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?

Thanks in advance.

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    $\begingroup$ 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 $\mathcal{O}((-2n) + (n-1))$. $\endgroup$ – Jason Starr Aug 15 '16 at 16:12
  • $\begingroup$ I'm not able to log myself into my account L. Kern. Thus I had to build a new one but no reputations = not able to answer in comments. But thanks for those quick answers, I think @Jason Starrs answer might help. $\endgroup$ – L_K666 Aug 17 '16 at 11:47

Here is a group theoretical solution which makes it is easy to compute the canonical bundle of any generalized flag variety.

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 $K=\mathcal O(-n-1)$.

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  • $\begingroup$ What is the best source to learn these things? $\endgroup$ – Vít Tuček Aug 16 '16 at 8:50

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