Suppose $n \ge 3$ and $X$ is a path-connected subset of $\Bbb{CP}^n$ under the manifold topology. If for every complex hyperplane $H \subset \Bbb{CP}^n$, $H \cap X$ is a degree $k$ algebraic hypersurface in $H$ (could be singular or reducible), is it true that $X$ itself is a degree $k$ algebraic hypersurface in $\Bbb{CP}^n$? This question arose when I was constructing an algebra-free definition of quardric surface (where $n=3$ and $k=2$ respectively), and I have no idea how to prove or disprove this.
Appendix: The $n=2$ case can be easily proved to be false as stated in the comment. However the reason might be that a "hypersurface" in $\Bbb{CP}^1$ is just a set of finite points, thus no enough rigidity occurs. I wonder whether things change in higher dimensions.
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1$\begingroup$ For $n=2$, wouldn't any subset $X$ that intersects any line in finitely many points (but at least one) be a counterexample? There certainly exist such subsets that are not algebraic hypersurfaces. $\endgroup$– Antoine LabelleCommented Jan 12, 2022 at 3:45
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2$\begingroup$ For $n=2$, the OP is requiring that the intersection with every line is a fixed number of points (possibly with multiplicity), say $k$. Is there a $X$ with this property that is not an algebraic curve of degree $k$? $\endgroup$– Francesco PolizziCommented Jan 12, 2022 at 6:32
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$\begingroup$ @AntoineLabelle Just as Francesco Polizzi said, a degree $k$ "hypersurface" in $\Bbb{CP}^1$ is the zero locus of a single degree $k$ polynomial, which is $k$ points with multiplicity. $\endgroup$– ZeroxCommented Jan 12, 2022 at 11:35
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1$\begingroup$ For an arbitrary subset of $\mathbb{CP}^2$, how are you defining "intersection multiplicity" for the intersection with a line at a point? I do not understand your hypothesis for subsets of $\mathbb{CP}^2$. $\endgroup$– Jason StarrCommented Jan 12, 2022 at 11:50
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1$\begingroup$ Then the result is wrong. Start with a plane curve and remove a subset. $\endgroup$– Jason StarrCommented Jan 12, 2022 at 11:55
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