# MOD 1 : N=4 SYM and the friendly cusps

Here I say "the" cusp, because a cusp is an equivalence class, but one should not forget that it contains many (an infinite number of) points. Let us recall the definition of a cusp. Given a subgroup $\Gamma$ of $SL(2,\mathbb{R})$ and a point $z \in \mathfrak{H} \cup \mathbb{R} \cup \{\infty\}$, we say that

- $z$ is an
*elliptic point*if it is the fixed point of some elliptic element of $\Gamma$; - $z$ is a
*cusp*if it is the fixed point of some parabolic element of $\Gamma$.

I recall that $\alpha \in \Gamma \subset SL(2,\mathbb{R})$ is elliptic (resp. parabolic) if and only if $|\mathrm{tr} \, \alpha| <2$ (resp. $|\mathrm{tr} \, \alpha| =2$). One can show that elliptic points are always in $\mathfrak{H}$ while cusps are always in $\mathbb{R} \cup \{\infty\}$. And it so happens that if we choose $\Gamma = SL(2,\mathbb{Z})$, the cusps are exactly the points in $\mathbb{Q} \cup \{\infty\}$, and they form only one conjugacy class of $SL(2,\mathbb{Z})$. This is why we say that the modular group has only one cusp, which is $\mathbb{Q} \cup \{\infty\}$.

Let us go back to the $\mathcal{N}=4$ theory. The points in the cusp are precisely the points where we can have a good Lagrangian description. For one of these points, $\{\infty\}$, the theory is weakly coupled, and for the other points, that is the elements of $\mathbb{Q}$, there is an equivalent description which is weakly coupled, obtained using a duality. For instance, the point $\tau = 0 \in \mathbb{Q}$ is related to $\{\infty\}$ by the standard $S$-duality.

Next time, we will begin an adventurous journey to the inland of the coupling space, even rubbing shoulders with the elliptic points, leaving behind the friendly cusps, which are somewhat paradoxically those singular points of the upper half-plane where things are under perturbative control.