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It is well known that for SU(2) $(g^{-1}dg)^3$ can be expressed as a total differential d(...). However in physics literature I could not find similar expressions for the orthogonal groups. Is $(g^{-1}dg)^3$ still a total differential for $g \in SO(p,q)$? If so there any closed formula for (...) in $d(...)=(g^{-1}dg)^3$? Thank you.

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Of course I meant Tr(g^{-1}dg)^3 (in matrix representation) –  Jerzy Kowalski-Glikman Nov 9 '11 at 11:45
You can correct mistakes by clicking on "edit" below the question. Moreover, if you put your TeX code inside dollar signs it will be much easier to read. –  Hans Lundmark Nov 9 '11 at 11:59
@JerzyKowalski-Glikman Are you referring to the fact that for the left invariant form $\omega = \operatorname{Tr}(g^{-1} \mathrm{d} g)$, the form $\omega \wedge \omega \wedge \omega$ is a differential of some other form ? This is a statement about de-Rham cohomology of $SU(2)$. See Nakahara's book. –  Sasha Nov 9 '11 at 13:29

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