modular forms

So why do we care about the modular group $G=SL_2(\mathbb{Z})/\left\{ -1,1\right\}$?

Well actually the answer has to do with complex lattices. The most important thing to know about complex lattices is the following fact:

Theorem. \[ \left\{ 1,z\right\}, \left\{ az+b, cz+d\right\} \] generate the same lattice, when $ad-bc=1$.

Proof. The first lattice is obviously contained in the second. The fact that the matrix abcd is invertible in $SL_2(\mathbb{Z})$ shows the other inclusion.

So what does this mean? Well, suppose that we consider $\left\{ (u,w)\; : \;u,w\in \mathbb{C} \right\}$ modulo we consider $(u,w)\equiv (u',w')$ if $u/w = u'/w'$. This is called homothety. Of course this is in correspondence with $\mathbb{C}$. But then there is an interesting and natural way to define an action of $SL_2(\mathbb{Z})$ on this equivalence class of vectors: matrix multiplication.

So FACT:

Theorem. Let $H$ be the upper half plane, $G$ the modular group, $R$ the set of complex lattices and $R/\mathbb{C}$ mean lattices modulo homothety. Then, $H/G R/. $

Ok, so lattices are actually something rather interesting and motivated so now it maybe makes a bit more sense why we care about the modular group.

Here’s an example of a modular form: \[ G_k(\Gamma) = \sum_{\gamma\in \Gamma}\frac{1}{|\gamma|^{2k}}. \]

Yeah so that’s a very nice object and it’s related to the modular group because we showed that the modular group is in some sense related to lattices.

boolean anlaysis

A trick for sampling things in a $p$-biased measure when you want them to have disjoint supports. You put them in buckets and then sample with $1/2$ measure.

coupling

Talked about coupling in probability class.

So, what is this why do we care. Well, there’s this notion of TV distance between distributions:

\[ d_{TV}(P,Q) = \frac{1}{2}\sum_x |P(x)-Q(x)| = \sup_{\text{events }A} P(A)-Q(Q) = \inf_{\text{couplings }X,Y} \Pr[P(X)\neq P(Y)].\]

So we are interested in analyzing mixing times of markov chains: how many steps do you have to go before the TV distance is small?

And a nice way to do this is by giving a coupling and arguing “locally” that the coupling causes $P(X)\neq P(Y)$ to decrease.