Let \(\eta^{a}\) be the basis of 1-forma so that \(\{\eta^{a},\eta^{b}\} = 0\). Let \(\iota_{a}\) be the dual basis such that \(\iota_{b}(\eta^{a}) = \delta_{b}^{a}\) and \(\{ \iota_{a} ,\iota_{b} \} =0\). I half suspect that \(1 \equiv \eta^{a}\iota_{a}\) but I'm probably being thick in that I can't prove it. Can someone either confirm or falsify this for me please?
when you wrote (na,nb)=0 did you mean to put that comma there making them corrdinates or no you meant (nanb) (I know the a and b powers I just can't write that) that is all I can read and actually my comma question doesn't matter because either way n=0
You've missed the point. Curly brackets indicate an anti-commutator, \(\{a,b \} = ab + ba\). The small letters that you see are summation indicies, so what AN really means by \(\eta^a \iota_a \) is \(\sum_a \eta^a \iota_a \). In physics when and index is repeated it is summed over. AN. Doesn't your condition \(\eta^a \iota_b = \delta_b^a \) imply that \(\eta^a \iota_a = d\) where d is the number of spacetime dimensions?
It's the other way around, \(\iota_{a}(\eta^{b})\). Strictly speaking the Clifford algebra Clif(d,d) is defined by the non-vanishing terms \({ \eta^{a},\iota_{b} } = delta^{a}_{b}\). Acting \(\eta^{1}\iota_{1}\) on \(eta^{a_{1}\ldots a_{p}}\), ie \(\eta^{1}\iota_{1}(eta^{a_{1}\ldots a_{p}})\) gives 0 if \(a_{n} \neq 1\) and \(\pm eta^{a_{1}\ldots a_{p}})\) otherwise. I get the feeling if you sum over all \(\eta^{a}\iota_{a}\) you end up with something nice but whether its 1 or not I'm not able to prove either way. Science man, these are not coordinates or numbers, they are something different. Have a Wiki for Grassmann numbers.
This is a recommended paper for students of Clifford Algebras and the connection to gauge theories (published by some Cambridge researchers). http://arxiv.org/abs/gr-qc/0405033 I think section 2.2 is relevant to AN's question (the forms of the equations look similar), but I'm a total noob here so please do not take my word for it.
There is more structure here than just a vector space, so there's an awful lot of freedom. You have a Clifford algebra generated by (V,g) (some vector space V and a quadratic form g), and you're at liberty to consider the associated Clifford algebra generated by (V*,g), but V and V* are going to be isomorphic in the finite dimensional case, so there's no additional structure. But remember here the "dual" aspect of the vector space only cares about the vector space - not any additional algebraic structure On a different, perhaps more natural outlook, it should be possible to define the "dual" of the element of a clifford algebra in much the same way as you do in the exterior calculus (i.e. Hoge dual), given the similarities. In fact, my scribbles suggest you can definitely do this. The bonus of this approach gives you an automatic algebraic structure in your "dual space" (in this context).
