Is orthogonal complement a rational map between Grasmannians? 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?
 A: One can define an orthogonal complement map without any usage of a metric metric or bilinear form,  just as they do in algebraic geometry. The  "orthogonal" complement of a subspace $L\subset V$ is the subspace $L^\perp \subset V^*$ consisting of linear functionals $v:V\to\mathbb{C}$ such that
$$v(x)= 0,\;\;\forall x\in L.  $$
We get a map
$$\mathcal{O}: Gr_k(V)\to Gr_{\dim V-k}(V^* ). $$
If we fix $L_0\in Gr_k(V)$, then all $k$ dimensional subspaces  of $V$ close to $L_0$  are  graphs of a linear map
$$T: L_0\to V/L_0, $$
i.e.,  a neighborhood of $L_0$ in $GR_k(V)$ can be identified with $ {\rm Hom}\;(L_0, V/L_0) $. Now observe that the dual of $V/L_0$   is naturally identified   with $L_0^\perp$ and we can view the   dual (transpose) of $T$ as a map
$$ L_0^\perp\to V^*/L_0^\perp $$
The map $\mathcal{O} $  in the above coordinates  by the   transpose map
$$ {\rm Hom}\;(L_0, V/L_0)\to {\rm Hom}\;(\; (V/L_0)^*, L_0^* )= {\rm Hom}\;( L_0^\perp, V^*/L_0^\perp). $$
Since the above map is linear, we conclude that  $\mathcal{O}$ is  rational.
A: 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.
A: Let's be more explicit. Let $k^N$ have orthonormal basis $e_1$, \ldots, $e_n$. Let $L$ be a $d$-plane with Plucker coordinates $p_I$. Let $q_J$ be the Plucker coordinates of $L^{\perp}$. Then $p_I = \pm q_{[n] \setminus I}$, where I am too lazy to work out the sign. In particular, this is a well defined morphism, not just a rational map. Here $[k]$ denotes $\{1,2,\ldots, k \}$.
This is probably more easily done by an example than a proof. Let's look at the $2$-plane in $5$-space given as the row span of
$$\begin{pmatrix} 1 & 0 & a & b & c \\ 0 & 1 & d & e & f \end{pmatrix}$$
Its Plucker coordinates are
$$p_{12} =1,\ p_{13} = d,\ p_{14}=e,\ p_{15}=f,\ p_{23}=-a,\ p_{24} = -b,\ p_{25} = -c,$$
$$\ p_{34} = ae-bd,\ p_{35} = af-cd,\ p_{45} = bf-ce.$$
The orthogonal complement is
$$\begin{pmatrix}
a & d & -1 & 0 & 0 \\
b & e & 0 & -1 & 0 \\
c & f & 0 & 0 & -1
\end{pmatrix}.$$
(Check it for yourself!) The Plucker coordinates of this matrix are
$$q_{345} = -1,\ q_{245} = d,\ q_{235} = -e,\ q_{234}=f,\ q_{145} = a,\ q_{135}=-b,\ q_{134} = c$$
$$q_{125} = - (ae-bd),\ q_{124} = af-cd,\ q_{123} = - (bf-ce).$$
It should be pretty clear how to redraw this example for a larger Grassmannian, up to getting the sign details right.
