Wow.. beautiful.. How on earth is idempotency equivalent to vector bundles?
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The idempotent arises since a vector bundle (over interesting spaces, like CW complexes) is a direct summand of a trivial bundle. The projection map from the trivial bundle unto V is an idempotent. (Bundles are finite dimensional here.)
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Ok, so this allows us to go from vector bundles to idempotents, but how do you create a vector bundle from an arbitrary idempotent in a ring?
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the projection on the base space? I'm also curious
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So an indempotent q^2=q generates the non invertible left/right ideals from which base vectors of the algebra can be extracted. Are you saying something about these being the frames basis of the vector bundle associated to the PFB?
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Plebanski's bivector chiral formulation of gravity (basis of anti-self dual two-forms Σ_AB) uses Hodge dual ∗ as an indempotent algebraic structure splitting local algebra of the complexified SO(1, 3)C→SO(4, C) gauge bundle over space-time, M into left and right-handed idealspic.twitter.com/bxDV8D8JHX
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