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Let $X\subset\mathbb{P}^N$ be a quadratic manifold. That is $I(X)$ is generated by quadratic polynomials $Q_1,...,Q_m$.

Let $\mathcal{I}_X$ be the ideal sheaf of $X$ and $\mathcal{I}_X/\mathcal{I}_X^2$ be the conormal bundle of $X$. Is it true that $H^0(X,\mathcal{I}_X/\mathcal{I}_X^2(2))$ has rank $m$ and that it is generated by $dQ_1,...,dQ_m$ ?

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No. The simplest example is the twisted cubic $C\subset \mathbb{P}^3$. The conormal bundle on $C\cong \mathbb{P}^1$ is $\mathcal{O}_{\mathbb{P}^1}(-5)^2$. Hence $H^0(C,\mathcal{I}_C/\mathcal{I}_C^2(2))$ has dimension 4, while $I(C)$ is generated by 3 quadrics.

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  • $\begingroup$ Sure, you are right. Thanks. Do you know if for some reason it is true for Grassmannians? $\endgroup$
    – user82886
    Commented Nov 20, 2015 at 19:55
  • $\begingroup$ It should be decidable, but I am not brave enough to do the computations. In general, you have an exact sequence $0\rightarrow K\rightarrow \mathcal{O}_X^m\rightarrow \mathcal{I}_X/\mathcal{I}_X^2(2)\rightarrow 0$, and the trouble comes from $H^1(X,K)$. If $X$ is a Grassmannian $K$ is a homogeneous vector bundle, so its cohomology should be computable. $\endgroup$
    – abx
    Commented Nov 21, 2015 at 5:24
  • $\begingroup$ For the Grassmannian $G$ with a universal SES, $$0\to S\to \mathcal{O}^{\oplus n}_G \to Q \to 0,$$ with $S$, resp. $Q$, locally free of rank $k$, resp. $n-k$, there is an increasing filtration on $\mathcal{I}_G/\mathcal{I}_G^2(2)$, $0 = F_0 \subset F_1 \subseteq \dots \subseteq F_{k-1} = \mathcal{I}_G/\mathcal{I}_G^2(2)$ with $F_r = \bigwedge^k(S^\vee)\otimes \bigwedge^{r-1}(S^\vee)\otimes \bigwedge^{k+1-r}(Q^\vee)$. Thus, when $k=2$, the sheaf is just $\bigwedge^2(S^\vee)\otimes \bigwedge^2(Q^\vee)$. By Borel-Weil-Bott, the global sections are zero or an irrep, hence surjectivity holds. $\endgroup$
    – Jason Starr
    Commented Nov 21, 2015 at 13:03
  • $\begingroup$ The argument above for $k=2$ might extend (I am not certain). Please confer Exercise 15.32, pp. 226-227 of W. Fulton, J. Harris, "Representation Theory", GTM 129, Springer-Verlag, NY, 1991. I am guessing that the formula they give there for $I(G)$ agrees with the formula from Borel-Weil-Bott for $H^0(G,\mathcal{I}_G/\mathcal{I}_G^2(2))$. $\endgroup$
    – Jason Starr
    Commented Nov 22, 2015 at 22:03

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