9

Many resources I have read prove Hua's identity more-or-less mechanically. I have seen there is more than one raison d'être for Hua's identity: e.g. its connection to the fundamental theorem of projective geometry, and also Jordan algebra theory. My impression, though, is that these two things are mostly application rather than inspiration. (I could be wrong, though.)

I would very much like to know how Hua's identity arose, hopefully with motivation/intuition as to how it was discovered.

I have intended to get ahold of the(?) original proof by Hua in hopes that it contained such information, but so far I haven't managed to lay my hands on the original citation(s). This would be a much-appreciated bonus to any solution.

If it turns out there is a good retroactive motivation/intuition for deriving the identity that beats the original, of course that would be welcome as well.

Happily I've seen the original paper now (thanks Martin). Surprisingly, the identity cited by all authors since the paper is different-looking from the original. I will have to compare the two versions and see if this version gives any more insight. No direct intuition about its origins are apparent, and indeed it is called "nearly trivial" although it seems a bit mystifying, IMO.

Jose Brox
  • 5,216
rschwieb
  • 160,592
  • Interesting question. The therem and its proof in Hua's paper "On the Automorphisms of a sfield" do not use "Hua's identity" from Wikipedia; also the theorem is different: Hua assumes $\sigma(aba)=\sigma(a)\sigma(b)\sigma(a)$, not just $\sigma(a^{-1})=\sigma(a)^{-1}$. Could it be that Cohn generalized Hua's Theorem and used the same name? By the way, the paper "Some properties of a sfield" by Hua contains another mysterious "trivial" identity. – Martin Brandenburg Jan 06 '16 at 22:01
  • 1
    It seems that Cohn has copied Artin's proof from "Geometric algebra" (1957), pp. 37-40. There, Artin writes "Hua has discovered a beautiful theorem ... ", without giving a reference. He also gives a motivated proof of "Hua's identity". Did you already see this? – Martin Brandenburg Jan 06 '16 at 22:12
  • @MartinBrandenburg Thank you very much for the original citation. I don't understand the advice about not using the Wiki version: the identity is exactly the same as the other five major authors specify, Artin included. Please understand that my question is about the identity and not the theorem. – rschwieb Jan 07 '16 at 03:34
  • @MartinBrandenburg Yes, the discussion in Geometric algebra is one of my resources. It is a pristine example of the problem I'm describing with most proofs of Hua's identity being unmotivated. It literally says "first we establish an identity" and then mechanically computes it. I'm interested in knowing how one would have even thought of it. Hopefully the answer is not just "because that's what you need to prove Hua's theorem..." I agree the theorem is better motivated there, though. – rschwieb Jan 07 '16 at 03:38
  • Hm, have you found a publication by Hua where "Hua's identity" is mentioned? – Martin Brandenburg Jan 07 '16 at 08:31
  • @MartinBrandenburg Few authors are so presumptuous, don't you think? No, and I did not expect to find it under that name from Hua himself. It is entirely possible someone like Jacobson or Artin slimmed down the original identity and the name has been a misattribution since. – rschwieb Jan 07 '16 at 11:10
  • @MartinBrandenburg Not so far: my access to his work is limited. – rschwieb Jan 07 '16 at 14:30
  • 1
    There is now a similar question on MathOverflow for anyone interested. – rschwieb Sep 11 '17 at 13:28
  • Someone else asked again – rschwieb Jan 28 '19 at 12:09
  • @rschwieb I have added some information on the MathOverflow question, may it be useful! – Jose Brox Jan 28 '19 at 12:25
  • @JoseBrox Cool, thanks. I still don't have intuition for it, but that's understandable since I have not thought about the area where it seems to be most relevant. – rschwieb Jan 28 '19 at 14:26
  • 1
    @rschwieb I have added an answer here motivating the development of the identity, instead of the necessity of finding it. – Jose Brox Jan 28 '19 at 16:12

1 Answers1

9

Well, I can only guess, but if I were Hua this is what I would have thought to derive the identity and say it is "almost trivial":

1) We want to generalize the theorem of Cartan and Dieudonné (now called Cartan-Brauer-Hua), so we want to express an element $a$ as a combination of sums, products and inverses of conjugates of $a,b$, for any other $b$ (such that $ab\neq ba$).

2) Immediately we think about proving our luck with the well known formula for multiplicative commutators in division rings, since it can readily give us $a$ as a factor so that we can solve it "as a fraction", and it is close to conjugations: If we denote $(a,b):=a^{-1}b^{-1}ab$ then $$a(a,b)=(a-1)(a-1,b)+1.$$

Therefore $$a((a,b)-(a-1,b))=1-(a-1,b).$$

3) If we expand we see we don't yet have conjugates due to the $b$ factors at the right of $(a,b)=a^{-1}b^{-1}ab$, etc.; but since it is the same for all terms, we add a right $b^{-1}$ to solve the problem, and get

$$a((a,b)-(a-1,b))b^{-1}=(1-(a-1,b))b^{-1}$$ $$a(a^{-1}b^{-1}a-(a-1)^{-1}b^{-1}(a-1))=b^{-1}-(a-1)^{-1}b^{-1}(a-1)$$

and now Hua's identity follows by solving for $a$.

Jose Brox
  • 5,216
  • 1
    Ah, there we go, that's more like what I was looking for. I didn't realize the connection with the Cartan–Dieudonné theorem. I guess if one is has blinders on for division rings, that would not be obvious, but since it works much more generally, that's a good observation. – rschwieb Jan 28 '19 at 16:42
  • Incidentally, have you ever heard of the generalization of the Cartan-Diudonné theorem to linear-fractional transformations? I know one source, but the curious thing is that I can't seem to find other references for it... – rschwieb Jan 28 '19 at 16:50
  • @rschwieb LOL, what I meant by Cartan-Dieudonné in this context was not the theorem on reflections, but the same as Hua in his paper, now known as the Cartan-Brauer-Hua theorem, which is the one that Hua wanted to generalize! But I suppose there is a connection (at least in $\mathbb{R}^3$) through quaternions... Is that what you are thinking? I don't know about the generalization to Möbius transformations, sorry... but sounds interesting! – Jose Brox Jan 28 '19 at 16:59
  • Yes, you have it now. I had no idea you meant the CBH theorem. I'm not sure about the connection to quaternions you're referring to. My original interpretation of what you said was to express something with more operators than just the product, which I thought was interesting in itself. – rschwieb Jan 28 '19 at 17:01