is there a bundle with any given Euler class? Given a cohomology class $\alpha \in H^{k+1}(X,\mathbb{Z})$, is there an (oriented) $S^k$ bundle over $X$ with Euler class $\alpha$?
 A: The answer is no.
In fact this question has a nice relationship to the classic Hopf-invariant-one problem, since the Hopf maps define sphere bundles of Euler class equal to the volume form of $S^n$ for $n = 2, 4, 8$ (there is also an $n = 1$ map but it does not define an orientable $S^0$ bundle). For $n$ which are not these, there is no Euler class equal to the volume form.
Further there is a relationship to the Steenrod problem since Euler classes may be represented in Poincar\'e duality by embedded manifolds (the zero locus of a suitably regular smooth section of the associated vector bundle). Thus there are lots of classes we expect cannot be represented this way.
On the other hand it was proven by L. Guijarro, T. Schick, and G. Walschap in ``Bundles with Spherical Euler Class" that if $X$ is an $n$-dimensional CW complex and $\alpha \in H^k(X,\mathbb{Z})$, $k$ even, there is a multiple of $\alpha$ which is an Euler class. This vibes with the previous comments since the obstructions to representing a class by an immersed manifold (not sure about embedded) are all torsion. Interestingly, the multiplication factor is a universal function of $n$ and $k$. There are also some more general results about representation by Pontryagin classes here.
That we need $k$ even can be seen by studying $S^1$. No multiple of the volume form is the Euler class of an $S^0$ bundle, since there are only two of these, the trivial and the Mobius bundle.
