Let $\mathfrak g$ be a Lie superalgebra.

If $\mathfrak a$ is not a grade subspace of $\mathfrak g$, then why does $[\mathfrak g, \mathfrak a]$ and $[\mathfrak a, \mathfrak g]$ are not same?

For me as sets they are linear span of $[a,x]$ and $[x,a]$ and hence they are same. But in book it is given they are different and the author has defined left and right ideal separately.

I am reading the book "Lie superalgebras and enveloping algebras by Ian M.Musson" Proposition 1.2.2.

Kindly help me with this.

Thank you.

  • $\begingroup$ @KonstantinosKanakoglou thank you. I have added the book name. $\endgroup$
    – GA316
    Aug 28, 2018 at 16:22
  • 2
    $\begingroup$ $[a,x]$ and $[x,a]$ are not collinear in general $\endgroup$
    – YCor
    Aug 28, 2018 at 16:25

1 Answer 1


The linear spans of $[a,x]$ and $[x,a]$, in a Lie superalgebra (i.e. a $\mathbb{Z}_2$-graded Lie algebra) are generally not the same (unlike the Lie algebras case):

Since $\mathfrak a$ is not a graded subspace of $\mathfrak g$, then in general its elements are not homogeneous. So for $a\in \mathfrak a$ we generally have: $a=a_0+a_1$, i.e. its decomposition into even and odd parts.
Thus, if we take for example an odd element $x\in \mathfrak g_1$: $$ [a,x]=[a_0,x]+[a_1,x]=-[x,a_0]+[x,a_1] $$ while $$ [x,a]=[x,a_0]+[x,a_1] $$ Consequently, the elements $[a,x]$ and $[x,a]$, are -in general- not collinear, as has already been indicated in the comments to the OP.

  • $\begingroup$ It would be nice to give an example: despite this, there remains a suspicion that if one generates the ideal (with nested brackets), the "left" result might still somehow engulf the "right" one (because of the (super)Jacobi identity)... $\endgroup$ Aug 29, 2018 at 4:21
  • $\begingroup$ @მამუკა ჯიბლაძე, i am just answering why the spans of $[a,x]$ and $[x,a]$ are not identical (as is incorreclty claimed in the OP) and thus why $[a,\mathfrak g]$ and $[\mathfrak g, a]$ are -in general- not the same even as vector subspaces. $\endgroup$ Aug 29, 2018 at 17:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.