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A Fermionic field $\psi$ is defined as a field over spacetime taking values in a Grassmann algebra $\mathcal{G}$. Why is the Grassmann algebra $\mathcal{G}$ usually taken to be infinite dimensional? Is there a particular vector space we choose when defining the Grassmann algebra of a field theory?

My first thought was that a field theory has an infinite degrees of freedom, but I think this is already taken into account by $\psi$ being dependent on $x \in \mathbb{R}^4$. My understanding is that the Grassmann algebra has no physical meaning other than to capture the anti-commuting nature of the fields, so couldn't we take the Grassmann algebra to be of any dimension since all Grassmann algebras are anti-commuting?

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1 Answers1

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  1. Ref. 1 does define a $2^N$-dimensional Grassmann algebra $\Lambda_N$ with finitely many anti-commuting generators $\xi^1,\ldots,\xi^N,$ and proceeds to write

    We shall usually, though not always, deal with the formal limit $N\to\infty$.

  2. For starters a finite $N<\infty$ implies that the soul $z_S$ of a Grassmann-even supernumber $$z~=~z_B+z_S\tag{1.1.2}$$ is automatically nilpotent $$z_S^{N+1}~=~0.\tag{1.1.4}$$ To avoid this one often takes $N=\infty$.

    (In contrast, a Grassmann-odd soul $z_S$ always squares to zero: $z_S^2=0$, which is a cherished feature.)

  3. Moreover, mathematical purists would like a universal construction of supernumbers that doesn't refer to a particular choice of generators $\xi^1,\ldots,\xi^N$, so they use a categorical definition with a so-called functor of points, see e.g. references in this Phys.SE post.

References:

  1. Bryce DeWitt, Supermanifolds, Cambridge Univ. Press, 1992; p.1.
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