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I am trying to prove that $\mathfrak{sl}(2,\Bbb C)$ is simple.

Since this takes the $[x,y]=xy-yx$ matrix commutator bracket, this is clearly non-abelian. So to prove it is simple, we need only show that it has no trivial ideals.

Now for it to have a non trivial ideal, this ideal would need to be spanned by one or two of the basis elements of $\mathfrak{sl}(2)$, i.e. we have the basis of $\mathfrak{sl}(2)$:

$e=\begin{bmatrix}0&1\\0&0\end{bmatrix},\quad h=\begin{bmatrix}1&0\\0&-1\end{bmatrix},\quad f=\begin{bmatrix}0&0\\1&0\end{bmatrix}$

and this non-trivial ideal must have basis:

1) $\{e,h\}$ 2) $\{e,f\}$ 3) $\{h,f\}$ 4) $\{e\}$ 5) $\{h\}$ 6) $\{f\}$

But we know from the commutation relations that: $[h,e]=2e,\quad[h,f]=-2f,\quad[e,f]=h$

and this does indeed eliminate all of these basis choices, since we don't have $[\mathfrak{i},\mathfrak{sl}(2,\Bbb C)]\subseteq \mathfrak{i}$ in any case.

Am I missing anything, or making any incorrect assumptions?

Also I would be interested to know if this is the only simple complex Lie algebra of dimension $3$ up to isomorphism.

glS
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So many hats
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    It is not true that those are the only possibilities for ideals. Any linear combination of $e, h, f$ could potentially also generate an ideal. – Qiaochu Yuan Aug 26 '15 at 00:04
  • Good point @QiaochuYuan, are there any other cases if I also consider the linear combinations? Also is there a much smarter way to handle this? – So many hats Aug 26 '15 at 00:40

2 Answers2

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It's not very hard to just do this explicitly, along the lines you started but keeping my comment in mind. Rather than $h, e, f$ I'll write $H, X, Y$. Suppose $I$ is a nonempty ideal. Then it contains some nonzero element $aH + bX + cY$. Applying $[H, -]$ to this element gives

$$b [H, X] + c [H, Y] = 2b X - 2c Y$$

from which we conclude that either $a = 0$ or $I$ contains $H$. If $I$ contains $H$, then it contains $[H, X] = 2X$ and $[H, Y] = 2Y$, and hence must be all of $\mathfrak{g} = \mathfrak{sl}_2(\mathbb{C})$.

Otherwise, $a = 0$. Now, applying $[X, -]$ to $bX - cY$ gives

$$-c [X, Y] = -c H$$

from which we conclude that either $c = 0$ or $I$ contains $H$. If $I$ contains $H$ then as above we are done; otherwise, $I$ contains $bX$, so it contains $[X, Y] = H$, and again we're done.

A slicker way to organize this argument is as follows. Any ideal of $\mathfrak{g}$ must in particular be invariant under the adjoint action of $H$. The adjoint action of $H$ is diagonalizable with eigenvectors $H, X, Y$, so in fact any ideal of $\mathfrak{g}$ must be a direct sum of the corresponding eigenspaces; that is, it actually turns out to be true that any ideal must have as a basis some subset of $\{ H, X, Y \}$.

Qiaochu Yuan
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    $b [H, X] + c [H, Y]$ is equal to $2b X - 2c Y$, not $2b X + 2c Y$. But you can apply $[H, -]$ twice, and you will get $4b X + 4c Y$. – Litho Aug 26 '15 at 05:44
  • Right, right. It doesn't really affect the structure of the argument. Note that the only fact used about $\mathbb{C}$ in this proof is that $2$ is invertible: in fact this proof shows that $\mathfrak{sl}_2(F)$ is simple for any field $F$ of characteristic not $2$. In characteristic $2$, though, $\mathfrak{sl}_2(F)$ is solvable. – Qiaochu Yuan Aug 26 '15 at 06:05
  • @QiaochuYuan I'm trying to show this in general and the step "[...] any ideal must be a direct sum of corresponding eigenspaces;" has eluded me for hours. Could you elaborate about it? – hjhjhj57 Sep 21 '15 at 03:40
  • @hjhjhj: the general fact is that if $T : V \to V$ is a diagonalizable linear operator (so $V$ has a basis of eigenvectors of $T$), then any $T$-invariant subspace also has a basis of eigenvectors of $T$. This is just saying that $T$ remains diagonalizable when restricted to a $T$-invariant subspace. – Qiaochu Yuan Sep 21 '15 at 06:58
  • @QiaochuYuan Great. Thank you. – hjhjhj57 Sep 21 '15 at 16:15
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Concerning the last question, whether $\mathfrak{sl}(2,\mathbb{C})$ is the only complex simple Lie algebra of dimension $3$, the answer is yes. There is just one connected Dynkin diagram for rank one. Over the real numbers, there are already $2$ different simple Lie algebras of dimension $3$, see here. And over $\mathbb{Q}$ the situation is much more complicated.

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Dietrich Burde
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