# What's the minimal embedding of orthogonal Grassmannian?

Suppose $X$ is the orthogonal Grassmanian. We know the Plücker embedding does not span the whole background $\mathbb{CP}^n,$ just a subspace $\mathbb{CP}^m.$

My question: is there an expression of the isometric embedding of $X$ into $\mathbb{CP}^m$?

By expression, I mean a way to write down explicitly the coordinates such as Plücker coordinates for the Grassmannian?

• I posted a similar question, but in the Lagrangian context. I'm sure that minor modifications will allow you to fit the answer(s) I got to the orthogonal context you're interested in. See: mathoverflow.net/questions/209058/… – Giovanni Moreno Sep 30 '15 at 10:13
• By the way, the analogous answer for the embedding in the case of the Lagrangian Grassmannian is this: take the Plucker embedding applied to a symmetric matrix and delete all redundant minors due to the symmetry. The target is now $\bigwedge^n \mathbb{C}^{2n} / \omega\wedge \bigwedge^{n-2} \mathbb{C}^{2n}$, the fundamental representation of the symplectic group with the appropriate highest weight. – Luke Oeding May 24 '17 at 5:14

The image under the Plücker embedding of the isotropic Grassmannian of $r$-planes in $\mathbb{C}^k$ (of course, $r\leq k/2$) spans the subspace generated under $SO(k)$ acting on the highest weight vector $v_1\wedge \cdots \wedge v_r$. This is the entirety of $\bigwedge^{r}\mathbb{C}^k$ with the sole exception of when $k=2r$. In that case, I believe it spans a half dimensional subspace, but to be honest, it is late, and I've forgotten the details.
• I also believe the case $k=2r$ is an exceptional one. In fact, I asked the very same question posted above, but in the skew-symmetric (i.e., symplectic/Lagrangian) context, and the answer turned out to be a little bit trickier than the easy "half dimensional subspace" guess. See: mathoverflow.net/questions/209058/… – Giovanni Moreno Sep 30 '15 at 10:10