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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 $L$?

Ronald
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    You have no guarantee that the Lie bracket of something from $K$ and something from $I$ would vanish. If they were both ideals, then you would get that immediately (because the intersection is trivial). This means that you cannot reconstruct the Lie bracket of $L$ from those of $I$ and $K$ alone. IOW you can only conclude that $[I,K]\subseteq I$. If we could conclude that $[I,K]\subseteq I\cap K$ we would, indeed, get a direct sum. – Jyrki Lahtonen Mar 21 '16 at 16:53
  • Could you please provide an example ? thanks – Ronald Mar 21 '16 at 16:57
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    Take $L$ to be upper triangular matrices, $K$ the diagonal matrices and $I$ the strictly upper triangular matrices. It's been 25 years since I really played with Lie algebras, so I'm rusty, but that should work. – Jyrki Lahtonen Mar 21 '16 at 17:05
  • Thanks Jyrki, so in general we require that $[i,k]=0$ whenever $i∈I$ and $k∈K $ in order to have a Lie algebra direct sum? (is that a defintion of Lie algebra direct sum? – Ronald Mar 21 '16 at 17:17

2 Answers2

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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 $L$.

Tsemo Aristide
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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.

ಠ_ಠ
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