why is there no non-degenerate 2-forms on 4-sphere? The question is in the title. I have been told that there are 
actually no non-degenerate 2-forms on $S^{2n}$ for $n \neq 1,3$.
I have found the following question:
No symplectic structure on $S^{2n},\ n>1$
but it only eliminates the possibility of existence a closed non-degenerate 2-form (which seems to matter in $S^6$ case, for example).
 A: The existence of a non-degenerate 2-form in a 2n-dimensional smooth manifold is tantamount to the existence of an almost-complex structure. The spheres $S²$ and $S⁶$ are almost-complex and the remaining spheres are not almost complex: $S⁴$ is not almost-complex by Riemann-Roch and the rest are not almost-complex because of the divisibility of the image of the Chern character in K-theory [see this MO question for details].
Non-degenerate 2-forms being homotopically the same as almost complex structure can be seen as follows: the first is a reduction of the structure group of the tangent bundle from Gl(2n,R) to the symplectic linear group Sp(2n,R), and the latter is a reduction to the unitary group U(n). Since this is a homotopy theory issue, only the homotopy type of the group is relevant; U(n) is a maximal compact subgroup in Sp(2n,R) and thus they are homotopy equivalent. This is the algebraic topology approach.
The differential geometry interpretation is the so-called 2-out-of-3 property, which given two elements in the set
{inner product, non-degenerate 2-form, almost complex structure}
corresponding to O(n), Sp(2n,R) and U(n), constructs the third via linear algebra. Since a Riemann metric exists thanks to partition of unity, we can assume that we have a fibrewise inner product and then non-degenerate 2-forms and almost complex structures become equivalent. Hope this helps.
