5
$\begingroup$

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.

$\endgroup$
4
  • $\begingroup$ Might it perhaps be somewhat easier to consider QSymm's dual NSymm which classifies Hasse-Schmidt derivations? I'm not sure. $\endgroup$ Aug 31, 2013 at 3:54
  • 1
    $\begingroup$ 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. $\endgroup$ Aug 31, 2013 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 $\endgroup$ Aug 31, 2013 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. $\endgroup$ Aug 31, 2013 at 4:15

1 Answer 1

3
$\begingroup$

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.

$\endgroup$
8
  • $\begingroup$ But is that specifically over $\mathbb{Q}$? Moreover, those variables don't commute, right? $\endgroup$ Aug 31, 2013 at 3:32
  • $\begingroup$ I'm a little confused. What is a diagonal power series? $\endgroup$ Aug 31, 2013 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. $\endgroup$ Aug 31, 2013 at 4:17
  • $\begingroup$ I edited to answer your questions. $\endgroup$
    – AFK
    Aug 31, 2013 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 $a_{\mathbf{s}}=f(M_{\mathbf{s}})$? $\endgroup$ Aug 31, 2013 at 15:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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