nLab
super vector bundle

Idea

An ordinary smooth vector bundle on a manifold $X$ may be identified with its sheaf of sections, which is a locally free sheaf of modules over its structure sheaf and all such localyy free module sheaves arise this way.

The definition of vector bundles in terms of sheaves of sections therefore immediately generalizes to every ringed space and in particular to supermanifolds.

Definition

A super vector bundle over a supermanifold $X$ is a locally free sheaf on the category of open subsets of $\mid X\mid$ of modules for the structure sheaf $O_X$.

Examples

super tangent bundle

The tangent bundle of an ordinary manifold has the sheaf of sections given by the derivations of the structure sheaf. The same definition works here:

the super tangent bundle $TX$ of a supermanifold $X$ is given by the sheaf $U\mapsto \mathrm{Der}{O}_X(U)$.

so a super tangent vector is a global section of this sheaf of derivations.

Example on the supermanifold ${ℝ}^{1\mid 1}$ with its canonical coordinates

$$t\in C^{\mathrm{\infty }}({ℝ}^{1\mid 1}{)}^{\mathrm{ev}}$$t \in C^\infty(\mathbb{R}^{1|1})^{ev}

$$\theta \in C^{\mathrm{\infty }}({ℝ}^{1\mid 1}{)}^{\mathrm{odd}}$$\theta \in C^\infty(\mathbb{R}^{1|1})^{odd}

there is the odd vector field

$$D:={\partial }_{\theta }+\theta \cdot {\partial }_t$$D := \partial_\theta + \theta \cdot \partial_{t}

whose super Lie bracket with itself vanishes

$$[D,D]=0.$$[D, D] = 0 \,.

Claim This odd vector field $D$ is left invariant with respect to the super translation group structure on ${ℝ}^{1\mid 1}$.

This means that $\mathrm{Lie}({ℝ}^{1\mid 1})$ is free on one odd generator.