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Let $L$ be a Lie algebra over field $F$, $I$ - ideal in this algebra. It's stated that $I^2$ (and so any item of central series) is also ideal in $L$.

1) For any $a, b \in I^2: [a,b] \in I^2$. True, since $a, b \in I$.

2) For any $k \in F, a \in I^2: ka \in I^2$. True, because $a = [a_1,a_2],$ $a_i \in I$, and $ka = k[a_1,a_2] = [ka_1,a_2]$. As $ka_1, a_2$ are from $I$, so $[ka_1,a_2]$ is from $I^2$.

3) For any $a, b \in I^2: a + b \in I^2$. Why is that? How could you represent $[a_1,a_2] + [b_1,b_2]$ as $[c_1, c_2]$ for some $c_1,c_2 \in I$?

Glinka
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  • Also asked here: http://mathoverflow.net/questions/164270/why-the-square-of-ideal-in-lie-algebra-is-also-ideal – Glinka Apr 24 '14 at 20:59
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    $I^2$ is the subspace generated by all brackets $[i,i']$ with $i,i'\in I$. By definition, it satisfies the subspace axioms. To show that it is an ideal, you need to prove that $[i'',x]\in I^2$ for all $i''\in I^2$ and $x\in L$. By bilinearity of the bracket, you only nedd to prove $[i'',x]\in I^2$ for $i''$ of the form $[i,i']$ with $i,i'\in I$. The result is a direct consequence of the Jacobi identity, and the fact that $I$ is an ideal. This actually shows that $[I,J]$, the subspace generated by all the $[i,j]$ with $i\in I$ and $j\in J$, for $I,J$ ideals, is an ideal. – Olivier Bégassat Apr 24 '14 at 21:05
  • You might as well have just posted that as an answer. – Seth Apr 24 '14 at 21:26
  • $I^2$ is a subspace by definition! That's the clue! Thank you very much! That's not for the first time with me: recently I tried to prove that the sum of two ideals is an ideal, but forgot that bilinearity of the bracket is given... – Glinka Apr 24 '14 at 21:34
  • You must add @username somewhere in your comment for the person you are responding to to be notified. – Olivier Bégassat Apr 29 '14 at 15:23

1 Answers1

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Let $I,J$ be two ideals of a Lie algebra $L$. The subspace $[I,J]$ generated by all brackets $[i,j]$ with $i\in I$ and $j\in J$ is an ideal of $L$. Indeed, for $x\in L$ and $i\in I,j\in J$, $$[x,[i,j]]=[\underbrace{[x,i]}_{\in I},j]+[i,\underbrace{[x,j]}_{\in J}]\in[I,J]$$ since $I$ and $J$ are ideals. In particular, $I^2=[I,I]$, the subspace of $L$ generated by all $[i,i']$, with $i,i'\in I$, is an ideal of $L$.

Olivier Bégassat
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