Why do we require that a simple Lie algebra be non-abelian?

Question

We say that a Lie $k$-algebra is simple if it is a simple object in the category of Lie algebras, and also nonabelian. The only simple object which we do not consider to be a simple Lie algebra under this definition is the line $k$. Is there any particular reason why $k$ would be problematic if were to consider it to be a simple Lie algebra?

Answer 1

I think it's mainly historical and practical. There is no deep reason why one agrees that groups of prime order are simple groups and 1-dimensional Lie algebras (resp. algebraic groups, Lie groups) are not simple Lie algebras (resp. algebraic groups, Lie groups).

One difference between the two contexts is that a finite-dimensional Lie algebra (at least in char 0) has a solvable radical and the quotient is a direct product of (non-abelian) simple Lie algebras. This "separates" the abelian part (the solvable radical, which is an iterated extension of abelian guys) and the semisimple part, which is made of non-abelian simple factors. In finite group theory there is no such separation. The simplest counterexample to such a result is the symmetric group on $\ge 2$ letters, which has no nontrivial solvable normal subgroup but has an abelian Jordan-Hölder factor. In this case there is a separation the other way round, but in general is just more tangled (complicated examples can be cooked up using wreath products).

In any case, there are many results for which one has to specify "non-abelian" simple groups. In Lie algebras, I guess that if by the sake of coherence, abelian ones were allowed as simple, one would often have to specify "non-abelian simple", more often than one has to write "simple or 1-dimensional abelian".

Answer 2

Getting rid of the abelian case just makes all the theorem statements nicer. $\mathbb{R}$ does not behave in any way like the other simple Lie algebras (let me work over $\mathbb{R}$ here): its finite-dimensional representation theory is not semisimple, its Killing form is not nondegenerate, it is not classified by a Dynkin diagram, etc. Just keep learning the theory and you'll see what happens.