# Possible orders of automorphisms for the Poincare homology sphere

Let $$M^3$$ denote the Poincare homology sphere. I am wondering what the possible orders of (smooth) automorphisms of $$M$$ are (I'm not sure if allowing arbitrary homeomorphisms changes things?). By presenting the space as $$-1$$-surgery on the trefoil, there is a automorphism of order 3 given by the 3-fold symmetry of the trefoil. Similarly, the framed link description of $$M$$ obtained by presenting $$M$$ as the 5-fold cyclic branched cover of $$S^3$$ branched over the trefoil has a 5-fold symmetry. Both of these symmetries can also be seen from the dodecahedron/icosahedron definition of $$M$$.

I believe that the $$5k$$-fold cyclic branched covers of $$S^3$$ branched over the trefoil are all $$M$$ for any $$k \geq 1$$ and therefore, by drawing the resulting framed link representations, we obtain automorphisms of order $$5k$$ for any $$k$$. I am not sure if any of these automorphisms are isotopic to the identity.

What are the possible orders of automorphisms of $$M$$? Which possible orders have an automorphism of that order that is not isotopic to the identity?

• The answer to your question for all spherical 3-manifolds is given by the Rubinstein-McCullough paper. In that paper they compute the isometry groups of spherical 3-manifolds. You can use geometrization to argue any finite-order automorphism is conjugate to such an isometry. Sep 21 '18 at 4:00

Let $$G$$ act smoothly and orientably. The quotient $$M/G$$ is a spherical 3-orbifold. Therefore, by the elliptization theorem, it may be given a metric of constant curvature 1; pulling back, then $$M$$ is given a constant curvature metric for which $$G$$ acts by isometries. (This was known to Thurston when the action of $$G$$ carried fixed points, and due to Perelman when $$G$$ acts freely.)

So we may just ask what the finite groups of isometries of $$S^3/I$$ is, where $$I$$ is the binary dihedral group. Consider the subgroup $$\text{Isom}_I(S^3)$$ of isometries of $$S^3$$ which normalize the (right) action of $$I$$, in the sense that $$f(xi) = f(x)i'$$ for some $$i' \in I$$. It is easy to see that every such function is a lift of an isometry of $$S^3/I$$, and there is a short exact sequence $$1 \to I \to \text{Isom}_I(S^3) \to \text{Isom}(S^3/I) \to 1$$.

Now recall that $$\text{Isom}(S^3) = SO(4) = S^3 \times_{\pm 1} S^3$$, the two factors acting by $$(q_1, q_2) \cdot v = q_1 v q_2^{-1}$$. The condition above on the action of $$(q_1, q_2)$$ becomes "For fixed $$i$$, there is an $$i'$$ so that $$q_1 v i q_2^{-1} = q_1 v q_2^{-1} i'$$ for all $$v$$." This is true if and only if $$q_2 i q_2^{-1} = i'$$ for every $$i$$. This implies $$q_2 \in N_{SU(2)}(I) = I$$.

Now we see that $$\text{Isom}(S^3/I) = (S^3 \times_{\pm 1} I)/(1 \times I)$$; the quotient is $$S^3/\pm 1 = SO(3)$$.

So every finite group of automorphisms is conjugate to the action of some subgroup of $$SO(3)$$, acting on $$S^3/I = SO(3)/(I/\pm 1)$$ by left multiplication. One concludes by listing all finite subgroups of $$SO(3)$$; this includes, of course, all cyclic groups.

The Smale conjecture (certainly known now for $$\Sigma(2,3,5)$$, but I don't know the history well enough to give the correct attribution) dictates that $$\text{Isom}^+(M) \to \text{Diff}^+(M)$$ is a homotopy equivalence when $$M$$ is a spherical 3-manifold. So $$M$$ has trivial mapping class group.

• I would be surprised if you couldn't extend this to the case of compact positive-dimensional group actions (it should be easier...), but I have put no effort into this.
– mme
Sep 20 '18 at 20:17