# nLab super module

Contents

### Context

#### Algebra

higher algebra

universal algebra

supersymmetry

# Contents

## Idea

A more general version of super vector space.

## Definition

### $\mathbb{Z}/2\mathbb{Z}$-graded $R$-modules

Given a commutative ring $R$, an $\mathbb{Z}/2\mathbb{Z}$-graded $R$-module is a $R$-module $V$ with decomposition functions $\mathcal{D}_0:V \to V$ and $\mathcal{D}_1:V \to V$, such that

• for all $v:V$, $v = \mathcal{D}_0(v) + \mathcal{D}_1(v)$
• for all $a:R$, $b:R$, $v:V$, and $w:V$, $\mathcal{D}_0(a v + b w) = a \mathcal{D}_0(v) + b \mathcal{D}_0(w)$
• for all $a:R$, $b:R$, $v:V$, and $w:V$, $\mathcal{D}_1(a v + b w) = a \mathcal{D}_1(v) + b \mathcal{D}_1(w)$
• for all $v:V$, $\mathcal{D}_0(\mathcal{D}_0(v)) = \mathcal{D}_0(v)$
• for all $v:V$, $\mathcal{D}_0(\mathcal{D}_1(v)) = 0$
• for all $v:V$, $\mathcal{D}_1(\mathcal{D}_0(v)) = 0$
• for all $v:V$, $\mathcal{D}_1(\mathcal{D}_1(v)) = \mathcal{D}_1(v)$

As a result, the image of the two decomposition functions $\im(\mathcal{D}_0)$ and $\im(\mathcal{D}_1)$ are $R$-modules and there exists a linear isomorphism $i:V \cong \im(\mathcal{D}_0) \otimes \im(\mathcal{D}_1)$, where $A \otimes B$ is the tensor product of modules.

The elements of $\im(\mathcal{D}_0)$ are called even elements or bosonic elements, and the elements of $\im(\mathcal{D}_1)$ are called odd elements or fermionic elements.

### Super modules

The tensor product of $\mathbb{Z}/2\mathbb{Z}$-graded $R$-modules $A \otimes B$ for $a:A$, $b:B$ is defined as the following:

$\mathcal{D}_0(a \otimes b) = \mathcal{D}_0(a) \otimes \mathcal{D}_0(b) + \mathcal{D}_1(a) \otimes \mathcal{D}_1(b)$
$\mathcal{D}_1(a \otimes b) = \mathcal{D}_0(a) \otimes \mathcal{D}_1(b) + \mathcal{D}_1(a) \otimes \mathcal{D}_0(b)$

This plus the linearity of the $\mathcal{D}_0$ and $\mathcal{D}_1$ functions result in the category of $\mathbb{Z}/2\mathbb{Z}$-graded $R$-modules to be a monoidal category.

A super module is an object of the category of $\mathbb{Z}/2\mathbb{Z}$-graded $R$-modules with the braiding for the tensor product $A \otimes B$:

$t_{A, B}:(A \otimes B) \to (B \otimes A)$

such that

$\mathcal{D}_0(t_{A, B}(a, b)) = \mathcal{D}_0(a) \otimes \mathcal{D}_0(b) - \mathcal{D}_1(a) \otimes \mathcal{D}_1(b)$
$\mathcal{D}_1(t_{A, B}(a, b)) = \mathcal{D}_0(a) \otimes \mathcal{D}_1(b) + \mathcal{D}_1(a) \otimes \mathcal{D}_0(b)$

## See also

Last revised on May 11, 2022 at 11:47:56. See the history of this page for a list of all contributions to it.