Spectral-gap equality conjecture for generalised pancake graphs

Counterexample at $m=6,\ n=3$, with a completely exact six-dimensional spectral certificate.

The original conjecture is Greaves–Zhu, Discrete Mathematics 349 (2026), 115102, Conjecture 2. The subsequent paper S. A. Blanco, arXiv:2608.15398, Section 5, posted August 15, 2026, explicitly states that this same full-graph versus Schreier-quotient spectral-gap equality remains open. Its counterexample to a different expander conjecture does not address the equality considered here.

Take the generalized pancake graph

$$\mathcal P_6(3) = \operatorname{Cay}\bigl((\mathbb Z/6\mathbb Z)^3\rtimes S_3, \{r_k^+,r_k^-:1\le k\le3\}\bigr).$$

It has $6^3\cdot3!=1296$ vertices, is $6$-regular, and has top adjacency eigenvalue $6$.

Let $\zeta=e^{2\pi i/6}$, and for a color word $c=(c_1,c_2,c_3)$ write

$$\chi_a(c,\sigma) = \zeta^{a_1c_1+a_2c_2+a_3c_3}, \qquad a\in(\mathbb Z/6\mathbb Z)^3.$$

These are functions on the full colored-permutation graph. If $R_ka$ reverses the first $k$ coordinates, direct application of the six prefix-flip generators gives

$$A\chi_a = \sum_{k=1}^3 2\cos\left(\frac{\pi}{3}\sum_{j=1}^k a_j\right) \chi_{R_ka}. \tag{1}$$

The orbit

$$(0,1,5),\ (0,5,1),\ (1,0,5),\ (1,5,0),\ (5,0,1),\ (5,1,0)$$

spans an invariant subspace. In this order, (1) acts by the integer symmetric matrix

$$M= \begin{pmatrix} 2&0&1&0&0&2\\ 0&2&0&2&1&0\\ 1&0&1&0&2&0\\ 0&2&0&1&0&2\\ 0&1&2&0&1&0\\ 2&0&0&2&0&1 \end{pmatrix}.$$

Its characteristic polynomial is exactly

$$\det(xI-M) = (x-3)(x+1)(x^2-5x+1)(x^2-x-7).$$

Every orbit character is orthogonal to constants. Therefore

$$L=\frac{5+\sqrt{21}}2,\qquad 4<L<5,$$

is a nontrivial full-graph eigenvalue, and

$$\lambda_2(\mathcal P_6(3))\ge L. \tag{2}$$

The conjectured equitable quotient records the color and position of one distinguished symbol. Its Fourier blocks are

$$B_t = \begin{pmatrix}0&0&0\\0&2&0\\0&0&4\end{pmatrix} + t\begin{pmatrix}1&1&1\\1&1&0\\1&0&0\end{pmatrix}, \qquad t\in\{2,1,-1,-2\}.$$

For $t=2$,

$$\det(xI-B_2)=x(x-4)(x-6),$$

so its nontrivial eigenvalues are $0$ and $4$.

For the other three blocks, use $L^2-5L+1=0$. The successive leading principal minors of $LI-B_t$ are

$$\begin{array}{c|ccc} t&\Delta_1&\Delta_2&\Delta_3\\ \hline 1&L-1&L+1&L-2\\ -1&L+1&5L-3&L+8\\ -2&L+2&7L-5&13-2L. \end{array}$$

All entries are strictly positive because $4<L<5$. Sylvester's criterion therefore shows that every eigenvalue of these blocks is strictly less than $L$. Consequently

$$\boxed{ \lambda_2\bigl(Q(\mathfrak P_{6,3})\bigr) < \frac{5+\sqrt{21}}2 \le \lambda_2\bigl(\mathcal P_6(3)\bigr). }$$

Since both graphs have top eigenvalue $6$,

$$6-\lambda_2\bigl(\mathcal P_6(3)\bigr) < 6-\lambda_2\bigl(Q(\mathfrak P_{6,3})\bigr).$$

The proposed spectral-gap equality is therefore false.