Are all Equivariant Bundles of a Total Flag Manifold Constructable from Line Bundles? As we all know, for any homogeneous space $G/H$ we have that the equivariant vector bundles over $G/H$ are characterized by the representations of $H$. Thus, for the the complex projective line $CP^1 \simeq SU(2)/U(1)$, it must hold that all its line bundles are indexed by the integers $L_k$, for $k \in Z$, and more generally, its rank-$k$ (equivarian) vector bundles are of the form
$$
L_{\bf z} = L_{z_1} \oplus \cdots \oplus L_{z_k}, {\text ~~~ for ~~~ } {\bf z} \in Z^k.
$$
Does this then extend to all the total flag manifolds $F(n)$, ie the spaces of the form 
$$
F(n) := SU(n)/(U(1)^{\otimes n-1}).
$$
Edit: I omitted the word equivariant by mistake and have now entered it as (equivariant)
 A: I think perhaps the confusion stems from the following.
It is true that the category of $G$-equivariant $G$-bundles on $G/H$ is equivalent to the category of representations of $H$. The flag variety $\mathcal F l_n$ can be realized as either $U(n)/U(1)^n$, or as $GL_n(\mathbb C)/B_n$, where $B_n$ is the group of upper triangular matrices.
Thus $GL_n(\mathbb C)$-equivariant vector bundles on $\mathcal Fl_n$ are equivalent to representations of $B_n$. This group is not reductive, and not every representation is a direct sum of 1-dimensional representations. Thus, not every equivariant vector bundle is a direct sum of line bundles.
On the other hand, the category of $U(n)$-equivariant principal $U(n)$-bundles is equivalent to representations of $U(1)^n$. This category is semisimple.
A: The answer is no. For example, if $n = 3$ then $F(3)$ is a divisor of bidegree $(1,1)$ in $P^2\times P^2$ and the pullback of the tangent bundle from any factor is an example of an equivariant bundle which is not a sum of line bundles.
On the other hand, any equivariant bundle on $F(n)$ can be obtained as an iterated extension of line bundles.
