On Bardakov's Question on the Linearity of Automorphism Groups of Artin Groups of Finite Type: A Reduction to Six Exceptional Types

Problem 17.16 of the Kourovka Notebook, posed by V. G. Bardakov, asks whether the automorphism group \(\operatorname{Aut}(A)\) of an Artin group \(A\) of finite type is linear, as \(A\) is by the theorems of Bigelow, Krammer, Digne and Cohen–Wales; the Notebook records the affirmative answer of Bardakov and Bryukhanov for braid groups. We give a partial answer: \(\operatorname{Aut}(A)\) is linear for every Artin group \(A\) of finite type without components of types \(E_6\), \(E_7\), \(E_8\), \(F_4\), \(H_3\), \(H_4\) (type \(D_5\) relying on an unrefereed preprint), and the question reduces to linearity of \(\operatorname{Aut}(A/Z(A))\) for these six irreducible types. First, \(\operatorname{Aut}(A)\) is linear over \(\mathbb{C}\) if and only if \(\operatorname{Aut}(A_i/Z(A_i))\) is linear over \(\mathbb{C}\) for every irreducible component \(A_i\) of \(A\). The proof assembles known facts (linearity of \(A\), a lifting argument of Charney and Crisp, parabolic closures after Cumplido, Gebhardt, González-Meneses and Wiest, a lemma of Cumplido and Paris); what we did not find in print is the reduction itself, for reducible groups and in the converse direction. Second, for the types \(A_n\), \(B_n\), \(D_n\) and \(I_2(m)\), \(\operatorname{Aut}(A)\) has a subgroup of finite index isomorphic to \(A/Z(A)\) or to \(A/Z(A)\times\mathbb{Z}\). This follows, by the argument of Bardakov and Bryukhanov, from the published computations of \(\operatorname{Aut}(A)\) by Dyer–Grossman, Charney–Crisp, Gilbert–Howie–Metaftsis–Raptis, Crisp–Paris, Bell–Margalit, Soroko, Castel–Paris and Paris–Soroko (for \(D_5\) a preprint of 2026), and may be regarded as folklore. For the six exceptional types, finiteness of \(\operatorname{Out}(A/Z(A))\) would suffice; we found no computation of it in print. This is an unrefereed note.