Recal that $\frak{sl}_2$ is the Lie algebra with basis elements $e,f,h$, and bracket $$ [e,f] = h, ~~~ [h,e] = 2e, ~~~ [h,f] = 2f. $$ For $M$ a $2n$complex manifold, the Lefschetz identities tell us that any Hermitian metric $h$ induces a representation $\pi_h$ of $ \frak{sl}_2 $ on the Dolbeault double complex $\Omega^{(\bullet,\bullet)}(M)$. More specifically, if $\omega$ is the fundamental form of $h$, $L$ the operator which wedges by $\omega$, $\Lambda$ the adjoint of $L$, and $H := \bigoplus_{k=0}^{2n} (kn)\Pi^k$ (where $\Pi^k$ is projection onto the $k$forms), then a presentation of $\frak{sl}_2$ is given by $$ \pi_h(e) = L, ~~~ \pi_h(f) = \Lambda, ~~~ \pi_h(h) = H $$ Does every representation $\rho :\frak{sl}_2 \to $End$[\Omega^{(\bullet,\bullet)}(M)]$, for which $\rho(h) = H$, arise in this way? In other words, can we always realise $\rho(e)$ as wedging by the fundamental form of some particular metric?
The answer to the newly edited question is now "no".
The reason is that $\Omega^{(\bullet,\bullet)}(M)$ is not an irreducible $\frak{sl}_2$module, even when you fix $\rho(h) = H$.
For example, you can do this: Choose two different Kähler forms $\omega$ and $\eta$ and let $L$ and $\Lambda$ be associated to $\omega$ on the sum of the components $\Omega^{(p,q)}(M)$ for which $p{+}q$ is even and let $L$ and $\Lambda$ be associated to $\eta$ on the sum of the components $\Omega^{(p,q)}(M)$ for which $p{+}q$ is odd. This will give a representation of $\frak{sl}_2$ by the above process, but it won't be the one associated to a single Kähler structure.

$\begingroup$ But can we always realise $L$ as wedging by a form? $\endgroup$ – Christian Fischmann Oct 20 '14 at 18:42

$\begingroup$ cf my edit to make the question clearer. $\endgroup$ – Christian Fischmann Oct 20 '14 at 18:44

$\begingroup$ I have rewritten the question to make it clearer, I hope this helps, and sorry for the confusion. $\endgroup$ – Christian Fischmann Oct 20 '14 at 19:12