It is well known that the saturated ideal of the 10-dimensional spinor variety $S^{10}\subset\mathbb{P}^{15}$ is generated by 10 quadrics. Is it possible to get $S^{10}$ as scheme-theoretic intersection of less than 10 quadrics of $\mathbb{P}^{15}$ ?
Thanks.
I think the answer is yes. Indeed, note that the 10-dimensional space $V$ of quadrics has a structure of a natural representation of the group SO(10). In particular, there is a natural quadric in this space. Let us take any hyperplane $V' \subset V$, which is not tangent to the quadric. Let us prove that the map $V'(-2) \to I_S$ is surjective, where $I_S$ is the ideal of the spinor variety.
First let us check that the map is surjective along $S$. For this we have to restrict the map $V'(-2) \to I_S$ to $S$. We will obtain the map $$ V'\otimes O_S(-2) \to I_S\otimes O_S = I_S/I_S^2. $$ But the conormal bundle $I_S/I_S^2$ of $S$ is isomorphic to $U^*(-2)$, where $U$ is the tautological rank 5 vector bundle on $S = OGr_+(5,V)$ (the one restricted from $Gr(5,V)$). Hence the above map after the $O(2)$-twist is $$ V'\otimes O_S \to U^*, $$ the composition of the embedding $V' \to V$ with the natural epimorphism $V\otimes O_S \to U^*$ (restricted from $Gr(5,V)$). If the map $V' \to U^*$ is not surjective at a point corresponding to a 5-dimensional isotropic subspace $U \subset V$, then $U \subset V'$, but then $V'$ should be tangent to the quadric. So if $V'$ is not then the map is surjective.
Now it follows that the support of the cokernel of the map $V'(-2) \to I_S$ is disjoint from $S$. Thus the zero locus of quadrics in $V'$ has 2 connected components, one is $S$ and the other is some other $Z \subset P^{15}$. Moreover, the dimension of $Z$ is greater or equal than $15 - 9 = 6$. Since the dimension of $S$ is 10, the intersection of $Z$ and $S$ cannot be empty, unless $Z = \emptyset$. Thus the map $V'(-2) \to I_S$ is surjective.
The ten quadrics $Q_i,Q^i, i =1, 2,\dots, 5$ defining the spinor variety $S_{10} \in \mathbb{P}^{15}$ can be chosen in such a way that they satisfy the relation
$Q_i Q^i = 0 \; (\star)$ [Mukai]. The ideal $I_9$ generated by the nine-quadrics $Q_i,Q^i, i =1, 2,\dots, 5$ and $Q_5 + Q^5$ also satisfies $Q_5 Q^5 = 0$ by $ (\star)$. Therefore $I_9$ also contains $(Q_5 - Q^5)^2$ and hence the saturation of $I_9$ is the defining ideal of the spinor variety $S_{10}.$ This argument is from section 6 of [arXiv:1807.03766]. Presumably it appears somewhere else much earlier.
