I first asked this question at [math.stackexchange](https://math.stackexchange.com/questions/4373073/contact-variety-is-equidimensional) with no success, so I decided to repost it here.


I am reading the paper "Weakly Defective Varieties" by L. Chiantini and C. Ciliberto, available [here](https://www.ams.org/journals/tran/2002-354-01/S0002-9947-01-02810-0/S0002-9947-01-02810-0.pdf). In it they define the contact variety as follows.


Let $X$ be a reduced irreducible non-degenerate projective variety in $\mathbb{P}^n\_{\mathbb{C}}$ and $P\_1,\ldots,P\_{k+1}\in X$ be general points. If $H$ is a general hyperplane tangent to $X$ at these $k+1$ points, then the *contact variety* of $H$ is the union $\Sigma$ of the irreducible components of $\text{Sing}(H\cap X)$ containing $P\_1,\ldots,P\_{k+1}.$


Then on page 2 the authors claim:



> 
> Since $P\_1,\ldots,P\_{k+1}$ are
> general points, an obvious monodromy argument shows that $\Sigma$ is equidimensional.
> 
> 
> 


Could someone please elucidate this 'obvious' argument? I honestly don't even know what monodromy even means in this context, I am only familiar with it in the context of automorphisms of covering spaces in topology.



Contact variety to projective variety is equidimensional
I think that the monodromy argument they refer to is not entirely obvious. It is presumably related to a version of the *uniform position principle* but in the context of tangent hyperplanes, which I guess involves some intricate technical details. Let me instead sketch a proof of their statement that avoids monodromy.


For any $k \geq 0$, let us denote by $S^k(X)$ the variety:
$$ S^k(X) = \overline{ \{z \in \mathbb{P}^n, \ \textrm{exists a generic $k$-uple} \ (x\_1, \ldots, x\_k) \in X^k, \ \textrm{with} \ z \in \langle x\_1, \ldots, x\_k \rangle \} }, $$
where $\overline{A}$ is the Zariski closure of $A$. The variety $S^k(X)$ is called the variety of $k$-secants $\mathbb{P}^{k-1}$ to $X$. The variety $S^{k}(X)$ is the closure of the projection of a $\mathbb{P}^{k-1}$-fibration over an open dense subset of $X^k$, hence it is irreducible.


Terracini's Lemma asserts that for generic $x\_1, \ldots, x\_k \in X$ and for generic $z \in \langle x\_1, \ldots, x\_k \rangle$, we have:
$$T\_{S^k(X),z} = \langle T\_{X,x\_1}, \ldots, T\_{X,x\_k} \rangle.$$


Consider $X^\*$ the dual variety of $X$ (that is the closure of the set of generic tangent hyperplanes to $X$). The variety $X^\*$ is the closure of the image in $\left(\mathbb{P}^n\right)^\*$ of the projectivization of the normal bundle of $X$ over the dense open subset $X\_{smooth}$. Hence it is irreducible. We also define $X^\*\_k$ to be:
$$ X^\*\_k = \overline{ \{ H \in X^\*,\ \textrm{exists a generic} \ (x\_1,\ldots, x\_k) \in X^k, \ \textrm{with} \ T\_{X,x\_1} \subset H, \ldots, T\_{X,x\_k} \subset H \}}$$


Terracini's lemma implies that $X^\*\_k = S^k(X)^\*$, where $S^k(X)^\*$ is the projective dual to $S^k(X)$. We let
$$p : \overline{\mathbb{P}(N^\*\_{X\_{smooth}/\mathbb{P}^n}(-1))} \longrightarrow \mathbb{P}^n$$
and $$ q : \overline{\mathbb{P}(N^\*\_{X\_{smooth}/\mathbb{P}^n}(-1))} \longrightarrow \left(\mathbb{P}^n \right)^\*$$
the natural projections, where $\overline{\mathbb{P}(N^\*\_{X\_{smooth}/\mathbb{P}^n}(-1))}$ is the closure of the incidence variety *hyperplane/tangent spaces* in $\mathbb{P}^n \times (\mathbb{P}^n)^\*$. Set $I\_{X^\*\_k} = q^{-1}(X^\*\_k)$. By hypothesis, we have:
$$ p(I\_{X\_k^\*}) = X.$$


Furthermore, Terracini's lemma implies that the generic fiber (over $x \in X$) of $p|\_{I\_{X^\*\_k}}$ is $S(T\_{X,x}, S^{k-1}(X))^\*$, which is irreducible, as the projective dual of an irreducible variety. As a consequence, there is only one irreducible component of $I\_{X^\*\_k}$ which dominates $X$. We denote this component by $Z$. Note that $Z$ dominates $X^\*\_k$. Moreover, since $Z$ is the only irreducible component of $I\_{X^\*\_k}$ which dominates $X$, the *contact locus* of $H$ with $X$ as defined in the paper you mention is:
$$\mathrm{Tan}(H,X) = q|\_{Z}^{-1}(H),$$
for generic $H \in X^\*\_k$.


The Theorem of the generic fiber guarantees that the generic fiber of the restriction $q|\_{Z} : Z \longrightarrow X^\*\_k$ is equidimensional.


