# Modular forms- definition of a cusp

• A
this is probably a stupid question but for the fundamental domain for SL2(Z), we say the cusp is only at infinity.

Compare say to hecke subgroups which are congruence subgroups where we say the equivalence classes are given by the points where the fundamental domain intercepts the real axis as well as infinity.

Definition of a cusp :
A pointed end where two curves meet.

So my stupid question is, the fundamental domain for sl2(z), isn't ##t= e^{\pm\frac{2\pi}{3}}## such a point ? Isn't this a cusp ?

Thanks

## Answers and Replies

Infrared
Science Advisor
Gold Member
It's been a while since I've looked at this (which is unfortunate considering my choice of avatar), but I think that in this context, a cusp is an orbit in $\mathbb{P}^1(\mathbb{Q})$. Since $SL(2,\mathbb{Z})$ acts transitively on $\mathbb{P}^1(\mathbb{Q})$, there is only one cusp, but the action isn't transitive for the congruence subgroups, so there are more cusps.

It's been a while since I've looked at this (which is unfortunate considering my choice of avatar), but I think that in this context, a cusp is an orbit in $\mathbb{P}^1(\mathbb{Q})$. Since $SL(2,\mathbb{Z})$ acts transitively on $\mathbb{P}^1(\mathbb{Q})$, there is only one cusp, but the action isn't transitive for the congruence subgroups, so there are more cusps.
Hey thank you for your reply.
I am a bit confused however.
First of, I can see that transitivity will fail if closure does not hold. To me it is not obvious that if ##a \in \Gamma_0 (N) ## and ##b \in \Gamma_0 (N) ## then ## a.b ## is also... whereas for sl2(z) closure is clear.

From what I can see, all the group properties are involved, at some point, in proving the three criteria of the equivalence relation, and, identity, inverse and associtivity will clearly still hold as we go from ## SL2(Z) \to \Gamma_0(N) ##, the only one I would be unsure of would be closure, which will mean transitivity does not hold.

However if this is the case :
1) what is the definition of ##\Gamma_0(N) ## equivalence if transitivity does not hold, by definition there is no equivalence relation ? Whereas my notes say we still have a notion of ##\Gamma_0(N) ## equivakence ...

2) without..erm...closure, the definition of the Hecke subgroup being a subgroup is also not true

Ta

Infrared
Science Advisor
Gold Member
Fortunately $\Gamma_0(N)$ is a subgroup of $SL_2(\mathbb{Z})$. You can just check closure directly.

Fortunately $\Gamma_0(N)$ is a subgroup of $SL_2(\mathbb{Z})$. You can just check closure directly.
But closure and associtivity are all that are needed to prove transitivty ?

Infrared
Science Advisor
Gold Member
Suppose a group $G$ acts on a set $X$. Consider the relation $\sim$ on $X$ defined by $x\sim y$ if there exists $g\in G$ such that $g\cdot x=y$. This is an equivalence relation.

I recommend reviewing some of this group theory before studying modular forms.

Suppose a group $G$ acts on a set $X$. Consider the relation $\sim$ on $X$ defined by $x\sim y$ if there exists $g\in G$ such that $g\cdot x=y$. This is an equivalence relation.

I recommend reviewing some of this group theory before studying modular forms.
huh? I know this.
And any equivalence relation is transitive, reflexive and symmetric. Where transitory says if a is equivalent to b, and b to c, then a must be equivalent to c.

You said transitivity no longer holds. I attempted a proof of this and it only used closure and associtivity. You said closure still holds, so I don't understand why transitivity fails.

Infrared
Science Advisor
Gold Member
A transitive group action is one in which there is exactly one orbit. This is not the same thing as a transitive relation.