Why does the Grassmann integral ∫dθ θ equal 1?

[FJ Rolfes]

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]

huh?

 

[FJ Rolfes]

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