The tilde species

Question

Consider a combinatorial species $F$, that is, an action of the symmetric group $\mathfrak S_n$ on a finite set $F[n]$. Recall that the elements of $F[n]$ are called structures. Furthermore, recall the notation $F[\sigma](s)$ for the structure in $F[n]$ obtained by acting with $\sigma\in\mathfrak S_n$ on $s$.

Joyal defined the species $\tilde F$ (roughly) as follows:

• let $\tilde F[n] = \{(s, \alpha)\in F[n]\times\mathfrak S_n | F[\alpha](s) = s\}$, that is, the set of pairs $(s, \alpha)$ such that $\alpha$ is an automorphism of $s$, and
• let $\tilde F[\sigma](s, \alpha) = (F[\sigma](\alpha), \sigma\circ\alpha\circ\sigma^{-1})$.

The main motivation to introduce this species is, as far as I know, the proof of the substitution formula, i.e., the computation of the character of $F\circ G$ (which, for 'molecular' species, is essentially the wreath product).

Playing with a new package I developed with Travis Scrimshaw and Mainak Roy for SageMath (https://github.com/sagemath/sage/pull/38446), I came to wonder what is known about $\tilde F$.

As pointed out on page 21 in Joyal, Andre, Une théorie combinatoire des séries formelles, Adv. Math. 42, 1-82 (1981). ZBL0491.05007, $\widetilde{F+G} = \tilde F + \tilde G$ and $\widetilde{F G} = \tilde F \tilde G$, so we can restrict attention to atomic species. Recall that a species is called atomic if the corresponding permutation group is indecomposable under direct product.

The number of isomorphism types of $\tilde F$ is the number of conjugacy classes of the corresponding permutation group.

It is not hard to see that $\tilde F = \frac{n!}{|F[n]|} F$ if $F$ is an atomic species corresponding to an abelian group of degree $n$.

Finally, for the species $E_n$ of sets, $\tilde E_n$ is, by definition, the species of permutations, i.e., the adjoint action of the symmetric group.

The two other 'classical' families of atomic species which do not correspond to an abelian group are

• $P_n$ be the species of polygons, corresponding to the dihedral groups, and
• $E^\pm_n$ be the species of oriented sets, corresponding to the alternating groups.

Here are two small tables ($C_n$ is the species of cycles, corresponding to the cyclic groups, $P^b_n$ is the species of polygons whose edges are alternately coloured red and blue). For the species of polygons I obtain

• $\tilde P_3 = C_3 + E_3 + XE_2$
• $\tilde P_4 = 2P_4 + C_4 + P^b_4 + E_2^2$
• $\tilde P_5 = P_5 + 2C_5 + XE_2(X^2)$
• $\tilde P_6 = 2P_6 + 2C_6 + 2P^b_6$
• $\tilde P_7 = P_7 + 3C_7 + XE_2(X^3)$
• $\tilde P_8 = 2P_8 + 3C_8 + P^b_8 + \{((1,2)(3,4)(7,8), (1,3)(2,4)(5,6))\}$

The expression $\{((1,2)(3,4)(7,8), (1,3)(2,4)(5,6))\}$ should be interpreted as the species corresponding to the permutation group with the given two generators. It appears that this (atomic) species cannot be written as a substitution.

For the species of oriented sets I obtain

• $\tilde E^\pm_2 = X^2$
• $\tilde E^\pm_3 = 3C_3$
• $\tilde E^\pm_4 = E^\pm_4 + P^b_4 + 2XC_3$
• $\tilde E^\pm_5 = E^\pm_5 + 2C_5 + X^2 C_3 + X P^b_4$
• $\tilde E^\pm_6 = E^\pm_6 + 2C_3^2 + 2XC_5 + \{((1,2)(3,4), (1,3)(5,6))\} + \{((1,2,3,4)(5,6))\}$

Questions:

1. Are there nice expressions for $\tilde P_n$ or $\tilde E^\pm_n$, or at least for their characters?
2. Is there an efficient way to compute the character of $\tilde F$?
3. Is there a reasonably efficient way to compute $\tilde F$?
4. Is this interesting at all?

Comments:

Apparently, the groups appearing in the molecular decomposition of $\tilde F$ are precisely the centralizer subgroups of the group corresponding to $F$. For the dihedral groups these are listed explicitly in the thesis Parthasarathy, Bhargavi, Conjugacy Classes of Centralizers in Groups, whereas for the alternating groups they are determined in Bhunia, Sushil; Kaur, Dilpreet; Singh, Anupam, $z$-classes and rational conjugacy classes in alternating groups, J. Ramanujan Math. Soc. 34, No. 2, 169–183 (2019). ZBL1469.20002.

Answer 1

Let $M_G$ be the molecular species with corresponding permutation group $G$. Let $C_G(g) = \{h\in G| gh = hg\}$ be the centralizer subgroup of $g\in G$. Finally, let $Conj(G)$ be the set of conjugacy classes of $G$.

Then $$ \tilde M_G = \sum_{g\in Conj(G)} M_{C_G(g)}, $$ where the sum is over a set of representatives of conjugacy classes.

Proof:

For $a\in Conj(G)$ let $\tilde M_G^a = \{\sigma\cdot (G, a) | \sigma\in\mathfrak S_n\}$. Then the map $\sigma C_G(a) \mapsto \sigma\cdot(G, a)$ is well defined and an isomorphism between $M_{C_G(a)}$ and $\tilde M_G^a$.

Moreover, $\bigcup_{a\in Conj(G)} \tilde M_G^a = \tilde M_G$, and the union is disjoint.

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I can imagine that the statement appears implicitly in the proof of the substitution formula by Joyal.