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?


1 Answer 1


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.

  • 1
    $\begingroup$ Nice it see it from a different viewpoint! Thanks for your answers. $\endgroup$ Feb 15, 2017 at 23:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.