TITLE: The Lefschetz operator
QUESTION [15 upvotes]: Let $\omega=\sum_{i=1}^n dx_i\wedge dy_i\in\bigwedge^2(\mathbb{R}^{2n})^*$ be a standard symplectic form. The following result is due to Lefschetz:

For $k\leq n$, the Lefschetz operator
$L^{n-k}:\bigwedge^k(\mathbb{R}^{2n})^*\to
 \bigwedge^{2n-k}(\mathbb{R}^{2n})^*$ defined by  $$
 L^{n-k}\alpha=\alpha\wedge\underbrace{\omega\wedge\ldots\wedge\omega}_{n-k}=\alpha\wedge\omega^{n-k}
$$ is an isomorphism.

This follows from Proposition 1.2.30 [1]. The proof is slick and it uses representation theory.

Question: What are other standard references for this result? I want to quote it properly and I would like to see a reference to a
proof by brute force. A proof by brute force is not difficult but ugly
and perhaps in some reference I can find an elegant presentation of
such a proof.

This result seems so fundamental that it should be available in many textbooks, but I am not aware of any except [1].
[1] D. Huybrechts, Complex geometry. An introduction.. Universitext. Springer-Verlag, Berlin, 2005.

(MathsciNet review).

REPLY [16 votes]: I find the proof mentioned by Robert Bryant so beautiful that I cannot resist and I have to include it here. The following proof is from [1] Proposition 1.1a.

Definition. Given $w\in\mathbb{R}^n$ and $1\leq k\leq n$, we define the  interior product $$
 \iota_w:\bigwedge\nolimits^k(\mathbb{R}^n)^*\to
 \bigwedge\nolimits^{k-1}(\mathbb{R}^n)^* \quad \text{by} \quad
 \iota_w(\xi)(v_1,\ldots,v_{k-1})=\xi(w,v_1,\ldots,v_{k-1}). $$

For the proof of the next lemma see Proposition 2.12 in [2].

Lemma. If $\xi\in \bigwedge\nolimits^k(\mathbb{R}^n)^*$, $\eta\in \bigwedge\nolimits^\ell(\mathbb{R}^n)^*$ and $v\in\mathbb{R}^n$, then
  $$
 \iota_v(\xi\wedge\eta)=\iota_v(\xi)\wedge\eta+(-1)^k\xi\wedge\iota_v(\eta). $$

This lemma and a simple induction imply that
$$
\iota_v(\omega^k)=k\omega^{k-1}\wedge\iota_v(\omega),
\quad
\text{for $v\in\mathbb{R}^{2n}$.}\ \ \ \ \ \ \ (*)
$$
Now we can prove the main result:

Theorem. For $1\leq k\leq n$, the Lefschetz operator $$ L^k:\bigwedge\nolimits^{n-k}(\mathbb{R}^{2n})^*\to
 \bigwedge\nolimits^{n+k}(\mathbb{R}^{2n})^*, \quad L^k(\xi)=\xi\wedge\omega^k
 $$ is an isomorphism.

Proof.
Since the spaces $\bigwedge\nolimits^{n-k}(\mathbb{R}^{2n})^*$ and $\bigwedge\nolimits^{n+k}(\mathbb{R}^{2n})^*$ have equal dimensions, it suffices to show that $L^k$ is injective. We will use a backwards induction starting with $k=n$. If $k=n$, then $\omega^n=n!dVol$ and it is obvious that
$$
L^n:\bigwedge\nolimits^0(\mathbb{R}^{2n})^*=\mathbb{R}\to \bigwedge\nolimits^{2n}(\mathbb{R}^{2n})^*
$$
is an isomorphism and hence injective. 
Suppose that the result is true for $k$. We need to prove it is true for $k-1$. Thus 
we need to prove that if
$L^{k-1}(\xi)=\xi\wedge\omega^{k-1}=0$ for some
$\xi\in\bigwedge\nolimits^{n-(k-1)}(\mathbb{R}^{2n})^*$, then $\xi=0$.
Clearly, $\xi\wedge\omega^k=0$. If $v\in\mathbb{R}^{2n}$, then the Lemma and (*) yield
$$
0=\iota_v(\xi\wedge\omega^k)=
\iota_v(\xi)\wedge\omega^k\pm \xi\wedge\iota_v(\omega^k)=
\iota_v(\xi)\wedge\omega^k\pm k\underbrace{\xi\wedge\omega^{k-1}}_{0}\wedge\iota_v(\omega)
$$
so $L^k(\iota_v(\xi) )=\iota_v(\xi)\wedge\omega^k=0$ and hence $\iota_v(\xi)=0$ by the injectivity of $L^k$ (induction hypothesis).
Since $\iota_v(\xi)=0$ for all $v\in\mathbb{R}^{2n}$, it follows that $\xi=0$. The proof is complete.
$\Box$
[1] R. Bryant, P. Griffiths, D. Grossman, Exterior differential systems and Euler-Lagrange partial differential equations. Chicago Lectures in Mathematics. University of Chicago Press, Chicago, IL, 2003.
[2] F. W. Warner, Foundations of differentiable manifolds and Lie groups. Corrected reprint of the 1971 edition. Graduate Texts in Mathematics, 94. Springer-Verlag, New York-Berlin, 1983.