# Equidistribution on $\mathrm{SU}_2$

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 $$$$\lim_{n\to\infty}\frac{1}{\mathrm{Card}(A_n)}\sum_{a\in A_n}f\big(\rho(a)\big)=\int_{\mathrm{SU}_2}fdv_{\text{Haar}},$$$$ where $$dv_{\text{Haar}}$$ is the normalized Haar measure with volume $$1$$.

I appreciate any help or reference.

## 2 Answers

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.

• 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. Feb 22 at 13:57
• 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 $f$). Feb 22 at 14:03
• @MikaeldelaSalle Thanks for your comment, dense image looks like a necessary condition for equidistribution, but how to prove the sufficiency? Feb 23 at 3:18

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.

• Thank you so much for the detailed explanation! Feb 23 at 6:50