariant.

The Exponential Map

Consider the exponential mapping

$$q : \tau \mapsto e^{2\pi i \tau}$$

$q$ takes for example, the reals into the unit complex numbers. This mapping is one-to-many, but given the action of $m_1$,

$$m_1\cdot q = e^{2\pi i (\tau + 1),}$$

we begin to see how $q$ and $SL_{2}\mathbb{Z}$ might be related.

For $\tau$ in the so-called upper-half plane, $\mathcal{H}$:

$$\mathcal{H} = \big\{ z \in \mathbb{C} | \mathsf{Im}\,z > 0 \big\},$$

$q$ maps $\tau$ in to the interior of the unit disk in $\mathbb{C}$ - the same disk bounded by the reals. As with the reals, the real part of $\tau\in\mathcal{H}$ also wraps angularly, in periodic fashion. This is basically the opposite streographic projection of the plane, where the north pole is mapped to the origin. These facts about $q$ - particularly as they involve $\mathcal{H}$, will play a prominent role in our discussions.

Before moving on, it's instructive to formally verify on of those claims we made.

Claim: $\mathcal{H}$ is closed under the action of $SL_{2}\mathbb{Z}$.

Proof:

Let $\gamma \in SL_{2}\mathbb{C}$ and $\tau\in\mathcal{H}$, where

$$\gamma = \left(\begin{array}{cc}a&b\\c&d\end{array}\right).$$

We compute,

$$\gamma \tau = \frac{a\tau + b}{c\tau + d} = \frac{(a\tau + b)(c\bar{\tau} + d)}{|c\tau + d|^{2}} = \frac{ac|\tau|^2 + db + ad\tau + bc\bar{\tau}}{|c\tau + d|^{2}}.$$

The imaginary part of $\gamma \tau$ is then

$$\mathsf{Im} \gamma\tau = \frac{(ad - bc)}{|c\tau + d|^2}\mathsf{Im} \tau.$$

But $\gamma$ is in $SL_{2}\mathbb{Z}$, so

$$\mathsf{Im} \gamma\tau = \frac{1}{|c\tau + d|^2}\mathsf{Im} \tau,$$

which implies that if $\tau\in \mathcal{H}$, then $\gamma\tau$ must be too. $\blacksquare$

I'll leave you today with one final thing to chew on, and we'll pick up with its implications next time.

Example 4: The derivative of $\gamma \tau$

Consider $\gamma^{\prime}(\tau)$:

$$\gamma^{\prime}(\tau) = \frac{d}{d\tau} \frac{a\tau + b}{c\tau + d} = \frac{a}{c\tau + d} - c\frac{a\tau + b}{(c\tau + d)^{2}}.$$

Simplifying,

$$\gamma^{\prime}(\tau) = \frac{a(c\tau + d) - c(a\tau + b)}{(c\tau + d)^2} = \frac{ad - bc}{(c\tau + d)^{2}}.$$

Again we find our friend $\det\gamma$.

Thus,

$$\gamma^{\prime}(\tau) = \frac{1}{(c\tau + d)^2}.$$

This pattern will become very familiar to us in future discussions. One curious fact is that it leads to what appears to be an obstruction to computing $SL_{2}\mathbb{Z}$-invariant counter integrals,

$$\oint dz f(z),\quad f: \mathcal{H}\rightarrow \mathbb{C}.$$

As anyone who's studied Riemannian geometry can attest, great pains are taken to ensure that the volume element of a volume integral of some scalar function is invariant. This of course extends to considerations of Stoke's theorem for surface, hypersurface and line integrals in $\mathbb{R}^{n}$. The technology of linear algebra used in vector analysis is more than sufficient for these purposes. For our present case, however, we're interesting in some class of functions that transform under $\gamma$ in a curious way,

$$f(\gamma(z)) = (c\tau + d)^{k}f(z),$$

for $k=2$. For any integer $k$, this observation quickly grows from a curiosity into a long and deep story.

Proof of the Proposition

At the top of this section, we claimed that $SL_{2}\mathbb{Z}$ is generated by two matrices,

$$m_{1} = \left(\begin{array}{cc}1&1\\0&1\end{array}\right),\quad m_{2} = \left(\begin{array}{cc}0&1\\-1&0\end{array}\right).$$ We now prove it.

Let $\Gamma$ be the subgroup of $SL_{2}\mathbb{Z}$ generated by $m_{1}$ and $m_{2}$.

Proposition:

For any $\alpha\in SL_{2}\mathbb{Z}$, there exists a $\gamma \in \Gamma$ such that $\alpha\gamma = \mathbb{1}$. Hence, $\alpha\in\Gamma$, and hence $\Gamma = SL_{2}\mathbb{Z}$.

