Characterizations of Amorphic Schemes and Fusions of Pairs

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Problem: Formalized question: Does there exist a finite symmetric association scheme $\mathcal X=(X,\{A_i\}_{i=0}^d)$, with primitive idempotents $\{E_i\}_{i=0}^d$, such that its fusing-relations graph $\Gamma_{\mathcal A}(\mathcal X)$ and fusing-idempotents graph $\Gamma_{\mathcal E}(\mathcal X)$, both on vertex set $\{1,\dots,d\}$, are not isomorphic?

Here $ij$ is an edge of $\Gamma_{\mathcal A}$ if fusing $A_i,A_j$ gives a fusion scheme, and $ij$ is an edge of $\Gamma_{\mathcal E}$ if fusing $E_i,E_j$ gives a fusion scheme.

Result: No such scheme exists. For every finite symmetric association scheme,

Proof. Let $P$ be the first eigenmatrix, with valencies $k_i=P_{0i}$, and write $p_r$ for row $r$ of $P$. By the Bannai-Muzychuk criterion, a pair $\{A_i,A_j\}$ fuses iff there is a unique pair $\{E_r,E_s\}$ such that, on the nontrivial coordinates,

Indeed, the relation partition has exactly one nonsingleton part $\{i,j\}$, so the corresponding idempotent partition must have exactly one nonsingleton part $\{r,s\}$. The constant row-sum condition is precisely equality outside $i,j$ and equality of the $i+j$ row sums.

Thus fusing relation-edges and fusing idempotent-edges are in canonical bijection:

Now use row orthogonality:

If two idempotent-edges $\{r,s\},\{r,t\}$ share $r$, put

Then

for the weighted inner product $\langle x,y\rangle=\sum_h x_hy_h/k_h$. Hence the supports of $\delta_s,\delta_t$, namely $\phi(\{r,s\})$ and $\phi(\{r,t\})$, must intersect. So $\phi$ preserves edge-incidence. Applying the same argument to the second eigenmatrix $Q$ gives the converse, so $\phi$ is an isomorphism of line graphs:

The only Whitney line-graph ambiguity is $K_3$ versus $K_{1,3}$. This ambiguity cannot occur here. If three idempotent-edges share $r$, their image supports are three pairwise intersecting 2-subsets. If they formed a triangle $\{a,b\},\{a,c\},\{b,c\}$, then for nonzero $x,y,z$,

and the product of their three pairwise weighted inner products would be

contradicting that each pairwise inner product equals $\langle p_r,p_r\rangle>0$. Thus stars map to stars, not triangles; dually, triangles cannot map to claws.

By Whitney’s line-graph theorem, the edge-containing components of $\Gamma_{\mathcal A}$ and $\Gamma_{\mathcal E}$ are therefore isomorphic componentwise. Both graphs have the same total vertex set size $d$, so the number of isolated vertices also agrees. Hence the whole graphs are isomorphic.

Therefore the requested non-isomorphic example does not exist.

Citation: No prior source is known to me for this exact conclusion. Definitions and the open question are from van Dam–Koolen–Xiong, arXiv:2404.00567. The proof uses the Bannai-Muzychuk fusion criterion and Whitney’s line-graph theorem: H. Whitney, “Congruent graphs and the connectivity of graphs,” Amer. J. Math. 54 (1932), 150–168.