The book Introduction to Elliptic Curves and Modular Forms by Koblitz says
We extend the usual topology on $\mathbb H$ to the set $\overline{\mathbb H}$ as follows. First, a fundamental system of open neighborhoods of $\infty$ is $N_C = \{z\in \mathbb H: \operatorname{Im} z > C\} \cup \{\infty\}$ for any $C > 0$. Note that if we map $\mathbb H$ to the punctured open unit disc $\mathbb D\setminus\{0\}$ by sending $$\phi: z\mapsto q = e^{2\pi iz} \tag{1.8}$$ and if we agree to take the point $\infty\in \overline{\mathbb{H}}$ to the origin under this map, then $N_C$ is the inverse image of the open disc of radius $e^{-2\pi C}$ centered at the origin, and we have defined our topology on $\mathbb{H}\cup \{\infty\}$ so as to make this map $(1.8)$ continuous.
$\mathbb H$ is the upper half plane, and $\overline{\mathbb H} := \mathbb H\cup \mathbb Q\cup \{\infty\}$.
- What is meant by a fundamental system of open neighborhoods of $\infty$?
- I presume the topology on $\mathbb H\cup\{\infty\}$ is defined so that the map $\phi$ is continuous, i.e., a subset $V\subset \mathbb H\cup\{\infty\}$ is open if and only if $\phi(V)$ is open in $\mathbb D$. What is the topology on $\mathbb H \cup \Bbb Q\cup \{\infty\}$?
The points $\Bbb Q\cup \{\infty\}$ are called cusps.
Next, near a cusp $a/c \in \Bbb Q \subset\overline{\Bbb H}$ we define a fundamental system of open neighborhoods by completing $a, c$ to a matrix $\alpha = \begin{pmatrix}a&b\\c&d \end{pmatrix} \in \Gamma = SL_2(\Bbb Z)$ and using $\alpha$ to transport the $N_C$ to discs which are tangent to the real axis at $a/c$. In other words, with this topology, to say that a sequence $z_j$ approaches $a/c$ means that $\alpha^{-1}z_j$ approaches $i\infty$, i.e., that $\operatorname{Im}\alpha^{-1}z_j$ approaches infinity in the usual sense. Notice, by the way, that the topology near the rational numbers $a/c$ does not agree with the usual topology on the real line, i.e., a sequence of rational numbers which approaches $a/c$ as real numbers will not approach $a/c$ in our topology.
- Why does a sequence $z_j$ approaching $a/c$ mean that $\alpha^{-1}z_j$ approaches $i\infty$, i.e.,$\lim_{j\to\infty} \operatorname{Im}\alpha^{-1}z_j = \infty$?
- Lastly, consider a sequence of rational numbers $r_j = \frac{a}{c} + \frac{1}{j}$. Then, $r_j \to a/c$ as $j\to\infty$. The author comments that $\{r_j\}$ will not approach $a/c$ in the topology defined on $\overline{\Bbb H}$ (if I'm not wrong); why is this true?
Thank you!
whene u have a point $x\in X$ ($X$ is a topological space), a fundemantal system of nbds is a subset $W$ of $V_{x}$(set of all nbds of $x$) such that ,for any open set $A$ containing $x$,there an open set $B$ which is an element of $W$ such that $B\subset A$,for example ,in the real line $\mathbb{R}$,$x=0$, the family $(\left ]\frac{-1}{n};\frac{1}{n} \right [)_{n\geq 1}$ is a fundemantal system of nbds of 0.Regarding your second question, the unit disc has already a topology,and the given map is continous from the upper half plane into the disc,he extends this map to $\mathbb{H}\cup \{\infty \}$ by defining the image of $\infty$ to be 0.Now the extended map is still continous if we endow $\mathbb{H}\cup \{\infty \}$ with the following topology:the nbds of every element of $\mathbb{H}$ are the usual euclidean nbds, whereas the nbds of $\infty$ are of the form $N_{C}$ (remark that $\lim_{Im(z)\to \infty} f(z)=0$ and u can see that $N_{C}=f^{-1}(B(0,e^{-2\pi c}))$.