# Definition

Let $\mathcal{A}$ be a commutative algebra with unit 1 and define a bilinear product $m:\mathcal{A}\otimes\mathcal{A}\to\mathcal{A}$ by

$m(a\otimes b) = ab.$

The universal differential envelope $\tilde\Omega(\mathcal{A})$ is the graded differential algebra

$\tilde\Omega(\mathcal{A}) = \bigoplus_{r=0}^n \tilde\Omega^r(\mathcal{A}),$

where

\begin{aligned} \tilde\Omega^0(\mathcal{A}) &= \mathcal{A}, \\ \tilde\Omega^1(\mathcal{A}) &= \Ker{m}, \\ \tilde\Omega^r(\mathcal{A}) &= \bigotimes_{r\:\text{times}} \tilde\Omega^1(\mathcal{A}). \end{aligned}

The derivation

$d:\tilde\Omega^r(\mathcal{A})\to\tilde\Omega^{r+1}(\mathcal{A})$

is given by the graded commutator

$d\alpha= [\tilde G,\alpha],$

where $\tilde G\in\mathcal{A}\otimes\mathcal{A}$ is the universal graph operator

$\tilde G = 1\otimes 1.$

Note that although $\tilde G$ is not in $\tilde\Omega^1(\mathcal{A})$, given any $a\in\tilde\Omega^0(\mathcal{A})$

$[\tilde G,a] = 1\otimes a - a\otimes 1$

is in $\tilde\Omega^1(\mathcal{A})$ since

$m(1\otimes a - a\otimes 1) = 0.$

Any differential graded algebra may be derived as a quotient of the universal differential envelope by a two-sided differential ideal.

# Example: Directed Graphs

Given a directed graph $G$, let $\mathcal{A}$ be the algebra of projections on $G_0$ in which case

$1 = \sum_v \pi_v.$

The universal graph operator is then given by

$\tilde G = \sum_{v,v'\in G_0} \pi_v\otimes\pi_{v'}.$

Interpreting the element $\pi_v\otimes\pi_{v'}$ as (being dual to) a directed edge $v\to v'$, the universal graph operator is seen to correspond to the complete directed graph with vertices $G_0$.

# Example: Smooth Manifolds

Given a smooth manifold $\mathcal{M}$, let $\mathcal{A}$ be the algebra of 0-forms on $\mathcal{M}$.

category: maths

Revised on September 25, 2009 at 04:00:47 by Eric Forgy