- #1

binbagsss

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- 11

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##.

##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.