Is there a notion of $\mathfrak{g}$-$\mathfrak{h}$-bimodules? Let $\mathfrak{g}$ and $\mathfrak{h}$ be two Lie algebras.

Is there a notion of $\mathfrak{g}$-$\mathfrak{h}$-bimodules such that $\mathfrak{g}$ becomes a $\mathfrak{g}$-$\mathfrak{g}$-bmodule via the adjoint actions of $\mathfrak{g}$ on itself from the left and from the right?

My thoughts so far:

*

*One could try to define a $\mathfrak{g}$-$\mathfrak{h}$-bimodule as a $\operatorname{U}(\mathfrak{g})$-$\operatorname{U}(\mathfrak{h})$-bimodule, or equivalently as a vector space $M$ together with a left action of $\mathfrak{g}$ on $M$ and a right action of $\mathfrak{h}$ on $M$, such that these two actions commute. But this does not apply to the adjoint actions, so it doesn’t make $\mathfrak{g}$ into a $\mathfrak{g}$-$\mathfrak{g}$-bimodule.

*There seems to exist the notion of a $\mathfrak{g}$-bimodule (coming from the more general notion of bimodules of a Leibniz algebra). For this, $\mathfrak{g}$ acts both from the left and from the right on $M$ (compatible with commutators) such that
$$
  (x \cdot m) \cdot y
  - x \cdot (m \cdot y)
  = [x, y] \cdot m \,.
$$
(Or some variation of this formula.)
But I don’t see how this definition could be generalized to a definition of $\mathfrak{g}$-$\mathfrak{h}$-bimodules.

 A: A left representation of $\mathfrak{g}$ is a vector space $V$ together with a linear map $\rho \colon \mathfrak{g} \to V$ such that $\rho([x, y]) = \rho(x) \rho(y) - \rho(y) \rho(x)$ for all $x, y \in \mathfrak{g}$. I would guess that a right representation of $\mathfrak{g}$ on $V$ is then a linear map $\pi \colon \mathfrak{g} \to V$ such that $\pi([x, y]) = \pi(y) \pi(x) - \pi(x) \pi(y)$.
If $\pi$ is a right representation, then $-\pi$ is a left representation, since $$(- \pi)([x, y]) = (-\pi)(x) (-\pi)(y) - (-\pi)(y) (-\pi)(x).$$
This shows that there is almost no difference between left and right representations of Lie algebras. Furthermore, the actions of two left representations $\rho, \eta \colon \mathfrak{g} \to V$ commute ($\rho(x) \eta(y) = \eta(y) \rho(x)$ for all $x, y \in \mathfrak{g}$) if and only if the actions of $\rho$ and $(-\eta)$ commute and conversely.
Therefore, any $(\mathfrak{g}, \mathfrak{h})$-bimodule $V$ with actions $(\rho, \pi)$ could just as well be considered a $\mathfrak{g} \times \mathfrak{h}$-module with the actions $(\rho, -\pi)$. So the answer to your question is that there is a notion of bimodules, but it is not more interesting than just considering modules over products of Lie algebras.
