Let $T$ be a finite symmetric set generating a Zariski dense subset of an algebraic group $G$ (specifically, $PSL\_2(\mathbb{C})$ or its subgroups). Is there an $\alpha>0$ such that the set $T^{\leq n}$ of words of length at most $n$ is not in any codimension-1 subvariety of degree $n^{\alpha}$?


"Escape from subvariety" arguments seem to prove similar results that are not polynomial.



Degree of balls in finitely-generated subgroups of SL_2(C)
In the case of $\text{PSL}(2,\mathbb{C})$, the set of words of length $O(n^3)$ is not contained in a subvariety of degree $n$. A polynomial of degree $n$ is a linear combination of matrix entries of the irreps of highest weight $\le n$. Let $A\_n$ be the direct sum of the corresponding matrix algebras; its dimension is a sum of consecutive squares which is then $O(n^3)$. The generating set $S$ yields a set of operators in $A\_n$, which then yields an algebra filtration of $A\_n$. Since $S$ generates a Zariski dense subgroup, some term of this algebra filtration is eventually all of $A\_n$. On the other hand, the filtration is generated by the term of degree 1, i.e., one can write
$$A\_n^{(k+1)} = A\_n^{(k)}A\_n^{(1)},$$
taking all linear combinations of all products of pairs on the right side. So the dimensions of the terms of the filtration have to keep going up by at least one until the filtration terminates, which then gives you the $O(n^3)$ bound.


There is a similar argument for any linear algebraic group, except that the algebra $A\_n$ is different for each one.


This is a really interesting question. Although this argument does give you the bound that you wanted, I have the feeling that the bound is not optimal.


