Is there a list of all 3-dimensional, connected Riemannian manifolds with 4-dimensional isometry group?



3-dimensional Riemannian manifolds with 4-dimensional isometry group
There is a uniform way to describe these Riemannian $3$-folds using the geometry of the Lie group of isometries, as YCor mentioned in his comment. The following description is essentially drawn from the classification of Bianchi:


Let $h\ge0$ and $k\not=h^2$ be real constants and consider the connected, simply-connected Lie group $G\_{h,k}$ of dimension $4$ that has a basis of left-invariant $1$-forms $(\alpha,\omega\_1,\omega\_2,\omega\_3)$ satisfying the structure equations
$$
\begin{aligned}
\mathrm{d}\omega\_1 &= -\alpha\wedge\omega\_2 
+ \phantom{2}h\,\omega\_2\wedge\omega\_3\\
\mathrm{d}\omega\_2 &= \phantom{-}\alpha\wedge\omega\_1 
+ \phantom{2}h\,\omega\_3\wedge\omega\_1\\
\mathrm{d}\omega\_3 &=\phantom{-\alpha\wedge\omega\_2}+ 2h\,\omega\_1\wedge\omega\_2\\
\mathrm{d}\alpha &= (h^2+k)\,\omega\_1\wedge\omega\_2\\
\end{aligned}
$$
Let $H\_{h,k}\subset G\_{h,k}$ denote the connected $1$-dimensional subgroup on which the $\omega\_i$ all vanish. Let $M^3\_{h,k} = G\_{h,k}/H\_{h,k}$ be the homogeneous quotient. Then there exists a unique Riemannian metric $g\_{h,k}$ on $M^3\_{h,k}$ that pulls back to $G\_{h,k}$ to become ${\omega\_1}^2+{\omega\_2}^2+{\omega\_3}^2$. The group $G\_{h,k}$ acts almost effectively on $M\_{h,k}$ preserving $g\_{h,k}$ and is the simply-connected cover of the identity component of the isometry group of $g\_{h,k}$.


Every connected, simply-connected Riemannian $3$-fold with $4$-dimensional isometry group is described this way for some unique $(h,k)$ with $h\ge0$ and $k\not=h^2$. The Ricci tensor of $g\_{h,k}$ pulls back to $G\_{h,k}$ to be the quadratic form
$$
(h^2{+}k)\,{\omega\_1}^2 + (h^2{+}k)\,{\omega\_2}^2 + 2h^2\,{\omega\_3}^2. 
$$
(Nothing goes wrong with the construction when $k=h^2$, but the resulting metric $g\_{h,h^2}$ has constant sectional curvature $h^2$ and hence its full isometry group has dimension $6$.)


