We will be working over an algebraically closed field of characteristic 0. We say that a projective variety $X\subset \mathbb{P}^n$ has projectively isomorphic plane sections if there is an open set $U\in(\mathbb{P}^n)^\vee$ such that the hyperplane sections $H\cap X,\ H\in U$ are all projectively isomorphic, i.e. for any two $H\_1,H\_2\in U$ there is some automorphism $f$ of $\mathbb{P}^n$ such that $f(H\_1)=H\_2, f(X\cap H\_1)=X\cap H\_2$.


Consider a smooth cubic hypersurface $X\subset \mathbb{P}^3.$ According to Theorem 2.12 of the paper "[Some Remarks on Varieties with Projectively
Isomorphic Hyperplane Sections](https://link.springer.com/article/10.1007/BF01263521)" by Rita Pardini, $X$ has projectively isomorphic hyperplane sections if and only if it's a projection of a normal rational ruled surface in $\mathbb{P}^4.$ I suspect that this is not the case in this situation.


Is there a way to see that $X$ does not have projectively isomorphic hyperplane sections? More generally, what are the ways to identify such projections of normal rational ruled surfaces of degree $d$ from $\mathbb{P}^{d+1}$ to $\mathbb{P}^d$?



Does a smooth cubic in $P^3$ have projectively isomorphic sections?
Try this: Consider a "Lefschetz pencil" of planes through a general line. I believe these cut the smooth cubic surface in a pencil of irreducible plane cubic curves that are generally smooth, and the special ones have only one node, in particular are stable curves. Hence the smooth ones converge to the nodal ones in the moduli space of stable curves of genus one. That space is hausdorff, so the smooth ones cannot all be isomorphic, much less projectively equivalent. Does this work?


