Timeline for How would you solve this tantalizing Halmos problem?
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20 events
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| S Jun 29, 2016 at 14:58 | history | suggested | A_S | CC BY-SA 3.0 |
Add some $...
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| Jun 29, 2016 at 14:40 | review | Suggested edits | |||
| S Jun 29, 2016 at 14:58 | |||||
| Jul 14, 2010 at 20:03 | history | edited | Bill Dubuque | CC BY-SA 2.5 |
Pers suggestion, reformulated so first couple lines show problem; deleted 11 characters in body
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| Jul 14, 2010 at 18:18 | comment | added | T.. | @Bill, yes, I believe that the relation to formal languages, rational automata, etc as studied by Schutzenberger and his collaborators was surmised and to some extent worked out, long before 1980. | |
| Jul 14, 2010 at 14:43 | comment | added | Bill Dubuque | @Richard: Searching on Krob reveals that it's part of Exercise 6.64 in v2 of Richard Stanley's book Enumerative Combinatorics, for which he refers to a 1960 paper of Schutzenberger. So perhaps the problem was known to some combinatorists long before Halmos popularized it in the Intelligencer, and before I may have discussed it with Rota (both circa '81). Richard: do you happen to recall any other approaches, perhaps from discussions with Rota around that time? If memory serves correct there are a couple other approaches that we have yet to discuss here - one with a model-theoretic flavor. | |
| Jul 14, 2010 at 14:41 | comment | added | Bill Dubuque | @Victor: I originally had a much more specific title in mind but on second thought I realized that it might make the problem look so elementary that it might not attract broad interest. One reason I asked the question is that I am trying to reconstruct the handful of approaches I knew a few decades ago as an MIT student (alas, my notes are long lost). I don't recall if I explicitly discussed it with Rota, but, if so, perhaps there was some discussion among combinatorists at the time and perhaps Richard Stanley's memory might be better than mine. | |
| Jul 14, 2010 at 4:43 | comment | added | Victor Protsak | Bill: Neither the title nor the first few lines of the question visible from the "questions" tab give any clue as to its content (in particular, "old chestnut" implied it's an unsolved elementary puzzle). It's sheer luck that I happened to read it at all. May I suggest using more descriptive titles in the future? | |
| Jul 14, 2010 at 4:27 | answer | added | Victor Protsak | timeline score: 81 | |
| Jul 13, 2010 at 1:37 | comment | added | Bill Dubuque | Many thanks to Richard Stanley for the links to Krob's 1991 paper. One approach I found circa '81 was based on earlier ('65-'75) work on rational identities etc by Amitsur, Bergman, and Cohn - which work I stumbled upon thanks to a reference from Rota on a related topic. As luck would have it that was around the same time '81 that Halmos's paper appeared, so the application was obvious. I haven't seen Krob's more recent work. Does anyone happen to know how it compares to said earlier work? | |
| Jul 12, 2010 at 21:31 | comment | added | Richard Stanley | The theorem of Krob I mention above is roughly the following. Any rational identity that holds in every ring (with identity) is "trivial", i.e., an algebraic consequence of $(1-a)^{-1}=1+a+a^2+\cdots$, and conversely. A nice example is $$ (1+x)(1-yx)^{-1}(1+y) = (1+y)(1-xy)^{-1}(1+x). $$ It is clearly true if we can expand $(1-yx)^{-1}=1+yx+(yx)^2+\cdots$ and similarly for $(1-xy)^{-1}$. Therefore it holds in every ring. (Of course it is assumed that the inverses exist.) It is rather tricky to prove this identity from scratch. | |
| Jul 12, 2010 at 21:05 | comment | added | Qiaochu Yuan | @Yemon: whoops, my apologies. I was thinking of a different and easier problem. Also, thanks for the counterexample. | |
| Jul 12, 2010 at 20:34 | comment | added | Yemon Choi | Also, I don't quite follow the assertion that "the power-series method is valid in any Banach algebra". It might be worth remarking here that we can have an invertible element $u$ in a Banach algebra such that the power series $\sum_{n\geq 0} (1-u)^n$ fails miserably to converge... | |
| Jul 12, 2010 at 20:31 | comment | added | Yemon Choi | @Qiaochu: I'm not quite sure what hypotheses are being put on your ring, so apologies if this comment misses the point: but it is well known that there are certain identities which cannot hold in Banach algebras but do hold in arbitrary complex algebras. The standard example I was given is $qp-pq=I$. | |
| Jul 12, 2010 at 20:13 | history | edited | Bill Dubuque |
edited tags
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| Jul 12, 2010 at 19:19 | answer | added | T.. | timeline score: 19 | |
| Jul 12, 2010 at 19:18 | comment | added | Richard Stanley | I believe that the paper to which Qiaochu refers is D. Krob, in Topics in invariant theory, (M.-P. Malliavin, ed.), Lecture Notes in Math., vol. 1478, Springer-Verlag, 1991, pp.215-243. A short discussion also appears in Section 8 of C. Reutenauer, A survey of noncommutative rational series, in Formal Power Series and Algebraic Combinatorics (New Brunswick, NJ, 1994), DIMACS Series Discrete Math. Theoret. Comput. Sci. 24, American Mathematical Society, 1996, pp. 159-169. | |
| Jul 12, 2010 at 19:15 | comment | added | Qiaochu Yuan | Question: it's not hard to see that the power-series method is valid in any Banach algebra. Suppose I take the subring of any ring consisting of any expression in 1, a, b, and inverses if they exist. Does this ring always embed into a Banach algebra? | |
| Jul 12, 2010 at 18:30 | history | edited | Bill Dubuque | CC BY-SA 2.5 |
added 35 characters in body
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| Jul 12, 2010 at 18:28 | comment | added | Qiaochu Yuan | There is a theorem to the effect that if you can prove a non-commutative identity by power series, it is true in an abstract ring. But I can never remember the reference. (My intuition is that this is analogous to proving that a polynomial identity holds over C via an analytical or topological argument to show that it must hold identically.) | |
| Jul 12, 2010 at 18:25 | history | asked | Bill Dubuque | CC BY-SA 2.5 |