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The plethysm $s_{\nu}[s_{\mu}]$ of two symmetric functions is the character of the composition of Schur functors $S^{\nu}(S^{\mu}(V))$. We know that this operation is linear and multiplicative in its first argument. But is there a way to develop

  1. $s_{\nu}[s_{\mu} + s_{\lambda}]$;

  2. $s_{\nu}[s_{\mu}s_{\lambda}]$;

in terms of plethysms $s_{\nu}[s_{\mu}]$ and $s_{\nu}[s_{\lambda}]$ ? I think I have heard of a formula for the first one, but I don't find it anymore!

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    $\begingroup$ If a symmetric function $f$ satisfies $\Delta\left(f\right) = \sum_{i=1}^k g_i \otimes h_i$ (where $\Delta$ is the comultiplication of the Hopf algebra of symmetric functions), then $f\left[u + v\right] = \sum_{i=1}^k g_i\left[u\right] h_i\left[v\right]$ whenever $u$ and $v$ are elements of a $\lambda$-ring (e.g., symmetric functions). Thus, any formula for $\Delta s_\nu$ (for example, the classical $\Delta s_\nu = \sum_{\lambda, \mu} c^\nu_{\lambda, \mu} s_\lambda \otimes s_\mu$) will give you a formula for $s_\nu\left[u + v\right]$. $\endgroup$
    – darij grinberg
    Commented Aug 31, 2020 at 19:16
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    $\begingroup$ The same applies to $f\left[uv\right]$, but this time you need the second comultiplication (i.e., the internal comultiplication, whose structure constants are the Kronecker coefficients) instead of $\Delta$. $\endgroup$
    – darij grinberg
    Commented Aug 31, 2020 at 19:16

1 Answer 1

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In principle one can develop (1) using the coproduct in the ring of symmetric functions. By the Littlewood–Richardson rule, $\Delta(s_\nu) = \sum_{\alpha}\sum_\beta c^\nu_{\alpha\beta} s_\alpha \otimes s_\beta$ where $c^\nu_{\alpha\beta}$ is a Littlewood–Richardson coefficient, and correspondingly

$$s_\nu[s_\lambda + s_\mu] = \sum_{\alpha}\sum_\beta c^\nu_{\alpha\beta} s_\alpha[s_\lambda] s_\beta[s_\mu].$$

Here the sum is over all partitions such that $|\alpha|+|\beta| = |\nu|$. Somewhat similarly,

$$s_\nu[s_\lambda s_\mu] = \sum_{\alpha}\sum_{\beta} k^\nu_{\alpha\beta} s_\alpha[s_\lambda] s_\beta[s_\mu]$$

where the sum is over all partitions $\alpha$ and $\beta$ of $|\nu|$ and $k^\nu_{\alpha\beta}$ is the Kronecker coefficient, most easily defined as the inner product $\langle \chi^\nu, \chi^\alpha \chi^\beta \rangle$ in the character ring of the symmetric group. Equivalently the$k^\nu_{\alpha\beta}$ are the structure constants for the internal product, usually denoted $\star$, on the ring of symmetric functions. These formulae can be found in MacDonald's textbook: see (8.8) and (8.9) on page 136, and hold replacing $s_\lambda$ and $s_\mu$ with arbitrary symmetric functions.

In practice, at least in my experience, this usually leads to a mess. One special case that's worth noting is when $\nu = (n)$, in which case the Littlewood—Richardson coefficient is non-zero only if $\alpha = (m)$ and $\beta = (n-m)$ for some $m \in \{0,1,\ldots, n\}$ and we get

$$s_{(n)}[s_\lambda + s_\mu] = \sum_m s_{(m)}[s_\lambda] s_{(n-m)}[s_\mu].$$

This is the symmetric function version of $\mathrm{Sym}^n (V \oplus W) = \sum_{m=0}^n \mathrm{Sym}^m V \otimes \mathrm{Sym}^{n-m} W$ for polynomial representations of $\mathrm{GL}_d(\mathbb{C})$. There is a corresponding rule for exterior powers and so for $s_{(1^n)}$.

