Wedge Products with the Symplectic Form

Question

Let $ \omega $ denote the symplectic form on $ \mathbb{R}^{2n} $, namely $ \omega = dx_1 \wedge dy_1 + \cdots + dx_n \wedge dy_n $. Then let $ T $ be the linear map from $ (n-1) $- forms to $ (n+1) $-forms given by $ T(\tau) = \omega \wedge \tau $. I believe that $ T $ is a linear isomorphism, but I need help finding the proof, please.

Answer 1

You are right, this map is an isomorphism. In fact, much more is true. $\mathbb{R}^{2n}$ (with its standard symplectic form, complex structure and scalar product) is a Kähler manifold. On any Kähler manifold $M^{2n}$ (with Kähler form $\omega$), each Lefschetz operator

$$L^k : \Omega^l_{dR}(M) \to \Omega^{l+2k}_{dR}(M) : \tau \mapsto \tau \wedge \omega^k$$

is an isomorphism when $l = n-k$. The proof I saw most often of this fact uses the 'Hodge adjoint' $\Lambda$ of $L$, the fact that the commutator $[L, \Lambda] = (k-n)Id$ and some induction. This is merely linear algebra, albeit not the most straightforward one.

There is certainly a more 'elementary' proof of this result in the case you are interested in, but it would probably be a reformulation and simplification of the above proof.

After all, one only needs to prove the injectivity of the map $T = L^1$, which would follow from the existence (for any $(n-1)$-form $\tau \neq 0$) of a $(n-1)$-form $\tau'$ such that $\tau \wedge \omega \wedge \tau' \neq 0$, and the above proof implies that such a $\tau'$ may be constructed by applying $L$ and $\Lambda$ on $\tau$.

For more information about this line of thinking, you might take a look at Huybrechts's book, or Voisin's book starting on p.139.