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.]

  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ @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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.