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I have studied that the Lie algebra $sl(2,\mathbb{Z})=\{A\in M_2(\mathbb{Z}):tr(A)=0\}$ is a perfect Lie algebra over the base ring $\mathbb{Z}$, i. e. $$sl(2,\mathbb{Z})=[sl(2,\mathbb{Z}),sl(2,\mathbb{Z})],$$ where $[sl(2,\mathbb{Z}),sl(2,\mathbb{Z})]=\ \langle AB-BA:A,B\in sl(2,\mathbb{Z})\rangle$. Now a basis of $sl(2,\mathbb{Z})$ is the set $\bigg\{\begin{pmatrix} 0 & 1\\ 0 & 0 \end{pmatrix}$, $\begin{pmatrix} 0 & 0\\ 1 & 0 \end{pmatrix}$, $\begin{pmatrix} 1 & 0\\ 0 & -1 \end{pmatrix}\bigg\}$.

My question is that if $sl(2,\mathbb{Z})$ is perfect then each of the basis element can be written in the form of $AB-BA$. I can write the matrix $\begin{pmatrix} 1 & 0\\ 0 & -1 \end{pmatrix}=[AB-BA]$, where $A= \begin{pmatrix} 0 & 0\\ 1 & -0 \end{pmatrix}$ and $B= \begin{pmatrix} 0 & 1\\ 0 & 0 \end{pmatrix}$. But I am not able to write the matrices $A$ and $B$ in this form. I request you to help me.

MANI
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    ${\mathbb Z}$ is not a field. – Moishe Kohan Jul 21 '22 at 04:11
  • this structure is also defined and some authors called it as Lie algebra, in any case if I am defining it over $\mathbb{Z}$, I think the question is valid. – MANI Jul 21 '22 at 04:20
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    It is not generally true in a perfect Lie algebra that each element can be written as a commutator, just as a sum of commutators. If I recall correctly, for $sl_2$ over a field it might be possible to write each element as a commutator, but as Moishe Kohan points out, your base ring $\mathbb Z$ is not a field. Just call it a base ring instead, then it's a good question. – Torsten Schoeneberg Jul 21 '22 at 04:28
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    @TorstenSchoeneberg I have edited the question accordingly. – MANI Jul 21 '22 at 05:23
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    $CB--BC=2A$ where $C=AB-BA$ You will not be able to get just $A$ as the sum of the entries in the commutator is even. – Ryszard Szwarc Jul 21 '22 at 05:45
  • Just for reference, qualifying my earlier comment: https://math.stackexchange.com/q/769881/96384, https://math.stackexchange.com/q/95537/96384. – Torsten Schoeneberg Jul 21 '22 at 05:58
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    Not sure what happens with Lie Algebras, but for groups not every element of the commutator subgroup needs to be a commutator. – calc ll Jul 21 '22 at 06:05
  • @Cpc I mean to say generated by, can we a finite sequence of commutators such that the property hold. – MANI Jul 21 '22 at 07:27

1 Answers1

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The commutator of two traceless matrices $\pmatrix{a_1 &a_2 \\a_3 &-a_1}$ and $\pmatrix{b_1 &b_2 \\b_3 &-b_1}$ over any commutative ring $R$ is $$\pmatrix{a_2b_3-a_3b_2 &2(a_1b_2-a_2b_1) \\2(a_3b_1-a_1b_3) &a_3b_2-a_2b_3},$$

in particular the set $\lbrace [A,B]: A, B \in \mathfrak{sl}_2(R) \}$ is contained in

$$\mathfrak{sl}_2(R) \cap \{ \pmatrix{R &2R \\ 2R & R} \}.$$

Hence, as soon as $2$ is not a unit in $R$ (e.g. $R$ a field of characteristic $2$, or your case $R= \mathbb Z$), the matrix $\pmatrix{0 &1 \\0 &0}$ is neither a commutator itself, nor is it contained in the $R$-submodule spanned by all commutators, which is the usual definition of the derived subalgebra. In particular, the Lie algebra $\mathfrak{sl}_2(\mathbb Z)$ is not perfect for any definition of "perfect" I would find reasonable here.

By the way, one can write it as a commutator in the bigger Lie algebra $\mathfrak{gl}_2(\mathbb Z)$. In fact,

$$\pmatrix{0 &a \\0 &0} = [\pmatrix{0 &0 \\0 &1}, \pmatrix{0 &-a \\0 &0} ]$$ in $\mathfrak{gl}_2(R)$ for any ring $R$ and $a \in R$.

Torsten Schoeneberg
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