8
$\begingroup$

Recall that the Kronecker product $s_\lambda * s_\mu$ of two Schur functions $s_\lambda$ and $s_\mu$ is the symmetric function whose expansion (in terms of Schur functions) is given by

\begin{equation} \sum_{\nu \, \vdash \, n} g_{\lambda \mu}^\nu \, s_\nu \end{equation}

where $\lambda$, $\mu$, and $\nu$ are partitions of $n$ and $g_{\lambda \mu}^\nu$ is the Kronecker coefficient, which famously counts the multiplicity of $V_\nu$ in the tensor product $V_\lambda \otimes V_\mu$ of the irreducible representations of the symmetric group $S_n$.

Switch now to the quasi-symmetric world: Given a composition $\alpha = (\alpha_1, \dots, \alpha_k)$ of $n$ let $L_\alpha$ be the fundamental quasi-symmetric functions defined by

\begin{equation} L_\alpha = \sum x_{\ell_1} \cdots \,x_{\ell_k} \end{equation}

where the sum is taken over all sequences $1 \leq \ell_1 \leq \cdots \leq \ell_k$ such that $\ell_i < \ell_{i+1}$ whenever $i = \alpha_1 + \cdots + \alpha_j$ for some $1 \leq j \leq k-1$.

The space of symmetric functions within the $\mathrm{QSym}_n :=\Bbb{Q}$-span of $\{ L_\alpha \, \big| \, \alpha \, \models \, n \}$ coincides with the $\mathrm{Sym}_n:= \Bbb{Q}$-span of $\{ s_\lambda \, \big| \, \lambda \, \vdash \, n \}$ the latter of which is endowed with the Kronecker $*$-product.

Question: Can the Kronecker $*$-product on $\mathrm{Sym}_n$ be extended to all of $\mathrm{QSym}_n$ so that there exist non-negative integers $\tilde{g}_{\alpha,\beta}^{\, \gamma}$ for each triple $\alpha$, $\beta$, $\gamma$ of compositions of $n$ satisfying

\begin{equation} L_\alpha * L_\beta = \sum_{\gamma \, \models \, n} \tilde{g}_{\alpha,\beta}^{\, \gamma} \, L_\gamma \quad \text{?} \end{equation}

p.s. Covertly, I am asking whether or not there is some kind of tensor product structure (as in a symmetric tensor category) on the projective indecomposable representations of the 0-Hecke algebra $H_n(0)$. Any thoughts on that would also be appreciated.

thanks, ines.

$\endgroup$
3
  • $\begingroup$ Sound right: Maybe this follows from the quasi-symmetric expansion $s_\lambda = \sum_{\text{Tableaux $T$}} L_{\mathrm{C}(T)}$ where $\mathrm{C}(T)$ is the descent composition ? $\endgroup$ Nov 27, 2020 at 23:49
  • $\begingroup$ Sorry, my formula was wrong. I misremembered the $s$-in-$L$ expansion. $\endgroup$ Nov 27, 2020 at 23:51
  • $\begingroup$ Maybe something like this: $g_{\lambda \mu}^\nu = \sum_{\stackrel{\scriptstyle \text{Tableaux $T''$}}{\text{of shape $\nu$}}} \sum_{\stackrel{\scriptstyle \text{Tableaux $T'$}}{\text{of shape $\mu$}}} \sum_{\stackrel{\scriptstyle \text{Tableaux $T$}}{\text{of shape $\lambda$}}} \tilde{g}^{\, \mathrm{C}(T'')}_{\mathrm{C}(T),\mathrm{C}(T')}$ ? $\endgroup$ Nov 28, 2020 at 0:38

2 Answers 2

5
$\begingroup$

In the world of symmetric functions, Kronecker coefficients give the structure constants for both the inner multiplication and the inner comultiplication. While the natural introduction of the inner multiplication uses the representations/characters of the symmetric group, the inner comultiplication has a very straightforward description: the structure constants are obtained via the expansion $$s_{\lambda}(x_1y_1,...,x_iy_j,...)=\sum_{\nu ,\mu\, \vdash \, n} g_{\nu \mu}^\lambda \, s_\nu (x)s_{\mu}(y).$$ Similarly we can ask whether $L_{\alpha}(xy)$ has such an expansion, and the answer is positive! There exist nonnegative integers $\tilde{g}_{\beta, \gamma}^{\, \alpha}$ such that $$L_{\alpha}(xy)=\sum_{\beta, \gamma \, \models \, n}\tilde{g}_{\beta, \gamma}^{\, \alpha}L_{\beta}(x)L_{\gamma}(y).$$

This gives a natural inner comultiplication for quasisymmetric functions. Moreover, by dualizing, these allow us to define a genuine inner multiplication for the dual $\widehat{\mathrm{QSym}_n}$, which is isomorphic to the descent algebra of Solomon.

