Real vs Complex Lie algebras

Question

I have taken a course on Lie algebras this semester and was reading a paper on classification of 3-dimensional Lie algebras. The main goal was to find a base of the Lie algebra with respect to which the structure constants of the bracket are orthonormal and such thing was achieved by working with the dimension of the derived algebra of the Lie algebra. However in class we are dealing with complex such algebras and so far I dont see why, so I wandered if the theory and results are richer when working over the field of complex numbers. Is there a way to "connect" a real Lie algebra with a complex one?

Answer 1

For the classification, also in dimension $3$, the field is important. This is one reason why in class other fields have been considered than real numbers. In particular complex numbers are nice because the field is algebraically closed.

Over the real numbers, for example, both Lie algebras $\mathfrak{so}_3(\Bbb R)$ and $\mathfrak{sl}_2(\Bbb R)$ are simple, but not isomorphic. Their complexifications however become both isomorphic to $\mathfrak{sl}_2(\Bbb C)$, which is the only complex simple Lie algebra in dimension $3$.

In general we can "connect" the classifications by considering Galois cohomology. This is used, for example, to obtain a classification of all $3$-dimensional Lie algebras over $\Bbb Q$ - see for example this post.

Answer 2

Yes. If $\mathfrak g$ is a real Lie algebra, you can complexify it, taking the tensor produc (over $\mathbb R$) $\mathfrak g_{\mathbb C}=\mathfrak g\bigotimes\mathbb C$. It then becomes a complex Lie algebra, with the bracket $[X\otimes\lambda,Y\otimes\mu]=[X,Y]\otimes(\lambda\mu)$.