4.1 Cumulant identities
For every \(\lambda \in \mathbb {R}^d\), \(T(\lambda )=\sum _{1\le i{\lt}j{\lt}k\le n}\lambda _{ij}\lambda _{ik}\lambda _{jk}\).
Expand \(\mathbb {E}[X_\lambda ^3]\) and observe that the only surviving multigraphs (every vertex with even multiplicity) are triangles. Each triangle contributes \(3!\) ordered triples, giving \(\mathbb {E}[X_\lambda ^3]=6\sum \lambda _{ij}\lambda _{ik}\lambda _{jk}\). Since \(T=\frac16\mathbb {E}[X_\lambda ^3]\), the claim follows.
\(Q(\lambda ) =-\frac{1}{12}\sum _e\lambda _e^4+\frac18\mathcal{C}_4(\lambda )\), where \(\mathcal{C}_4(\lambda )=\sum _{\text{ordered 4-cycles}}\lambda _{i_1 i_2}\lambda _{i_2 i_3}\lambda _{i_3 i_4}\lambda _{i_4 i_1}\).
Expand \(\mathbb {E}[X_\lambda ^4]=\sum _{e_1,\dots ,e_4}\lambda _{e_1}\cdots \lambda _{e_4} \mathbb {E}[\chi _{e_1}\cdots \chi _{e_4}]\). The expectation vanishes unless every vertex index has even multiplicity. Three types survive: (i) all four edges equal: \(\sum _e\lambda _e^4\); (ii) two distinct pairs: \(6\sum _{e{\lt}f}\lambda _e^2\lambda _f^2\); (iii) four distinct edges forming a 4-cycle: \(3\mathcal{C}_4(\lambda )\). Thus \(\mathbb {E}[X_\lambda ^4]=\sum _e\lambda _e^4+6\sum _{e{\lt}f}\lambda _e^2\lambda _f^2 +3\mathcal{C}_4\). Since \(3s(\lambda )^2=3\sum _e\lambda _e^4+6\sum _{e{\lt}f}\lambda _e^2\lambda _f^2\), subtracting gives \(\kappa _4=-2\sum _e\lambda _e^4+3\mathcal{C}_4\), and dividing by \(24\) yields the formula for \(Q\).
Let \(A\) be a symmetric zero-diagonal \(n\times n\) matrix, \(B\) the \((n{-}1)\times (n{-}1)\) submatrix deleting row/column \(n\), \(x\in \mathbb {R}^{n-1}\) the last column, and \(M(B)=B^2-\operatorname {diag}(B^2)\). Then
The diagonal quartic part splits as \(\sum _{i{\lt}j\le n}A_{ij}^4=\sum _{a{\lt}b\le n-1}B_{ab}^4+\sum _a x_a^4\). Cycles through vertex \(n\) correspond to \(x^\top M(B)x\); cycles avoiding \(n\) are \(\mathcal{C}_{4,n-1}(B)\). Combine with the \(1/8\) and \(-1/12\) coefficients.