I Congruence Subgroups and Modular Forms Concept Questions

1. Dec 27, 2016

binbagsss

1) Defintion :

A congruence subgroup of level $N$ is one that contains the principal subgroup at level $N$ which is defined as $(a b c d) \in SL_2(Z) : a,d\equiv 1 (mod N), b,c \equiv 0 (mod N)$ (apologies $( a b c d)$ is a 2x2 matrix.)

The Hecke group is one such example given by $T_{0}(N) = ( *, *) ( 0, *)$ (apologies again that's the first row and second row of the matrix respecitvely).

So is my understanding of these definitions correct: that $T(N) \in T_0(N)$ since can choose $( *, *) = (1,0)$ and $(0,d) = (0,1)$ to give the required $a,d \equiv 1 (mod N), b,c \equiv 0 (mod N)$ ?or eg $(*,*)= (cN+1, cN)$, $c$ is a constant, and then both of these elements divide by $N$? (that's the top row of the matrix $\in$ $SL_2(Z)$ , apologies again.

2) In my notes I have that the 'corresponding' condifiton for a function $f(t)$ to be modular for $T_0(N)$ differs to the $SL_2(Z)$ condition that must be holomorphic at $\infty$,( from $f(t)$can be written as an expansion as $f(t)= \sum\limits^{\infty}_{n=0} a_{n} q^n$, to :

$f$ is 'holomporphic at all cusps of $T_0 (N)$ , that is the limit
$lim _{q \to 0} (ct+d)^{-k} f( \alpha t)$ exists for all $\alpha$ \in $SL_2(Z)$.

And that this only needs to be checked at finitely many $\alpha$, for those which map the (inequivalent) cusps to $\infty$.

So i don't really understand why this is the condition, nor why it is sufficient to only check the $\alpha$ that map the cusps to $\infty$.

Here's what I know:

- In $SL_2(Z)$ all cusps are $T$-equivalent to $\infty$, since all rational numbers are, and so this is why it suffices to only check holomorphicity at $\infty$ for $SL_2(Z)$ ?

- Whereas for $T _0 (N)$ fewer points are $T$-equivalent and so $T_0$ can have other cusps at the rational numbes, that is, the cusps can no longer be mapped to $\infty$. So we check the expansion of $f$ mapped to $\infty$ from these inequivalent cusps - however I don't really under where the condition comes from. Why is the idea to map to $\infty$? Why is the condition working with maps $\in SL_2(Z)$ , how is this ok?

3) This is proabably a stupid question but in my notes it says:

$T _0 (N)$ touches the real axis at the point $0$. In fact, such a behavior will happen for any $T_0(N)$. Any fundamental domain for$T_0(N)$ will touch the the real axis in ﬁnitely many rational points, and we call these points (together with the inﬁnite cusp $∞$), the (equivalence classes) of cusps of $T_0(N)$.

I don't understand how the cusp at $0$ and $\infty$ are called equivalence classes, since $0$ is mapped to $\infty$ by $S$ , however, for e.g $T_0 (p)$ $s \notin T_0(p)$ and so the points $0$ and $\infty$ are not $T$-equivalent...

I'm pretty confused.

Any clarification what so ever greatly appreciated, ta.

2. Jan 1, 2017

bump.