Does the Hilbert polynomial determine the weight of the Galois representation associated to a variety - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T18:03:13Z http://mathoverflow.net/feeds/question/116886 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/116886/does-the-hilbert-polynomial-determine-the-weight-of-the-galois-representation-ass Does the Hilbert polynomial determine the weight of the Galois representation associated to a variety Mike Lowrey 2012-12-20T17:35:35Z 2012-12-20T19:41:57Z <p>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.</p> <p>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.</p> <p>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.</p> <p>Is the dimension of the representation bounded if we fix the Hilbert polynomial?</p> http://mathoverflow.net/questions/116886/does-the-hilbert-polynomial-determine-the-weight-of-the-galois-representation-ass/116893#116893 Answer by Will Sawin for Does the Hilbert polynomial determine the weight of the Galois representation associated to a variety Will Sawin 2012-12-20T19:02:43Z 2012-12-20T19:02:43Z <p>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.</p> <p>The divisor $6 H - 2 e_1$ on $\mathbb P^2$ blown up at a single point is very ample, and satisfies $D^2=32$, $D\cdot K= 16$.</p> <p>The divisor $7 H - 4 e_1 - e_2$ on $\mathbb P^2$ blown up at two points is very ample, and satisfies $D^2=32$, $D\cdot K= 16$.</p> <p>But $H^2$, or the weight $2$ Galois representation, is $2$-dimensional for the first surface and $3$-dimensional for the second.</p> <p>There is probably an easier example. This is just the first one I came up with.</p>