Bounds on Artin conductors over function fields Let $L/K$ be a geometric Galois extension of function fields over $\mathbb F_q$. Let $\chi$ be a non-trivial irreducible character of $\text{Gal}(L/K)$. According to Michael Rosen's Number Theory in Function Fields p.131, Weil's proof of the Riemann Hypothesis for curves over finite fields shows that the Artin L-function associated to $\chi$ is a polynomial in the variable $u := q^{-s}$ of degree $$\chi(1)(2g_K-2) + \deg_K \mathfrak{f}(\chi),$$ where $g_k$ is the genus of $K$ and $\mathfrak{f}(\chi)$ is an effective divisor of $K$ called the Artin conductor of $\chi$.

Question : I would like to have bounds on the previous degree, so for example on $\deg_K \mathfrak{f}$, in terms of $\chi(1)$ and possibly other invariants related to $L/K$, like $\deg_K D_{L/K}$ where $D_{L/K}$ is the discriminant of $L/K$.

In the number field case, I know there are for example the Odlyzko bounds on Artin conductors, and I was wondering if such bounds could be found in the function field case. Thanks in advance for any answer.
 A: It's not completely clear from your question if you want lower or upper bounds.
For lower bounds, the Odlyzko bounds come from analysis of the $L$-function. The lower bound we get by analyzing the $L$-function in the function field case is
$$ \deg_K \mathfrak{f}(\chi) \geq  (2 -2g_K) \chi(1) $$ coming from the obvious fact that the degree of a polynomial is nonnegative. This is sharp, which we can see by, for instance, taking $f$ to be a generic polynomial of degree $d$, $K = \mathbb F_q ( f(T))$, $L$ to be the Galois closure of $\mathbb F_q(T)$ over $K$, $\chi$ to be the $d-1$-dimensional standard representation of the Galois group $S_{d-1}$ of, in which case this $L$-function is indeed a polynomial of degree zero.
For upper bounds, the conductor-discriminant formula gives
$$ \deg_K D_{L/K} = \sum_{\chi} \chi(1) \deg_K \mathfrak{f}(\chi)$$ where the sum is over irreducible $\chi$ so $$\deg_K \mathfrak{f}(\chi) \leq \frac{  \deg_K D_{L/K} }{\chi(1) } $$ because all these degrees are nonnegative.
This is sharp for a few Galois groups.

From the definition of the Artin conductor, we can see that $$\deg \mathfrak{f} (\chi) = \frac{1}{|G|} \sum_{g \in G} i(\sigma) ( \chi(e) - \chi(\sigma) ) =\frac{1}{|G|} \sum_{g \in G} i(\sigma) \operatorname{Re}( \chi(e) - \chi(\sigma) )  $$ where $i(\sigma)$ is the maximum $i$ such that the $i$'th ramification group includes $\sigma$ and note that $Re( \chi(e) - \chi(\sigma)) > 0 $ because $\chi$ is a faithful representation of $g$. Combining this with the conductor-discriminant formula, we get
$$\deg_K D_{L/K} =  \sum_{ \sigma \in G, \sigma \neq e} i(\sigma)$$ and thus
$$ \frac{ \deg \mathfrak f(\chi) } { \deg D_{L/K}}\geq \min_{ \sigma \in G, \sigma \neq e}  \frac{\operatorname{Re} ( \chi(e) -\chi(\sigma)) } {|G|} .$$
In fact we can do slightly better because we can average in the numerator over all conjugate $\sigma$, where $\sigma$ and $\sigma'$ are conjugate if $\sigma$ is a power of $\sigma'$ and vice versa. This has the effect of averaging over all Galois conjugates of $\chi(e) -\chi(\sigma)$, which upon consideration of roots of unity means that the numerator is at least $1/2$.
Now $|G| = [L:K]$ so this gives $$\deg \mathfrak f(\chi) \geq \frac{ \deg D_{L/K}} { 2[L:K]}.$$
