# Differential modulo another differential

## Main Question or Discussion Point

Hello,

One says that an odd derivation d is a differential modulo another differential d' if this two things hold:

[d,d']=0
d^2=-[d',d"] for some d"

- The first condition says that d defines a derivative on H*(d'), but how? To have a well-defined action of this derivation on H*(d') we must have:
dx belongs to H*(d') whenever x belongs to H*(d') i.e: d'(dx)=0 & dx#d'y for every y
- d'(dx)=-dd'x=0 OK
- How to show the second one?

- The second condition says that d is indeed a differential i.e: d^2=0 in H*(d') but for x belongs to H*(d') we have:
d^2x=-d'd"x-d"d'x=-d'd"x ????????, How does it vanish?
Thanks for every help....

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dextercioby
Homework Helper
Off-topic: do you study BRST ?

On-topic, see the book of Henneaux and Teitelboim "Quantization of gauge systems", chapter # 8, page # 173.

Hi,
Yes I'm studying BRST
This book is precisely what I read now but I did not understand this section!
The problem is how the derivation d becomes a derivation on H*(d') by means of the first condition? an after that becomes nilpotent by means of the seconde condition?