On a Question of Freslon, Gerontogiannis and Skalski about Quantum Isometries of Cuntz–Krieger Algebras

For a primitive 0–1 matrix A, Freslon, Gerontogiannis and Skalski computed the quantum isometry group G_A^∞ of the log-Laplacian spectral triple on the Cuntz–Krieger algebra O_A. They showed that the quantum automorphism group QAut(A) is a quantum subgroup of a larger quantum group G_A^1 and asked if it is one of G_A^∞ when QAut(A) ≠ Aut(A). We show that the answer is negative in general. For the Cuntz algebras O_N and for A = J_N − I_N, with N ≥ 4, the quantum permutation group S_N^+ = QAut(A) is not a quantum subgroup of G_A^∞. For every primitive A with QAut(A) ≠ Aut(A), the canonical inclusion of QAut(A) in G_A^1 does not extend to G_A^∞, and the natural action of QAut(A) on O_A is not isometric for this spectral triple. Quantum groups such as S_M^+, M ≥ 4, cannot sit in G_A^∞ through an embedding that fixes a vertex. On the other hand, for two explicit primitive 4 × 4 matrices, the non-classical QAut(A) (the dual of the infinite dihedral group, and the hyperoctahedral quantum group H_2^+) is a quantum subgroup of G_A^∞. For general A it remains open whether S_M^+, M ≥ 4, can sit in G_A^∞ through an embedding that fixes no vertex. This is an unrefereed note.