Let $C\subset\mathbb{P}^4$ a rational normal curve. Then the variety of secant lines in $\mathbb{G}(1,4)$ is isomorphic to the second symmetric product of $C$, hence a $\mathbb{P}^2$. Is there a small flat deformation of this $\mathbb{P}^2$ in $\mathbb{G}(1,4)$ to something that is not a variety of secant lines?
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3$\begingroup$ The morphism from $\mathbb{P}^2=\text{Sym}^2(C)$ to $\text{Grass}(1,\mathbb{P}^4)$ is unramified. Thus we have a short exact sequence of locally free sheaves from the tangent bundle of $\mathbb{P}^2$ to the pullback of the tangent bundle of $\text{Grass}(1,\mathbb{P}^4)$ to the "normal bundle." By the long exact sequence of cohomology, the higher cohomology of the "normal bundle" vanishes. So the Hilbert scheme parameterizing closed subschemes of $\text{Grass}(1,\mathbb{P}^4)$ is smooth at this point with dimension equal to the "expected dimension." $\endgroup$– Jason StarrCommented Jun 18, 2023 at 21:55
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$\begingroup$ I don't know the notation that you are using. Please state which field your projective 4-space is defined over. Is your G(1,4) just another notation for P^3 (over the same field)? Thanks. $\endgroup$– Daniel AsimovCommented Jun 20, 2023 at 1:19
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$\begingroup$ @DanielAsimov I work over the complex numbers and I am looking at the Grassmannian of lines in $\mathbb{P}^4$. Anyways, I think I can work with what Jason Starr wrote (thanks Jason!) $\endgroup$– HansCommented Jun 26, 2023 at 8:23
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