The algebra of balanced dessins

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Problem: In the paper’s notation, $\mathcal D_{\mathbb Q}$ is the rational Burnside algebra of finite transitive $F_2$-sets, i.e. irreducible dessins, with multiplication induced by Cartesian product. For $D\in\mathcal D$,

$$\psi_D=\pi_{S_3}(D)-\pi_{S_3}\pi_{\mathcal G}(D),$$

and $\widetilde{\mathcal A}$ is generated by $1$ and all $\psi_D$.

Conjecture 6 is naturally read as follows: the partial factor choice promised in Conjecture 5—choosing for each $\psi_D$ the unique irreducible factor $Z_{\psi_D}$ of its minimal polynomial whose degree equals $|\mathcal G\psi_D|$—has a unique coherent extension to every $\psi\in\widetilde{\mathcal A}$.

Result: The conjecture is false: the preceding “partial choice” does not exist.

Indeed, every element of $\mathcal D_{\mathbb Q}$, hence every element of $\widetilde{\mathcal A}$, has minimal polynomial split completely into linear factors over $\mathbb Q$.

To see this, let $\Gamma=F_2$. Any finite set of dessins appearing in an element $\psi\in\mathcal D_{\mathbb Q}$ factors through some finite quotient $G=\Gamma/N$. Thus $\psi$ lies in the rational Burnside ring $B_{\mathbb Q}(G)$. For finite $G$, the mark homomorphism

$$B_{\mathbb Q}(G)\to \prod_{[H]\le G}\mathbb Q,\qquad X\mapsto (|X^H|)_H$$

is an injective ring map and, after tensoring with $\mathbb Q$, an isomorphism. Hence $B_{\mathbb Q}(G)\cong \mathbb Q^r$. Therefore every element satisfies a polynomial of the form

$$\prod_i (x-q_i),\qquad q_i\in\mathbb Q,$$

so all irreducible factors of its minimal polynomial are linear.

Now choose a connected dessin $D$ whose field of moduli has degree $7$ over $\mathbb Q$; such dessins exist by Birch’s theorem that every number field occurs as a field of moduli of a dessin. Then $|\mathcal G D|=7$. Since an $S_3$-orbit has size at most $6$, $\pi_{S_3}(D)$ cannot be $\mathcal G$-fixed. Therefore

$$\psi_D=\pi_{S_3}(D)-\pi_{\mathcal G}\pi_{S_3}(D)$$

has nontrivial $\mathcal G$-orbit, so $|\mathcal G\psi_D|>1$.

But $P_{\psi_D}(x)$ has only linear irreducible factors. Thus there is no irreducible factor of degree $|\mathcal G\psi_D|>1$. The factor required by Conjecture 5 does not exist for this $D$, so the “partial choice” mentioned in Conjecture 6 cannot have any coherent extension, unique or otherwise.

Verification audit: no extra hypothesis on $\widetilde{\mathcal A}$ was used; the argument applies to the ambient algebra $\mathcal D_{\mathbb Q}$. The counterexample uses a genuine irreducible/connected dessin, and the obstruction is exactly at the factor-degree condition invoked by the target conjecture.

Citation: A. J. Birch, “Noncongruence subgroups, covers and drawings,” in The Grothendieck Theory of Dessins d’Enfants, LMS Lecture Note Ser. 200, Cambridge Univ. Press, 1994.
The Burnside-ring mark isomorphism over $\mathbb Q$ is standard; the proof above includes the needed form.