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?
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2$\begingroup$ Well, this requires the data of an inner product on $V$. If you want something that doesn't require an inner product (so works over any field) you should take a subspace $W \subset V$ to its annihilator in $V^{\ast}$. $\endgroup$– Qiaochu YuanMar 4, 2012 at 22:42
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$\begingroup$ Ok, yes of course I mean there is a given inner product on V. Then why is the map algebraic? $\endgroup$– JerryMar 5, 2012 at 0:11
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$\begingroup$ I edited your post a little. It's customary not to write "thanks" or to sign your questions, since your username is automatically displayed below. $\endgroup$– Ryan ReichMar 5, 2012 at 0:29
3 Answers
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.
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$\begingroup$ Thanks Ryan! The only thing bothering me is that how can I see that the entries in the matrix (why are not uniquely determined) correspond to the canonical coordinates (the k-th exterior product) in the Grassmannian in an algebraic way? $\endgroup$– JerryMar 5, 2012 at 0:33
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$\begingroup$ The matrix entries are indeed non-unique, though if you column-reduce it you can get a somewhat-canonical form (depending on which rows you choose to put the 1's in). The coordinates in $\bigwedge^k (\mathbb{C}^n)$ are just the set of all $k\times k$ minors (determinants) from the matrix $M$. $\endgroup$ Mar 5, 2012 at 0:38
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.
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$\begingroup$ This is very enlightening! When we take $V=\mathbb{C}^n$, is this map biholomorphic? (I'm sure it is a diffeomorphism when we take $V=\mathbb{R}^n$) $\endgroup$ Jul 17, 2021 at 2:12
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.
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$\begingroup$ I find this answer a lot more satisfying than the checked answer (sorry Ryan!). $\endgroup$ Mar 21, 2012 at 14:16