- #1

Korybut

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- TL;DR Summary
- Is supercommutativity is necessary?

Hello!

I have some troubles with the definition of the so called

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

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

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 followsSuppose 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**.