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FJ Rolfes
Oct6-11, 03:59 PM
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 !

mathwonk
Oct9-11, 08:04 PM
huh?

FJ Rolfes
Oct10-11, 11:18 AM
Why is ∫dθ θ = 1 for a Grassmann number?