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:


*

*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 $\mathcal{O}_E^{\oplus 6}$).

*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. 
Thanks in advance! 
 A: 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 $G^* \simeq F$.
