The algebra of balanced dessins

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Problem: Reconstructed statement: in Fine’s notation, let $\mathcal D$ be the set of irreducible/connected dessins, equivalently transitive finite $F_2$-sets up to isomorphism, and let $\mathcal D_{\mathbb Q}$ be their rational Burnside-type algebra. Let $\mathcal G=\operatorname{Gal}(\overline{\mathbb Q}/\mathbb Q)$ and $S_3$ act on $\mathcal D_{\mathbb Q}$, and define

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

For $\psi=\psi_D$, let $P_\psi(x)\in\mathbb Q[x]$ be its minimal polynomial in $\mathcal D_{\mathbb Q}$. The conjecture asserts that $P_\psi$ has a unique irreducible factor $Z_\psi(x)$ with

$$\deg Z_\psi=|\mathcal G\psi|.$$

Here “prime factor” means irreducible factor over $\mathbb Q$.

Result: The conjecture is false.

First, a stronger structural fact holds: every element of $\mathcal D_{\mathbb Q}$ has a minimal polynomial split into linear factors over $\mathbb Q$.

Indeed, identify irreducible dessins with transitive finite $F_2$-sets $F_2/H$, where $H\le F_2$ has finite index. For each finite-index subgroup $K\le F_2$, define the mark

$$m_K(X)=|X^K|.$$

This extends to a $\mathbb Q$-algebra homomorphism

$$m_K:\mathcal D_{\mathbb Q}\to\mathbb Q.$$

The combined mark map is injective: if

$$a=\sum_i c_i[F_2/H_i]\neq 0,$$

choose $H_j$ of minimal index among those with $c_j\neq 0$. If $m_{H_j}(F_2/H_i)\neq 0$, then $H_j\le gH_ig^{-1}$ for some $g$, hence $[F_2:H_j]\ge [F_2:H_i]$. By minimality equality holds, so $H_i$ is conjugate to $H_j$. Thus

$$m_{H_j}(a)=c_j\,|N_{F_2}(H_j):H_j|\neq 0,$$

a contradiction.

Now for any $a=\sum_i c_i X_i$, the set of possible values $m_K(a)$ is finite, because each $m_K(X_i)$ lies between $0$ and $|X_i|$. Hence

$$q_a(x)=\prod_{\lambda\in\{m_K(a)\}}(x-\lambda)$$

annihilates $a$. Therefore the minimal polynomial $P_a$ divides $q_a$, so all irreducible factors of $P_a$ are linear.

It remains to produce $D$ with $|\mathcal G\psi_D|>1$. Such a $D$ exists. The absolute Galois action on connected dessins is faithful. If every $\mathcal G$-orbit on $\mathcal D$ had size at most $6$, then $\mathcal G$ would embed as a closed subgroup of a product of groups $S_n$ with $n\le 6$, hence every finite continuous quotient of $\mathcal G$ would have order divisible only by $2,3,5$. But $\mathcal G$ has a quotient $C_7$, for example from the degree-$7$ cyclic subfield of $\mathbb Q(\zeta_{29})$. Therefore some $D\in\mathcal D$ has

$$|\mathcal G D|>6.$$

For this $D$, set $\psi=\psi_D$. If $|\mathcal G\psi|=1$, then for every $\sigma\in\mathcal G$,

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

so $\pi_{S_3}(\sigma D)=\pi_{S_3}(D)$. Since these are formal averages over $S_3$-orbits in the basis $\mathcal D$, equality implies $\sigma D\in S_3D$. Thus

$$\mathcal G D\subseteq S_3D,$$

so $|\mathcal G D|\le |S_3D|\le 6$, contradiction. Hence $|\mathcal G\psi|>1$.

But $P_\psi(x)$ has only linear irreducible factors, so no prime factor of $P_\psi$ has degree $|\mathcal G\psi|>1$. Thus the asserted factor does not exist.

Verification audit: the argument uses Fine’s exact algebra $\mathcal D_{\mathbb Q}$, the stated $S_3$- and $\mathcal G$-averages, and the minimal polynomial before quotienting. No extra hypothesis is added. The counterexample is nondegenerate: it has genuinely nontrivial $\mathcal G$-orbit after balancing.

Citation: Fine, “The algebra of balanced dessins,” arXiv:1802.04531, Conjecture 5. Faithfulness of the absolute Galois action on dessins is classical; see Lando–Zvonkin, Graphs on Surfaces and Their Applications, Springer, 2004, Chapter 2. The mark-homomorphism argument is the standard Burnside-ring argument; a self-contained proof is included above.