Here is a topological question which seems quite elementary. The answer to this question may be useful e.g. in estimating the orders of the automorphism groups of some algebraic varieties and in computing the cohomology of some moduli spaces.

Let $K$ be a compact Lie group, $T$ a maximal torus of $K$ and $V$ a complex vector space of dimension $d$. (If one wishes, instead of $K$ and $T$ one can consider a complex reductive group and its Borel subgroup.) From a representation $R:T\to GL(V)$ we can construct a homogeneous vector bundle with total space

$$E=K\times_T V.$$

(I.e., we identify $(kt,v)$ with $(k,R(t)v$ for all $k\in K, t\in T, v\in V$.)

Set $E_0$ to be $E$ minus the zero section. This space is fibered over $K/T$ with fiber $V$ minus the origin.

Suppose the Euler (=top Chern) class of the vector bundle is zero. Then $H^{2d-1}(E_0,\mathbf{Z})\cong \mathbf{Z}$, because $K/T$ has only even-dimensional cells. Take a generator $a$ of $H^{2d-1}(E_0,\mathbf{Z})$. We can identify the integral cohomology of $E_0$ with $H^*(K/T,\mathbf{Z})\otimes\Lambda(a)$ where $\Lambda$ stands for exterior $\mathbf{Z}$-algebra.

We have the natural group action map $K\times E_0\to E_0$. The cohomology of $E_0$ has no torsion, so we can identify $H^*(K\times E_0,\mathbf{Z})$ with $H^*(K,\mathbf{Z})\otimes H^*(E_0,\mathbf{Z})$ using the projections. The latter can be written as $$H^*(K,\mathbf{Z})\otimes H^*(K/T,\mathbf{Z})\otimes\Lambda(a).$$

The pullback of $a$ under the action map is $1\otimes a+x\otimes 1$ for some $x\in H^{*}(K,\mathbf{Z})\otimes H^{*}(K/T,\mathbf{Z})$. Question: find $x$. The weights of the representation are assumed to be known; for simplicity one can assume that $K=U(n)$ or $SU(n)$.

For real cohomology I know how to reduce the problem to linear algebra using Lie algebra cohomology. But the result I am able to get in this way is not very illuminating. And moreover, I have no idea how to extract the integral structure out of it.

For projectivized (and not spherized) bundles, the question becomes trivial. But this does not seem to help.