Let $X$ be a curve or an abelian variety (over a finite field). Then the Galois representation associated to $X$ via the etale cohomology of $X$ (in degree $1$) is integral of weight $1$ and its dimension is determined by the Hilbert polynomial of $X$. This is a theorem of Weil.


Let $X$ be a variety with fixed Hilbert polynomial $h$. Are the dimension and weight of the Galois representations associated to $X$ via etale cohomology determined by $h$? Note that these representations have a well-defined dimension and weight by Deligne's proof of the Riemann hypothesis over finite fields.


Edit: Will Sawin points out that the dimension of the representation doesn't only depend on the Hilbert polynomial. Thus, I would like to ask the following weaker question.


Is the dimension of the representation bounded if we fix the Hilbert polynomial?



Does the Hilbert polynomial determine the weight of the Galois representation associated to a variety
No. By Hirzebruch-Riemann-Roch, the Hilbert polynomial of a surface embedded in $\mathbb P^1$ with hyperlane class $D$ is determined by the invariants $\chi(O\_X)$, $D \cdot D$, and $D \cdot K$. There is no reason to expect two surfaces with the same arithmetic Euler characteristic, and that each have a divisor with a fixed set of intersection numbers, to have the same Betti numbers.


The divisor $6 H - 2 e\_1$ on $\mathbb P^2$ blown up at a single point is very ample, and satisfies $D^2=32$, $D\cdot K= 16$.


The divisor $7 H - 4 e\_1 - e\_2$ on $\mathbb P^2$ blown up at two points is very ample, and satisfies $D^2=32$, $D\cdot K= 16$.


But $H^2$, or the weight $2$ Galois representation, is $2$-dimensional for the first surface and $3$-dimensional for the second.


There is probably an easier example. This is just the first one I came up with.


