Assume you have a function $f:S^1 \rightarrow \mathrm{GL}_d(\mathbb{C})$ whose coefficients are Laurent polynomials $f_{i,j}(q)\in \mathbb{C}[q^{\pm 1}].$
I am interested in getting conditions for the spectral radius of the matrix $$F_n= f(e^{\frac{2i\pi (n-1)}{n}})\ldots f(e^{\frac{2i\pi}{n}})f(1)$$ to be exponentially growing with $n.$
If $d=1,$ meaning that $f(q)$ is a single Laurent polynomial, you can easily find that $F_n=e^{n m(f)+o(1)},$ where $m(f)$ is the Mahler measure of $f.$ If we further assume $f(q)\in \mathbb{Z}[q^{\pm 1}]$ then $F_n$ is exponentially growing as long as $f$ is not a product of cyclotomic polynomials.
When $d>1,$ you can think of using the determinant, but that doesn't give you any information if $f$ takes values in $\mathrm{SL}_d(\mathbb{C}).$
Of course there are many examples where the spectral radius of $F_n$ doesn't grow exponentially: if $f$ takes value in $\mathrm{SU}_d$ so does $F_n$ and $F_n$ has spectral radius 1. Same thing would apply for upper triangular matrices with diagonal $1.$
Ideally I would want those counterexamples to be general, that is the spectral radius of $F_n$ grows exponentially or the image of $f$ lives in a smaller Lie group than $\mathrm{SL}_d(\mathbb{C}).$ But any sufficient condition for the spectral radius to grow exponentially would be interesting to me.
