I'm reading a few papers which deal with spinors and p-forms and they keep using a notation I've not seen before and I'm wondering if anyone else has come across it. If M is a 2n dimensional space then you can define the p-form spaces \(\Lambda^{p}(M)\) in the usual way from the wedge product of elements in T*M. Fine. However, the papers then make p-form valued spinors by considering something like \(\Lambda^{p}(M) \otimes \sqrt{\Lambda^{n}(M)}\). Note that the term in the square root is an n-form and M is 2n dimensional, which seems to mean something as they don't consider general p-forms inside the root but I'm not sure what. Is this some short hand for spinor, in that their phase transformations are only half of what you'd expect from a vector or p-form and hence the square root, or is there something more to it? Why only consider \(\Lambda^{n}(M)\) for M being 2n dimensional? Is this because if you're splitting a spin bundle \(S \to S^{\pm}\) via some kind of chirality or whatnot when you consider spinors? Any of this ringing any bells with anyone?
No takers I see.... For what it is worth I was mistaken about the particular expression, it is a spin bundle decomposition of the form : \(S^{\pm} = \Lambda^{\pm}V* \otimes \sqrt{\Lambda^{n}V}\) where \(\Lambda^{\pm}V*\) are the sum of even (for +) and odd (for -) \(\Lambda^{p}V*\) and you can regard V as n dimensional, not 2n dimensional. That makes the 1 dimensional nature of \(\sqrt{\Lambda^{n}V}\) clearer. The most relevant paper is http://arxiv.org/abs/math.DG/0209099 , specifically Section 3.2. Another paper, http://arxiv.org/abs/0807.4527 (Section 2.3) talks about making \(\Lambda^{\pm}V* \otimes \sqrt{\Lambda^{n}V}\) into something of the form \( \Lambda^{\pm}V* \otimes L\), where L is a line bundle fibre. If V is n dimensional then \(\Lambda^{n}V\) is obviously 1 dimensional and is akin to some kind of volume form, right? Is the root there to provide it with spinor transformation properties since otherwise it has vector transformation properties, ie \(e^{a} \to \Lambda^{a}_{b}e^{b}\) implies the volume form goes as \(\mu_{vol} \to Det(\Lambda) \mu_{vol}\)?
