Let $C_1$ and $C_2$ be two rationally equivalent curves in $\mathbb{P}^3$. Is it true that the dimension of $H^0(\mathcal{N}_{C_1|\mathbb{P}^3})$ equal to that of $H^0(\mathcal{N}_{C_2|\mathbb{P}^3})$?
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1$\begingroup$ No, that is not true. Even if $C_1$ and $C_2$ have the same Hilbert polynomial, which is stronger than being rationally equivalent in $\mathbb{P}^3$, that is not true. Even if the two curves are parameterized by points in the same irreducible component of the Hilbert scheme, that is not true. Let $C_1$ be a twisted cubic, and let $C_2$ be a specialization of a twisted cubic whose reduced curve is a planar, nodal cubic, but with a spatial embedded point at the node. $\endgroup$– Jason StarrCommented Nov 15, 2013 at 15:34
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1 Answer
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No, even if you assume $C_1$ and $C_2$ smooth irreducible. A rational equivalence class of curves in $\Bbb{P}^3$ is determined by its degree. Take for $C_1$ a complete intersection of 2 quadrics, you get $\dim H^0(\mathcal{N}_{C_1|\mathbb{P}^3})= 16$; and for $C_2$ a plane quartic curve, you find $\dim H^0(\mathcal{N}_{C_2|\mathbb{P}^3})= 17$.
