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Question about Grassmann Integral

  1. Oct 6, 2011 #1
    I can find various derivations of ∫ dθ = 0 which are satisfactory, but none of ∫dθ θ =1.

    Cheng and Li says it's just a normalization convention, of course that assumes that the integral is finite.

    Is this just a matter of definition, or is there a better reason that that?

    And would any of this relate to the exterior calculus, since I believe Grassmann algebra is an example of (or is) exterior albegra.

    Thanks very much !
     
  2. jcsd
  3. Oct 9, 2011 #2

    mathwonk

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  4. Oct 10, 2011 #3
    Why is ∫dθ θ = 1 for a Grassmann number?
     
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