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 ?

  • 2
    $\begingroup$ If $X$ is a hypersurface with one double point, $\dim X^{\vee}=\dim S^{\vee}=n-1$. $\endgroup$
    – abx
    Jan 18, 2021 at 15:15
  • 2
    $\begingroup$ Maybe a non-strict inequality was meant? $\endgroup$
    – Will Sawin
    Jan 18, 2021 at 16:05

1 Answer 1


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

$$\dim S^{\vee}=1<2=\dim X^{\vee}.$$


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