Suppose $X$ is a smooth projective variety of dimension $k$ in $\mathbb{P}^n$. Let $X^*$ be its dual variety in ${\mathbb{P}^n}^*$.

Question: What is the dimension of $X^*$ ?

  • 4
    $\begingroup$ In most cases $X^*$ is a hypersurface, but there are exceptions. See L. Ein, Varieties with small dual varieties I, Invent. Math. 86 (1986), no. 1, 63-74. $\endgroup$
    – abx
    Apr 5, 2022 at 3:58

1 Answer 1


The natural number $\delta(X)=n-1-\dim X^*$ is called the defect of $X$. A standard application of the incidence variety shows that $\delta(X) \geq k$ if and only if, for every $x \in X$ and for every hyperplane $H \supset T_xX$, we have $$\dim_x \{y \in X \; | \; H \supset T_yX \} \geq k.$$

In particular, $X$ is a hypersurface (namely, $\delta(X)=0$) if and only if, for every point $x \in X$ and for every hyperplane $H \supset T_xX$, there are only finitely many $y \in X$ such that $H$ is contained in $T_yX$. As explained in abx's comment, this is the usual situation, but there are exceptions.

A formula by Katz expresses the defect of $X$ in terms of the rank of a certain Hessian matrix, see Chapter 6 of the book Projective Dual varieties by E. Tevelev.


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