# Canonical bundle of the Lagrangian Grassmannian

I work through the paper On branched coverings of some homogeneous space of Kim and Manivel and I came across the definition of the canonical bundle of the Lagrangian Grassmannian $\mathbb{LG}_n$, the set of $n$-dimensional lagrangian subspaces of a $2n$-dimensional symplectic vector space. In the paper it is called 'a fact' that the canonical bundle $K_{\mathbb{LG}_n} = \mathcal{O}_{\mathbb{LG}_n}(-n-1)$, but I'm not able to verify it. Can someone help me in this?

• The canonical bundle of the Grassmannian of $n$-dimensional subspaces of a $2n$-dimensional vector space equals $\mathcal{O}(-2n)$. Thus you need to compute the normal bundle of the Lagrangian Grassmannian as a subvariety of the classical Grassmannian. Denoting by $\mathcal{O}^{\oplus 2n}\to S^\vee$ the universal quotient that is locally free of rank $n$, then the Lagrangian Grassmannian is the zero locus of a section of $\bigwedge^2 S^\vee$. This has first Chern class $(n-1)c_1(\mathcal{O}(1))$. Thus the canonical bundle is $\mathcal{O}((-2n) + (n-1))$. – Jason Starr Aug 15 '16 at 16:12
The Lagrangian Grassmannian is the homogeneous space $G/P$ where $G=Sp(2n)$ and $P$ is the maximal parabolic subgroup corresponding to the set of simple roots $\{\alpha_1,\ldots,\alpha_{n-1}\}$. Then $Pic(G/P)$ equals $\Xi(P)$, the character group of $P$. The latter is cyclic and its generator $\eta:=\sum_{i=1}^n\epsilon_i$ corresponds to $\mathcal O(-1)$. The canonical bundle corresponds to the determinant character of $(\mathfrak g/\mathfrak p)^*\cong\mathfrak p_u$ and therefore to the sum of roots in $\mathfrak p_u$: $$\sum_{i<j}(\epsilon_i+\epsilon_j)+\sum_i(2\epsilon_i)=(n+1)\eta.$$ Thus $K=\mathcal O(-n-1)$.