Let $(V,\omega)$ be a $2n$-dimensional linear symplectic space, and $(\mathbb{P}V,\theta\_\omega)$ the corresponding $(2n-1)$-dimensional contact manifold.


Given a smooth $(n-1)$-dimensional smooth projective variety $X\subset \mathbb{P}V$, I can define its **orthogonal**
$$
X^\perp:=\{ \pi\in \mathbb{P}V^\*\mid\pi=v^\perp\textrm{ for some }[v]\in X \}\, ,
$$
as well as its **dual**
$$
X^\ast:=\{ \pi\in \mathbb{P}V^\*\mid\pi\supset T\_v\widehat{X}\textrm{ for some }[v]\in X \}\, ,
$$
where $\widehat{X}$ is the cone over $X$.


Here's the point: I'm trying to define a "companion" $X\_\bullet$ of $X$ via the equation
$$
X\_\bullet^\ast=X^\perp\, .\quad\quad (1)
$$
*Example.* If $D<V$ is a linear $n$-dimensional subspace, and $X=\mathbb{P}D$, then $(1)$ unambiguously defines $X\_\bullet$, which turns out to be $\mathbb{P}D^\perp$. In particular, if $D$ is $\omega$-Lagrangian, then $X\_\bullet=X$ is also $\theta\_\omega$-Legendrian.



> 
> QUESTION: how to characterise the $X$'s possessing a well-defined (by means of $(1)$) "companion" $X\_\bullet$? In particular, does this property hold for arbitrary (i.e., not necessarily linear) $\theta\_\omega$-Legendrian $X$'s? Is it true that in this case $X=X\_\bullet$?
> 
> 
> 


I'm sure that this problem has been already addressed somewhere: any reference/keyword will be most welcome.


**EDITED after Libli's and Robert Bryant's comments**


Of course, by using the *reflexivity theorem* (see, e.g., [this paper by Tevlev, Theorem 1.7a](http://arxiv.org/abs/math/0112028v1)), one can solve $(1)$ and obtain
$$
X\_\bullet=(X^\perp)^\*\, .
$$
So, the answer to my question is "for all $X$". Nevertheless (recall that I'm assuming $X$ to be smooth), I can still ask, how to characterise those $X$ whose $X\_\bullet$ enjoys some extra properties, like being smooth and (which this is **what really I'm interested in**) having the same dimension as $X$ - or even being $X$ itself (the only purpose of the symplectic orthogonality is to have $X$ and its dual sitting in the same space, and compare them).



Symplectic orthogonality and projective duality: how do they work together?
**Corrected (Partial) Answer**


Given a smooth variety $X\subset \mathbb{P}V$ that is the projectivization of a smooth punctured cone $\hat X\subset V\setminus\{0\}$, the space $X\_\bullet$ that the OP defines is a variety in $\mathbb{P}V$ described as
$$
X\_\bullet 
= \bigl\{[v]\in\mathbb{P}V\ \bigl|\ \exists w\in \hat X,\ \ \omega(v,T\_w\hat X) = 0\ \bigr\}
\subset \mathbb{P}V.
$$
The OP's question then is: When is $X\_\bullet = X$? (In this case, say $X$ is $\omega$-self-dual.) [Note that the question of when $X\_\bullet$ is smooth or has the same dimension as $X$ is not really a symplectic question, since the classical dual $X^\*\subset\mathbb{P}V^\*$ is equivalent to $X\_\bullet$ under the correlation defined by $\omega$, and $X^\*$ does not depend on the symplectic structure.]


I don't have a complete answer, but here are a couple of observations:


1. If $X$ is $\omega$-self-dual, we must have $\dim X \ge n{-}1$, 
and, if $\dim X = n-1$, then $X = \mathbb{P}L$, 
where $L\subset V$ is a Lagrangian subspace of $V$.
The reason for this is that, if $\dim X = k$, then the
dimension of $T\_w\hat X$ is $k+1$, 
so its symplectic orthogonal $(T\_w\hat X)^\perp$
has dimension $2n-k-1$ 
and so $\mathbb{P}\bigl((T\_w\hat X)^\perp\bigr)$ has
dimension $2n-k-2$ and is contained in $X\_\bullet$.
Thus, if $X\_\bullet = X$ (or, more generally, if they just have the same dimension), then $k\ge 2n-k-2$, so $k\ge n{-}1$.
If equality holds then $X = X\_\bullet 
= \mathbb{P}\bigl((T\_w\hat X)^\perp\bigr)$, so $X = \mathbb{P}L$
for the Lagrangian space $L = (T\_w\hat X)^\perp = T\_w\hat X$.
2. If $\dim X = 2n{-}2$ (i.e., $X$ is a hypersurface), 
then $X$ must be a quadric hypersurface 
(since those are the only smooth hypersurfaces whose duals are smooth), 
but not every quadric hypersurface is $\omega$-self-dual when $n>1$:
The set of smooth quadric hypersurfaces in $\mathbb{P}V$ 
has dimension $n(2n{+}1)-1$ while the space of $\omega$-self-dual 
quadric hypersurfaces only has dimension $n(n{+}1)$.
In fact, it is not hard to show
that if $V = L\_+\oplus L\_-$ is a splitting of $V$ 
into two transverse Lagrangian subspaces $L\_+$ and $L\_-$, 
then the quadric hypersurface defined by
$$
X = \bigl\{ [v\_+ + v\_-]\in\mathbb{P}V\ \bigl|\ \omega(v\_+,v\_-) = 0\ \bigr\}
$$
is $\omega$-self-dual, and every $\omega$-self-dual quadric
hypersurface is of this form for some splitting $V = L\_+\oplus L\_-$.


