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I am asked to prove that given two ideals $I,J\subset \mathfrak{g}$ in a Lie algebra, then the set $$ [I,J]=\{[x,y]:x\in I,y\in J\} $$ is another ideal of $\mathfrak{g}$. It is somewhat easy to see that due to the Jacobi identity, for any $a\in \mathfrak{a},x\in I,y\in J$: $$ [a,[x,y]] = -[x,[y,a]]-[y,[a,x]], $$ so now I have only to check that $[I,J]$ is closed under sums (since each term in RHS is in $[I,J]$). I searched for previous posts and I don't even think this is true (see this answer): I think there's a typo in my book and that $[I,J]$ should be defined as the linear space spanned by all possible $[x,y]$ where $x\in I, y\in J$.

Can anyone either confirm that the definition is wrong or either give a proof that shows $ [x,y]+[a,b] \in \{[a',b']:a'\in I, b'\in J\} $ for $x,a\in I, y,b\in J$?

topolosaurus
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1 Answers1

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It is correct that $I$,$J$ and $[I,J]$ are assumed to be subspaces of the vector space $\mathfrak{g}$. So the definition of $[I,J]$ is, it is the subspace generated by all brackets $[a,b]$ with $a\in I$ and $b\in J$.

There are counterexamples for your definition, e.g., for $I=J=\mathfrak{g}$, i.e., a Lie algebra where the commutator subalgebra $[\mathfrak{g},\mathfrak{g}]$ does not consist only of commutators $[a,b]$.

When is the set of commutators of a Lie Algebra not equal to its derived Lie Algebra?

Dietrich Burde
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