TITLE: spin structures on full flag manifolds
QUESTION [5 upvotes]: 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?

REPLY [7 votes]: $G/T$ 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".)