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Assume that $k$ is a field of characteristic zero. Show that the Lie algebra $\mathfrak{sl}(2)$ has no ideals but $\{0\}$ and the algebra itself. Deduce that $\mathfrak{sl}(2) = [\mathfrak{sl}(2), \mathfrak{sl}(2)].$

For the first part, I know how to exactly do it still I do not know how to deduce the second part, could someone clarify this to me, please? will I expand the commutator? but then I will get zero not $\mathfrak{sl}(2),$ am I right? or there is something that I do not know?

luxerhia
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Emptymind
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    Since $[\mathfrak{sl}(2),\mathfrak{sl}(2)]$ is a nonzero ideal, and $\mathfrak{sl}(2)$ is simple, we obtain equality. This is what the hint was suggesting to you. Note that for a Lie algebra $L$, the commutator subalgebra $]L,L]$ is an ideal by the Jacobi identity - see this duplicate. – Dietrich Burde Apr 22 '23 at 09:43
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    The main work is to show that $\mathfrak{sl}(2)$ is simple, i.e., has only two ideals. – Dietrich Burde Apr 22 '23 at 10:38
  • @DietrichBurde, Emptymind: I think that in the special case $I=J=L$ which we have here, unlike in the generality of the duplicate, one does not even need the Jacobi identity to show that $[L, L]$ is an ideal in $L$. It's a one-line proof at most: The commutator of a general element and an element of the space in question is ... well, a commutator. – Torsten Schoeneberg Apr 22 '23 at 16:51
  • @TorstenSchoeneberg Yes, you are right. I wanted to give the more general link, because this is also very useful. For the commutator alone there is no link. – Dietrich Burde Apr 22 '23 at 16:58

2 Answers2

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To show the second part from the first, just note that $[\mathfrak{sl}_2, \mathfrak{sl}_2]$ is an ideal in $\mathfrak{sl}_2$. And that it is not $=\{0\}$.

Torsten Schoeneberg
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  • how can I show the definition of an ideal on it? – Emptymind Apr 22 '23 at 05:21
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    You can show it like here. – Dietrich Burde Apr 22 '23 at 10:36
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    @Emptymind: For that, I suggest to empty your mind from anything about the special case $\mathfrak{sl}_2$ here, and convince yourself that for any Lie algebra $L$, the so-called derived subalgebra $[L, L]$, which by definition is the span of all commutators inside $L$, is always an ideal. That proof should not take more than one line. – Torsten Schoeneberg Apr 22 '23 at 16:48
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To show that $\mathfrak{sl}(2)=[\mathfrak{sl}(2),\mathfrak{sl}(2)]$, we need to prove two things:

  1. $\mathfrak{sl}(2)\subseteq [\mathfrak{sl}(2),\mathfrak{sl}(2)]$: any element of $\mathfrak{sl}(2)$ can be written as a linear combination of the three matrices $$ h=\begin{pmatrix}1&0\\0&-1\end{pmatrix}, \qquad e=\begin{pmatrix}0&1\\0&0\end{pmatrix}, \qquad f=\begin{pmatrix}0&0\\1&0\end{pmatrix}, $$ which satisfy the relations $$ [h,e]=2e, \qquad [h,f]=-2f, \qquad [e,f]=h. $$ Since $[\mathfrak{sl}(2),\mathfrak{sl}(2)]$ is the linear span of commutators $[x,y]$ with $x,y\in\mathfrak{sl}(2)$, it suffices to show that $[h,e]$, $[h,f]$, and $[e,f]$ are in $[\mathfrak{sl}(2),\mathfrak{sl}(2)]$. This is easy to see, since \begin{align*} [h,e] &= 2e = [e,f]-h, \\ [h,f] &= -2f = -(e,f)+h, \\ [e,f] &= h = [h,e]+[h,f], \end{align*} and all of these commutators are linear combinations of commutators in $[\mathfrak{sl}(2),\mathfrak{sl}(2)]$.

  2. $[\mathfrak{sl}(2),\mathfrak{sl}(2)]\subseteq \mathfrak{sl}(2)$: this follows from the fact that $[\mathfrak{sl}(2),\mathfrak{sl}(2)]$ is a Lie subalgebra of $\mathfrak{sl}(2)$ (i.e., it is closed under the Lie bracket operation), and $\mathfrak{sl}(2)$ is a simple Lie algebra, which means that it has no nontrivial proper Lie subalgebras.

Therefore, we conclude that $\mathfrak{sl}(2)=[\mathfrak{sl}(2),\mathfrak{sl}(2)]$.

brooke phoenix
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    I can not see in the first part why $[h,e] = [e,f] - h$ could you clarify this please? – Emptymind Apr 22 '23 at 05:15
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    Also in the second line you are using parantheses and not square brackets ..... is this really what you meant? – Emptymind Apr 22 '23 at 05:16
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    Also, in the third line $h \neq [h,e] + [h,f]$ – Emptymind Apr 22 '23 at 05:19
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    the third one needs multiplication by half I think – Emptymind Apr 22 '23 at 05:33
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    The last three equations in part 1 are a) wrong and b) not needed. That some commutators are in $[\mathfrak{sl}_2, \mathfrak{sl}_2]$ is trivial, but also not needed for this direction. What one should actually show in part 1 is that $h, e$ and $f$ are in $[\mathfrak{sl}_2, \mathfrak{sl}_2]$, and that follows from the first couple of computations and the fact that $1/2$ is in the base field $k$. – Torsten Schoeneberg Apr 22 '23 at 16:43
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    Also, part 2 is trivial, as the left hand side is very obviously a subset of the right hand side. Nothing about simplicity or subalgebras is needed. Which is good, because the statement that "a simple Lie algebra has no nontrivial proper Lie subalgebras" is plain wrong. -1 for this answer. – Torsten Schoeneberg Apr 22 '23 at 16:44