Outer automorphism group of Brieskorn homology sphere?

Question

In this post, it is discussed how a Brieskorn homology sphere $\Sigma(a_1,a_2,a_3)$ with $\displaystyle \frac{1}{a_1}+ \frac{1}{a_2}+ \frac{1}{a_3} < 1$ is an aspherical manifold with a superperfect fundamental group with non-trivial center. Would anyone know what the outer automorphism groups of their fundamental groups are? I'm looking to do semidirect products with these groups as the kernel group with another group ($\mathbb{Z} \times \mathbb{Z}$) as the quotient group, so I need to know the outer automorphism groups of these groups.

EDIT: Would SnapPy be able to help with this? https://www.math.uic.edu/t3m/SnapPy/

Answer 1

The answer is that the outer automorphism group is $\mathbb{Z}_2$. (Compare Ian Agol's answer to What is the order of the isotopy group of the Brieskorn homology 3-sphere?). This is obtained by cobbling together several standard facts. Much of this can be found in Boileau and Otal, Scindements de Heegaard et groupe des homéotopies des petites variétés de Seifert. (French) [Heegaard splittings and homeotopy group of small Seifert manifolds] Invent. Math. 106 (1991), no. 1, 85–107.

The first is that the outer automorphism group of the fundamental group coincides with the diffeomorphism group modulo isotopies. The second is that any self-diffeomorphism of a Brieskorn sphere preserves orientation; see Neumann-Raymond, Seifert manifolds, plumbing, μ-invariant and orientation reversing maps. Boileau-Otal show that up to isotopy, any orientation preserving diffeomorphism preserves the fibers (setwise) and hence gives an automorphism of the 2-orbifold quotient by the circle action.

Because of the relatively prime condition (easy to verify by homology calculations) there is only one automorphism of that orbifold; it is induced from a reflection in the 2-sphere and so flips the orientation. The lift to the Brieskorn sphere also reflects the fiber so is orientation preserving.