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 "polar dual" for projective varieties?
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.


