## Does the Hilbert polynomial determine the weight of the Galois representation associated to a variety

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?

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 I am not sure I completely understand your question. What cohomology do you consider exactly. Is it something like $H^i (X \times \bar k, \mathbb Z_l)$ for some fixed $i$? Now, if you fix the degree $i$, then the weight of the Galois representation on that space is $i$ by Deligne, for $X$ proper and smooth. Do you consider some non-proper or non-smooth $X$? – Joël Dec 20 at 19:53 Thank you for your comment. It actually answered my question. I didn't know Deligne had proved the weight was $i$. I just thought it was some integer. – Mike Lowrey Dec 20 at 20:16 Okay. Just so you know, if you take $X$ non-smooth or non-proper, the action of Galois (i.e. of Frobenius) on the $H^i$ above may have several different weights -- one says it is mixed --. If $X$ is smooth but not proper, all the weights appearing are at most $i$. If $X$ is proper but not smooth, at leads $i$. – Joël Dec 20 at 21:47

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.

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Ok. So the dimensions change. But the weight is the same apparently. But probably there's an example where the weights are different too, no? – Mike Lowrey Dec 20 at 19:15
Also, do you think that the dimension is "bounded" if we fix the Hilbert polynomial? This is weaker than the question I originally asked. – Mike Lowrey Dec 20 at 19:31
1. What do you mean by "the weight is the same"? For a smooth projective variety, $H^n$ always has weight $n$. Do you want to consider non-smooth varieties? 2. The dimension is clearly bounded if you fix the space it's embedded in, because the Hilbert scheme will be Noetherian, and the dimensions of the Dolbeaut cohomology groups are semicontinuous by the semicontinuity theorem, so their is a maximum to them, so there is a maximum to the Betti cohomology groups. – Will Sawin Dec 20 at 19:53
I had not completely understood Deligne's theorem. You have completely answered my question. Thank you very much! – Mike Lowrey Dec 20 at 20:17