and the claimed inequality follows.

Inside the dual projective space $\PP\bigl(\bigwedge^2 H^0(C, E)\bigr)$, we identify $\PP\bigl(\mbox{Ker } \lambda\bigr)$ with the space of hyperplanes in $\PP\bigl(\bigwedge^2 H^0(C, E)^{\vee}\bigr)$ containing the span $\langle \phi_L(C)\rangle$. We set $$\GG:=G\bigl(2, H^0(C, E)^{\vee}\bigr),\ \ \PP:=\PP\bigl(\bigwedge^2 H^0(C, E)^{\vee}\bigr)\  \mbox{ and }\ \Lambda:=\mbox{Im } \PP(\lambda^{\vee})\subset \PP.$$

Let us assume that $h^0(C, E)=4$. Lemma \[PRlemma\] implies that $\mbox{dim}(\mbox{Im } \PP(\lambda^{\vee})) \geq 4$, and $$Q_E:=\PP\bigl(\lambda^{\vee}\bigr)^{-1} \bigl( G(2, H^0(C, E)^{\vee})\bigr)\in \mbox{Sym}^2 H^0(C, L)$$ is a quadric of rank at most $6$ containing $\phi_L(C)$. In particular, the multiplication map $$\nu_2(L):\mbox{Sym}^2 H^0(C, L)\rightarrow H^0(C, L^{\otimes 2})$$ is not injective. Equivalently $K_{1, 1}(C, L)=\mathrm{Ker}\ \nu_2(C, L)\neq 0$.

More generally, diagram (\[diag\]) induces a pull-back morphism at the level of quadrics $$\mathrm{res}_{C}: H^0\bigl(\PP, \I_{\GG/\PP}(2)\bigr)\rightarrow \mathrm{Ker } \ \nu_2(C, L).$$ To link the geometry of $E$ to a syzygy type statement, we estimate the rank of $\mbox{res}_C$.

\[restrquadrics\] Assume $E$ is a globally generated rank $2$ vector bundle on $C$, without sub-pencils and with $h^0(C, E)\leq 5$. Then the map $\mathrm{res}_C$ is injective.

We begin with a *Plücker quadric* $Q\in H^0(\PP, \I_{\GG/\PP}(2))$, that is, a rank $6$ quadric corresponding to a $4$-dimensional quotient of $H^0(C, E)^{\vee}$. The dual $Q^{\vee}\subset \PP\bigl(\bigwedge^2 H^0(C, E)\bigr)$ is $4$-dimensional and contained 