This is a follow-up to the question

Rigidity/flexibility of Nil-, Sol-, $\widetilde{\rm SL}_2$- structures on closed 3-manifolds

From the answers/comments there and from an excellent survey by Bonahon ("Geometric Structures on 3-Manifolds") I now figured out what are the deformation spaces in the cases of Seifert geometries. What still remains unclear to me is the case of Sol-geometry.

So, let $M$ be a closed 3-manifold admitting Sol-geometry. Thereby, it is a torus bundle over circle with Anosov monodromy. We say that two Sol-structures on $M$ are equivalent if they differ by a diffeomorphism isotopic to identity. Can there be non-equivalent Sol-structures on $M$? If yes, what is known about the deformation space?

(Note that the respective section of Bonahon's survey about rigidity/flexibility is called "Seifert geometries and Sol", however, unfortunately, it seems to me that despite the title, the case of Sol-geometry is omitted there.)

  • $\begingroup$ I'd like to see the question restated in group-theoretic terms. Let $G$ be the maximal isometry group of SOL (it is 3-dimensional with 8 components, with maximal compact subgroup $K$ finite of order 8). Is the question equivalent to whether there are two non-conjugate torsion-free lattices $\Gamma_1,\Gamma_2$ in $G$ with $\Gamma_1\backslash G/K$ and $\Gamma_2\backslash G/K$ diffeomorphic? $\endgroup$
    – YCor
    Feb 26 at 9:58
  • $\begingroup$ @YCor: Since Sol is contractible and homotopy-equivalent solve-manifolds are known to be homeomorphic (Johannson?), I think the question can be phrased as the exact analogue of Mostow rigidity: if two lattices in $G$ are isomorphic, must they be conjugate? (Looking at Sam Nead's answer, I guess the answer is no, but only up to a 1-parameter ambiguity.) $\endgroup$
    – HJRW
    Feb 26 at 10:35
  • 1
    $\begingroup$ @HJRW Thanks. The automorphism group can maybe viewed as the wreath product $G=(\mathbf{R}\rtimes\mathbf{R}^*)\wr(\mathbf{Z}/2\mathbf{Z})$, in which the maximal isometry group $H$ can be identified to the kernel of the homomorphism $G\to\mathbf{R}_{>0}$, $((x_1,t_1),(x_2,t_2),\varepsilon)\mapsto |t_1t_2|$. Then one might wonder if isomorphic torsion-free lattices of $H$ are conjugate inside $G$ — indeed confirming this 1-parameter ambiguity. One might consider only lattices of $H$ meeting the abelian normal subgroup as covolume 1 subgroup to wonder if isomorphic implies conjugate in $H$. $\endgroup$
    – YCor
    Feb 26 at 10:46

1 Answer 1


Let's look at the case of Solv manifolds $M$ which are torus bundles over the circle. (All Solv manifolds are finitely covered by such bundles.) The square of the length element is as follows:

$$ds^2 = e^{2t} dx^2 + e^{-2t}dy^2 + dt^2$$

(See the bottom of page 470 of Peter Scott's article The geometries of 3-manifolds.) The directions $\partial/\partial x$ and $\partial/\partial y$ give the tangent plane to the two-torus fibre $T \subset M$. Thus the fibering of $M$ is deduced from the metric. Let $A \colon T \to T$ be the resulting monodromy. We may assume that the monodromy $A$ fixes a point of $T$ (which we can call the "origin" of the $xy$-plane). By taking a cover (done above) we have ensured that $A$ does not rotate the eigendirections at the origin. It follows that the length of the closed loop through the origin depends (only) on (the eigenvalues of) $A$. The only parameter left is the area of $T$. Scaling this gives a one-parameter family of Solv metrics on $M$


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