next up previous
Next: exterior derivation on 1-forms Up: Topics in Non commutative Previous: exterior algebras

exterior derivation

Let $ A$ be a commutative ring. Let $ B$ be a commutative $ A$ -algebra. We have already defined the exterior derivation

$\displaystyle d:B\to \Omega^1_{B/A}.
$

We define

$\displaystyle \Omega^\bullet_{B/A}=\wedge_{B} \Omega^1_{B/A}.
$

2

We would like to extend this to a map

$\displaystyle d:\Omega^\bullet_{B/A} \to \Omega^\bullet_{B/A}
$

which satisfies the following rules.

(EXT1) $\displaystyle d^2=0$

(EXT2) $\displaystyle d(\alpha\wedge \beta)=(d\alpha)\wedge \beta + (-1)^k\alpha\wedge ...
...d
(\forall \alpha \in \Omega^k_{B/A} ,\forall \beta \in \Omega^\bullet_{B/A})
$

It is easy to see that $ d$ is uniquely determined by the



Subsections

2012-02-29