# Is there a uniform solution of the Ruziewicz problem?

For any integer $$n\geq 2$$ there is one and only one (up to rescaling) rotation-invariant, finitely-additive measure on the Lebesgue $$\sigma$$-algebra of $$S^n$$.

The proof of this statement I'm aware of treats two cases separately: $$n\geq 4$$ and $$n=2, 3$$ (the latter in a bizarre turns of events ends up relying on the Ramanujan conjecture). I wonder if there is an argument that is uniform in $$n$$.

## 1 Answer

First, you don't need the full Ramanujan Conjecture: property $$(\tau)$$ is enough, so in that sense you can get a uniform proof (it's just that for $$n\geq 4$$ we have also property (T), which is easier to prove and stronger). Second, you can in fact induce property $$(\tau)$$ up the ladder (see Burger--Sarnak). Thirdly, his book  Sarnak makes this explicit, showing how to convert the spectral solution for the problem for $$O(n)$$ to a solution for $$O(n+1)$$.

 Sarnak, Peter, Some applications of modular forms, Cambridge Tracts in Mathematics, 99. Cambridge: Cambridge University Press. x, 111 p. \textsterling 17.50; \$ 29.95 (1990). ZBL0721.11015.