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An elliptic point of a modular curve is the image of a point of the complex upper half-plane with a nontrivial stabilizer subgroup in the effective modular subgroup. For the full modular group the stabilizer subgroup orders are two and three, represented by $i$ and $e^{2\pi i/3}$. A local disk coordinate $u$ upstairs descends as $u^e$ when the stabilizer subgroup order is $e$. The central matrices $\pm I$ are factored out before counting this order.

The closed standard fundamental domain of the modular group isIts vertical sides are paired by $T:z\mapsto z+1$, and its circular sides by $S:z\mapsto -1/z$. It is a fundamental region with boundary identifications. To obtain exactly one representative of each orbit, use $-1/2<x\le 1/2$ and, on $|z|=1$, retain the half with $x\ge 0$. We will include both vertices when discussing the closure.

To prove existence of a representative, fix $z\in \mathbb{H}$. Among primitive integer pairs $(c,d)$ choose one minimizing $|cz+d|$. The minimum exists: below any fixed bound, $|c|y\le |cz+d|$ bounds $c$, and then the real part bounds $d$, so only finitely many pairs occur. Extend this primitive pair to the bottom row of a determinant-one integer matrix $\gamma$. Sincethis makes the imaginary part maximal over the orbit. Translate by an integer to put the real part in $[-1/2,1/2]$. If the resulting point had modulus below one, inversion by $S$ would increase its imaginary part, a contradiction. Thus every orbit meets $\mathcal{F}$.

For uniqueness in the interior, note that $y\ge \sqrt{3}/2$ throughout $\mathcal{F}$ and that $|cz+d|\ge 1$ for every primitive pair. If $|c|\ge 2$, then $|cz+d|\ge |c|y\ge \sqrt{3}>1$. For $|c|=1$, the minimum over integer $d$ is attained at $d=0$ or, at an endpoint, at a neighboring integer: $|z|\ge 1$ and $|z\pm 1{|}^2=|z{|}^2\pm 2x+1\ge 1$. All larger $|d|$ give a strict inequality. If $c=0$, primitivity gives $d=\pm 1$.

If two points in $\mathcal{F}$ are equivalent, applying this inequality to the transformation and its inverse shows their imaginary parts are equal. For two interior points equality forces $c=0,d=\pm 1$, hence an integer translation. Their real parts lie in an interval of length one, so the translation is zero. The only matrices acting trivially are $\pm I$. On the boundary, the equality cases give precisely the stated vertical and circular identifications. This establishes the fundamental-domain claim, including its boundary convention.

An elliptic point of a modular curve means a fixed point of a nonidentity element of the effective modular group $PS{L}_2(\mathbb{Z})$; the central matrix $-I$ is not counted, since it acts trivially everywhere. The equality cases above show that the only such points in the closed region areTheir stabilizer subgroups in $S{L}_2(\mathbb{Z})$ arewhereIndeed $S^2=-I$, $(ST{)}^3=(TS{)}^3=-I$, and their fractional-linear fixed-point equations give the listed points. To see that these are the full stabilizer subgroups, equality in $|cz+d|\ge 1$ leaves only bottom rows $(0,\pm 1),(\pm 1,0)$, with the extra neighboring rows at the vertices. The fixed-point equation then fixes the top row. This yields exactly four matrices at $i$, six at each vertex, and just $\{\pm I\}$ elsewhere. These are the elliptic stabilizers of the modular group. After passing to $PS{L}_2(\mathbb{Z})$ their effective orders are two and three. The two vertices represent the same elliptic orbit.