Direct sum as vectors but not Lie algebra direct sum

Question

Let $L$ be a Lie algebra and let $I$ be an ideal in $L$ and $K$ be a subalgebra in $L$ such that $L=I\oplus K $. Why this sum is direct as vector subspaces but not Lie algebra direct sum? Can't we define a Lie bracket on $I\oplus K $ in someway to meet the Lie bracket on $L$?

Answer 1

Consider $L=Vect_R(u,v)$ endowed with the Lie bracket $[u,v]=u$ , $Vect(u)$ is an ideal, and $Vect(v)$ a subalgebra. $L$ is the sum of $Vect(u)$ and $Vect(v)$ has a vector space. The Lie algebra $Vec(u)\oplus Vect(v)$ is commutative but not $L$.

Answer 2

Suppose you have a short exact sequence of Lie algebras

$$0 \rightarrow I \hookrightarrow L \twoheadrightarrow K \rightarrow 0$$

such that the homomorphism $L \twoheadrightarrow K$ has a splitting $ K \hookrightarrow L$. Then $K$ is identified with a subalgebra of $L$ by the splitting and $I$ is an ideal in $L$. As Lie algebras, we have $L \cong K \ltimes I$.

However, if we forget the Lie algebra structure and just think of these as vector spaces and linear maps, then it is always true that $L \cong K \oplus I$ as vector spaces.

Keep in mind that the category of Lie algebras is semiabelian (like the category of not-necessarily-abelian groups), not abelian like the category of vector spaces or modules, so exact sequences behave somewhat differently.