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In proving the Cartan-Brauer-Hua theorem, Hua uses an obscure identity. That is: $$ a=(b-(a-1)^{-1}b(a-1))(a^{-1}ba-(a-1)^{-1}b(a-1))^{-1} $$ for $a$ and $b$ such that $ab \neq ba$. All these elements belong to a division ring. This identity confused me.I want to know some motivation or intuition about this identity. Thanks for any help.

yuan
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    Can you provide some more detail about what kind of objects $a$ and $b$ are? If everything commutes, this is not a well-defined statement. Otherwise it may be but it is still unclear which operations are permitted. – quarague Jan 28 '19 at 10:33
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    Conjugation in a group is not "obscure". – Dietrich Burde Jan 28 '19 at 10:48
  • I think whatever this identity is, it's probably equivalent to Hua's identity? If so, you're re-asking this question of mine, just with a different formulation of the identity. – rschwieb Jan 28 '19 at 12:07
  • @rschwieb,does this identity could be implied by Hua's identity? – yuan Jan 28 '19 at 12:13
  • @rschwieb I think Hua's identity could be deduced from $(1-ba)^{-1}=b(1-ab)^{-1}a+1$. This identity could be described that if $1-ab$ is invertible then so is $1-ba$. But I don't know why he choose the initial form. – yuan Jan 28 '19 at 12:33
  • @yuan It's been a long time since I looked into it, but I do think I remember Hua using a different version of the identity later attributed to him. I think later authors simplified it to what I wrote. I haven't attempted to prove their equivalence yet... – rschwieb Jan 28 '19 at 14:28
  • @yuan The reason I asked my question is I thought it is a little distasteful to juggle identities without clues to why they should work. Unfortunately, I haven't been able to rectify this much :( – rschwieb Jan 28 '19 at 14:44
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    I have added an answer to this question in https://math.stackexchange.com/questions/1602573/what-was-the-genesis-of-huas-identity – Jose Brox Jan 28 '19 at 16:16
  • Related (a proof of the identity) – rschwieb Dec 06 '20 at 13:46

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