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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).

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Dima Sustretov
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    But he claim is for $n\neq 1,3$. Non-degenerate means that restriction of the form to any fibre of the ctangent bundle is non-degenerate as a bilinear form, i.e. induces an isomorphism $T_x M \to T_x^*M$. – Dima Sustretov Sep 26 '14 at 19:45

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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.

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Casals
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