# Rank 3 Lagrangian vector bundles on an elliptic curve

Let $$k$$ be an algebraically closed field of characteristic zero (feel free to assume $$k= \mathbb{C}$$) and $$E$$ an elliptic curve over $$k$$ with identity $$P \in E(k)$$.

I am interested in certain morphisms from $$E$$ to $$LG(3,6)$$, the Lagrangian Grassmannian of $$3$$-dimensional Lagrangian subspaces of a $$6$$-dimensional symplectic vector space over $$k$$, namely those morphisms $$E \rightarrow LG(3,6)$$ such that the pullback of $$\mathcal{O}(1)$$ on $$LG(3,6)$$ (coming from the Plücker embedding) to $$E$$ along this morphism is isomorphic to $$\mathcal{O}_E(6P)$$.

More concretely, I am interested in rank $$3$$ vector bundles $$V$$ on $$E$$ with the following two properties:

1. There exists a surjection $$\mathcal{O}_E^{\oplus 6} \twoheadrightarrow V$$ whose kernel is a Lagrangian subvectorbundle of $$\mathcal{O}_E^{\oplus 6}$$ (where we put the standard symplectic form on $$\mathcal{O}_E^{\oplus 6}$$).
2. We have an isomorphism $$\text{det}\, V \simeq \mathcal{O}_E(6P)$$.

An example of such a $$V$$ is given by $$\mathcal{O}_E(2A)\oplus \mathcal{O}_E(2B) \oplus \mathcal{O}_E(2C)$$ where $$A,B,C \in E(k)$$ sum to zero, i.e. the divisor $$A+B+C$$ is linearly equivalent to $$3P$$.

Question: Is every $$V$$ of this form?

I believe that if $$V$$ is a direct sum of line bundles then it must necessarily be of the form described above. So we could equivalently ask: is every $$V$$ satisfying the above two conditions a direct sum of line bundles?

The variety $$LG(3,6)$$ is a homogenous space for the algebraic group $$Sp_6$$, but I haven't been able to find results in the literature which treat this specific case.

• Reversing the problem (since you are using a Mukai--style construction), you may ask for the existence of an homogeneous bundle $F$ on $SG(3,6)$ of rank 5 and $c_1(F)=4$. The only non--degenerate example I can come up with is $F= \bigwedge^2 R^{\vee} \oplus \mathcal{O}(1) \oplus \mathcal{O}(1)$, where $R$ is the rank 3-tautological. This corresponds to an elliptic curve as double (multi)-linear section of $(\mathbb{P}^1)^3$. Otherwise you can use $R^{\vee}$ (or $Q$) as a bundle, but then the embedding is degenerate in $SG(3,6)$, and you should embed $E$ directly in a 3-dimensional quadric. – Enrico Mar 26 at 13:46
• Can you explain why the problem you pose is related to the one given here, and what $SG(3,6)$ stands for? I'm not really familiar to Mukai-style constructions. – Jef Mar 27 at 9:29
• The Lagrangian Grassmannian $LG(k, 2k)$ is a special case of the symplectic Grassmannian $SG(k,n)$ (basically I typed the comment in a hurry using the notation I am most used to, sorry). – Enrico Mar 27 at 11:24
• By Mukai-style I mean the vector bundle method (see for example section 3 of library.msri.org/books/Book28/files/mukai.pdf or section 5 of "Algebraic Geometry V"). Finding a rank 3 bundle with 6 sections on $E$ gives you a morphism to $Gr(3,6)$. Then your original $E$ can be recovered by taking zero locus of an appropriate number of sections of $\bigwedge^i R^{\vee}$. Note that $LG(3,6)$ is itself the zero locus of $\bigwedge^2 R^{\vee}$ on $Gr(3,6)$. – Enrico Mar 27 at 11:32

I think your guess is correct and one can proceed as follows (some details are missing though). Let $$V$$ be a six dimensional symplectic vector space and $$F$$ be a rank three-vector bundle on $$E$$ wiht an exact sequence $$0 \longrightarrow G \longrightarrow V \otimes \mathcal{O}_E \longrightarrow F \longrightarrow 0,$$ where $$G$$ is a Lagrangian subbundle of $$V \otimes \mathcal{O}_E$$.

Assume furthermore that $$\det(F) = \det(G^*) = \mathcal{O}_{E}(6P)$$.

Let $$W \subset V$$ be a generic Lagrangian subspace and consider the map:

$$\phi : G \longrightarrow V/W \otimes \mathcal{O}_{E}.$$

The genericity of $$W$$ implies that it is generically on E an ismorphism. Furthermore, $$\phi$$ is (globally) injective as $$G$$ is torsion free. We denote by $$Z \subset E$$ the subscheme corresponding to the degeneracy locus of $$\phi$$. Since $$\det(G^*) = \mathcal{O}_{E}(6P)$$, we have the linear equivalence $$Z \sim 6P$$.

