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Today I was reading Jean-Louis Loday's classic paper, "Realization of the Stasheff polytope", in which he produces a simple and very pretty realization of the associahedra as convex polytopes. He introduces the permutohedra like so:

Let us recall the definition of the permutohedron (alias zylchgon).

I had not heard this alternate name for permutohedra before, and tried to figure out the origin, but couldn't find anything. Sadly, Loday is no longer alive, so I can't ask him. Does anyone know the origin of "zylchgon"?

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  • $\begingroup$ This is almost assuredly not the reason, but: the Chow polytope of a 0-dimensional variety (i.e., a set of points) is a Minkowski sum of standard simplices, i.e., a generalized permutohedron. So possibly 0-dimensional = "zilch"? $\endgroup$
    – Sam Hopkins
    Commented Oct 19, 2018 at 18:11
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    $\begingroup$ Aha: it seems to have originated in section 6 of Carlsson and Milgram's chapter, "Stable homotopy and iterated loop spaces", maths.ed.ac.uk/~v1ranick/papers/carlmilg.pdf . "We now introduce a family of combinatorial cells which do just this, the Zilchgons, (also called permutahedra by combinatorialists), $C(n)$." I'm still not sure where the "zilchgon" terminology comes from. $\endgroup$
    – Nathaniel Bottman
    Commented Oct 19, 2018 at 18:25
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    $\begingroup$ in physics the "zilch current" is a concept from electromagnetism, so called because it has "zilch" (zero) physical significance; the name originated in a 1964 paper (as discussed here on page 241) --- who knows, it may have inspired Milgram... $\endgroup$
    – Carlo Beenakker
    Commented Oct 19, 2018 at 20:07
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    $\begingroup$ incidentally, the first instance of "zilchgon" I have found is a 1974 paper by Milgram, Unstable Homotopy from the Stable Point of View . Milgram cites an earlier 1966 paper as the origin of the concept, but I did not find the name there. $\endgroup$
    – Carlo Beenakker
    Commented Oct 19, 2018 at 20:18
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    $\begingroup$ Hi Nate! Have you considered emailing Gunnar? Sometimes he writes back, and he's as likely to know the answer as anyone else: math.stanford.edu/~gunnar $\endgroup$
    – Vidit Nanda
    Commented Oct 19, 2018 at 22:36

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