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Dec 16, 2021 at 16:57 history closed Johannes Hahn
LeechLattice
Ben McKay
coudy
Dima Pasechnik
Needs details or clarity
Dec 8, 2021 at 18:02 comment added AlexArvanitakis @WillieWong: I think F^2 means F\wedge F. Hence $dx\wedge dy$ has $F^2=0$. But also what is the question? Is it whether the claim is true (seems so, it's been referenced) or whether it's obscure?
Dec 8, 2021 at 14:50 comment added Willie Wong What do you mean by $F^2$? And can you clarify your statement? For example, the two-form $dx\wedge dy$ has Lorentzian inner product with itself non-zero, but it is not the sum of two orthogonal simple bivectors (unless you admit the 0 bivector as simple).
Dec 8, 2021 at 13:55 history edited Amir Sagiv CC BY-SA 4.0
tex edit
Nov 25, 2021 at 16:25 review Close votes
Dec 16, 2021 at 16:57
S Nov 25, 2021 at 15:35 review First questions
Nov 25, 2021 at 16:01
S Nov 25, 2021 at 15:35 history asked Fabrice Pardo CC BY-SA 4.0