# Interaction of plethysm with other operations

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!

• 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 $s_\nu\left[u + v\right]$. Aug 31 '20 at 19:16
• 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 $\Delta$. Aug 31 '20 at 19:16

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.

• 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. Sep 1 '20 at 1:46
• 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. Sep 1 '20 at 6:54
• @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). Sep 1 '20 at 11:14
• 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 $f$. Sep 1 '20 at 15:13
• @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 $p_2(1,0,0,\dots)-p_2(q,0,0,\dots)=1-q^2$. Sep 6 '20 at 0:57