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We will work over $\mathbb C$ and the notation will be coherent with the paper of Ottaviani (see [Ott]).

Consider a $n$-dimensional quadric hypersurface $Q_n \subset \mathbb P^{n+1}$. We have embeddings of the quadric in a Grassmannian, according to the parity of $n$: $$ s:Q_{2k+1} \hookrightarrow G\left(\mathbb P^{2^k-1}, \mathbb P^{2^{k+1}-1}\right)=G\left(2^k,2^{k+1}\right), $$ $$ s':Q_{2k} \hookrightarrow G\left(\mathbb P^{2^{k-1}-1},\mathbb P^{2^k-1} \right)=G\left(2^{k-1},2^k\right), \qquad s'':Q_{2k} \hookrightarrow G\left(\mathbb P^{2^{k-1}-1},\mathbb P^{2^k-1} \right)=G\left(2^{k-1},2^k\right). $$ Let $U$ be the universal bundle over the Grassmannian, fitting in the short exact sequence $$ 0 \to U \to V \otimes \mathcal O \to Q \to 0. $$ We define the spinor bundle over $Q_{2k+1}$ as $S=s^* U$, its rank is $2^k$. On the other hand, we call the spinor bundles over $Q_{2k}$ to be $S'=(s')^* U$ and $S''=(s'')^* U$, both of rank $2^{k-1}$. If $f: Q_{2k} \to Q_{2k}$ is an automorphism exchanging the two families of $\mathbb P^k$, then $f^* S' \simeq S''$ and $f^* S'' \simeq S'$.

Moreover, one can prove that these vector bundles are homogeneous ones.

In Section 2 of [Ott], we have the following result (Theorem 2.8):

Theorem.

  1. Consider the spinor bundle $S \to Q_{2k+1}$. Then there is a short exact sequence $$ 0 \to S \to \mathcal O^{\oplus 2^{k+1}} \to S(1) \to 0 $$ and $S^\vee \simeq S(1)$.
  2. Consider the spinor bundles $S',S'' \to Q_{2k}$. Then there are two short exact sequences $$ 0 \to S' \to \mathcal O^{2^k} \to S''(1) \to 0, \qquad 0 \to S'' \to \mathcal O^{2^k} \to S'(1) \to 0. $$ If $k \equiv 0 \pmod 2 $, we have $(S')^\vee \simeq S'(1)$ and $(S'')^\vee \simeq S''(1)$. Otherwise, if $k \equiv 1 \pmod 2$, we have $(S')^\vee \simeq S''(1)$ and $(S'')^\vee \simeq S'(1)$.

Finally the question. Consider the orthogonal Grassmannian $OG(m+1,n+2)$, which parametrize $\mathbb P^m \subset Q_n$ (note that $Q_n$ is obtained by $m=0$). Can we generalize the construction of spinor bundles to some bundles over $OG(m+1,n+2)$? In particular, is there an embedding of $OG(m+1,n+2)$ into a Grassmannian $G$ like $s,s',s''$?

At this point, one defines the spinor bundles over $OG(m+1,n+2)$ as the pullback of the universal bundle over the Grassmannian $G$ and the rank is clear.

Second question. Having defined the spinor bundles over the orthogonal Grassmannians, can we produce some short exact sequences like the ones in the theorem above?

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    $\begingroup$ See Section 6 in [Kuznetsov, Alexander Exceptional collections for Grassmannians of isotropic lines. Proc. Lond. Math. Soc. (3) 97 (2008), no. 1, 155–182]. $\endgroup$
    – Sasha
    Jul 12, 2022 at 9:59
  • $\begingroup$ @Sasha: that's a perfect reference, thanks! Just a more question, if you can reply to me. In the paper that you cite, one can define the (generalized) spinor bundle starting from the $\mathcal O(1)$ of one of the two spinor variety, let's say $\mathfrak F^+_m$ (I am now using the notation contained in the paper). However, the definition of spinor bundle is not given for $\mathfrak F^-_m$, can one generalize the incidence construction a little more to include this case? $\endgroup$
    – Bobech
    Jul 13, 2022 at 13:42
  • $\begingroup$ $$ \mathfrak F^-_m \longleftarrow \mathfrak F^\pm_m \longrightarrow \mathfrak F^+_m$$ and one define the spinor bundle to be $(\pi_-)_*(\pi_+)^* \mathcal O(1)$? What is its rank? $\endgroup$
    – Bobech
    Jul 13, 2022 at 13:43
  • $\begingroup$ I can compute such a bundle using a different technique ($\mathbb C^*$-actions, smooth drums...), but I want to understand if such a definition of (generalized) spinor bundle can be extended in this direction $\endgroup$
    – Bobech
    Jul 13, 2022 at 14:02
  • $\begingroup$ Sorry, but I don't understand the question. What is the variety in the middle of your diagram. What are the maps $\pi_\pm$? What are smooth drums? $\endgroup$
    – Sasha
    Jul 13, 2022 at 19:27

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