Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Differential modulo another differential

  1. Mar 22, 2008 #1

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

    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....
  2. jcsd
  3. Mar 22, 2008 #2


    User Avatar
    Science Advisor
    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.
  4. Mar 23, 2008 #3
    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?
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook