Question

It is known that any full flag manifold $G/T$ is a spin manifold. For example, we can prove it using that $G/T$ is a complex manifold, by computing its 1st Chern class as follows: For full flag manifolds we have that the first Chern class is given by $c_{1}(G/T)= 2\delta_{G} \cdot $ generator of $H^{2}(G/T, \mathbb{Z})$, where $$\delta_{G}=\frac{1}{2}\sum_{\alpha\in R^{+}}\alpha.$$ Here $R^{+}$ are the postive roots of $G$. It is well known that $\delta_{G}=\Lambda_{1}+\cdots+\Lambda_{\ell}$, where $\Lambda_{i}$ are the fundamental weights and $\ell=\dim T= {\rm rank}G$. Then $$ c_{1}(G/T)= 2\delta_{G} \cdot g =2(\Lambda_{1}+\cdots+\Lambda_{\ell}) \cdot g,$$ where $g\in H^{2}(G/T, \mathbb{Z})$ is the generator. Therefore $c_{1}(G/T)$ is even and since for a complex manifold $M$ the second Whitney class $w_{2}(M)$ is the reduction $\mod 2$ the first Chern class $c_{1}(M)$, we conclude that $$w_{2}(G/T)=0 \ \ thus \ \ G/T \ \ is \ a \ spin \ manifold.$$

Question How can we give a proof of this fact by a different way, for example, by using arguments from homology or cohomology theory?

Answer 1

$G/T$ is a co-adjoint orbit in $\mathfrak g^*$. The normal bundle to the inclusion $G/T\rightarrow \mathfrak g^*$ is trivial, so the tangent bundle of $G/T$ is stably trivial. This implies its Stiefel-Whitney and Pontryagin classes vanish.

(Argument stolen from Dan Freed "Flag Manifolds and Infinite Dimensional Geometry" in MSRI proceedings volume "Infinite Dimensional Groups with Applications".)

Answer 2

I suppose that $G$ is a compact Lie group and $T$ is a maximal torus, right? Let $N$ be the dimension of $G/T$; passing if necessary to a double cover of $G$, you may assume that the isotropy representation $T\rightarrow SO(N)$ lifts to $T\rightarrow Spin(N)$; this gives you a lift of the frame bundle $G\times_T SO(N)$ of $G/T$, to a $Spin(N)$- bundle.