This also gives one indication that (2) is even harder: one related question was asked on MathOverflow. Example 3 on page 137 of MacDonald gives the special case for $\nu = (n)$, when $\chi^{(n)}$ is the trivial character, and so $\langle \chi^{(n)}, \chi^{\alpha}\chi^{\beta}\rangle = \langle \chi^{\alpha}, \chi^\beta\rangle = [\alpha=\beta]$. Hence

$$s_{(n)}[s_\lambda s_\mu] = \sum_{\alpha} s_\alpha[s_\lambda] s_\alpha[s_\mu]. $$

Great care is needed when extending these rules to arbitrary symmetric functions. For instance, $s_\nu[-f] = (-1)^{|\nu|} s_{\nu'}[f]$ for any symmetric function $f$ and, as Richard Stanley points out in a comment below, the expression $s_\nu[f-f]$ should be interpreted as a plethystic substitution using the alphabets for $f$ and $-f$, not as $s_\nu[0]$; correctly interpreted, it can be expanded using the coproduct rule and the rule for $s_\nu[-f]$ just given.

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    $\begingroup$ It is not true that $s_\nu[0] = s_\nu[f-f]$. This illustrates the subtlety of plethysm notation. The expression $f-f$, at least when $f$ is a sum of monomials, means $f+(-f)$, where the $+$ refers to a union of alphabets and $-f$ to a negative alphabet. $\endgroup$
    – Richard Stanley
    Commented Sep 1, 2020 at 1:46
  • $\begingroup$ The expansion of $s_{(n)}[s_\lambda+s_\mu]$ can be interpreted nicely in terms of (very slightly generalized) combinatorial species: $s_{(n)}$ is the (cycle index series of the) species of sets with $n$ elements, so the expansion says: any set of $n$ items, some of which are $\lambda$ime coloured and some are $\mu$agenta coloured, is a set of $n-m$ $\lambda$ime coloured items together with a set of $m$ $\mu$agenta coloured items. In general, combinatorial species are quite a good language to phrase plethystic identities in. $\endgroup$
    – Martin Rubey
    Commented Sep 1, 2020 at 6:54
  • $\begingroup$ @RichardStanley: Really? Can you give a counterexample? I'm pretty sure that $f\left[u-u\right] = f\left[0\right]$ for any symmetric function $f$ (a consequence of the defining axiom of the antipode in a Hopf algebra). $\endgroup$
    – darij grinberg
    Commented Sep 1, 2020 at 11:14
  • $\begingroup$ What a minefield. I think darij is correct. By P2 in the survey article by Loehr and Remmel, $g \mapsto p_m \circ g$ is an algebra homomorphism. Hence $p_m[-u] = -p_m[u]$. Using $\Delta[p_m] = p_m \otimes 1 + 1 \otimes p_m$, we get $p_m[u-u] = p_m[u]1[-u] + 1[u]p_m[-u] = p_m[u] + p_m[-u] = p_m[u] - p_m[u] = 0$. By P1 in the survey article, for any $h$, the map $f \mapsto f \circ h$ is an algebra homomorphism. Hence $p_\mu[u-u] = \prod p_{\mu_i}[u-u] = 0$. Since the $p_\mu$ span, P3 implies that $f[u-u] = 0$ for all $f$. $\endgroup$
    – Mark Wildon
    Commented Sep 1, 2020 at 15:13
  • $\begingroup$ @darijgrinberg: oops, you are right. I was thinking of the ambiguity of notation such as $p_2(1-q)$, which could mean either $p_2(1-q,0,0,\dots)=(1-q)^2$ or $p_2(1,0,0,\dots)-p_2(q,0,0,\dots)=1-q^2$. $\endgroup$
    – Richard Stanley
    Commented Sep 6, 2020 at 0:57

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