Let $X$ be a Fano threefold of index $1$ and genus $8$ (i.e. degree $14$). In this (top of page 5) paper, the authors state that the rank $2$ Gieseker stable vector bundle $E_0$ with $c_1=1$ and $c_2 =5$ giving the embedding $X \hookrightarrow \mathrm{Gr}(2, 6)$ is defined by the Serre construction for any elliptic quintic $C$ on $X$. They link this paper as a reference for this, but having looked through I can't seem to find the relevant part of the paper.
Would I be correct in assuming then, that $E_0$ fits into the short exact sequence
$$ 0 \to \mathcal{O}_X \to E_0(H) \to I_C(2H) \to 0 $$
where $H$ is the choice of polarisation on $X$, and $C \subset X$ is an elliptic quintic. I'm mimicking the Serre construction on cubic threefolds to obtain instanton bundles here, but I'm not sure whether we should use the same short exact sequence or a different one for my Fano threefold case?
Thanks.
Edit: I've had some more thoughts as follows: $X$ is a threefold and $C$ is a codimension $2$ subscheme which is a locally complete intersection, and $C$ can be regarded as the locus of a section $s_0$ of the vector bundle $E_0$. Then we should have a short exact sequence
$$ 0 \to\mathcal{O}_X \xrightarrow{s_0} E_0 \to I_C \otimes \Lambda^2E_0 \to 0 . $$
I guess my question becomes: what is $\Lambda^2E_0 = \det (E_0)$? My Fano threefold $X$ is anticanonically embedded, and since the embedding is "given" by $E_0$, I guess we should have $\det(E_0) = – K_X = \mathcal{O}_X(H)$, the second equality because $X$ is index $1$. I don't know if this is actually the correct thinking though.
Best Answer
The bundle $E_0$ is defined by the exact sequence $$ 0 \to \mathcal{O}_X \to E_0 \to I_C(H) \to 0, $$ where $H = -K_X$ is the anticanonical polarization. You can find this sequence on page 14 of the paper.