Is there a graph manifold (https://en.wikipedia.org/wiki/Graph_manifold) that doesn't admit an orientation reversing involution? If so, what would be a simple example?
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2$\begingroup$ There's plenty. You can answer this question completely with a little leg work. The start of the inductive process is to answer it for Seifert-fibered spaces. So why not start with something simple like lens spaces? Automorphisms of lens spaces preserve their genus $1$ Heegaard splitting, up to isotopy. $\endgroup$– Ryan BudneyCommented Nov 2, 2021 at 2:20
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$\begingroup$ Thanks @RyanBudney , I started with the most trivial case - $S^1$-bundles over surfaces, and they all have such an involution. Following your suggestion, I've just tried with lens spaces, but don't see why they will give an example. Indeed, a lens space is a quotient $S^3/\mathbb Z_n$ where $\mathbb Z_n$ is acting linearly on $\mathbb C^2$. But it should be possible to find a $\mathbb Z_2$ extension of this action that would also reverse orientation. So your suggestion for lens spaces doesn't seem to work (am I correct?)? If you have a simple example that works, I would be grateful to see it. $\endgroup$– aglearnerCommented Nov 4, 2021 at 9:41
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$\begingroup$ Orientation-reversing involutions can be subtle. There are lens spaces that do not admit them. For the lens space denoted $L_{p,q}$ where $p$ is the order of the fundamental group, usually the condition is written $q^2 \neq -1 \ mod \ p$. $\endgroup$– Ryan BudneyCommented Nov 4, 2021 at 15:39
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1 Answer
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$\mathbb{N}il^3$-manifolds and $\widetilde{SL}$-manifolds are orientable and do not have orientation-reversing self homotopy equivalences. This is most easily seen algebraically. The fundamental group $\Gamma$ of the $S^1$-bundle over the torus $T$ with Euler class a generator of $H^2(T;\mathbb{Z})$ is a central extension of $\mathbb{Z}^2$ by $\mathbb{Z}$. (This is the Heisenberg group of upper triangular matrices in $SL(3,\mathbb{Z})$.) Direct calculation of the automorphisms of $\Gamma$ shows that an automorphism which induces $A\in{GL(2,\mathbb{Z})}$ on the central quotient $\mathbb{Z}^2$ acts by $\det{A}$ on the centre, and hence is orientation-preserving on the extension.
In general, the fundamental groups of such 3-manifolds are virtually central extensions by $\mathbb{Z}$ of orientable surface groups $S$, with non-zero extension class in $H^2(S;\mathbb{Z})$. Moreover, the centre of the group is $\mathbb{Z}$, and so the extension is preserved under any automorphism of the group. Considering the effect of an automorphism on the centre and the quotient shows that the extension class is preserved if and only if the induced automorphisms of the centre and the central quotient $S$ are both orientation-preserving or both orientation reversing.
An analogous situation holds for $\mathbb{S}^3$-manifolds. However the argument breaks down for some lens spaces (as observed by Ryan), as these admit self-maps which do not preserve the Seifert fibration.
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$\begingroup$ I don't understand the remark "(This is the Heisenberg group of upper triangular matrices...)" It seems like you saying that the only group appearing in this way is the Heisenberg group? or are you saying that the Heisenberg group is an example of such a group? $\endgroup$– Nick LCommented Nov 8, 2021 at 13:39