*This is a copy from MSE where the question did not attract much attention.*

I'm working over $\mathbb{C}$ here. Let $G=\mathrm{SO}(2n+1)$ be the odd orthogonal group, and $P$ be the maximal parabolic corresponding to the $1$st node in the Type $B_n$ Dynkin diagram, following Bourbaki notation- I mean the endpoint which is adjacent to a doubled edge. (This is a *minuscule* node.) Then $G/P$ should be what is called the *(maximal) orthogonal Grassmannian* $\mathrm{OG}(n,2n+1)$: these are the isotropic subspaces (with respect to a nondegenerate symmetric bilinear form) of maximal dimension in $\mathbb{C}^{2n+1}$.

The Borel-Weil theorem says that the $m$th homogeneous component of the coordinate ring of $G/P=\mathrm{OG}(n,2n+1)$ should be isomorphic to the irreducible representation $V^{m\omega_1}$, where $\omega_1$ is the corresponding fundamental weight. This should hold at least at say the level of representations of the Lie algebra $\mathfrak{g}=\mathfrak{so}(2n+1)$. Actually, it might be that we get the *contragredient* representation $(V^{m\omega_1})^*$ this way (because we're acting on functions). But in Type B negation belongs to the Weyl group so I think we should have $(V^{\lambda})^*\simeq V^{\lambda}$ for any irreducible representation.

So in particular, the linear part of the coordinate ring of $\mathrm{OG}(n,2n+1)$ is the $\mathfrak{g}$ representation $V^{\omega_1}$. Now, the linear part of this coordinate ring also seems like a perfectly good $G$ representation to me. And I would guess that it is the irreducible representation $V^{\omega_1}$. But that can't be right: $V^{\omega_1}$ *should not be realizable* as an $\mathrm{SO}(2n+1)$ representation, because of the fact that $\mathrm{SO}(2n+1)$ is not simply connected; to get this representation we are supposed to have to take the simply connected double cover $\widetilde{\mathrm{SO}}(2n+1)$, which is also called the *spin group* $\mathrm{Spin}(2n+1)$. (This representation $V^{\omega_1}$ is often called the *spin representation*.)

**Question**: where am I getting confused here? What is (the linear part of) the coordinate ring of the orthogonal Grassmannian as a representation of the special orthogonal group?

isdefined at the level of $G$) as you describe, and that if you wish to recover $V^{\omega_1}$ you need to move to the $2$-fold covering $\mathrm{Spin}(2n+1)/P$ of $\mathrm{SO}(2n+1)/P$ to get a square root of the line bundle? (Disclaimer: I didn't give this much thought, so maybe this is stupid.) $\endgroup$ – Gro-Tsen May 18 '20 at 14:16