The Lefschetz operator

There is an elementary proof in our 2003 book Exterior Differential Systems and Euler-Lagrange Partial Differential Equations (Bryant, et al, University of Chicago Press). It does not use any representation theory and is not 'brute force'; it only takes a couple of paragraphs using elementary facts about exterior algebra. See Proposition 1.1a, with the proof on page 13.

I learned that proof from Eugenio Calabi more than 30 years ago, and he told me that he had found it sometime back in the 50s.

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.