# Super tangent bundles

## Idea

The tangent bundle of an ordinary manifold is a vector bundle whose sheaf of sections? is given by the derivations of the structure sheaf. The same idea applies to a supermanifold to produce a super vector bundle.

## Definition

The super tangent bundle $TX$ of a supermanifold $X$ is given by the sheaf $U↦\mathrm{Der}{O}_{X}\left(U\right)$.

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

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

there is the odd vector field

$D≔{\partial }_{\theta }+\theta \cdot {\partial }_{t}$D \coloneqq \partial_\theta + \theta \cdot \partial_{t}

whose super Lie bracket with itself vanishes

$\left[D,D\right]=0\phantom{\rule{thinmathspace}{0ex}}.$[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 $\mathrm{Lie}\left({ℝ}^{1\mid 1}\right)$ is free on one odd generator.

Revised on December 7, 2011 02:06:32 by Toby Bartels (64.89.53.157)