Link between the Grassmann algebra and spinors What is the exact link between spinors and the Grassmann algebra? I'm pretty sure there's one, based on the following: 


*

*The Berezin integral in path integrals is done over the Grassmann algebra of $\Bbb C$

*There's a mapping from the tangent bundle of a supermanifold (which is locally $\Bbb R^n \times $ the Grassmann algebra), and the spinor bundle of a spin manifold

*On the other hand, spinors are defined just as the vector space associated to a representation of the spin group.


I'm not quite sure what the relation is. If we deal with spinors larger than basic Dirac spinors $\Bbb C^2$ (like a product of two spinors), are they automatically described by a member of the Grassmann algebra? Can it be shown that only those transform properly under the spin group?
 A: There is no direct link between the Graßmann algebra and spinors. The reason both are connected in physics is the spin-statistic theorem: The fields/operators associated to fermionic objects, i.e. Dirac spinors, anticommute, which is basically the defining property of the Graßmann algebra.
The algebra that actually controls the spinor representation - whose irreducible representation actually essentially defines what a Dirac spinor is - is the Clifford algebra, not the Graßmann algebra. Quantization links the two, see my and David Bar Moshe's answers to this question. The gist is that the defining relation of the Graßmann algebra is $\theta^2 = 0$, while that of the Clifford algebra is $\theta^2 = \hbar \eta(\theta,\theta)$ for the metric $\eta$. For a "fermionic"/anticommuting classical variable, the $\theta^2$ is the super-Poisson bracket. So precisely like the quantization of bosonic variables deforms the Poisson bracket by factors of order $\hbar$ and higher, the quantization of fermionic variables deforms the anticommutator, and consequently the Graßmann algebra into the Clifford algebra.
