Grassmann variables were introduced to make path-integral formalism easier to handle fermionic (anti-commutating) fields.

Mathematically they represent the exterior algebra of forms (or exterior derivatives).

Furthermore they can be represented (matrix representation theory) as matrices of dimension $2^n \times 2^n$.

However a similar concept is found in complex numbers which have both a matrix representation and a geometric one (as $2$-dimensional vectors).

What is the geometric representation/interpretation of a Grassmann number (even in higher dimension if needed)?

More specificaly what would be the geometric intepretation of $\theta^2=0$?

For example a mixed number of the form $a+b\theta$ (in geometric analogy to complex number)

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    $\begingroup$ If you know they are essentially the exterior algebra - what more than the geometric interpretation than that of the $k$-th exterior power as $k$-planes (and thus of $\theta$ as a "line" that cannot span a plane, hence $\theta^2= 0$) do you think there is? $\endgroup$ – ACuriousMind Nov 5 '14 at 2:44
  • $\begingroup$ @ACuriousMind, yes, i was thinking sth along the lines of vectors similar to complex numbers. for example a mixed number of the form $a+b\theta$ as a vector. in effect i want to see the geometric analogy with complex numbers, if you want i can ask a converse question about complex numbers in geometric analogy with grassman numbers $\endgroup$ – Nikos M. Nov 5 '14 at 2:51
  • $\begingroup$ Why do you think that Grassmann numbers are analogous to complex numbers? Though you can obtain both by formally adjoining an element $\mathrm{i}$ resp. $\theta$ to $\mathbb{R}$, the fact that one imposes the relation $\mathrm{i}^2 = -1$ and the other $\theta^2 = 0$ means they are fundamentally different - for instance, the complex numbers are a field extension of the reals, while the Grassmann numbers are not. $\endgroup$ – ACuriousMind Nov 5 '14 at 3:02
  • $\begingroup$ @ACuriousMind, this is the question, more about geometry that number theory $\endgroup$ – Nikos M. Nov 5 '14 at 3:04
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    $\begingroup$ Related: mathoverflow.net/q/22247/13917 and physics.stackexchange.com/q/5005/2451 $\endgroup$ – Qmechanic Nov 5 '14 at 11:57

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