Non-homogeneous line bundles over a homogeneous space Let $G$ be a compact Lie group and $G/K$ a connected homogeneous space. A homogeneous vector bundle over $G/K$ is a vector bundle is one that is isomorphic to a vector bundle of the form
$$
G \times_{\rho} V \to G/K, ~~~ (g,v) \mapsto [g],
$$
where $(V,\rho)$ is a $K$-module. Can there exist line bundles over $G/K$ that are not homogeneous? What are some typical examples?
 A: Yes. This happens whenever $G$ admits nontrivial vector bundles $E$ which can be equipped with an equivariant structure for the $K$-action. Then $E$ descends in the same way to $G \times_{\rho} E \to G / K$. The homogenous ones are exactly the case that $E$ is a trivial $G$-bundle.
A trivial example is $G = S^1$ and $K = 1$ then take $E$ to be the Mobius bundle. This gives the Mobius bundle on $G/K$ which is not homogeneous according to this presentation. Of course, it is homogeneous if identify $G$ with $G / \{ \pm 1 \}$ and equip the trivial line bundle with the flip $\{ \pm 1 \}$-action.
Less trivial examples arise from the observation that if $K$ is connected then $\pi_1(G) \to \pi_1(G/K)$ is a surjection and therefore $H^1(G/K, \mathbb{Z}/2\mathbb{Z}) \to H^1(G, \mathbb{Z}/2 \mathbb{Z})$ is injective. These are the spaces parametrizing real line bundles via the first Stiefel–Whitney class. Its injectivity means that if $K$ is compact and $G/K$ admits any nontrivial line bundle, it stays nontrivial after pulling back to $G$ so cannot be homogenous.
For example, take $G = U(n)$ and $K = SU(n)$ then $G/K = S^1$ then the Mobius bundle on $S^1$ is not homogenous.
However, in a better sense, every line bundle is homogeneous if you choose the right presentation. Every line bundle $L$ defines via its Stiefel–Whitney class, a map $w_1(L) : \pi_1(G) \to \mathbb{Z}/2\mathbb{Z}$ giving a finite index normal subgroup. This defines a $\mathbb{Z}/2\mathbb{Z}$-covering group $G' \to G$ so that $L$ is trivialized on $G'$. Then $L$ is homogeneous for the pair $(G', \mathbb{Z} / 2 \mathbb{Z})$. Hence for each line bundle $L$ on $G/K$ we can form $(G', K')$ with $K'$ the preimage under $G' \to G$ such that $L$ is homogeneous for the new pair $(G', K')$ noting that $G'/K' = G/K$.
Now if you meant complex line bundle the story is more interesting. Complex line bundles $L$ are classified by their first Chern class $c_1(L) \in H^2(X, \mathbb{Z})$. Just as before we are interested in whether the class is sent to zero under $H^2(G/K, \mathbb{Z}) \to H^2(G, \mathbb{Z})$. It is an interesting fact that every compact Lie group has $\pi_2(G) = 0$. Therefore, if $\pi_1(G) = 0$ then $H^2(G, \mathbb{Z}) = 0$ so every complex line bundle on the homogeneous space will again be homogeneous.
