Question about Grassmann Integral

FJ Rolfes · Oct 6, 2011

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 · Oct 9, 2011

FJ Rolfes · Oct 10, 2011

Why is ∫dθ θ = 1 for a Grassmann number?

Question about Grassmann Integral

Thread 'Derivation of equations of stress tensor transformation'

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