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 $H := \bigoplus\_{k=0}^{2n} (k-n)\Pi^k$ (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?



Which $\frak{sl}_2$-Representations Arise From Hermitian Metrics
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. 


