Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 137 / 3

Let $S=\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)$ and $T=\left(\begin{array}{cc}1& 1\\ 0& 1\end{array}\right)$. The modular group acts on the complex upper half-plane by fractional linear transformations, andFor fixed $z=x+iy$, the values $|cz+d|$ with primitive integer pairs $(c,d)$ have a positive minimum: the conditions $|cz+d|\le R$ bound both $|c|\le R/y$ and $|d|\le R+|cx|$, leaving finitely many pairs. By Bezout identity, a minimizing pair is the bottom row of a matrix in $S{L}_2(\mathbb{Z})$. Applying that matrix gives a point of maximal imaginary part in its orbit.

Translate it by a power of $T$ into $-1/2<x\le 1/2$. It must have modulus at least one, since otherwise $S$ would increase its imaginary part. If it is on the unit circle with negative real part, apply $S$, which there changes $x$ to $-x$. This gives an element of the specified half-open standard fundamental domain of the modular group. It proves existence by reduction to the standard modular region.

For uniqueness, if $z$ lies in the closed standard region then $y\ge \sqrt{3}/2$ andfor every primitive pair. If $c=0$, this follows from $d=\pm 1$; if $|c|\ge 2$, it follows from $|c|y\ge \sqrt{3}$. If $|c|=1$, then $d=0$ uses $|z|\ge 1$, while $|d|\ge 1$ givesIf both $z$ and $z^{\prime }=\gamma z$ lie in the region, applying this inequality to $\gamma$ and ${\gamma }^{-1}$ forces their imaginary parts to agree, hence $|cz+d|=1$.

The equality cases describe every possible boundary identification. For $c=0$ the map is an integer translation; the half-open vertical strip retains one representative. For $|c|=1,d=0$, equality requires $|z|=1$, and the map is an inversion followed by an integer translation. Inversion exchanges the two halves of the circle, so retaining only the right half removes the duplication; the only translation that returns the exchanged endpoint is the one identifying the two corners. The remaining equality cases have $|c|=|d|=1$, $|z|=1$ and $x=\pm 1/2$, and give only those same corner identifications. Our region retains the right corner alone. Therefore $z^{\prime }=z$, proving

The equality analysis also gives the elliptic stabilizers of the modular group. The central matrices $\pm I$ fix every point. The point $i$ is fixed by $S$, with $S^2=-I$, and the retained corner ${\rho }_{+}=e^{i\pi /3}$ is fixed by $R=TS$, with $R^3=-I$. There are no other noncentral stabilizers, and the possible bottom rows in the equality cases give exactlyThe corresponding stabilizer orders in $PS{L}_2(\mathbb{Z})$ are two, three and one; the requested group is $S{L}_2(\mathbb{Z})$, so the center must be included.

Use the convention that a weak modular form is a holomorphic function $f$ on the complex upper half-plane satisfyingwith no condition at infinity. In particular $f(z+1)=f(z)$, so it has a Laurent series $f(z)={\sum }_{n\in \mathbb{Z}}{a}_n{q}^n$ on $0<|q|<1$, where $q=e^{2\pi iz}$. It is a modular form when it is holomorphic at a cusp, here the unique level-one cusp at infinity, equivalently $a_n=0$ for $n<0$. It is a cusp form when additionally $a_0=0$. If one uses the narrower weakly holomorphic convention, one also requires finitely many negative terms; the derivative argument below holds in that subspace as well. The matrix $-I$ forces odd-weight forms to be zero.

Put $(a{)}_r=a(a+1)\cdots (a+r-1)$, with $(a{)}_0=1$, andWe prove the derivative transformation of a weak modular form in the uniform formFor $\ell =0$ this is the weight-$k$ transformation under $S$. Differentiating the formula and multiplying by $z^2$, since $d(-1/z)/dz=z^{-2}$, gives the recurrencewhere out-of-range coefficients are zero. The displayed explicit coefficients satisfy it. Indeed, for interior $j$, extract the common polynomial $(k+j{)}_{\ell -j}$; the remaining identity isThis follows from Pascal's identity and $j\left(\genfrac{}{}{0ex}{}{\ell }{j}\right)=(\ell -j+1)\left(\genfrac{}{}{0ex}{}{\ell }{j-1}\right)$. The endpoint coefficients satisfy the recurrence directly. This polynomial proof also applies when a factor is zero, avoiding division by that factor. As $C_{\ell ,\ell }=1$, the result isThe letter $l$ in the printed exponent is the same derivative order $\ell$.

For integer $k<0$, take $\ell =1-k$. If $j<\ell$, the product runs from $k+j\le 0$ through $k+\ell -1=0$, and therefore vanishes. Consequently $f^{(1-k)}$ has weight $k+2(1-k)=2-k$ under $S$. It is also invariant under $T$, by differentiating $f(z+1)=f(z)$, and $S,T$ generate the modular group. Its holomorphy is preserved by differentiation. With $D=(2\pi i{)}^{-1}d/dz=qd/dq$, termwise differentiation of the locally convergent Laurent series gives the Bol identity for modular forms:For negative indices the exponent $1-k$ is a positive integer, and the constant term is killed, so every term is unambiguous.