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I have been trying to do this by finding the $\Gamma_0(4)$ orbit of $\infty$, then finding an element not in here (namely $0$) and considering the orbit of this. But I feel like there is a slight error in my computation as I suspect there are in fact 3 non-equivalent cusps.

Here is what I have done thus far:

$\Gamma_0(4)\cdot\infty = \{\left( \begin{array}{cc} a & b \\ 4c & d \end{array} \right)\infty : ad - 4bc = 1\} = \{\frac{a}{4c} : gcd(a,4c) = 1\} = \{\frac{p}{q} : gcd(p, q) = 1, 4|q\}$

and then similarly for $0$. This method would give just the two inequivalent cusps which I am pretty sure is wrong. I feel like the error is probably in my final equality but I cannot see why this would be incorrect.

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  • $\begingroup$ The calculation for the orbit of $\infty$ seems to be right. Why do you think it is wrong? $\endgroup$
    – u1571372
    Commented Nov 10, 2016 at 16:27
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    $\begingroup$ The missing cusp is $1/2$. $\endgroup$
    – davidlowryduda
    Commented Nov 10, 2016 at 16:29
  • $\begingroup$ That cusp is in the orbit of $0$. A similar calculation shows that the orbit of $0$ is $\{\frac{p}{q}: 4 \nmid q\}$, which contains $\frac{1}{2}$. $\endgroup$
    – u1571372
    Commented Nov 10, 2016 at 16:32
  • $\begingroup$ Yes, see this is what I thought but know it to be false so feel that the orbit of $0$ cannot actually be that. $\endgroup$
    – user291678
    Commented Nov 11, 2016 at 12:32

1 Answer 1

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I guess you know how to compute the number of inequivalent cusps of $\Gamma_{0}(N)$: $\sum_{d\mid N}\phi(\gcd(d,N/d))$. So the number of cusps of $\Gamma_{0}(4)$ has three inequivalent cusps. We know that the cusps $\infty$ and 0 are inequivalent. I claim that $1/2$ is not equivalent to neither $\infty$ nor $0$.

If $1/2$ were equivalent to $0$, then there would exist $j\in\mathbb{Z}$ such that $2j+1\equiv 0\pmod{4}$ (please see p. 99 of 'A first course in modular forms' written by Diamond and Shurman), which is absurd. We conclude that $1/2$ is not equivalent to $0$. Similarly, $1/2$ is not equivalent to $\infty$.

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  • $\begingroup$ This logic makes sense to me, but am I not right in saying that a similar computation for the orbit of $0$ is: $\Gamma_0(4)\cdot0 = \{\frac{p}{q}: 4\nmid q, gcd(p,q) = 1\}$ which would contain \frac{1}{2}, would it not? $\endgroup$
    – user291678
    Commented Nov 11, 2016 at 12:29
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    $\begingroup$ I'm sorry for misunderstanding. The similar computation for the cusp $0$ show that the $\Gamma_{0}(4)$-orbit of $0$ is the set $\{\frac{b}{d}:\gcd{b,d}=1,d\nmid 4\}=\{\frac{b}{d}:\gcd{b,d}=1,d\equiv 1\pmod{4}\}$. So the missing part is the case when the denominator is 2. The formula for the number of cusps indicates that this case appears. $\endgroup$
    – user262841
    Commented Nov 11, 2016 at 12:42
  • $\begingroup$ Sorry, there is a typo. $\gcd(b,d)=1$ not $\gcd b,d=1$. $\endgroup$
    – user262841
    Commented Nov 11, 2016 at 12:51
  • $\begingroup$ Ooooh okay right, I get it. I just wasn't explicitly understanding what the orbit of $0$ actually was. Wonderful, thanks a lot. $\endgroup$
    – user291678
    Commented Nov 11, 2016 at 12:56

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