Modular Forms Part III, Michaelmas 2019

$$\def\NN{\mathbb{N}} \def\ZZ{\mathbb{Z}} \def\QQ{\mathbb{Q}} \def\RR{\mathbb{R}} \def\CC{\mathbb{C}} \def\FF{\mathbb{F}}$$

Notes on lecture 5 (22 October)

In the lecture I remarked that the mentioned the fact that the quotient $SL_2(\RR)/\{\pm1\}$ is commonly and traditionally called $PSL_2(\RR)$, but that this notation is somewhat confusing. For the interested, here is the explanation of this remark. (Warning: algebraic geometry lurks below!)

The matrix group $GL_n$ is in fact an affine algebraic group, one of the simplest examples of algebraic variety which is a group. For a field $K$, $GL_n(K)$ is the set of $K$-valued points of $GL_n$. Likewise, $SL_n$ is an affine algebraic group.

One basic fact in algebraic geometry is that Galois theory holds for points: if $L/K$ is a Galois extension of fields, $V$ is an algebraic variety over $K$, and $x\in V(L)$ is an $L$-valued point, then $x\in V(K)$ iff $x$ is fixed by the Galois group of $L/K$.

Why is the notation $PSL_2(\RR)$ confusing? It suggests (to anyone who has taken a course in algebraic geometry!) that it is the set of $\RR$-valued points of an algebraic group $PSL_2$. But $SL_2(\RR)/\{\pm1\}$ is not the set of elements of $SL_2(\CC)/\{\pm1\}$ fixed by complex conjugation. Indeed, consider the matrix $g=\begin{pmatrix}0 & i \\ i & 0\end{pmatrix} \in SL_2(\CC).$ Since $\bar g = -g$, the image of $g$ in $SL_2(\CC)/\{\pm I\}$ is fixed by complex conjugation, but doesn't belong to $SL_2(\RR)/\{\pm I\}$. This means that there is no algebraic group $PSL_2\text{''}$ whose set of $K$-valued points, for a field $K$ of characteristic zero, is $SL_2(K)/\{\pm I\}$.

[Further remarks: let $Z_n\subset GL_n$ be the subgroup of scalar matrices. One can prove that for a Galois extension $L/K$ of fields, $PGL_n(K)=GL_n(K)/Z_n(K)$ is the set of Galois invariant elements of $PGL_n(L)$, and is in fact the set of $K$-valued points of an affine algebraic group $PGL_n$, which is the quotient (in the sense of group schemes) $GL_n/Z_n$.

One could then try to form the quotient of $SL_n$ by its subgroup $SL_n\cap Z_n$ of scalar matrices (which is zero-dimensional, given by $n$-th roots of unity). But this quotient turns out to be $PGL_n$ as well, because for any algebraically closed field $K$, the map $SL_n(K)/(SL_n(K)\cap Z_n(K)) \to GL_n(K)/Z_n(K)$ is a bijection.]

The integral in the proof of KLF (lecture 18, 21 November)

In the lecture I used the fact that for any $a\gt 0$ and $b\ge0$, $I= \int_0^\infty e^{-\pi(ax^2+b/x^2)} dx = \frac{1}{2\sqrt{a}} e^{-2\pi\sqrt{ab}}$ (which can be found on the Wikipedia page "List of definite integrals", without proof or reference). Here is a possible proof. First put $x=y/\sqrt{a}$ and $c=\sqrt{ab}$ to get $I = \int_0^\infty e^{-\pi(y^2+c^2/y^2)} \frac{dy}{\sqrt{a}} = \frac{1}{\sqrt{a}}e^{-2\pi c} J,\qquad J = \int_0^\infty e^{-\pi(y-c^2/y)^2} dy.$ Split $J$ into two integrals over $[c,\infty)$ and $(0,c]$, and in the second of these substitute $z=c^2/y$ (a familiar ruse from the lectures). This gives $J = \int_c^\infty e^{-\pi(y-c^2/y)^2} \biggl(1+\frac{c^2}{y^2}\biggr) dy =\int_0^\infty e^{-\pi t^2} dt = \frac{1}{2}$ putting $t = y-c^2/y$.