If $X\subset \mathbb{P}^{n}$ is a cubic hypersurface that is not normal, what's the easiest way to see that the nonnormal locus is a linear subspace of dimension $n-2$?

As for a reference, there is a paper that classifies the nonnormal cubic hypersurfaces that gives a proof by expressing the hypersurface as a cone over the projection of a rational normal scroll of degree 3 (http://www.sciencedirect.com/science/article/pii/S0022404910002872), but just the fact about the nonnormal locus was already stated an earlier paper without proof (third paragraph of page 6 of https://arxiv.org/pdf/math/0005146v1.pdf), which makes me guess there might an easy way to prove this fact if we aren't after a classification.


1 Answer 1


Singularities of a hypersurface, supported in codimension 2, are normal. So, the question is, what are the $(n-2)$-dimensional components of the singular locus of $X$. Denote the union of these components by $Z$.

Note that the secant variety $Sec(Z)$ of $Z$ is contained in $X$ (indeed, any secant of $Z$ intersects $X$ with multiplicity 4 or greater, hence is contained in $X$). The dimension of $Sec(Z)$ is at least $n-1$, unless $Z = \mathbb P^{n-2}$. Further, $\dim Sec(Z) = n-1$ only if $Sec(Z)$ is a hyperplane in $\mathbb P^n$, and $Z$ is a hypersurface in $Sec(Z)$. So, if $X$ is irreducible, the second case is impossible, and we conclude that $Z = \mathbb P^{n-2}$.

In case when $X = H \cup Q$ is reducible, the nonnormal locus is $H \cap Q$ is a quadric of dimension $n-2$. So, irreducibility is necessary for linearity of the nonnormal locus.


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