Are Grassmann numbers a concept of graded Lie algebras or is something specific to superalgebras? What are they (i.e: how are they defined, important properties, etc.)? Is there a reasonable introduction to them?

I think that what makes me wonder a little is, since there does not seem to be a sensible constructivist approach to these entities (other than accepting them as the entities that satisfy the required properties) is there nothing that stops someone from going into 'constructing' meta-superalgebras by defining 'numbers' $\kappa_{i}$, such that, e.g.,

$$\kappa_{i} \kappa_{j} = \theta_{k} \quad (\leftarrow \text{Grassmann odd}),$$ $$\kappa_{i} \kappa_{j}\kappa_{m} \kappa_{n} = \theta_{p}\quad (\leftarrow \text{Grassmann even}).$$

So I define such numbers as 'square-roots' of grassmann $a$-numbers. It seems nothing stops this process ad infinitum. Maybe there is some property I'm missing that will allow the algebra to be closed but I don't know what that could be.

Btw, I think this is a great reference Phys.SE question regarding this topic: "Velvet way" to Grassmann numbers.


1 Answer 1


Here I will just make a couple of general remarks.

1) Graded algebras usually refer to $\mathbb{Z}$- or $\mathbb{N}$-graded algebras, while superalgebras are $\mathbb{Z}_2$-graded algebras.

2) Grassmann numbers are oddly graded supernumbers.

Please click on the links for further information, important properties and references.


1) Bryce De Witt, "Supermanifolds", Cambridge Univ. Press, 1992.

2) Deligne, Pierre and John W. Morgan, "Notes on Supersymmetry (following Joseph Bernstein)". Quantum Fields and Strings: A Course for Mathematicians (1999). American Mathematical Society. pp. 41–97. ISBN 0-8218-2012-5.

Concerning v3 of the question. The $\kappa_i$'s correspond to a $\mathbb{Z}_4$ grading, and there are indeed research works in that direction. However, many properties of numbers and supernumbers do not generalize easily to $\mathbb{Z}_n$-grading with $n>$2. For instance, I think that already Berezin showed that it is not possible to define a useful notion of (Berezin) determinant of matrices with $\mathbb{Z}_n$-graded entries if $n>$2.

  • 1
    $\begingroup$ thanks!, that property $(\theta_{i})^2 = 0$ seems to suggest these 'numbers' have zero divisors, so i'm wondering if there is a representation of these numbers using sedenions or some other algebra with zero divisors? although the infinite expansion seems to imply an infinite dimensional algebra $\endgroup$
    – lurscher
    May 17, 2011 at 18:52

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