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?