B Is Super Commutativity Essential in Defining a Super Lie Module?

[Korybut]

TL;DR

Is supercommutativity is necessary?

Hello!

I have some troubles with the definition of the so called super Lie module. In Alice Rogers' textbook "Supermanifolds theory and applications" definition goes as follows

Suppose that $\mathbb{A}$ is a super algebra and that #\mathfrak{u}# is a super Lie algebra which is also a super $\mathbb{A}$ module such that
$[AU_1,U_2]=A[U_1,U_2]$
for all $A$ in $\mathbb{A}$ and $U_1,U_2$ in $\mathfrak{u}$. Then $\mathfrak{u}$ is said to be super Lie module over $\mathbb{A}$.

According to this definition I assume that $AU\in \mathfrak{u}$ for $A\in \mathbb{A}$ and $U\in \mathfrak{u}$. However if one considers chain of transformations
$[AU_1,BU_2]=A[U_1,BU_2]=-(-)^{|U_1|\, (|B|+|U_2|)}A[BU_2,U_1]=...$

$-(-)^{|U_1|\, (|B|+|U_2|)}AB[U_2,U_1]=(-)^{|B|\, |U_1|}AB[U_1,U_2]$
On the other hand one can do it differently
$[AU_1,BU_2]=-(-)^{|AU_1|\, |BU_2|}[BU_2,AU_1]=...=(-)^{|A| \, |B|} (-)^{|B|\, |U_1|}BA[U_1,U_2]$
If someone adds supercommutativity of the algebra #\mathbb{A}# in definition than everything is fine.

Book also provides an example

Suppose that $\mathfrak{u}$ is a super Lie algebra, and that $\mathbb{A}$ is a super algebra. Then $\mathbb{A}\otimes \mathfrak{u}$ is a super Lie module over $\mathbb{A}$, with bracket defined by
$[AX,BY]=(-)^{|B|\, |X|} AB[X,Y].$

This example is not clear also due to same issue.

In definition of left(right) super #\mathbb{A}#-module algebra is supposed to be super sommutative (which may be relaxed I suppose). However this not required or written explicitly in definition or example of super Lie module.

 

[fresh_42]

I get the following equations

\begin{align}
[A.U_1,B.U_2]&=A.[U_1,B.U_2]\\&=(-1)^{|U_1||B.U_2|}A.[B.U_2,U_1]\\&=(-1)^{|U_1||B.U_2|}A.B.[U_2,U_1]\\&=(-1)^{|U_1||B.U_2|}(-1)^{|U_2||U_1|}A.B.[U_1,U_2]\\
&=(-1)^{|U_1||B|}A.B.[U_1,U_2]\\[6pt]
[A.U_1,B.U_2]&=(-1)^{|A.U_1||B.U_2|}[B.U_2,A.U_1]\\&=(-1)^{|A.U_1||B.U_2|}(-1)^{|U_2||A|}B.A.[U_2,U_1]\\
&=(-1)^{|A||B|+|U_1||B|}B.A.[U_1,U_2]\\&=(-1)^{|U_1||B|}A.B.[U_1,U_2]
\end{align}

but have difficulties checking yours. I used the first result from (24) to (25) and the fact that we are only allowed to pull out the "scalar" at the first position of the bracket.

 

[Korybut]

I get the following equations

\begin{align}
[A.U_1,B.U_2]&=A.[U_1,B.U_2]\\&=(-1)^{|U_1||B.U_2|}A.[B.U_2,U_1]\\&=(-1)^{|U_1||B.U_2|}A.B.[U_2,U_1]\\&=(-1)^{|U_1||B.U_2|}(-1)^{|U_2||U_1|}A.B.[U_1,U_2]\\
&=(-1)^{|U_1||B|}A.B.[U_1,U_2]\\[6pt]
[A.U_1,B.U_2]&=(-1)^{|A.U_1||B.U_2|}[B.U_2,A.U_1]\\&=(-1)^{|A.U_1||B.U_2|}(-1)^{|U_2||A|}B.A.[U_2,U_1]\\
&=(-1)^{|A||B|+|U_1||B|}B.A.[U_1,U_2]\\&=(-1)^{|U_1||B|}A.B.[U_1,U_2]
\end{align}

but have difficulties checking yours. I used the first result from (24) to (25) and the fact that we are only allowed to pull out the "scalar" at the first position of the bracket.

I do get the same results as you actually. But moving from line (8) to (9) you have used supercommutativity $AB=(-1)^{|A|\, |B|}BA$ but according to the textbook's definition algebra $\mathbb{A}$ is not necessarily super commutative it is just any super algebra. Perhaps author forgot to add this fact in the definition.

 

[fresh_42]

I think super commutativity is inevitable for compatibility reasons. The calculation uses the Lie multiplication to switch the order between $A.B.U$ and $B.A.U$. This has to be leveled.

 

[Korybut]

I think super commutativity is inevitable for compatibility reasons. The calculation uses the Lie multiplication to switch the order between $A.B.U$ and $B.A.U$. This has to be leveled.

One more time thanks for your help!