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

• B
• Korybut
In summary, the author is having trouble understanding the definition of a so-called super Lie module, and is unsure whether or not super commutativity is needed for compatibility reasons.
Korybut
TL;DR Summary
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.

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
fresh_42 said:
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.

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
fresh_42 said:
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!

## 1. What is a Super Lie module?

A Super Lie module is a mathematical structure that combines the properties of a Lie algebra and a module. It is a vector space equipped with a bilinear operation that satisfies the Jacobi identity and a module action that satisfies certain compatibility conditions.

## 2. How is a Super Lie module different from a regular Lie module?

A Super Lie module differs from a regular Lie module in that it has an additional grading or Z2-grading, which assigns elements to either an even or odd degree. This grading is used to define the compatibility conditions between the Lie algebra and the module action.

## 3. What is the significance of a Super Lie module in mathematics?

Super Lie modules have important applications in the study of supersymmetry, a concept in theoretical physics that extends the standard model of particle physics. They also have connections to other areas of mathematics such as representation theory and algebraic geometry.

## 4. Can you give an example of a Super Lie module?

One example of a Super Lie module is the space of differential forms on a supermanifold, which is a mathematical object that extends the concept of a manifold to include odd coordinates. The exterior derivative and Lie derivative operations on these forms make them into a Super Lie module.

## 5. How are Super Lie modules studied and classified?

Super Lie modules are studied using tools from representation theory, homological algebra, and category theory. They can be classified by their properties, such as the dimension of the underlying vector space, the type of Lie algebra and module involved, and the grading structure. Classification results can provide insight into the structure and behavior of Super Lie modules.

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