Originally this was shown in Gessel's paper, where you will find the details and combinatorial interpretations

I. Gessel, "Multipartite P-partitions and inner products of skew Schur functions", Contemporary Mathematics, 34:289–317, 1984

You can also read about it in

C. Malvenuto, C. Reutenauer "Duality between quasi-symmetric functions and the Solomon descent algebra" J. Algebra, 177 (1995), pp. 967-982


Now, as far as a natural inner multiplication for $\mathrm{QSym}$, this is an open problem, at least according to Hazewinkel in

M. Hazewinkel, "Symmetric functions, noncommutative symmetric functions, and quasisymmetric functions", Acta Appl. Math. 75 (2003), 1-3, 55–83

The same paper mentions that there is no known structure on $H_n(0)$ that induces the inner comultiplication from above, so the answer to the last question in the postscript is also missing at the moment.

$\endgroup$
8
  • 1
    $\begingroup$ No I think the construction Gjergji suggests is correct. I just don’t think these coefficients precisely generalize Kronecker coefficients (at least, I don’t see how you recover Kronecker coefficients as a special case…) $\endgroup$ Dec 22, 2022 at 2:47
  • 2
    $\begingroup$ If I understand your second comment correctly, then, yes, a different basis would give you different structure constants and you can play around with that, but the product structure you are defining would be the same. $\endgroup$ Dec 22, 2022 at 3:39
  • 1
    $\begingroup$ @JeanneScott, I looked at the link again. On page 34 they are talking about the outer coproduct of QSym which is obtained by expanding $f(x_1,x_2,\dots, y_1, y_2, \dots)$ rather than the inner coproduct. $\endgroup$ Jan 8, 2023 at 21:01
  • 1
    $\begingroup$ @GjergjiZaimi, It sounds like the generalized Kronecker coefficients $\hat{g}^\alpha_{\beta,\gamma}$ are just the structure constants for Solomon's descent algebra $\Sigma_n$ for compositions $\alpha, \beta, \gamma \models n$ ? $\endgroup$ Jan 15, 2023 at 10:49
  • 1
    $\begingroup$ @JeanneScott, Yes, that is one of the results in Gessel's paper mentioned above. $\endgroup$ Jan 15, 2023 at 20:18
2
$\begingroup$

One silly (or super wishful thinking approach) is to use the formula $$ g_{\lambda \mu \nu} = \frac{1}{n!} \sum_{\sigma \in S_n} \chi^{\lambda}(\sigma) \chi^{\mu}(\sigma) \chi^{\nu}(\sigma). $$

Perhaps some version of $$ g_{\alpha, \beta,\gamma} = \frac{1}{n!} \sum_{\tau \in S_n} \chi^{\alpha}(\tau) \chi^{\beta}(\tau) \chi^{\gamma}(\tau) $$ where now $\chi^{\alpha}(\tau)$ are the coefficients which show up when the quasisymmetric power sums are expanded in terms of the Gessel quasisymmetric functions.

Here, I suppose that in the sum, one does not only consider the cycle type, but decide to order the cycles by smallest element, and then let $\alpha$ be the integer composition given by the cycle lengths in that order.

If the coefficients are non-negative by some miracle, then, well, you have something cool to start with.

Ballantine, Cristina; Daugherty, Zajj; Hicks, Angela; Mason, Sarah; Niese, Elizabeth, On quasisymmetric power sums, J. Comb. Theory, Ser. A 175, Article ID 105273, 36 p. (2020). ZBL1442.05241.

$\endgroup$
7
  • $\begingroup$ So, for example the permutation $\tau \in S_6$ with cycle-decomposition $(235) \cdot (14) \cdot (6)$ would correspond to the composition $\alpha = (2,3,1)$ following your recipe ? $\endgroup$ Nov 27, 2020 at 21:33
  • $\begingroup$ Presumably these "character" values $\chi^\alpha(\tau)$ (would) have some meaning in terms of the 0-Hecke algebra ? $\endgroup$ Nov 27, 2020 at 21:48
  • $\begingroup$ @InesInstitoris yes, there is such an interpretation $\endgroup$ Nov 28, 2020 at 14:45
  • 1
    $\begingroup$ There are two versions of quasi-symmetric power sum functions (and consequently two versions of "character" values) in the paper you reference. Is a this somehow related to the fact that the principal quasi-symmetric function $F_\alpha$ of a composition $\alpha$ is related to the {\it simple} irreducible module $S_\alpha$ of $H_n(0)$ while the corresponding (non-commutative) {\it ribbon Schur-function} $R_\alpha$ is related to the indecomposable projective module $P_\alpha$ of $H_n(0)$ ? $\endgroup$ Nov 28, 2020 at 16:47
  • 1
    $\begingroup$ Right, there are two versions, but one of them is much better well-behaved imho... $\endgroup$ Nov 28, 2020 at 16:53

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.