I'm trying to read *Convergence conditions for random quantum circuits* by Emerson, Livine, Llyod (https://doi.org/10.1103/PhysRevA.72.060302), arXiv version: (https://arxiv.org/abs/quant-ph/0503210) which as I understand is a sort of "Central Limiting Theorem" for products of random unitaries.

Stated in not so rigorous, i.e. physics language it is the statement that if you compose unitaries drawn independently from a universal gate set (could be continuous, or discrete) then the distribution of such long products (i.e., "deep circuits") tends towards the Haar distribution.

In more mathematical language the measure of a product of unitaries $g = g_m \cdots g_2 g_1$ where each unitary $g_i$ is drawn independently from a probability measure $f$ over the full unitary group $G$, which generates the full group, converges exponentially fast in $m$ to the constant function w.r.t. the Haar measure on the group.

Broadly speaking I understand the logic is that they analyze the distribution of the product of unitaries, which is given by multiple convolutions of the underlying distribution $f$. A Fourier transform then turns the analysis of convolutions into an analysis of products of the Fourier components of the underlying distribution. The Fourier components $\hat{f}^s$ of $f$ are organized in terms of unitary irreps $s$ of $G$, where the trivial representation $s=0$ has dimensionality $d_0 = 1$. The claim is then (I guess for suitably well-behaved $f$s) Fourier components for irreps $s \neq 0$ upon raising them to a high power are subleading to the $\hat{f}^{s=0}$ component, so only this component survives, which is Fourier coefficient of the constant function.

The crux seems to be showing that Fourier components $\hat{f}^s$ for $s\neq0$ has norm less than dimension $d_s$ so that $(\hat{f}^s/d_s) < 1$ and hence $(\hat{f}^s/d_s)^m \to 0$ as $m\to \infty$ (where $m$ is the number of convolutions/products) --- this is Eq. (9).

Indeed, it is stated that for all vectors $x$ in the $s$ representation, \begin{align} |\hat{f}^s x| = |d_s \int_G d\mu(g) f(g) \overline{D}^{(s)}(g) x|_2 \leq d_s \int_G d\mu(g) f(g) \| \overline{D}^{(s)}(g) \| |x|_2 = d_s |x|_2 \end{align} where $d\mu(g)$ is the Haar measure on $G$ and $\overline{D}^{(s)}(g)$ is (as I understand) is the representation of element $g \in G$ in the $s$-irrep (so that $\| \overline{D}^{(s)}(g)\|=1$). Note the 1st term within the modulus is the (matrix-valued) Fourier coefficient $\hat{f}^s$ acting on vector $x$, and the 2nd term within the modulus without $x$ is the definition of $\hat{f}^s$ -- it is just extracting the Fourier component in the $s$-irrep like in standard Fourier analysis $\hat{f}(k) = \int dx f(x)e^{-ikx}$.

It is also stated (top of page 3 of the arXiv version) that equality holds iff $D(g)x = \xi(g) y$ is true for all $g$ for which $f(g) > 0$ where $\xi(g) \in \mathbb{C}$. And that since any $g \in G$ can be generated by the support of $f$, $D(g) x = \xi(g) y$ must hold also for all $g \in G$. This implies that we have a 1d rep of $G$ embedded in the irrep $s$.

Here is where I do not understand:

Why does equality hold iff $D(g)x = \xi(g) y$ is true for all $g$ for which $f(g) > 0$? If I plug this into the integral, I get $\hat{f}^s x = d_s \int_G d\mu(g) f(g) \overline{\xi}(g) \overline{y}$. But I don't see why the RHS has norm $d_s \|x\|_2$. I also don't see the other direction.

The second statement that since any $g \in G$ can be "generated by the support of $f$" the relation must hold for all $g \in G$ is cryptic to me. Could someone explain?

Lastly, why does equality imply we have a 1d rep of $G$ embedded in the irrep $s$?

Thank you.