If $X$ is a variety and $X_p$ is its projection from a point $p$ not on $X$, can we relate the Hilbert polynomial of $X_p$ to the one of $X$?

$\begingroup$ It's smaller... $\endgroup$ – Allen Knutson Jan 27 '14 at 20:43

$\begingroup$ It depends on how far the map $X \to X_p$ is from being an isomorphism. $\endgroup$ – Sasha Jan 27 '14 at 20:47
The projection induces a map $\pi :X\rightarrow X_p$, which is finite and birational (at least in characteristic 0, or for $p$ general enough, and codim$(X)>1$ in its ambient projective space). There is an exact sequence $$0\rightarrow \mathcal{O}_{X_p}\rightarrow \pi _*\mathcal{O}_X\rightarrow \mathcal{Q}\rightarrow 0$$ where $\mathcal{Q}$ is supported on the locus where $\pi $ is not an isomorphism. The Hilbert polynomial $P_X$ of $X$ is given by $P_X(m)=\chi (\mathcal{O}_X(m))$, and same for $X_p$; so the exact sequence gives $P_{X_p}(m)=P_X(m)\chi (Q(m))$. If you have some control on $\mathcal{Q}$ (e.g. if you know that $\pi $ is an isomorphism outside finitely many points, or even better an isomorphism), you'll be able to say something; in general, however, it may be rather complicated.

$\begingroup$ are you saying that if $\pi$ is an isomorphism, then the Hilbert Polynomials are identical ? $\endgroup$ – aginensky Jan 28 '14 at 3:36

$\begingroup$ Yes. In that case $\pi _*\mathcal{O}_X=\mathcal{O}_{X_p}$, so $\chi (\mathcal{O}_{X_p}(m))=\chi (\mathcal{O}_{X}(m))$. $\endgroup$ – abx Jan 28 '14 at 6:26

$\begingroup$ Be careful, if $X$ is an hypersurface $\pi$ is far from being birational! $\endgroup$ – IMeasy Jan 28 '14 at 8:20

$\begingroup$ Right! I was assuming (implicitely...) codim($X)\,\geq 2$. $\endgroup$ – abx Jan 28 '14 at 8:26
