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In the case of the $(1, -1, -1, -1)$ metric, a bivector (or a 2-form) $F$ decomposes uniquely as the sum of two orthogonal simple bivectors if $F^2 \ne 0$.

I have the impression that it is very little known. Am I wrong?

References found:

  • Clifford Numbers and Spinors, Marcel Riesz, Springer, 1993, p. 206. The corresponding lecture, given in 1959, was edited by E. Folke Bolinder and Pertti Lounesto.

  • On the canonical form of the electromagnetic field, Luigi Stazi, Annali dell’Università di Ferrara 52: 127–135 (2006), doi:10.1007/s11565-006-0011-8

Strangely, the editor of Riesz's book only considers the euclidean case, completely different and less interesting, in his book

  • Clifford Algebra and Spinors, 2nd ed, Pertti Lounesto, Cambridge Univ. Press, 2001, p. 87.
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  • $\begingroup$ 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). $\endgroup$
    – Willie Wong
    Commented Dec 8, 2021 at 14:50
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    $\begingroup$ @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? $\endgroup$
    – AlexArvanitakis
    Commented Dec 8, 2021 at 18:02

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