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Jun 2, 2011 at 11:02 comment added Spiro Karigiannis @Benjamin: I think you said it best yourself, that you should think of spinors as "square roots" of classical geometry. There is another notion from calibrated geometry that explains how a calibration (such as the Kahler form) is a "square" of a spinor (see the book by Reese Harvey). Also, manifolds which admit spinor fields that satisfy a nice differential equation (harmonic spinors, Killing spinors, etc) tend to have nice properties on their Riemannian metrics (Ricci-flat, Einstein, etc)
Jun 2, 2011 at 4:57 comment added Benjamin I would like to add that spinor geometry plays a very important role in the Atiyah-Singer index theorem (I could make this statement more precise), so perhaps this gives other people a lead to my question, since a lot of spinor geometry is used in this theorem.
Jun 2, 2011 at 4:52 comment added Benjamin @r0b0t: Thanks, I forgot this! To define the square root of the Laplace-Beltrami operator I need spinors to get the Dirac operator. More generally my teacher always said that spinors are the "square root" of usual geometry. I also know that Cartan thought like this, and it seems very natural to him.
Jun 2, 2011 at 4:42 vote accept Benjamin
Jun 2, 2011 at 4:58
Jun 1, 2011 at 23:51 history answered Vít Tuček CC BY-SA 3.0