Every 1-dimensional or simple complex Lie algebra admits an invariant, symmetric and non-degenerate bilinear form. This form is unique up to multiplication by a nonzero constant (which in Yang-Mills theories plays the role of the coupling constant). 


Are there other Lie algebras with such a unique scalar product?



Lie algebras with unique invariant scalar product
Let $\mathfrak g$ be a finite dimensional Lie algebra with an invariant, symmetric and non-degenerate bilinear form $\langle \cdot , \cdot \rangle $.


**Claim : $\mathfrak g$ admits only one such a bilinear form if and only if $\mathfrak g $ is either simple or abelian and one dimensional.**


*Proof* : Denote the Levi decomposition of $\mathfrak g$ by $\mathfrak r + \mathfrak s$, then for any $\lambda \in \mathbb R $ the form defined by
$$\langle r\_1 + s\_1 , r\_2 + s\_2 \rangle\_\lambda = \langle r\_1 , r\_2 \rangle + \langle s\_1 , r\_2 \rangle + \langle r\_1 , s\_2 \rangle +\lambda \langle s\_1 , s\_2 \rangle $$
Is invariant.
For some $\lambda$ around 1, the form is also definite because it is an open property. 


Having only one such form implies that $\mathfrak g$ is either solvable and equal to $\mathfrak r$ or semi-simple and equal to $\mathfrak s$.
We know that in the semi-simple case, only simple ones have only one such form.
Assume from now on, that $\mathfrak g = \mathfrak r$ is solvable.
We write $\mathfrak g = [\mathfrak g , \mathfrak g] + A$ as a direct sum of vector spaces and as before for some $\lambda$ the bilinear form defined by
$$\langle g\_1 + a\_1 , g\_2 + a\_2 \rangle\_\lambda = \langle g\_1 , g\_2 \rangle + \langle g\_1 , a\_2 \rangle + \langle a\_1 , g\_2 \rangle +\lambda \langle a\_1 , a\_2 \rangle $$
is symmetric and definite.


