Let $\Sigma(p,q,r)$ be the Brieskorn homology 3-sphere with $\frac{1}{p}+\frac{1}{q}+\frac{1}{r}<1$ (so not the 3-sphere or the Poincare sphere). The fundamental group is given by $$ \pi_1(\Sigma(p,q,r)=\langle a,b,c\, |\, a^p=b^q=c^r=abc \rangle $$, the centrally extended triangle group for $(p,q,r)$. The isotopy group $\pi_0(Diff(\Sigma(p,q,r)))$ must be finite as well as the outer automorphisms of the fundamental group $Out(\pi_1(\Sigma(p,q,r))$. Furthermore the map $$\pi_0(Diff(\Sigma(p,q,r)))\to Out(\pi_1(\Sigma(p,q,r))$$ is injective ( I found this result in D. McCullough, Virtually geometrically finite mapping class groups of 3-manifolds, J. Differential Geom. 33 (1991), no. 1, 1–65.) Now to my questions:

  1. Are there new results to show that this map is an isomorphism? Or, what is the order of the isotopy group?
  2. In particular, I'm interested in the example $\Sigma(2,5,7)$ (bounding a contractable smooth 4-manifold).
  3. I do not find any result about the outer automorphism of tre centrally extended triangle group. What is known?

In fact, much more is known: $Diff(\Sigma(p,q,r))\simeq Isom(\Sigma(p,q,r))$ when $\frac1p+\frac1q+\frac1r <1$ by a result of McCullough and Soma. The metric is a homogeneous metric on $\Sigma(p,q,r)$ modeled on the homogeneous space $\widetilde{SL_2(\mathbb{R})}$. In this case, the isometry group preserves the Seifert fibering, so descends to isometries of the quotient 2-orbifold $S^2(p,q,r)$, which is made from two hyperbolic triangles with angles $\pi/p,\pi/q,\pi/r$. There is a reflection isometry fixing $(p,q,r)$, which lifts to an action on the fiber by orientation reversal. The quotient of $\Sigma(p,q,r)$ by such an isometry is an orbifold whose underlying singular locus is a Montesinos link.

If $p=q\neq r$, there is also an orientation preserving isometry given by an involution which exchanges the two orbifold points and the two isometric triangles. This isometry lifts to $\Sigma(p,q,r)$ preserving the fibers. If $p=q=r$, then there is a dihedral group action, together with the reflection.

Thus, the identity component of $Isom(\Sigma(p,q,r))\cong S^1$, obtained by rotating along the fibers. Also, there is an extension by a finite group of isometries of $S^2(p,q,r)$. In the case of $\Sigma(2,5,7)$, this is just an extension by $\mathbb{Z}/2\mathbb{Z} \cong \pi_0(Diff(\Sigma(2,5,7)))$.

  • $\begingroup$ Thanks a lot Ian. The answers is right what I expected. Is the isotopy group $\mathbb{Z}/2\mathbb{Z}=\pi_0(Diff(\Sigma(2,5,7)))$ the number of components of the isometry group for the $\tilde{SL_2(\mathbb{R})}$ isometry group (I read it in Scott article about the geometric structures on 3-manifolds for this geometry.) $\endgroup$ – Torsten Asselmeyer-Maluga Jan 8 '14 at 20:51
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    $\begingroup$ @TorstenAsselmeyer-Maluga: I think that's correct: there are two components of isometries of $\widetilde{SL_2(\mathbb{R})}$, determined by the action on the orientation of the fibers. $\endgroup$ – Ian Agol Jan 8 '14 at 21:20

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