Let M be a compact one-manifold, and let Diff 1+bv + (M ) denote the group of C 1 orientation preserving diffeomorphisms of M whose first derivatives have bounded variation. We prove that if G is a group which is not virtually metabelian, then (G × Z) * Z is not realized as a subgroup of Diff 1+bv + (M ). This gives the first examples of finitely generated groups G, H Diff ∞ + (M ) such that G * H does not embed into Diff 1+bv + (M ). By contrast, for all countable groups G, H Homeo + (M ) there exists an embedding G * H → Homeo + (M ). We deduce that many common groups of homeomorphisms do not embed into Diff 1+bv + (M ), for example the free product of Z with Thompson's group F . We also complete the classification of right-angled Artin groups which can act smoothly on M and in particular, recover the main result of a joint work of the authors with Baik [3]. Namely, a right-angled Artin group A(Γ) either admits a faithful C ∞ action on M , or A(Γ) admits no faithful C 1+bv action on M . In the former case,where Gi is a free product of free abelian groups. Finally, we develop a hierarchy of right-angled Artin groups, with the levels of the hierarchy corresponding to the number of semi-conjugacy classes of possible actions of these groups on S 1 .for the relevant manifolds (cf. [26,30])?Returning to the original motivation, Theorem 1.1 combined with results of Farb-Franks and Jorquera completes the classification of right-angled Artin groups admitting actions of various regularities on compact one-manifolds. Before stating the result, we define some terminology. We write Γ for a finite simplicial graph with vertex set V (Γ) and edge set E(Γ). The right-angled Artin group (or RAAG, for short) on Γ is defined as
SANG-HYUN KIM AND THOMAS KOBERDAAfter some set theoretic computation, one sees the following:Note that we used bd ⊆b ∪d, and alsoc ∩d = ∅. It now suffices for us to prove the following claim:Claim. We have the following:cTo see the first part of the claim, let us consider x ∈ X satisfying cb(x) ∈c ∩ cbd.Then we have x ∈d and cb(x) ∈c. Sincec ∩d = ∅ and b(x) ∈ c −1c =c, we see x = b(x). In particular, we have x ∈b and cb(x) ∈ cb(b ∩d). This proves the first part of the claim. The second part follows by symmetry. Lemma 3.7. If b, c, d ∈ Diff 1 + (I) are given such that supp c ∩ supp d = ∅,Proof. As in the proof of Lemma 3.4, we let B = π 0 supp b and C = π 0 supp c. Letfor each B ∈ B. By Lemma 3.6, we have thatMoreover, for each B ∈ B we note thatClaim. The following set is a finite collection of intervals:We will employ the C 1 -hypothesis for this claim. Let us write