Proof:

We begin the proof with a series of intermediate results. Let

$$\alpha = \left(\begin{array}{cc}a&b\\c&d\end{array}\right),$$

and in what follows $a^{\prime},b^{\prime}$, etc simply refer to a generic matrix element in $\mathbb{Z}$.

Lemma 1:

For any $\alpha\in SL_{2}\mathbb{Z}$, there exists a $\gamma \in \Gamma$ such that

$$\alpha\gamma = \left(\begin{array}{cc}a^{\prime} &b^{\prime}\\c^{\prime}&d^{\prime}\end{array}\right),$$

where $|c^{\prime}| = c^{\prime}$ and $|d^{\prime}| = -d^{\prime}$.

Proof:

If $\mathsf{sign}(c) = \mathsf{sign}(d)$, then let $\gamma = m_{2}$. Computing,

$$\alpha\gamma = \left(\begin{array}{cc}a^{\prime} &b^{\prime}\\c^{\prime}&d^{\prime}\end{array}\right)\left(\begin{array}{cc}0&1\\-1&0\end{array}\right) = \left(\begin{array}{cc}-b^{\prime} &a^{\prime}\\-d^{\prime}&c^{\prime}\end{array}\right). \quad\blacksquare$$

Lemma 2:

For any $\alpha\in SL_{2}\mathbb{Z}$, there exists a $\gamma \in \Gamma$ such that

$$\alpha\gamma = \left(\begin{array}{cc}a^{\prime} &b^{\prime}\\c^{\prime}&d^{\prime}\end{array}\right),$$

where $|c^{\prime}| < |d^{\prime}|$.

Proof:

If $|c^{\prime}| \geq |d^{\prime}|$, let $\gamma = m_{1}^{n}$. Computing,

$$\alpha m_{1}^{n} = \left(\begin{array}{cc}a^{\prime} &b^{\prime}\\c^{\prime}&d^{\prime}\end{array}\right)\left(\begin{array}{cc}0&1\\-1&0\end{array}\right) = \left(\begin{array}{cc}a^{\prime} &na^{\prime}+b^{\prime}\\c^{\prime}&nc^{\prime} + d^{\prime}\end{array}\right).$$

Clearly there is some natural number $n$ such that

$$|c| < |nc + d|. \quad\blacksquare$$

We now begin an algorithm whose starting point requires the above two lemmas.

Claim:

For any $\alpha\in SL_{2}\mathbb{Z}$, there exists a $\gamma \in \Gamma$ such that

$$\alpha\gamma = \left(\begin{array}{cc}a^{\prime} &b^{\prime}\\ 0 &d^{\prime}\end{array}\right).$$

Proof:

Without loss of generality, suppose $\alpha\in SL_{2}\mathbb{Z}$ such that $|c| = c, |d| = -d,$ and $|c|\leq |d|$. If $c$ is not zero, define $m$ and $r$ by

$$d = mc + r,$$

where the remainder $r$ satisfies $0\leq r < c$. We can then consider

$$\alpha m_{1}^{m} = \left(\begin{array}{cc}a& mb + a\\ c & cm + d \end{array}\right) = \left(\begin{array}{cc}a& mb + a\\ c & r \end{array}\right).$$

Now act with $m_{2}^{3}$,

$$\alpha m_{1}^{m}m_{2}^{3} = \left(\begin{array}{cc}a& mb + a\\ c & r \end{array}\right)\left(\begin{array}{cc}0&1\\ -1 & 0 \end{array}\right) = \left(\begin{array}{cc}mb + a& -a\\ r & -c \end{array}\right).$$

If $r = 0$, we're done. If not, notice that $\alpha m_{1}^{m}m_{2}^{3}$ has the same form as required by the hypothesis. So we can interate this procedure until $r$ vanishes. $\blacksquare$

Because all of the matrices under consideration are elements of $SL_{2}\mathbb{Z}$, they have unit determinant. This fact let's us fix $d = 1$, at least up to action by $m_{2}^{2}$. To see this, let $\gamma \in \Gamma$ be as above. So

$$\alpha\gamma = \left(\begin{array}{cc}a&b\\ 0 & d \end{array}\right),$$

and so

$$\det\big(\alpha\gamma\big) = \det\alpha \det\gamma = 1 \Rightarrow ad = 1.$$

Since $a,d\in\mathbb{Z}$, we have that either both $a$ and $d$ are equal to one or minus one. Therefore, up to an action of, $m_{2}^{2}$, we have

$$\alpha\gamma = \left(\begin{array}{cc}1&b\\ 0 & 1 \end{array}\right).$$

Thus, there is some $\gamma^{\prime}\in \Gamma$, where

$$\alpha \gamma^{\prime} = \mathbb{1},$$

and that $\gamma^{\prime}$ is given by

$$\gamma^{\prime} = \gamma (m_{1})^{-b},$$

where if $b>0$ it's understood that $m_{1}^{-1}$ is given above.