Let $P\_{\alpha}$ be the principal series representation of $GL\_n(\mathbb{F}\_q)$, where $\alpha = ( \alpha\_1, \alpha\_2, \cdots, \alpha\_n)$ and $\alpha\_i : \mathbb{F}\_q^\* \rightarrow \mathbb{C}^\*$. 


The character value of $P\_{\alpha}$ on the unipotent conjugacy class ( Jordan form has eigenvalues $x$) are of the form $f(q) \prod\_i \alpha\_i(x) $, where $f$ is a polynomial in $q$. Is it possible to say what is the degree of this polynomial ( or maybe a non trivial bound on its degree)?


(p.s. There is already an explicit formula for the character value of the cuspidal representations on unipotent conjugacy classes, and if we do a bases change to $\mathbb{F}\_{q^n} $, the cuspidal representation becomes a principal series, hence I'm inclined to think that the degrees match for the two of them, but don't know how to proceed with this argument.)


(p.p.s. if $\alpha\_i = \alpha\_j$ for all $i,j$, then the principal series is reduced to a direct sum of unipotent representations, and if one can say anything about the maximum possible degree of the polynomial occurring in unipotent representation for each conjugacy class, then that gives a non-trivial bound. This should give a bound on the degrees occurring in a given unipotent class, and here again I don't know how to prove it for all unipotent representations)



Character values of principal series representations of $GL_n(\mathbb{F}_q)$
By the standard formula for the character of an induced representation, the value f(q)/|B| is equal to the number of g for which gug-1∈B (here u is my unipotent element).


Rewriting this condition as u∈g-1Bg, we see that f(q) is equal to the number of Borel subgroups containing u, i.e. the number of points in the Springer fibre.


Thus the degree of f(q) is equal to the dimension of the Springer fibre, which is known. So if u has Jordan type λ, then 
$$\operatorname{deg}(f)=\sum\_i {\lambda^t\_i\choose 2}$$


