ideals of Lie algebras

Question

I have a question about ideals of Lie Algebras and the quotient Lie Algebra.

Why are ideals of Lie algebras defined the way they are? I am assuming this has something to do with the quotient algebra, and it being well defined.

What if we constructed a quotient algebra by using subalgebras instead of ideals?

Answer 1

I want to address something you asked in the comments.

Why don't we define the quotient algebra with a different Lie bracket, so that we only need subalgebras?

The point of the quotient construction is not just as a way of making new Lie algebras out of old ones. In some sense, the most important feature of the quotient construction is that you get a homomorphism $f:L\rightarrow L/I$. That means you can learn something about $L$ by understanding $L/I$.

So, I would like to refine your question to the following:

Why don't we define the quotient algebra with a different Lie bracket, for which the projection map is a homomorphism, so that we only need subalgebras?

Then I claim that there is NO such definition.

To see this, consider the Lie algebra $L = \operatorname{span}_\mathbb{R}\{x,y\}$ with bracket given by $[x,y]_L = y = -[y,x]_L$ (with, of course, $[x,x]_L = [y,y]_L = 0$.)

Let $A = \operatorname{Span}_\mathbb{R}\{x\}$. Note that $A$ is $1$-dimensional, so the only choice for Lie bracket on $L/A$ is the trivial one: $[a,b]_{L/A} = 0$ for any $a,b\in L/A$.

Now, $\pi:L\rightarrow A$ maps $y\in L$ to something non-trivial in $L/A$ (because $y\notin A$). Further, $\pi(x) = 0\in L/A$ since $x\in A$. But now we have \begin{align*} 0 &\neq \pi(y)\\ &= \pi([x,y]_{L})\\ &= [\pi(x), \pi(y)]_{L/A}\\ &= 0.\end{align*}

More generally, we have the following:

Suppose $A\subseteq L$ is an algebra and that some Lie bracket is chosen on $L/A$ so that the natural projection map $\pi:L\rightarrow L/A$ is a homomorpism. Then $A$ is an ideal and the bracket on $L/A$ is the usual one: $[x+A, y+A]_{L/A} = [x,y] + A$.

Proof: First, assume for a contradiction that $A$ is not an ideal, which means there is some $x\in L$ and some $a\in A$ for which $[x,a]_L\notin A$. Because $[x,a]_L\notin A$, $\pi([x,a]_L) \neq 0\in L/A$. Also, since $a\in A$, $\pi(a) = 0 \in L/A$. But then \begin{align*} 0 &\neq \pi([x,a]_L)\\ &= [\pi(x), \pi(a)]_{L/A}\\ &= [\pi(x), 0]_{L/A}\\ &= 0. \end{align*} Thus, we have a contradiction, so $A$ must be an ideal.

Now, on $L/A$ we know we have one bracket for which $\pi$ is a homomorphism: $[a+A, b+A]_0 = [a,b]+A$. But why is it the only one? Because it's determined by $\pi$. Concretely, because $\pi$ is surjective for any $a,b\in L/A$, we can find preimages $a',b'\in L$. Then $$[a,b]_{L/A} = [\pi(a'), \pi(b')]_{L/A} = \pi([a',b']_L) = [\pi(a'),\pi(b')] + A = [a,b]+A.$$ $\square$