# Hilbert polynomial of the graph and boundedness of Hom scheme?

Let $$X,Y$$ be two projective variety over an algebraically closed field, then we know $$Hom(X,Y)$$ has a scheme structure by considering the embedding as a open subscheme of $$Hilb(X\times Y)$$ using the graph.

We know for each fixed polynomial $$P$$, the Hilbert scheme with Hilbert polynomial $$P$$ is projective hence finite type. And it seems that the graph of any $$f:X\longrightarrow Y$$ is isomorphic to $$X$$ hence has the same Hilbert polynomial viewed as a closed subscheme of $$X \times Y$$. However, Hom scheme can have infinitely many component (for example consider $$End(E)$$ where $$E$$ is an elliptic curve). So where does my intuition goes wrong? If $$X,Y$$ are both smooth curves, can we decide all of the connected components?

• Sasha already answered your question perfectly. Let me just add that your intuition is wrong, as can be already seen in the following example. Let $X$ be isomorphic to the projective line. Then, one can linearly embed $X$ into $\mathbb{P}^2$. Such an embedding makes $X$ into a degree one subvariety of $\mathbb{P}^2$. However, you could have also embedded $X$ into $\mathbb{P}^2$ as a conic. This then makes $X$ into a degree two subvariety of $\mathbb{P}^2$. So, subvarieties of $\mathbb{P}^2$ (or the Hilbert scheme for that matter) which are abstractly isomorphic could have different degree. Jan 20, 2019 at 15:05

To define a Hilbert polynomial you need to choose an ample line bundle. A reasonable choice on $$X \times Y$$ is $$L_X \boxtimes L_Y$$, where $$L_X$$ and $$L_Y$$ are line bundles on $$X$$ and $$Y$$. Then if $$f \colon X \to Y$$ is a morphism, the restriction of $$L_X \boxtimes L_Y$$ to the graph $$\Gamma(f)$$ of $$f$$ corresponds to the line bundle $$L_X \otimes f^*L_Y$$ on $$X$$, hence the Hilbert polynomial of $$\Gamma(f)$$ equals the Hilbert polynomial of $$X$$ with respect to $$L_X \otimes f^*L_Y$$. In particular, it does depend on $$f$$.