I will denote by $\Lambda$ the ring of symmetric functions, and by $\mathrm{QSym}$ the ring of quasisymmetric functions (both in infinitely many variables $x_1$, $x_2$, $x_3$, ..., both over $\mathbb Z$). See Victor Reiner, Hopf algebras in combinatorics, chapters 2 and 5, respectively, for the definitions.

It is known that one can define a plethysm $f\circ g \in \mathrm{QSym}$ for any $f\in\Lambda$ and any $g\in\mathrm{QSym}$. This is described, e. g., in Claudia Malvenuto, Christophe Reutenauer, Plethysm and conjugation of quasi-symmetric functions, Discrete Mathematics 193, 225-233 (1998). In a nutshell, if $g$ is a sum of monomials in the $x_1$, $x_2$, $x_3$, ..., one can construct $f\circ g$ by substituting these monomials as indeterminates into $f$. It takes some more work (and is less intuitive) to define $f\circ g$ when $g$ has negative coefficients, but the above should give some feeling for what $f\circ g$ is. Notice that $e_1\circ g = g$ (where $e_1$ is the $1$-st elementary symmetric function). For $g \in \Lambda$, the plethysm $f\circ g$ becomes the usual plethysm in $\Lambda$.

In section 3 of the preprint Michiel Hazewinkel, Explicit generators for the ring of quasisymmetric functions over the integers, it is claimed that this construction extends to all $f\in \mathrm{QSym}$, where the monomials are substituted into $f$ in lexicographic order. I cannot follow this claim, because it seems to me that the "addition formula"

(1) $f \circ \left(g+h\right) = \sum\limits_{(f)} \left(f_{(1)}\circ g\right) \left(f_{(2)}\circ h\right)$ (using Sweedler notation, where $\sum\limits_{(f)} f_{(1)} \otimes f_{(2)}$ is the first coproduct of $f$)

is no longer satisfied for general $f\in\mathrm{QSym}$, and the definition of $f\circ g$ outside the case of $g$ being a sum of monomials hinges on this formula (of course, there are better definitions in the $f \in \Lambda$ case which don't depend on this formula, but they don't look generalizable at all).

What I want to know, apart from whether or not my doubts on this definition are justified, is whether there is any reasonable definition of a plethysm of two elements of $\mathrm{QSym}$ known, or whether there are good reasons no such beast exists in nature.

[EDIT: At a second glance, if we take Hazewinkel literally, he isn't claiming this all; he is only defining $f\circ g$ for $f$ and $g$ being monomial quasisymmetric functions, which (I guess) he can do as he pleases. But I think he is trying to define $f\circ g$ for all $f\in\mathrm{QSym}$ and $g\in\mathrm{QSym}$. If I interpret his definition of $f\circ g$ as being only formulated for the monomial quasisymmetric functions, and then try to extend it using (1) to all $f\in\mathrm{QSym}$ and $g\in\mathrm{QSym}$, then I think I obtain a contradiction due to the non-cocommutativity of $\mathrm{QSym}$.]

[Let me remark that Hazewinkel's proof of the polynomial freeness of $\mathrm{QSym}$ does not depend on this kind of plethysm. He only ever uses it for $f\in\Lambda$. A clean version of his proof can be found in Chapter 6 of Michiel Hazewinkel, Nadiya Gubareni, V.V. Kirichenko, Algebras, Rings and Modules, Volume 3, and the only gap in it (the unproven footnote 13) can be filled in using Section 2 of David E. Radford, A natural ring basis for the shuffle algebra and an application to group schemes.]

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    $\begingroup$ Can you give an example where the addition formula doesn't seem to hold? $\endgroup$ May 1, 2013 at 0:18
  • $\begingroup$ Try the simplest example: $f = \sum\limits_{i < j} x_i^2 x_j$ and $g = h = x_1 + x_2 + x_3 + ...$. Then, the left hand side of the addition formula is $4 \sum\limits_{i < j} x_i^2 x_j + \sum\limits_i x_i^3$, while the right hand side is $3 \sum\limits_{i < j} x_i^2 x_j + \sum\limits_{i < j} x_i x_j^2 + \sum\limits_i x_i^3$. The difference stems from the fact that the order in which the monomials are substituted into $f$ matters (I'll now clarify this in the OP). $\endgroup$ May 1, 2013 at 0:34
  • $\begingroup$ Well, $f\circ(g+h)=\sum(f_{(1)}\circ g)(h\circ f_{(2)}$ fixes this, and seems like a more natural formula anyway. $\endgroup$
    – David Hill
    May 1, 2013 at 0:50
  • $\begingroup$ Hmm. Setting $h=0$, we want the right hand side to be $f\circ g$. But $0\circ u=0$ for every $u$... $\endgroup$ May 1, 2013 at 0:55
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    $\begingroup$ @darij I believe you want to interpret $g + h$ as $x_1 + x_2 + \cdots + x_1 + x_2 + \cdots$ even though the total order on the monomials occuring in $g$ and $h$ suggests that one should write $g+h = x_1 + x_1 + x_2 + x_2 + \cdots$. You might want to look at the 2011 J. Alg. Comb. paper by Loehr and Remmel. They take some care in discussing when plethysm (for $f$ symmetric) can be thought of as substitution and when you really need to work with the ring homomorphisms determined by $g$ and $h$ $\endgroup$ May 1, 2013 at 2:51

1 Answer 1


As per @darij's suggestion regarding my above comment: Loehr and Remmel's A computational and combinatorial exposé of plethystic calculus is a good resource for a combinatorial point of view of plethysm; I think it clarifies what one should do when both functions are quasisymmetric. Edit: Apparently the paper is Open Access.


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