Let $V=\mathbb C^{2n}$ with the standard basis $\{e_1,e_2, \cdots , e_{2n}\}$ and let $\sigma$ be the involution $e_i \mapsto e_{2n+1i}$. This induces an involution of the Grassmannian $G(n,2n)$ of $n$ dimensional subspaces of $\mathbb C^{2n}$. Then what are the fixed points of this involution ? Does it have a nice structure as a projective variety ?

$\begingroup$ The involution is just flipping signs of $n$ coordinates. $\endgroup$ – Fan Zheng Apr 3 '17 at 17:19

$\begingroup$ Also posted on MSE 15 hours ago. $\endgroup$ – Michael Albanese Apr 3 '17 at 17:21
In general, the fixed points of any map of the Grassmannian induced by a diagonalizable linear transformation is just a union of products of Grassmannians in the eigenspaces. Any subspace $V$ fixed by a diagonalizable transformation $A$ is the sum of the intersections of $V$ with the eigenspaces of $A$ (since the projection to each eigenspace is a polynomial in $A$). If we fix the dimension of each of these intersections, we get a map to the product of Grassmannians of the eigenspaces, which is obviously an isomorphism.
In this case, $\mathbb{R}^{2n}$ is the sum of the 1 and 1 eigenspaces, both having dimension $n$. Thus, the fixed points are a disjoint union of $Gr(k,n)\times Gr(nk,n)$ for all $0\leq k\leq n$.

2$\begingroup$ Could you please give a reference of the 1st line of your answer or a hint to prove this ? Thanks. $\endgroup$ – Mark Apr 4 '17 at 0:43

2$\begingroup$ @Mark The rest of the answer is an explanation of how to prove it. The map is given by looking at the intersection with the eigenspaces. $\endgroup$ – Ben Webster♦ Apr 4 '17 at 13:34

1$\begingroup$ The above involution is induced by the Dynkin automorphism $\alpha_i \mapsto \alpha_{2ni}$ and the parabolic $P_n$ is stable under this involution. So isn't the fixed locus same as one of the Symplectic Grassmannian ? $\endgroup$ – jack Apr 4 '17 at 14:12

1$\begingroup$ @jack I understand why you might think that, but if it's the case, please explain the map and tell me the problem with my argument above. Checking any example would have shown you this guess is wrong. The fact that the map doesn't use a symplectic form should be a dead giveaway. If you want the symplectic Grassmannian, pick a symplectic form, and look at the involution given by taking symplectic orthogonal. $\endgroup$ – Ben Webster♦ Apr 4 '17 at 15:25

$\begingroup$ Your arguments are fantastic. I have no problem with that. Of course my guess is wrong since the fixed locus is disconnected in this case. I thought the map is compatible with the standard symplectic form after some change of coordinates. $\endgroup$ – jack Apr 4 '17 at 15:53
Let us analyze in details the simplest nontrivial case, namely $n=2$. Let $$x_{ij} :=x_i \wedge x_j, \quad i <j$$ be the Plücker coordinates of $\mathbb{P}^5$. Since your involution $\sigma$ exchanges $e_1$ with $e_4$ and $e_2$ with $e_3$, it is easy to check that the action on the Plücker coordinates is
\begin{equation*}
\begin{split}
[x_{12}: \, x_{13}: \, x_{14}: x_{23}: \, x_{24}: \, x_{34}] \mapsto & [x_{34}: \, x_{24}: \, x_{14}: \, x_{23}: \, x_{13}: \, x_{12}] \\
=& [x_{34}: \, x_{24}: \, x_{14}:\, x_{23}: \, x_{13}: \, x_{12}].
\end{split}
\end{equation*}
The fixed locus $\Sigma$ of such an involution in $\mathbb{P}^5$ is given by a disjoint union $$\Sigma = \Sigma_1 \sqcup \Sigma_2,$$
where $\Sigma_1$ is the line of equation
$$x_{12}+x_{34}=x_{13}+x_{24}=x_{14}=x_{23}=0,$$
whereas $\Sigma_2$ is the $2$plane of equation
$$x_{12}x_{34}=x_{13}x_{24}=0.$$
Now recall that the Grasmannian $\mathbb{G}(1, \, 3)=G(2, \, 4)$ of lines in $\mathbb{P}^3$ (or, equivalently, of $2$planes in $\mathbb{C}^4$) is the quadric hypersurface $Q \subset \mathbb{P}^5$ whose Plücker equation is $$x_{12}x_{34}x_{13}x_{24}+x_{23}x_{14}=0.$$
Such a quadric is $\sigma$invariant, as expected, and the fixed locus $\Sigma_Q$ of the involution $\sigma \colon Q \to Q$ is given by intersecting $Q$ with $\Sigma$. Then we obtain a disjoint union
$$\Sigma_Q = \Sigma_{Q1} \sqcup \Sigma_{Q2},$$
where $\Sigma_{Q1}:=Q \cap \Sigma_1$ consists of the two points
$$[1: \, 1: \, 0: \, 0: \, 1: \, 1], \quad [1: \, 1: \, 0: \, 0: \, 1: \, 1],$$
whereas $\Sigma_{Q2}:=Q \cap \Sigma_2$ is a smooth linear section of dimension $2$, namely a smooth quadric surface.
Summing up, the fixed locus of the involution induced by $\sigma$ on the Grasmannian $\mathbb{G}(1, \, 3)$ consists of the disjoint union of two distinct points and a copy of $\mathbb{P}^1 \times \mathbb{P}^1$, and this agrees with Ben Webster's answer.

2$\begingroup$ Is it just two points or union of two points along with $\mathbb P^1 \times \mathbb P^1$ as the other answer suggests ? $\endgroup$ – jack Apr 3 '17 at 19:30