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Let $j:\mathbf{H}\to \mathbf{C}$ be the $j$-invariant. It's a modular function for $\Gamma(1) = \textrm{PSL}_2(\mathbf{Z})$.

For $\epsilon>0$ small, let $B(\epsilon)$ be the image of the strip $$\{x+iy: 0\leq x <1 , y> \frac{1}{\epsilon}\}\subset \mathbf{H}$$ under the quotient map $\mathbf{H}\longrightarrow X(1)$. Note that $B(\epsilon)$ is an open disc around the cusp $\infty$.

Is the absolute value of the $j$-invariant bounded from below (by a positive real number depending on $B(\epsilon)$ in terms of \epsilon$) on $\epsilon$?B(\epsilon)$?

The answer to this is probably no. But what if we take an annulus?

Is the absolute value of the $j$-invariant bounded from below on $B(\epsilon) - B(\epsilon/2)$ ?in terms of $\epsilon$?

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Is the absolute value of the j-invariant bounded from below on an annulus

Let $j:\mathbf{H}\to \mathbf{C}$ be the $j$-invariant. It's a modular function for $\Gamma(1) = \textrm{PSL}_2(\mathbf{Z})$.

For $\epsilon>0$ small, let $B(\epsilon)$ be the image of the strip $$\{x+iy: 0\leq x <1 , y> \frac{1}{\epsilon}\}\subset \mathbf{H}$$ under the quotient map $\mathbf{H}\longrightarrow X(1)$. Note that $B(\epsilon)$ is an open disc around the cusp $\infty$.

Is the absolute value of the $j$-invariant bounded from below on $B(\epsilon)$ in terms of $\epsilon$?

The answer to this is probably no. But what if we take an annulus?

Is the absolute value of the $j$-invariant bounded from below on $B(\epsilon) - B(\epsilon/2)$?

show/hide this revision's text 1

Is the absolute value of the j-invariant bounded from below

Let $j:\mathbf{H}\to \mathbf{C}$ be the $j$-invariant. It's a modular function for $\Gamma(1) = \textrm{PSL}_2(\mathbf{Z})$.

For $\epsilon>0$ small, let $B(\epsilon)$ be the image of the strip $$\{x+iy: 0\leq x <1 , y> \frac{1}{\epsilon}\}\subset \mathbf{H}$$ under the quotient map $\mathbf{H}\longrightarrow X(1)$. Note that $B(\epsilon)$ is an open disc around the cusp $\infty$.

Is the absolute value of the $j$-invariant bounded on $B(\epsilon)$ in terms of $\epsilon$?