Polarity

Remarks

A correlation sends subspaces of dimension $k$ to subspaces of dimension $n + 1 - k$, so it exchanges points and hyperplanes. For a polarity $\delta$, the hyperplane $\delta(P)$ is the polar of the point $P$, and $P$ is the pole of that hyperplane.

Example: let $F$ be a field, $V = F^{n+1}$ and $\delta(U) = U^\perp = \set{y : x \cdot y = 0 \text{ for all } x \in U}$, with the dot product $x \cdot y = \sum_i x_i y_i$. If $U \subseteq W$ then $W^\perp \subseteq U^\perp$; since the dot product is a non-degenerate symmetric bilinear form, $\dim U^\perp = n + 1 - \dim U$ and $(U^\perp)^\perp = U$. So $\delta$ is a bijection that reverses inclusion and has $\delta^2 = \id$: it is a polarity.

Non-example: on $F^3$ let $M$ be the matrix with rows $(1, 1, 0)$, $(0, 1, 0)$, $(0, 0, 1)$, and let $\delta(U) = \set{y : x^T M y = 0 \text{ for all } x \in U}$. This is a correlation, but $\delta(\delta(U)) = M^{-1} M^T U$. For $U = \langle e_1 \rangle$ one gets $M^T e_1 = (1, 1, 0)$ and $M^{-1} (1, 1, 0) = (0, 1, 0)$, so $\delta^2(\langle e_1 \rangle) = \langle e_2 \rangle \ne \langle e_1 \rangle$, and $\delta$ is not a polarity.

Every correlation of $\mathbb{P}(V)$ comes from a non-degenerate sesquilinear form $f$ by $\delta(U) = \set{y : f(x, y) = 0 \text{ for all } x \in U}$. It is a polarity exactly when $f$ is reflexive, that is $f(x, y) = 0$ implies $f(y, x) = 0$; up to a scalar, such forms are symmetric, alternating or Hermitian.