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Let us fix the setup over $\mathbb C$. Let $V$ be a $n$-dimensional vector space endowed with a non-degenerate symmetric bilinear form $q: V \times V \to \mathbb C$. We have that $$ Q^{n-2}=OG(1,V)=\{\mathbb P W \subset \mathbb P V : \dim W=1 \text{ and } W \text{ is }q-\text{isotropic}\} \subset \mathbb P V $$ is the $(n-2)$-dimensional smooth quadric. We denote by $\mathcal S$ the tautological line bundle $\mathcal O_{Q}(-1)$ over the quadric, and by $\mathcal S^\perp$ the bundle over the quadric such that over a point $[W]$ it puts the vector space $[W^\perp]$ where $$ W^\perp=\{v \in V: q(v,w)=0 \text{ for all }w \in W\}. $$ Since $W \subset W^\perp$, everything is well-defined and we can define the quotient $\mathcal S^\perp / \mathcal S$. At this point we define the projective bundle $$ \mathbb P(\mathcal S^\perp/ \mathcal S) \to Q^{n-2} $$ which puts over $[W]$ the projective space given by $\mathbb P(W^\perp/W)$. This projective bundle comes with a line bundle $\mathcal O(1)$ that gives us a map $\mathbb P(\mathcal S^\perp/\mathcal S) \to \mathbb P^N= \mathbb P(H^0(\mathcal O(1))^\vee)$.

I cannot understand what is this $\mathbb P^N$. My hope is that $\mathbb P^N= \mathbb P(\wedge^2 V)$...any hint on how to proceed?

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    $\begingroup$ If you twist your vector bundle $\mathcal{S}^{\perp}/\mathcal{S}$ by $\mathcal{O}_Q(1)$ (which of course does not change the projective bundle), then this is isomorphic to the tangent bundle to the quadric hypersurface. Hence $H^0(\mathcal{O}(1))$ - isomorphic to the space of global sections of the tangent bundle - is therefore isomorphic to the adjoint representation of the Lie algebra $\mathfrak{so}(n)$, which is what are you hoping for. $\endgroup$
    – Enrico
    Aug 25, 2021 at 16:58
  • $\begingroup$ @Enrico: You right! I think it is a corollary of an exercise in chapter 4 of the book you suggest me in the other question. I think that this argument can be generalized for $OG(k,V)$ and the target of the map given by the $\mathcal O(1)$ of $\mathbb P(\mathcal S^\perp/\mathcal S)$ is then $\mathbb P( \wedge^{k+1} V)$, am I right? $\endgroup$
    – Bobech
    Aug 26, 2021 at 6:42
  • $\begingroup$ But I cannot use directly the tangent bundle of $OG(k,V)$ $\endgroup$
    – Bobech
    Aug 26, 2021 at 6:56

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