We have an exact sequence:

$$0 \longrightarrow G \longrightarrow V/W \otimes \mathcal{O}_E \longrightarrow \mathcal{F} \longrightarrow 0,$$ where $$\mathcal{F}$$ is scheme theoretically supported on $$Z$$.

The vector space $$W \subset V$$ is generic and $$E$$ is a curve, so that the corank of $$\phi$$ is exactly $$1$$ on $$Z$$. As a consequence $$\mathcal{F}|_{Z}$$ is a line bundle on $$Z$$.

Let $$Z_{red} = \{P_1, \ldots, P_l\}$$ with the $$P_i$$ distincts. We write:

$$\mathcal{F} = \bigoplus_{i=1}^{l} \mathcal{F}_i,$$

where $$\mathcal{F}_i$$ is the restriction of $$\mathcal{F}$$ to the connected components of $$Z$$ corresponding to $$P_i$$.

For any subbundle $$F$$ of $$V \otimes \mathcal{O}_E$$ whose quotient is a vector bundle, we denote by $$F^{\perp} = (V/F)^*$$.

We have han exact sequence: $$0 \longrightarrow G^{\perp} \longrightarrow V^*/(W^{\perp}) \otimes \mathcal{O}_E \longrightarrow \mathcal{H} \longrightarrow 0,$$ where $$\mathcal{H}$$ is scheme theoretically supported on a subscheme of $$E$$ linearly equivalent to $$6P$$.

We similarly split $$\mathcal{H}$$ as $$\bigoplus_{i=1}^q \mathcal{H}_i$$, where the $$\mathcal{H}_i$$ correspond to the various connected component of the support of $$\mathcal{H}$$.

The bundles $$G$$ and $$W \otimes \mathcal{O}_E$$ being Lagrangian, the skew-symmetric form $$\sigma : V \longrightarrow V^*$$ induces isomorphisms:

$$\sigma_{G} \ : \ G \stackrel{\sim}\longrightarrow G^{\perp} \ \textrm{and} \ \sigma_{V/W} \ : \ V/W \stackrel{\sim}\longrightarrow V^*/(W^{\perp})$$ which are compatible with the maps:

$$G \longrightarrow V/W \ \textrm{and} \ G^{\perp} \longrightarrow V^*/(W^{\perp}).$$

We deduce that $$\mathcal{H}$$ and $$\mathcal{F}$$ are equal and that up to a reordering of the we have $$\mathcal{H}_i = \mathcal{F}_i$$, for all $$i$$.

For all $$i \in \{1, \ldots, l\}$$, the skew symmetric isomorphism $$\sigma$$ induces a skew-symmetric isomorphism:

$$\sigma_i : \mathcal{F}_i \stackrel{\sim}\longrightarrow \mathcal{F}_i,$$ which lifts to a skew-symmetric isomorphism:

$$h^0(\sigma_i) \ : \ H^0(E,\mathcal{F}_i) \stackrel{\sim}\longrightarrow H^0(E, \mathcal{F}_i).$$

The skew-symmetry of the isomorphism $$h^0(\sigma_i)$$ forces the dimension of the vector spaces $$H^0(E,\mathcal{F}_i)$$ to be even. As a consequence, of the Riemman-Roch formula on $$E$$, the multiplicity of $$P_i$$ as a connected component of $$Z$$ must always be even.

The generic situation (that is when $$E \longrightarrow LG(3,6)$$ is a generic point in a component of $$Hom(E, LG(3,6))$$ should correspond to the case:

$$Z_{red} = \{A,B,C\}$$ with $$A,B,C$$ distincts and $$Z = \{2A,2B,2C\}$$ as a subscheme of $$E$$.

Now I would like to deduce from this that we have a map:

$$\mathcal{O}_E(-2A) \oplus \mathcal{O}_E(-2B) \oplus \mathcal{O}_E(-2C) \longrightarrow G$$ which is generically an isomorphism (I have a vague idea why this should be true, but I don't have a precise argument to offer, perhaps someone else will find).

If we have such a map which is generically an isomorphism, then it must be an isomorphism, owing to the relation $$\det(G) = \det(\mathcal{O}_E(-2A) \oplus \mathcal{O}_E(-2B) \oplus \mathcal{O}_E(-2C))$$.

We conclude that $$F \simeq \mathcal{O}_E(2A) \oplus \mathcal{O}_E(2B) \oplus \mathcal{O}_E(2C)$$ as $$G^* \simeq F$$.