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A non-simply laced simple root system can be constructed from the simply-laced root system by folding the Dynkin diagram and hence the corresponding non-simply-laced Lie algebra can be constructed by taking the fixed points of a non-trivial diagram automorphism (outer automorphism). Then how are their Grassmannians related ? Specifically, if $\sigma$ is an outer automorphism of $SL_{2n}$ and $P$ is a maximal parabolic, then is it true that $(G/P)^{\sigma}$ is a Grassamanian of $Sp_{2n}$ ? Is every Grassmannian of $Sp_{2n}$ obtained in this way ?

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Most maximal parabolics of $SL_{2n}$ are not $\sigma$-invariant, not even up to conjugation. So $(G/P)^\sigma$ does not make sense. The correct statement is: Let $I\subseteq\{1,\ldots,2n-1\}$ be a symmetric subset, i.e., with $i\in I\Leftrightarrow 2n-i\in I$. Let $P_I\subseteq SL_{2n}$ be the corresponding parabolic. Then $J:=I\cap\{1,\ldots,n\}$ corresponds to a parabolic $P_J\subseteq Sp_{2n}$. Then $\sigma(P_I)=P_I$ and $(SL_{2n}/P_I)^\sigma=Sp_{2n}/P_J$. In particular, the $i$-th Grassmannian of $Sp_{2n}$ corresponds to the submaximal parabolic $P_I$ with $I=\{1,\ldots,2n-1\}\setminus\{i,2n-i\}$.

In geometric terms this corresponds to the following simple fact: the involution $\sigma$ induces an involution $U\mapsto U^\perp$ on subspaces of $\mathbb C^{2n}$. The $i$-th Grassmannian of $Sp_{2n}$ consists of $i$-dimensional isotropic subspaces $U$, i.e., with $U\subseteq U^\perp$. Thus, such a $U$ corresponds to a $\sigma$-invariant partial flag $U\subseteq V$ with $\dim U=i$ and $\dim V=2n-i$.

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    $\begingroup$ If we take $I=\{n\}$ then $SL_{2n}/P_I=Gr(n,2n)$, where $P_I$ is the maximal parabolic corresponding to $I$. Then $(SL_{2n}/P_I)^{\sigma}=Sp_{2n}/P_n$ the Lagrangian Grassmannian. But the answer here suggests that the fixed point set is disconnected. mathoverflow.net/questions/266274/fixed-points-of-an-involution I am wondering what am I missing here ? $\endgroup$ – jack May 13 '17 at 3:03

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