Definition of invariant bilinear form Given a bilinear form $\beta $ on a Lie algebra $ \mathfrak{g} $, what does it mean for $ \beta $ to be $ \mathfrak{g} $-invariant?
 A: It means $\beta([x, y], z) = \beta(x, [y, z])$.
It is not at all obvious that this is the correct definition. It's written to look nice but you may rightfully ask why this condition deserves the name "invariant." It may make a little more sense to rewrite it as
$$\beta(\text{ad}_y(x), z) + \beta(x, \text{ad}_y(z)) = 0;$$
this reveals that $\mathfrak{g}$-invariant means invariant with respect to the tensor action of $\mathfrak{g}$ on $\mathfrak{g} \otimes \mathfrak{g}$, where $\mathfrak{g}$ carries the adjoint representation. (Invariant with respect to a Lie algebra action means annihilated.) If you've never done this calculation, it would be good to check that

*

*the tensor action of a Lie algebra $X \mapsto X \otimes 1 + 1 \otimes X$ exponentiates to the tensor action of a Lie group $g \mapsto g \otimes g$, and

*exponentiation identifies "invariant with respect to $\mathfrak{g}$" and "invariant with respect to $G$" in the following sense: a vector $v \in V$ in a $\mathfrak{g}$-representation is $X$-invariant in the sense that $Xv = 0$ iff it is $\exp(tX)$-invariant in the sense that $\exp(tX) v = v$ for all $t \in \mathbb{R}$.

This means that a bilinear form on $\mathfrak{g}$ is $\mathfrak{g}$-invariant iff it is $\text{Inn}(\mathfrak{g})$-invariant in the obvious sense ($\beta(gx, gy) = \beta(x, y)$), where $\text{Inn}(\mathfrak{g})$ is the connected Lie subgroup of $\text{Aut}(\mathfrak{g})$ with Lie algebra the Lie algebra of inner derivations $\text{ad}_y(-)$.
