A "polar dual" for projective varieties? Given a projective variety $X$ (over $\mathbb{C}$, say) with an affine paving $X=\sqcup_i C_i$, one can construct a poset $P_X$ on the set of cells $\{C_i\}$ by saying $C_i \leq C_j$ whenever $C_i \subseteq \overline{C_j}$.  For example, doing this for Schubert cells in the flag variety gives the Bruhat order.
The dual $P^*$ of a poset $P$ is obtained by "flipping $P$ upside down": $x \leq_{P^*} y$ if and only if $y \leq_P x$.
Is there a natural construction of a "polar dual" variety $X^*$, along with an affine paving, such that $P_{X^*}\cong P_X^*$?  Ideally $X^{**}$ should be "the same" stratified variety as $X$.  I am willing to assume any niceness conditions on the decomposition $X=\sqcup_i C_i$ that you like.
The reason I use the name "polar dual" is that if one views a convex polytope $Q$ as the union of its open faces, and defines the poset $P_Q$ using the closure relation as above, then $P_{Q^{\circ}}=P_Q^*$, where $Q^{\circ}$ is the polar dual polytope.
 A: The answer to your question is no.  If you have an irrdeucible projective variety $X$ of dimension $d$ with an affine paving, then for all $i < d/2$, the number of $i$-cells is less than or equal to the number of $(d-i)$-cells.  This is the main idea of this paper by Bjorner and Ekedahl:
https://arxiv.org/pdf/math/0508022.pdf
The proof is as follows:  the Hard Lefschetz Theorem for intersection cohomology tells you that multiplication by the $(d-2i)$th power of the ample class gives an isomorphism from $I\!H^{2i}(X)$ to $I\!H^{2(d-i)}(X)$.  The existence of an affine paving implies that cohomology sits inside of intersection cohomology, so the above isomorphism restricts to an injection from $H^{2i}(X)$ to $H^{2(d-i)}(X)$.  But the dimension of $H^{2i}(X)$ is equal to the number of $i$-cells, so we win.
The upshot is that you can only hope to find a polar dual variety in the sense that you describe if the Poincare polynomial is palindromic, and this does not always happen.
For example, let $V$ be a generic 3-dimensional subspace of $\mathbb{C}^4$, and let $X$ be its closure inside of $(\mathbb{P}^1)^4$.  Then $X$ admits an affine paving in which the cells are indexed by looking at what subset of coordinates is equal to infinity.  There is one 0-cell (all coordinates equal to infinity), four 1-cells (three coordinates equal to infinity), six 2-cells (two coordinates equal to infinity), and one 3-cell (no coordinates equal to infinity).  (It's easy to convince yourself that it is not possible for exactly one coordinate to equal infinity.)  Since 4 is not equal to 6, $X$ cannot have a polar dual.
