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Let $F_{a_1,a_2}$ be the free group with a free generating set $\{a_1,a_2\}$ of two elements, and for any $n\in\mathbb{N}$, set $A_n=\{\text{reduced words in } F_{a_1,a_2} \text{with length} \leqslant n\}$.

Can we construct a representation $\rho\colon F_{a_1,a_2}\to \mathrm{SU}_2$, such that the image is equidistributed in the following sense: for any continuous function $f$ on $\mathrm{SU}_2$, we have \begin{equation} \lim_{n\to\infty}\frac{1}{\mathrm{Card}(A_n)}\sum_{a\in A_n}f\big(\rho(a)\big)=\int_{\mathrm{SU}_2}fdv_{\text{Haar}}, \end{equation} where $dv_{\text{Haar}}$ is the normalized Haar measure with volume $1$.

I appreciate any help or reference.

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In the article "On the spectral gap for finitely-generated subgroups of SU(2)" by Jean Bourgain and Alex Gamburd (Invent. Math. 171, No. 1, 83-121 (2008)), they show that free subgroups of $SU_2$ generated by elements with algebraic entries have a spectral gap property. This implies that the averaging operator that appears in the LHS of your equation converges exponentially fast to the RHS.

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    $\begingroup$ Correct, but it is a much more elementary fact that given $\rho$, the convergence holds if (and only if) $\rho$ has dense image. Indeed, your argument is that $A$, the average of the generators and their inverses acting on $L_2(\mathrm{SU}_2)$ has spectrum contained in $[-1+\varepsilon,1-\varepsilon]\cup\{1\}$ with simple eigenvalue $1$, and therefore $A^n$ converges in norm to the projection on the eigenspace for eigenvalue $1$. But if you only want convergence pointwise, all you need is that $A$ is self-adjoint of norm $\leq 1$ and $-1$ is not an eigenvalue. $\endgroup$
    – Mikael de la Salle
    Commented Feb 22 at 13:57
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    $\begingroup$ The soft argument I provide does not give an a priori speed of convergence, so Bourgain and Gamburd's result gives much more under suitable assumptions of $\rho$ (but I have to admit that I do not see how your argument does, without any assumption on the modulus of uniform continuity of $f$). $\endgroup$
    – Mikael de la Salle
    Commented Feb 22 at 14:03
  • $\begingroup$ @MikaeldelaSalle Thanks for your comment, dense image looks like a necessary condition for equidistribution, but how to prove the sufficiency? $\endgroup$
    – Local
    Commented Feb 23 at 3:18
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It was asked in the comments that I provide some details. I prove slightly more: if $\mu_n$ denotes uniform probability on the sphere of radius $n$ and if $\rho:F_2 \to \mathrm{SU}_2$ is a homomorphism with dense image, then for any continuous function $f$ on $\mathrm{SU}_2$, $$ \lim_n \int f(\rho(a)) d\mu_n(a) =\int f.$$

For the proof, consider the unitary representation $\pi$ on $L_2(\mathrm{SU}_2)$, $\pi(a) f(x) = f( \rho(a^{-1}) x)$. The assumption that $\rho$ has dense image gives that the $\pi(F_2)$-invariant functions are the constant functions. By strict convexity of $L_2$, this can be translated to: the eigenvectors of $A:=\pi(\mu_1) = \int \pi(a) d\mu_1(a)$ with eigenvalue $1$ are the constant functions. By similar argument, $A$ does not have $-1$ has an eigenvalue.

Now a classical computation shows that $\mu_1 \ast \mu_n = \frac{3}{4} \mu_{n+1} + \frac 1 4 \mu_{n-1}$ for every $n \geq 1$, and therefore we can write $\mu_n = P_n(\mu_1)$ where $P_n$ is the polynomials defined by $P_0=1$, $P_1=X$ and $X P_n = \frac{3}{4} P_{n+1} + \frac 1 4 P_{n-1}$. Solving the recurrence we see that $P_n(1)=1$ and $\lim_n P_n(x)=0$ for every $x \in (-1,1)$. By the spectral theorem, we obtain that $\lim_n \pi(\mu_n)f = \int f$ in $L_2(\mathrm{SU}_2)$ (and in particular in probability) for every $f \in L_2(\mathrm{SU}_2)$.

Finally, if $f$ is continuous, the family of functions $\pi(\mu_n)f$ is uniformly equicontinuous, so convergence in probability implies uniform (and in particular pointwise at the identity) convergence, QED.

As explained by Lucas Kaufmann, in some situations it is known that $A$ has spectral gap, so the convergence of $\pi(\mu_n)$ to the orthogonal projection on the constants happens in operator norm, and not only pointwise.

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    $\begingroup$ Thank you so much for the detailed explanation! $\endgroup$
    – Local
    Commented Feb 23 at 6:50

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