I'm looking for a particular description of the Hopf algebra structure on the ring of quasisymmetric functions.

Let me illustrate by giving this kind of description for the Hopf algebra of symmetric functions.

Fix a ground field $k$.

Edit: I'm happy to assume $k=\mathbb{Q}$. As darijgrinberg pointed out in the comments, $\mathrm{QSym}$ only has the Lyndon monomial basis when $\mathrm{char}(k)=0$.

Let $\Lambda$ be the ring of symmetric functions over $k$. Putting a Hopf algebra structure on a $k$-algebra $R$ is the same as putting a group structure on $\mathrm{Hom}_{k-alg}(R,S)$ for each $k$-algebras $S$ such that postcomposition is a group homomorphism. Since $\Lambda$ is a polynomial algebra in the elementary symmetric functions $e_1,e_2,\ldots$, we can identify $\mathrm{Hom}_{k-alg}(\Lambda,S)$ with the set of sequences $(s_1,s_2,\ldots)$ of elements of $S$ (where $f:\Lambda\to S$ is identified with $(f(e_2),f(e_2),\ldots)$). If we further identify the sequence $(s_1,s_2,\ldots)$ with the power series $1+s_1T+s_2T^2+\ldots\in 1+TS[[T]]$, then the group structure (coming from the Hopf algebra structure on $\Lambda$) is multiplication of power series.

I am hoping for something similar for the ring of quasisymmetric functions. Specifically, I would like, for each $k$-algebra $S$, a group $G(S)$ (which won't generally be abelian) and a bijection $\mathrm{Hom}_{k-alg}(\mathrm{QSym}_k,S)\leftrightarrow G(S)$, such that the group multiplication $G(S)\times G(S) \to G(S)$ is induced by the comultiplication on $\mathrm{QSym}$. I'm hoping the description of $G(S)$ is as explicit as possible (like the group $1+TS[[T]]$ in the case of symmetric functions).

It is known that $\mathrm{QSym}$ is a polynomial algebra. One possible free generating set is the monomial quasi-symmetric functions $$ M_{(s_1,\dots,s_k)}:=\sum_{i_1<\ldots<i_k}x_{i_1}^{s_1}\ldots x_{i_k}^{s_k}, $$ where $(s_1,\ldots,s_k)$ runs through the Lyndon words (a Lyndon word is one that is lexicographically smaller than all of its proper right subwords). This means we can identify $\mathrm{Hom}_{k-alg}(\mathrm{QSym},S)$ with the set of set maps $\{\text{Lyndon words}\}\to S$. However, under this identification it looks like a huge mess to describe the group structure, since (as far as I know) it is not easy to read off the comultiplication of $M_{(s_1,\ldots,s_k)}$ as a sum of simple tensors of other monomial quasisymmetric functions indexed by Lyndon words.

  • $\begingroup$ Might it perhaps be somewhat easier to consider QSymm's dual NSymm which classifies Hasse-Schmidt derivations? I'm not sure. $\endgroup$ – Jonathan Beardsley Aug 31 '13 at 3:54
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    $\begingroup$ The monomial quasisymmetric functions of the Lyndon words are not in general a free generating set unless $\mathrm{char} k = 0$. To get a free generating set over any commutative ring, you need a more complicated construction, such as the one done in arXiv:math/0410366v1 by Hazewinkel. You can play around with Hazewinkel's basis (or one of his bases, in fact) in Sage since patch #15131 ( trac.sagemath.org/ticket/15131 ), but I have not found any particularly useful patterns while doing so. $\endgroup$ – darij grinberg Aug 31 '13 at 4:10
  • $\begingroup$ @darijgrinberg Oh interesting, I hadn't realized this. The case I'm most interested in is $k=\mathbb{Q}$, so maybe I'll modify the question to ask just about this case $\endgroup$ – Julian Rosen Aug 31 '13 at 4:13
  • $\begingroup$ @JonBeardsley: I think asking for bialgebra homomorphisms from NSymm into a (generally noncommutative) algebra and asking for algebra homomorphism from QSymm into a commutative algebra are two rather different questions. In the case of NSymm, generators are crystal clear. $\endgroup$ – darij grinberg Aug 31 '13 at 4:15

Not sure that it helps but if I'm not mistaken, the $k$-Hopf algebra $Qsym$ you are looking for is the topological dual completed Hopf algebra $k \langle\!\langle y_i , i \geq 1 \rangle\! \rangle$ of power series in an infinity of non commutative variables with coproduct $\Delta_\star (y_n) = \sum_{p+q = n} y_p \otimes y_q$ (with $y_0 = 1$).

As a result, $Hom_{k\textrm{-alg}}(QSym,A)$ is identified with diagonal power series $\Phi_\star = \sum a_s y^s \in A\langle\!\langle y_i , i \geq 1 \rangle\! \rangle$. Diagonal meaning $\Delta_\star \Phi_\star = \Phi_\star \widehat{\otimes} \Phi_\star$ and $\varepsilon(\Phi_\star) = 1$.

Here's a detailled article on the subject: http://www1.mat.uniroma1.it/people/malvenuto/Duality.pdf . Hope it clarifies things for you.

  • $\begingroup$ But is that specifically over $\mathbb{Q}$? Moreover, those variables don't commute, right? $\endgroup$ – Jonathan Beardsley Aug 31 '13 at 3:32
  • $\begingroup$ I'm a little confused. What is a diagonal power series? $\endgroup$ – Julian Rosen Aug 31 '13 at 3:51
  • $\begingroup$ I think YBL is confused by the notation $\mathrm{Hom}$ being used for algebra homomorphisms rather than additive group homomorphisms here. I must say this also confused me a bit. $\endgroup$ – darij grinberg Aug 31 '13 at 4:17
  • $\begingroup$ I edited to answer your questions. $\endgroup$ – AFK Aug 31 '13 at 13:00
  • $\begingroup$ Just to make sure I understand: this bijection takes $f:\mathrm{QSym}\to A$ to the diagonal power series $\sum a_\mathbf{s} y^\mathbf{s}$ with $a_{\mathbf{s}}=f(M_{\mathbf{s}})$? $\endgroup$ – Julian Rosen Aug 31 '13 at 15:31

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