super vector bundle

**superalgebra**
and
**supergeometry**
## Formal context ##
* superpoint
* super ∞-groupoid, smooth super ∞-groupoid, synthetic differential super ∞-groupoid
* synthetic differential supergeometry
## Superalgebra ##
* super vector space, SVect
* super algebra
* Grassmann algebra
* Clifford algebra
* superdeterminant
* super Lie algebra
* super Poincare Lie algebra
## Supergeometry ##
* supermanifold, SDiff
* super Lie group
* super translation group
* super Euclidean group
* NQ-supermanifold
* super vector bundle
* complex supermanifold
* Euclidean supermanifold
* integration over supermanifolds
* Berezin integral
## Structures
* supersymmetry
* division algebra and supersymmetry
## Applications ##
* supergravity
* supersymmetric quantum mechanics
* geometric model for elliptic cohomology

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 locally 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.

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

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** $T X$ of a supermanifold $X$ is given by the sheaf $U \mapsto Der O_X(U)$.

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

On the supermanifold $\mathbb{R}^{1|1}$ with its canonical coordinates

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

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

there is the odd vector field

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

whose super Lie bracket with itself vanishes

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

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

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

Revised on April 5, 2013 13:42:24
by Anonymous Coward
(134.76.82.221)