On the Structure and Generators of the nth-order Chromatic Algebra

NEW

Problem: Reconstructed statement: over $K=\mathbb C((Q))$, let $\mathcal C_n$ be Liu’s $n$-th chromatic algebra and

$$E_n=\{e^n_{i,j}:1\le i<j\le n\},$$

where $e^n_{i,j}$ connects all top and bottom boundary points in positions $i,\dots,j$ to one inner vertex and leaves all other positions vertical. Conjecture 5.12 asserts that every multiplicative generating set of the unital $K$-algebra $\mathcal C_n$ has cardinality at least

$$|E_n|=\binom n2 .$$

Result: The conjecture is false already for $n=3$. Write

$$a=e^3_{1,2},\qquad b=e^3_{2,3},\qquad c=e^3_{1,3}.$$

Then $E_3=\{a,c,b\}$, so $|E_3|=3$. I will show $c\in K\langle a,b\rangle$; since Liu proves that $E_3$ generates $\mathcal C_3$, it follows that $\{a,b\}$ generates $\mathcal C_3$.

Use bottom labels $b_1,b_2,b_3$ and top labels $t_1,t_2,t_3$. Let

$$\begin{aligned} P&=\{\{b_1,b_2\},\{t_1,t_2\},\{b_3,t_3\}\},\\ R&=\{\{b_1,t_1,t_2\},\{b_2,b_3,t_3\}\},\\ S&=\{\{b_1,b_2,t_1\},\{b_3,t_2,t_3\}\},\\ U&=\{\{t_1,t_2\},\{b_1,b_2,b_3,t_3\}\},\\ V&=\{\{b_1,b_2\},\{b_3,t_1,t_2,t_3\}\}. \end{aligned}$$

Direct stacking and use of the chromatic relations $G=G/e-G\setminus e$, loop deletion $G=(Q-1)G\setminus e$, and suppression of $2$-valent vertices gives:

$$\begin{aligned} a^2&=(Q-2)a+P,\\ ab&=c-R,\\ ba&=c-S,\\ Pb&=U,\\ bP&=V,\\ aba&=V-P+(Q-3)c+U+a. \end{aligned}$$

For example, $ab$ has one internal edge; contracting it gives $c$, deleting it gives $R$.

Now compute:

$$\begin{aligned} &(1-Q)a+a^2+(Q-2)(ab+ba)-a^2b+aba-ba^2\\ &=(Q-3)c. \end{aligned}$$

Since $Q-3$ is invertible in $\mathbb C((Q))$, we obtain

$$c=\frac{(1-Q)a+a^2+(Q-2)(ab+ba)-a^2b+aba-ba^2}{Q-3}\in K\langle a,b\rangle .$$

Thus $\{e^3_{1,2},e^3_{2,3}\}$ generates $\mathcal C_3$, a generating set of cardinality $2<3=|E_3|$. Hence Conjecture 5.12 is false.

Audit: the counterexample uses the same algebra, field, basis convention, and unital-generation convention as the paper. No extra hypothesis is introduced; $Q-3$ is invertible in $\mathbb C((Q))$. This is not a boundary case such as $n=1$; it is the first case where $E_n$ has three elements.

Citation: Definitions and the theorem that $E_n$ generates $\mathcal C_n$: Ethan Yi-Heng Liu, “On the Structure and Generators of the $n$th-order Chromatic Algebra,” arXiv:2401.06095, Theorem 2 and Conjecture 5.12.