# On the dimension of the dual variety of a singular hypersurface

I was primarily interested in the following question. Let $$n\geq 3$$, and let $$X\subset \mathbb{P}^n$$ be a degree $$d$$ hypersurface. Assume that its singularity locus $$S$$ (with reduced structure) is irreducible and smooth of dimension $$k$$. (As pointed out by @abx , this should not be a strict inclusion) Is it true that $$\dim S^\vee \geq \dim X^\vee ?$$ The statement is true for quadratic hypersurfaces and hypersurfaces with isolated singularities. I am wondering if in general this holds.

Also, is there any criteria on when this inequality is strict ?

I was reading the book Discriminants, Resultants, and Multidimensional Determinants by Gelfand-Kapranov-Zelevinksy. In the first chapter they introduce the Katz dimension theorem, which computes the dimension of the dual varieties via local coordinates. But I found it difficult to use, since it's quite hard to write down the local coordinates. Is there any comments on the Katz dimension theorem, especially on how to use it ?

• If $X$ is a hypersurface with one double point, $\dim X^{\vee}=\dim S^{\vee}=n-1$.
– abx
Commented Jan 18, 2021 at 15:15
• Maybe a non-strict inequality was meant? Commented Jan 18, 2021 at 16:05

No. Consider the cubic surface $$X$$ defined by the equation $$x_0^2x_2+x_1^2x_3=0.$$
Then $$X$$ is singular along the line $$x_0=x_1=0.$$ Then $$X^{\vee}$$ is a hypersurface (actually isomorphic to $$X$$) and $$S^{\vee}\cong \mathbb P^1$$ (consisting of the hyperplanes containing the line). So
$$\dim S^{\vee}=1<2=\dim X^{\vee}.$$