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?



spin structures on full flag manifolds
$G/T$ 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".)


