I am reading a significant paper of Bourgain and Gamburd on "Uniform expansion bounds for Cayley graphs of $\mathrm{SL}\_2(\Bbb{F}\_p)$". They have proven a proposition which in their new papers they call it "$L^2$-Flattening Lemma" (Indeed this proposition says more). 
Let me state their proposition:


Suppose $v$ is a symmetric probability measure on $\mathrm{SL}\_2(\Bbb{F}\_p)$; that is 
$$
v(g)=v(g^{-1}),
$$
satisfying the following three properties for fixed positive $\gamma$, $0<\gamma<3/4$:


$$
p^{-\frac{3}{2}+\gamma}<|| v ||\_2 < p^{-\gamma},
$$
and for any proper subgroup $G\_0$ we have


$$
v\ast v[G\_0]:=\sum\_{a\in G\_0}v \ast v(a)< p^{-\gamma}.
$$


Then for some $\epsilon=\epsilon(\gamma)> 0$, for all sufficiently large $p$:
$$
||v\ast v ||\_2< p^{-\epsilon}||v||\_2.
$$


**Question**:


What is a motivation for this proposition? Why this proposition is called "Flattening Lemma"? Is it because of the $L^2$-norm of $v\ast v$ is smallet than the $L^2$-norm of $v$? Or it has another reason. 


Thank you in advance. 



L^2-Flattening Lemma 
Perhaps I can add a little to Denis' answer. I have read Bourgain and Gamburd's wonderful paper several times - I don't pretend to fully understand it but I did [write some notes](http://users.mct.open.ac.uk/ng3636/expanders2.pdf) on it which you may find useful.


A facetious explanation as to why the $L\_2$-flattening lemma is important is that it is a crucial step in proving the main theorem of the paper. This is not really helpful but, in a sense, it is true. One should consider what Bourgain and Gamburd are trying to do:


We have a set $A\subset SL\_2(\mathbb{Z})$; we consider the Cayley graph $\mathcal{G}( SL\_2(\mathbb{Z}/p\mathbb{Z}), A\_p)$ which is obtained by taking the image of elements of $A$ mod $p$ for some prime $p$. BG assume that one has a lower bound on the girth of this Cayley graph, from which it is easy enough to show that, for $p$ large enough, the graph is connected.


So, now, one has to show that $\mathcal{G}( SL\_2(\mathbb{Z}/p\mathbb{Z}), A\_p)\_{p\to \infty}$ is a family of expanders. To do this, some notation:


Let $\mu\_p: SL\_2(\mathbb{Z}/p\mathbb{Z})\to \mathbb{R}^+$ be the indicator probability measure for the set $A\_p$, i.e. it takes the value $1/|A\_p|$ for elements of $A\_p$ and $0$ elsewhere. There is a standard notion of **convolution** of measures which is very important in this area: write $\mu\_p\ast \mu\_p$ for the convolution of $\mu\_p$ with itself, and write $\mu\_p^{(l)}$ to mean the measure obtained by doing that $l$ times.


Now the following identities are crucial (and easy):


$$N W\_{2l} = {\rm tr}(Adj^{2l}) = \sum\limits\_{j=0}^{N-1} \lambda\_j^{2l}$$


$$\mu\_p^{(2l)}(1) = \frac{W\_{2l}}{(2k)^{2l}}$$


where $N=|SL\_2(\mathbb{Z}/p\mathbb{Z})|$, $k=|A|=$ the valency of the Cayley graph, $Adj$ is the adjacency matrix of the Cayley graph and $\lambda\_0,\dots, \lambda\_{N-1}$ are the eigenvalues of $Adj$. Recall that to prove that we have an expander family you have to show an eigenvalue gap between $\lambda\_0$ and $\lambda\_1$, so these identities connect the eigenvalues to the $2l$-fold convolution of measure $\mu\_p$. 


In particular the $L^2$-flattening lemma is used crucially to prove the following



> 
> Prop. For every $\epsilon >0$, there exists $C$ such that for $l\geq C\log\_{2k}(p)$, 
>  $$\|\mu\_p^{(l)}\|\_2 < p^{-\frac32+\epsilon}.$$
> 
> 
> 


Combining this with the identities above one obtains that
$$\frac{W\_{2l}}{(2k)^{2l}} = \mu\_p^{(2l)}(1) = \|\mu\_p^{(l)}\|\_2^2 < p^{-3+2\epsilon}.$$


One closes out the argument by appealing to Frobenius' classical lower bound for the dimension of a non-trivial irreducible representation of $SL\_2(\mathbb{Z}/p\mathbb{Z})$. We know that this dimension is at least $\frac12 (p-1)$, hence an easy argument shows that the multiplicity of all eigenvalues of $Adj$ is at least $\frac 12(p-1)$. One obtains immediately that
$$\sum\limits\_{j=0}^{N-1} \lambda\_j^{2l} > \frac12(p-1)\lambda\_1^{2l}.$$
Rearranging, and combining inequalities one obtains an upper bound for $\lambda\_1$:
$$\lambda\_1^{2l} \leq 3\frac{(2k)^{2l}}{p^{1-2\epsilon}}.$$
Since $\lambda\_0=k$, we have our eigenvalue gap, and the job is done. END OF PROOF SKETCH.


**Final remarks**: So the point of the $L^2$-flattening lemma, then, is that it allows us to prove the above proposition, a bound on convolutions which turns out to be enough to prove expansion. The idea that the $L^2$-flattening lemma implies the given proposition is, I hope, plausible just by the form of the statements. 


What makes the $L^2$-flattening lemma so significant is that it turned out to be the right statement for connecting growth results to convolution bounds. BG's aim was to use Helfgott's growth results for the set $A\_p$ in $SL\_2(\mathbb{Z}/p\mathbb{Z})$ to prove expansion. One can then exploit the Balog-Szemeredi-Gowers result to derive a bound on the additive energy of $A\_p$, or equivalently on the convolution of the measure $\mu\_p$. This is the starting point for proving the $L^2$-flattening lemma.


