*This is a copy from [MSE](https://math.stackexchange.com/questions/3677192/confusion-over-spin-representation-and-coordinate-ring-of-maximal-orthogonal-gra) 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?



Confusion over spin representation and coordinate ring of orthogonal Grassmannian
Let me first treat the case $n = 1$. Then $G\cong PSL(2,\mathbb C),P\cong (GL(1,\mathbb C)/\mathbb Z/2)\ltimes \mathbb C$ embedded as upper triangular matrices, $G/P\cong \mathbb{CP^1}$, and $\omega\_1$ defines the line bundle $O(1)$ over $\mathbb{CP}^1$. The root of your confusion is that this line bundle is not $G$-equivariant; however, after passing to the covers $\widetilde G = SL(2,\mathbb C),\widetilde P = GL(1,\mathbb C)$, it is $\widetilde G$-equivariant since it is the associated bundle construction $\widetilde G\times\_{\widetilde P,\omega\_1}\mathbb C$ on which $\widetilde G$ acts from the left. And indeed, $V^{\omega\_1}$ is the 2-dimensional representation of $\widetilde G$, which is not a $G$-representation.


In the general case, we have $P = GL(n,\mathbb C)\ltimes \mathbb \{A\in \mathbb C ^{n\times n}\mid A^T = A\}$. The weight $\omega\_1$ does not define a character of $P$, but of a double cover $\widetilde P = ML(n,\mathbb C)\ltimes \mathbb \{A\in \mathbb C ^{n\times n}\mid A^T = A\}$, where $ML(n,\mathbb C) = \{U\in GL(n,\mathbb C),z\in \mathbb C^\*\mid z^2 = \det U\}$ is the *metalinear group* (the character is $(U,z)\mapsto z$ or $z^{-1}$). This means, again, that the line bundle $L^{\omega\_1}$ and its holomorphic sections $V^{\omega\_1}$ only carry representations of the universal cover $Spin(2n+1,\mathbb C)$. This works just as well for the line bundles $L^{k\omega\_1}$ whose holomorphic sections define the higher degree components of the coordinate ring: The action of $G$ is well-defined precisely if $k$ is even.


More generally, the action of a group $G$ on a projective variety $X$ only gives rise to a projective representation of $G$ on the (linear part of the) coordinate ring of $X$, i.e. a representation of a central extension, with the kernel of the extension acting by scalars in each degree. In this case, the central extension is precisely the spin group.


