Any symplectic manifold $(M,\omega)$ carries a representation of $\frak{sl}_2$: Define the maps $$ L: \Omega^\bullet \to \Omega^{\bullet}, ~~~~~~~~~ \Lambda: \Omega^\bullet \to \Omega^{\bullet}, ~~~~~~~~~ H: \Omega^\bullet \to \Omega^{\bullet}, $$ as follows $$ L(\nu) := \omega \wedge \nu, ~~~~~ \Lambda(\nu) := \sum \omega_{ij}^{-1} i_{\partial_i} i_{\partial_j}(\nu), $$ where $i$ denotes the interior product, and finally, for any $\mu \in \Omega^k$, we define $$ H(\mu) = (n-k) \mu. $$ This can be shown by direct calculation to satisfy $$ [L,\Lambda] = H, ~~~~~~ [L,H] = 2L, ~~~~~~ [\Lambda,H] = -2 \Lambda. $$ Is there a conceptual, or at least non-computational, proof of this result that anyone can direct me to. Also, I am happy to hear of more philosophical reasons for this to be true.

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    $\begingroup$ Hm, there is indeed such a representation, but as I know, you need your manifold to be Hermitian, or, more generally, you need a Riemannian metric together with a compatible almost complex structure (you will then get an almost symplectic structure). The operator $\Lambda$ is then defined in a coordinate-free way as the adjoint to $L$. Look Proposition 1.2.26 and Corollary 1.2.27 in D. Huybrechts, Complex Geometry: An Introduction, Springer (2005). Besides, is there an easy way to see that your definition of $\Lambda$ does not depend on the choice of coordinates? $\endgroup$ Dec 10, 2018 at 12:49
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    $\begingroup$ There is no need for Hermitian, it works for symplectic. See the paper of Yau $\endgroup$ Dec 10, 2018 at 15:00
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    $\begingroup$ arxiv.org/pdf/0909.5418.pdf $\endgroup$ Dec 10, 2018 at 15:00
  • $\begingroup$ Tseng and Yau - Cohomology and Hodge theory on symplectic manifolds (abs link, rather than PDF). $\endgroup$
    – LSpice
    Dec 11, 2018 at 21:42


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