I am wondering whether the bijection that takes a $k$-dimensional subspace $W\subset V$ in an $n$-dimensional space $V$ to its orthogonal complement $W^\bot$ is a rational (algebraic) morphism between the Grassmanians $Gr\_k(V)$ and $Gr\_{n-k}(V)$. And how to see this?



Is orthogonal complement a rational map between Grasmannians?
To expand on what Qiaochu wrote in his comment, there is a slight ambiguity about the notion of "inner product" when you are working over $\mathbb{C}$. You could use a nondegenerate, symmetric bilinear form, which gives essentially the same thing he wrote and works for any field. Or you could use a Hermitian inner product, which requires conjugation and thus only works for $\mathbb{C}$.


The upshot is that the bijection, using a Hermitian inner product, is rational (algebraic) as a map from $\operatorname{Gr}\_k(\mathbb{C}^n) \to \operatorname{Gr}\_{n - k}(\mathbb{C}^n)$, but only as varieties over $\mathbb{R}$. When using a general $\mathbb{C}$-bilinear form, it is rational over $\mathbb{C}$, and $\mathbb{C}$ can be replaced by any field as well.


To see this, there is a standard way of understanding $\operatorname{Gr}\_k(\mathbb{C}^n)$, namely as the set of $n \times k$ matrices $M$ of full rank $k$, modulo column operations. The columns represent basis vectors in one of the $k$-dimensional subspaces $V$ of $\mathbb{C}^n$, and clearly, $V^\perp$ (with respect to the standard symmetric bilinear form) is the kernel of $M^t$. Since a basis for the kernel can be computed using row reduction, which is algebraic, there's your isomorphism.


