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Nov 17, 2015 at 17:34 vote accept Saal Hardali
Nov 17, 2015 at 15:28 comment added Saal Hardali That's enough for me to give up the digging for now. Thanks!
Nov 17, 2015 at 15:27 vote accept Saal Hardali
Nov 17, 2015 at 15:48
Nov 17, 2015 at 15:26 comment added Denis Nardin Honestly I don't know. I never found those pictures of the exterior powers convincing (expecially because the grassmannian is only a subset of $\mathbb{P}\Lambda^kM$ and not the whole thing) so I'm probably the wrong person to ask.
Nov 17, 2015 at 15:23 comment added Saal Hardali I see your point. I was searching for a geometrical picture for why the determinant is involved. Top exterior powers of finitely generated sub modules corresponds locally to tangent hyperplanes and the transformation between them induced by the orthogonal projection using the bilinear form is preciesly the determinant which is a very reasonable generalization since it corresponds to the volume of a projected unit box. Just as the inner product correspods to the length of a projected unit vector. Does this sound right to you?
Nov 17, 2015 at 15:20 history edited Denis Nardin CC BY-SA 3.0
Reordered the argument so to make it more clear
Nov 17, 2015 at 15:12 comment added Denis Nardin I think that the definitions are the same, but I always have an hard time parsing everything with more than one index, so don't trust me too much on this. But I do not understand in which sense my definition is local. I only localize to verify that the pairing is perfect, all the rest is done globally. I guess that my point is that you do not need all those messy computations at all: everything is a simple algebraic fact.
Nov 17, 2015 at 15:10 comment added Saal Hardali My global constructions via maps between exterior powers was motivated by this precise definition. Since both of these agree locally (their localizations agree) Doesn't it mean that my geometric definition is the same as your "local" one?
Nov 17, 2015 at 15:06 history answered Denis Nardin CC BY-SA 3.0