# Lagrangian Grassmannian as a Spin Manifold

I am trying to better understand this nice answer to a question of mine, which states

Spin structures on a compact complex manifold $(M^{2n},J)$ are in bijective correspondence with isomorphism classes of holomorphic line bundles ${\cal L}$ such that ${\cal L}\otimes {\cal L} = {\cal K}$ where ${\cal K}$ is the canonical line bundle of $(M,2n)$.

Now in an answer to another question on the canonical bundle of the Lagrangian Grassmannians, it is stated that

(paraphrased) The Picard group of each Lagrangian Grassmannian $Sp(2n)/P$ (where $P$ is the maximal parabolic subgroup) is cyclic with generator ${\cal O}(-1)$. Moreover, its canonical bundle is $\cal{O}(-n-1)$.

Now since the Lagrangian Grassmannian is a flag manifold, and hence a complex manifold, these two facts seem to me to imply that

The Lagrangian Grassmannian is a spin manifold if and only if $n$ is odd.

Is my reasoning correct?

The complex Lagrangian Grassmannian $M=G/K=Sp(n)/U(n)$ is an isotropy irreducible Hermitian symmetric space, hence it admits a unique invariant complex structure. It occurs by painting black in the Dynkin diagram of $Sp(n)$ the last simple root $\alpha_{n}$, hence the second Betti number of $M$ equals to 1. Now, $M$ admits a spin structure if and only if $n$ is odd, since for example its first Chern class is given by $c_{1}(M)=(n+1)\Lambda_{n}$, where $\Lambda_{n}$ is the fundamental weight corresponding to the painted black simple root $\alpha_{n}$ and it can be thought of as the generator of $H^{2}(M; \mathbb{Z})$. When such a spin structure exists, then it will be invariant and unique.

Note: Recall that a complex manifold is spin, if and only if its first Chern class is divisible by 2, i.e. even.

• Nice it see it from a different viewpoint! Thanks for your answers. – Janos Erdmann Feb 15 '17 at 23:32