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 '18 at 16:22
  • 2
    $\begingroup$ $[a,x]$ and $[x,a]$ are not collinear in general $\endgroup$ – YCor Aug 28 '18 at 16:25

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.

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  • $\begingroup$ 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)... $\endgroup$ – მამუკა ჯიბლაძე Aug 29 '18 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. $\endgroup$ – Konstantinos Kanakoglou Aug 29 '18 at 17:51

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