Elliptic points are isolated?

Question

I'm reading A First Course in Modular Forms by Diamond and Shurman and am confused on a small point in Chapter 2. Let $\Gamma$ be a congruence subgroup of $\operatorname{SL}_2(\mathbb Z)$. $\gamma \in \mathscr H$ is called an elliptic point for $\Gamma$ if the stablizer of $\gamma$ in $\operatorname{PSL}_2$ is nontrivial.

Proposition 2.1.1 Let $\tau_1, \tau_2 \in \mathscr H$ be given. There exist open neighborhoods $U_i$ of $\tau_i$ in $\mathscr H$ such that if $\gamma \in \operatorname{SL}_2(\mathbb Z), \gamma(U_1) \cap U_2 \neq \emptyset$, then $\gamma(\tau_1) = \tau_2$.

Corollary 2.2.3 Let $\Gamma$ be a congruence subgroup of $\operatorname{SL}_2(\mathbb Z)$. Each point $\tau \in \mathscr H$ has a neighborhood $U$ in $\mathscr H$ such that $\gamma \in \Gamma, \gamma(U) \cap U \neq \emptyset$ implies $\gamma \in \operatorname{Stab} \tau$. Such a neighborhood has no elliptic points except possibly $\tau$.

Taking $\tau = \tau_1 = \tau_2$ and $U = U_1 \cap U_2$ in the proposition implies everything in the corollary except for the last sentence. How do we know that we can choose $U$ small enough to exclude all elliptic points? In other words, how do we know that the elliptic points in $\mathscr H$ form a discrete set?

Answer 1

Let $\tau$ and $U$ be as in the Corollary. If $\tau' \in U$ is an elliptic point different from $\tau$, with nontrivial stabilizer $\gamma \in \text{PSL}_2(\mathbb{Z})$, then $\gamma$ also fixes $\tau$ by the first part of the Corollary. It is thus sufficient to show that if $\gamma \in \text{PSL}_2(\mathbb{R})$ fixes two distinct points of $\mathscr{H}$, then $\gamma$ is the identity. This is immediately verified (say, by assuming that one of the points is $\sqrt{-1}$).

Answer 2

Since each elliptic fixed point of an element of $\Gamma$ is an elliptic fixed point of an element of $SL_2(\mathbb{Z})$, it suffices to prove the stronger statement that the elliptic fixed points of $SL_2(\mathbb{Z})$ form a discrete set. And in fact what I'll prove is even stronger, they form a discrete closed set.

Under the action of $SL_2(\mathbb{Z})$ on $\mathscr H$, there are exactly two orbits of elliptic points: the orbit of $z_0=0+1i$ which is the elliptic fixed point of $f_0(z)=-1/z$; and the orbit of $z_1=\frac{1}{2} + \frac{\sqrt{3}}{2}i$ which is the fixed point of $f_1(z) = \frac{z-1}{z}$. You can check this by verifying for yourself that if the trace of $f \in SL_2(\mathbb{Z})$ equals $0$ then it is conjugate to a power of $f_0$; if the trace is $\pm 1$ then it is conjugate to a power of $f_1$; and if the trace has absolute value $\ge 2$ then it is not elliptic.

Suppose that the elliptic fixed points did not form a discrete closed subset of $\mathscr H$. There would exist, therefore, a point $p \in \mathscr H$ and a sequence of infinitely many distinct elliptic fixed points $q_1,q_2,q_3,\ldots$ converging to $p$. Passing to finite index subsequence, we may suppose that there exists $i \in \{0,1\}$ such that each of $q_1,q_2,q_3,\ldots$ is in the orbit of $z_i$ (and so is fixed by a conjugate of $f_i$). Passing to a further subsequence, we may suppose that each $q_n$ has distance $\le 1/2$ from $p$. We therefore obtain a sequence of infinitely many distinct nontrivial elements of $SL_2(\mathbb{Z})$, which I'll denote $g_1,g_2,g_3,\ldots$ (each of which is conjugate to $f_i$) such that for each $n$ the point $q_n$ is fixed by $g_n$, and furthermore we have \begin{align*} d(g_n(p),p) &\le d(g_n(p),g_n(q_n)) + d(g_n(q_n),p) \\ &= d(p,q_n) + d(q_n,p) \\ & \le 1 \end{align*} The subset of all $g \in SL_2(\mathbb{R})$ such that $d(g(p),p) \le 1$ is compact. Therefore, the sequence $g_n \in SL_2(\mathbb{Z})$ has a convergent subsequence in $SL_2(\mathbb{R})$. But that is a contradiction, because $SL_2(\mathbb{Z})$ is a discrete subgroup of $SL_2(\mathbb{R})$: $\mathbb{Z}$ is discrete in $\mathbb{R}$, and so no sequence of distinct elements of $SL_2(\mathbb{Z})$ has a convergent subsequence in $SL_2(\mathbb{R})$.