fermions - Number of Grassmann generators for Dirac field?

How many Grassmann generators are sufficient for the description of a Dirac spinor in 4 dimensions? i.e. The Dirac field is a map to $\Lambda_N$, the space of supernumbers with $N$ real Grassmann generators. What is $N$?

This is a follow up question to my previous question Grassmann Paradox Weirdness. I am following Prakash's book Mathemaical Perspectives on Theoretical Physics, where they say a supernumber $z\in\Lambda_N$ can be thought of the extension of complex numbers by the addition of $N$ Grassmann generators $\zeta^1,\, \zeta^2,\,\ldots \zeta^N$. The most general supernumber is written $$z = z_0+z_i\zeta^i+\textstyle\frac{1}{2!}z_{ij}\zeta^i\zeta^j+\ldots,$$ where $z_i$, $z_{ij}$, $\ldots$, are complex-valued and antisymmetric. The odd part of this is anticommuting and is used to describe Fermion fields. The book says that for finite $N$, it takes $2^{N-1}$ complex numbers to specify an anticommuting number.

How do I figure out how many Grassmann generators $\zeta^i$ I need to specify a Dirac spinor in 4 dimensions.

Answer

For the Dirac field you need an uncountable infinite number of Grassmann generators. What we want is the following property $$ \psi_\mu(x)\psi_\nu(x')=-\psi_\nu(x')\psi_\mu(x)$$ for any $x,x',\mu$ and $\nu$, where $\psi_\mu(x)\psi_\nu(x')=0$ only for $x=x'$ and $\mu=\nu$. You can clearly not achieve this using a finite number of Grassmann generators.

Generate a Grassmann algebra with a generator for each point $x$ and component $\mu$ $$ \Omega(\mathcal M) = \langle\zeta_\mu(x)|x\in\mathcal M,\mu=1,\dots n\rangle,$$ where the notation $\langle a,b,c,\dots\rangle$ stands for a Grassmann algebra generated by elements $a,b,c,\dots$ and $\mathcal M$ is the manifold your spinor lives on. A general element of this space is thus $z\in\Omega(\mathcal M)$: $$ z= z_0 + \int_{\mathcal M}\text dx\; z_\mu(x)\zeta_\mu(x) + \frac 1{2!}\int_{\mathcal M}\text dx\text dy\; z_{\mu,\nu}(x,y)\zeta_\mu(x)\zeta_\nu(y) +\dots.$$

The spinor $\psi_\mu(x)$ then lives in the subspace $\Omega_1(\mathcal M)\subset\Omega(\mathcal M)$ with only one generator. In other words $$ \psi_\mu(x) = z_\mu(x)\zeta_\mu(x) $$ for some complex number $z_\mu(x)\in\mathbb C$. This will ensure the property we want. For conjugate field $\bar{\psi}_{\mu}(x)$, you need to extend the algebra with a new set of generators $\bar \zeta_\mu(x)$ and similarly extend the algebra for each new set of fermions in the theory.