Which $F_3$-by-$\mathbb{Z}$s are CAT(0)?
Leo Delage
TL;DR
This work resolves which free-by-cyclic groups $F_3\rtimes_\phi Z$ with a unipotent-polynomially-growing automorphism $\phi$ fixing a rank-2 factor admit a CAT(0) suspension, correcting a claim in Sam06. It proves a precise criterion: the suspension is CAT(0) iff $\phi$ is the identity or the twisting word $w(a,b)$ lies outside the commutator subgroup $[F_{\{a,b\}},F_{\{a,b\}}]$, and it connects these results to known non-CAT(0 obstructions of Gersten-type. The paper then develops CAT(0) structures for new $F_3$-by-$Z$ examples using a thickened Bridson tree-of-spaces construction, yielding both positive CAT(0) cases and explicit obstructions when $w$ lies in the commutator. Overall, it extends the catalog of CAT(0) free-by-cyclic groups and clarifies when polynomially growing automorphisms fail to produce CAT(0) suspensions, with implications for cocompact cubulation questions.
Abstract
In this note we point out a mistake in theorem 4.4 of [Sam06], which states that a semidirect product $F_3\rtimes_φ\mathbb{Z}$ whose defining automorphism $φ$ is unipotent-polynomially-growing and fixes a free factor of rank $2$ is a CAT(0) group. We give and prove the corrected statement: such a group is CAT(0), if and only if $φ$ is the identity or if the element of $F_2$ twisting the non-fixed generator is not in the commutator subgroup of $F_2$. This gives new examples of free-by-cyclic groups that cannot act properly by semisimple isometries on a CAT(0) space, that are similar to {Gersten}'s examples [Ger94]. We also construct CAT(0) structures for new examples of $F_3$-by-$\mathbb{Z}$s by thickening the strips in Bridson's tree of spaces construction [BH99].
