Computing characteristic numbers of homogeneous spaces I apologize in advance if this is a bad question.
I would like to prove or refute a conjecture about the vanishing of characteristic numbers of homogeneous spaces, and to this end am looking (and currently failing) to understand some simple examples.
The conjecture works for the Euler characteristics of partial flag manifolds. It also works for the Pontryagin and Stiefel–Whitney numbers of real Grassmannians, for basically dimensional reasons; in particular, I haven't yet calculated any nonzero such numbers.
To understand some more cases, I'd like to be able to compute some Pontryagin and Stiefel–Whitney classes (or at least numbers) for some partial flag manifolds, but I don't know how. 
I know how to decompose the tangent bundle of a partial flag manifold, 
say that of flags $\mathbb{R}^{k_1} < \mathbb{R}^{k_1 + k_2} < \mathbb{R}^{k_1 + k_2 + k_3}$,
into a direct sum of tensor products of pullbacks of tautological bundles over the Grassmannians $G(k_j,k_1+k_2+k_3)$, but I don't know how to use this decomposition, because I don't understand what tensor product does to characteristic classes, and because I think I would also need to understand the pullback maps on the cohomology of these spaces. 
Indeed, I don't even know the cohomology. I know the cohomology with $\mathbb{Z}/2$ and $\mathbb{Q}$ coefficients of the infinite-dimensional Grassmannians $G(k,\infty)$ are well-known, though the 2-torsion of the cohomology with $\mathbb{Z}$ coefficients is somewhat subtle, but understand the situation for the finite-dimensional Grassmannians to be rather more complicated, possibly too complicated for me to understand much in general.
How can I learn to calculate characteristic numbers? 
 A: First: notation.  $H\subseteq G$ are both connected compact Lie groups, $EG$ is a contractible space on which $G$ acts freely and $BG = EG/G$.  We will first relate the cohomology (with $\mathbb{Z}/2\mathbb{Z}$ coefficients) of $G/H$ to that of $G$ and $H$, and then compute Stiefel-Whitney classes.  So, I'm assuming that $H^\ast(G;\mathbb{Z}_2)$ and $H^\ast(H;\mathbb{Z}_2)$ are known.  From here on, all cohomology is assumed to be with $\mathbb{Z}/2\mathbb{Z} = \mathbb{Z}_2$ coefficients.
To start with, we have a universal bundle $G\rightarrow EG\rightarrow BG$.  Since $EG$ is contractible, the Serre spectral sequence associated to this bundle must converge to trivial cohomology.  This lets you compute $H^\ast(BG)$ in terms of $H^\ast(G)$.
Now, we also have an $H$-principal bundle $H\rightarrow G\rightarrow G/H$ which is classified by a map $\phi_H:G/H\rightarrow BH$.  Now, one proves that, up to homotopy, $G\rightarrow G/H\rightarrow BH$ is a fibration and that, if $i:H\rightarrow G$ is the inclusion, then $G\rightarrow G/H\rightarrow BH$ is the pull back of the universal bundle $G\rightarrow EG\rightarrow BG$ by the map $Bi$.
By naturality of spectral sequences, the the differentials in the spectral sequence for $G\rightarrow G/H\rightarrow BH$ are given as pull backs by $Bi^\ast$ of the differentials in $G\rightarrow EG\rightarrow BG$.  So, if you can compute $Bi^\ast$, you can compute the spectral sequence for $G\rightarrow G/ H \rightarrow BH$, and hence, the cohomology of $G/H$ (modulo extension problems).
The computation of $Bi^\ast$ follows methods of Borel and Hirzebruch (or was it Borel and Serre?).  Let $Q_G\subseteq G$ be a maximal 2-group, that is, a maximal subgroup of $G$ isomorphic to $Z_2^n$ for some $n$.  Likewise, pick $Q_H\subseteq H$ in such a way that $Q_H\subseteq Q_G$.  (Just like all maximal tori in a Lie group are conjugate, so are all maximal $2$-groups, so we can always arrange that $Q_H\subseteq Q_G$).
Now, $BQ_G = (\mathbb{R}P^\infty)^n$, so we understand its cohomology.  The inclusion $i:Q_H\rightarrow Q_G$ induces a map $Bi^\ast: H^\ast(BQ_G)\rightarrow H^\ast(BQ_H)$ which we can write down in terms of $i^\ast:H^\ast(Q_G)\rightarrow H^\ast(Q_H)$ (here, I mean the cohomlogy of the discrete space $H^\ast(Q_G)$, ignoring the group aspect of it).
Now comes the kicker.  The inclusion maps $Q_H\subseteq H$ and $Q_G\subseteq G$ induce injective maps $H^\ast(BH)\rightarrow H^\ast(BQ_H)$ and $H^\ast(BG)\rightarrow H^\ast BQ_G)$.  So, if we can determine the image of these maps, we can compute $Bi^\ast$ on $H^\ast (BG)$ by computing it on $H^\ast(BQ_G)$ and restricting.  Borel and the other guy compute the image in the case of all the classical groups - I'll try to dig up a reference for it.
All of this just allows you to compute the cohomology rings with mod 2 coefficients (modulo extension problems) for $G/H$.  To actually compute Stiefel-Whitney classes, we use the following formula:
$$w(G/H) = \phi_H^\ast\left(Bi^\ast\left[ \prod_{\alpha\in \Delta_2 G}(1+\alpha) \right]\cdot\prod_{\beta\in \Delta_2 H}(1+\beta)^{-1}\right).$$
I will spend the rest of the post trying to make sense of the formula.  First, one computes $\phi_H^\ast$ as the edge homomorphism is the spectral sequence for $G\rightarrow G/H\rightarrow BH$, which we have already seen how to compute above.
Moving inside, the next confusing thing is the $\Delta_2 G$ notation.  This denotes the set of $2$-roots of $G$.  To compute these, consider that adjoint action of $G$ on $\mathfrak{g}$ and restrict it to $Q_G$.  Since $Q_G$ is abelian, $\mathfrak{g}$ must break into a sum of 1-dim irreducible reps, the $2$-root spaces.  The corresponding eigenvalues for these root spaces are the $2$-roots of $G$.
Every such eigenvalue is really a linear functional $f:Q_G\rightarrow \mathbb{Z}_2$, so we can think of it as a function $f:H_0(Q_G)\rightarrow \mathbb{Z}_2$, that is, $f\in H^0(Q_G)$.  From the fibration $Q_G\rightarrow EQ_G\rightarrow BQ_G = (\mathbb{R}P^\infty)^n$, we may identify $H^1(BQ_G)$ with $H^0(Q_G)$, which allows us to think about $f\in H^1(BQ_G)$.  This is the interpretation we want.
One final note.  Then map $\phi_H^\ast Bi^\ast$ is actually the $0$-map on everything but $H^0$, so one could leave out that product over $\Delta_2 G$.  However, pratically, its often the case that a lot of canceling happens between the $Bi^\ast$ product and the other product, so one often leaves in the $\Delta_2 G$ part in hopes of not having to compute too many inverses.
