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?

  • $\begingroup$ Isn't it simply that you get $V^{2\omega_1}$ (the Cartan square of the spin representation, which is defined 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
  • $\begingroup$ Is $\mathrm{Spin}(2n+1)/P$ not isomorphic to $\mathrm{SO}(2n+1)/P$? $\endgroup$ – Sam Hopkins May 18 '20 at 14:21
  • $\begingroup$ @SamHopkins Could you please cite the version of Borel-Weil theorem you are using? $\endgroup$ – Vít Tuček May 21 '20 at 7:28
  • $\begingroup$ @VítTuček: Uh, I'm not sure of a precise reference- I thought this was classical. It is mentioned in some other MO questions too: mathoverflow.net/questions/23426/…, mathoverflow.net/questions/319903/… $\endgroup$ – Sam Hopkins May 21 '20 at 12:41

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.

  • $\begingroup$ Thanks, this is extremely helpful. Is it true that there are no issues with what I said at the level of Lie algebras? $\endgroup$ – Sam Hopkins May 18 '20 at 14:28
  • $\begingroup$ Yes, for at least two reasons: A priori we get a projective representation of $\mathfrak{so}(2n+1)$, but since it is semisimple it has no nontrivial central extensions; and more simply, the Lie algebra doesn't see the difference between $SO$ and $Spin$. However, the construction of the spin representation works more or less the same way for the infinte-dimensional space $V = L^2(S^1,\mathbb C^n)$, giving rise to a projective representation of the loop group $LU(n)$ which defines a nontrivial central extension already at the Lie algebra level (see Segal-Pressley's book on loop groups). $\endgroup$ – Bertram Arnold May 18 '20 at 14:36

This got too long for a comment.

The formulation of the Borel-Weil theorem I am familiar with identifies global sections of associated bundles with highest weight $G$-representations. So to get this folklore formulation that the OP invokes, one has to think about what associated bundles appear in the coordinate ring. Since higher degrees are given by symmetric powers (right?) and all the isomorphisms are natural, one has that $m$-degree component is isomorphic to global sections of $m$-th symmetric power which has the weight $m\lambda,$ where $\lambda$ is the highest weight corresponding to the degree one component.

So I think the question actually boils down to he part of proof of the "folklore Borel-Weil theorem" that identifies this degree one component.

In this specific case the Levi part has Lie algebra $\mathfrak{gl}(n, \mathbb{C})$ and I think that the representation that would give $V^{\omega_1}$ as the space of global sections, would need associated bundle from representation which does not integrate to $GL(n, \mathbb{C})$ but rather to it's double cover which was described in another answer. In other words, I think that in this case the "folklore Borel-Weil" says that the the ring of regular functions is isomorphic as $G$ representation to $\bigoplus_{m=1}^\infty V^{2\omega_1}.$

  • $\begingroup$ Interesting. Note that Bertram's answer basically says that the issue is that the action of $G$ on this projective variety $G/P$ only leads to a projective representation on the (homogenous) coordinate ring. Whereas you are saying the issue is more identifying the highest weight for the linear part. But, as discussed above, I am pretty sure that the weight for the linear part is correct at the level of $\mathfrak{so}(2n+1)$-representations, so I'm not sure how that could square with what you're saying. $\endgroup$ – Sam Hopkins May 21 '20 at 15:05
  • $\begingroup$ Maybe a precise statement for what I'm thinking of is Theorem 3.1 of: pdfs.semanticscholar.org/e07e/… $\endgroup$ – Sam Hopkins May 21 '20 at 15:32
  • $\begingroup$ I guess the only issue is whether $SO(2n+1)/P$ is the same as $Spin(2n+1)/P$, but I thought this was the case (see mathoverflow.net/questions/330999/… for a discussion of similar thing) $\endgroup$ – Sam Hopkins May 21 '20 at 15:40

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