20 October 2007

Homophony part two

As at least two readers have pointed out, my claim that the triviality of the homophony group was part of the unwritten folklore is false. Take a look at the paper Quotients Homophones des Groupes Libres/Homophonic Quotients of Free Groups, by Mestre, Schoof, Washington, and Zagier. They use a lot of the same relations that I did in my proof; in particular damn = damn, damned = dammed, barred = bard, bass = base, jeans = genes, ruff = rough, phase = faze. (The last three of these seem to me to be the ones that most people will find.) Like my proof, v is the last generator to fall; they use chivvy = chivy or leitmotif = leitmotiv. I'm not sure how I feel about these; in both cases these seem like alternate spellings, not different words. veldt = felt, as suggested by a commenter on the earlier entry, feels "right" to me; these are quite clearly two different words with different meanings, veldt being a certain kind of open space in Africa, felt being the stuff out of which the tips of markers are made. (I think this is the one I came up with the first time I saw this problem.) They don't seem satisfied with their proofs for v, either; they introduce the space as a 27th generator, and use avowers = of ours to show the triviality of v.

The paper in question is bilingual; the proof that the English homophony group is trivial is given in French, and the proof that the French homophony group is trivial is given in English. I particularly like the acknowledgements:

The third and fourth authors were partially supported by NSF and (by the results of this paper) numerous other government agencies.

A related problem is as follows: consider the group on twenty-six letters A, B, ..., Z with the relations that two words are equivalent if they are permutations of each other and each appears as a word in some dictionary of choice. Identify the center of the group. According to Steven E. Landsburg, "The Jimmy's Book", The American Mathematical Monthly, Vol. 93, No. 8 (Oct., 1986), pp. 636-638, much work was done on this problem at Chicago in the seventies.

To get started: post = pots = stop = opts = spot = tops. Since post = pots, s and t commute; since pots = opts, o and p commute. This is clearly much harder than the homophony group problem. By the way, elation = toenail. (I've been playing lots of Scrabble recently; that set of seven letters seems to come up a lot.)

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