3-dimensional Riemannian manifolds with 4-dimensional isometry group Is there a list of all 3-dimensional, connected Riemannian manifolds with 4-dimensional isometry group?
 A: 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 $6$.)
A: Let me rather address the question as follows: what are the connected Riemannian 3-folds whose isometry group has dimension $\ge 4$.
Robert has proved in a comment that the isometry group is transitive (and hence so is its unit component). Therefore, this essentially reduces to classifying those pairs $(G,K)$, with $G$ a connected Lie group of dimension $4$, $K$ a compact subgroup of codimension $3$ with trivial core (i.e., $G$ acts faithfully on $G/K$). 
I say essentially because a remaining part is to classify $G$-invariant Riemannian metrics on $G/K$ up to isometry (and except in the flat case there are continuum many, just by considering, the supremum of the sectional curvature; also in most case there are probably many up to homothety too).
Then $K$ acts faithfully on the tangent space at the basepoint, hence appears as subgroup of $\mathrm{O}(3)$. Hence its dimension is $0,1,3$. Dimension $0$ ($\dim G=3$)is excluded by assumption. Dimension $3$ ($\dim G=6$) implies that the sectional curvature is constant and this implies that the manifold is covered by a homothetic of $\mathbb{H}^3$, $\mathbb{E}^3$ or $\mathbb{S}^3$. Using (global) homogeneity, it is actually is homothetic to either $\mathbb{H}^3$, $\mathbb{E}^3$ or $\mathbb{S}^3$, or $\mathbb{P}^3$. I found it not natural to exclude this case even if it's the easiest part (indeed some constant curvature cases still appear in $\dim(G)=4$ in the non-simply connected case).
Next, assume $\dim(G)=4$. First case (1): $K$ is maximal compact. Since $G$ is connected, so is $K$.
Suppose (1a) that $G$ is not solvable. Then $G$ is locally isomorphic to $\mathrm{SL}_2(\mathbf{R})\times\mathbf{R}$, that is, quotient of $\widetilde{\mathrm{SL}_2(\mathbf{R}}\times\mathbf{R}$ by a central subgroup $Z$, such that the maximal compact subgroup of the quotient by $Z$ has dimension 1 
and trivial core. So $Z$ is infinite cyclic, and up to automorphism this yields two cases, denoting by $\xi$ one generator of the cyclic center of $\widetilde{\mathrm{SL}_2(\mathbf{R}}$: $Z$ is generated by $(\xi,0)$ or $(\xi,1)$. In the first case, we get $(\mathrm{PSL}_2(\mathbf{R})\times\mathbf{R},\mathrm{PSO}(2))$, so that the quotient can be viewed as $\mathbb{H}^2\times\mathbb{E}^1$ (I'm not sure whether it carries exotic metrics, i.e., not isometric to a product of constant curvature metric on factors), and in the second case we get the "$\widetilde{\mathrm{SL}_2(\mathbf{R}}$" geometry (I guess it carries non-homothetic invariant metrics but am not sure). 
(1b) $G$ is solvable. Then it can be written as $N\rtimes K$ with $N$ solvable; since $K$ is maximal compact, $N$ is simply connected. Classification of 3-dimensional real solvable Lie algebras with a positive-dimensional compact group of automorphism yields the following list:


*

*$N=\mathbf{R}^3$. In that case $G/K$ is actually 3-dimensional Euclidean;

*$N=$ Heisenberg;

*$N=$ $\mathbf{R}\rtimes\mathbf{R}$ with action by homotheties; then for at least one invariant metric, $G/K$ is homothetic to $\mathbb{H}^3$, but I guess that for most invariant metrics, it does not have constant curvature.


In all cases above except 1., $K$ is a maximal connected compact subgroup in $\mathrm{Aut}$ and they are all conjugate so there is essentially no choice; in case 1. it still holds that all 1-dimensional tori of automorphisms are conjugate so again the choice does not matter. 
Next suppose (2) $G$ is compact. If $G$ is a torus, then $K$ cannot have a trivial core. Otherwise, $G$ is locally isomorphic to $\mathrm{SO}(3)\times\mathbf{R}$. This gives three possibilities for $G$: $\mathrm{SU}(2)\times\mathrm{SO}(2)$, the same modulo a diagonal element of order $2$, $\mathrm{SO}(3)\times\mathrm{SO}(2)$. In each case, fix an injective homomorphism $i:\mathrm{SO}(2)\to S$, where $S$ is the simple factor; note that in the second case the killed element is $(i(-1),-1)$. Then for every $n\ge 0$ we can choose $K=K_n$ as $\{(i(z),z^n):z\in\mathrm{SO}(2)\}$. In the first case it does not have trivial core, but in the other two cases it does. When $n=0$ we obtain the product cases ($\mathbb{S}^2\times\mathbb{E}^1$ and $\mathbb{P}^2\times\mathbb{E}^1$). When $n\ge 1$, we obtain in both the second and the third case, $\pi_2(G/K_n)=0$. In the second case, it yields $\pi_1(G/K_n)\simeq C_n$ (cyclic of order $n$); in the third case we get $\pi_1(G/K_{n})\simeq C_{2n}$. These are probably lens spaces but I haven't checked further. The simply connected case is $G/K_1$ in the second case, which yields a 3-sphere (probably not always with constant curvature metrics): this might be the Berger spheres mentioned in Bullet51's (partial) answer. In the third case, $G/K_1$ corresponds to metrics on $\mathbb{P}^3$ (including constant curvature ones, but probably not only).
There are further cases in this (2), namely considering when $K$ is not connected. Hence, in the above cases, we have to consider the normalizer of $K$. It is 2 dimensional and probably has 1 or 2 components. Then one has to determine 1-dimensional subgroups of the normalizer with trivial core. Probably there is 1 or 2, I haven't looked in details at the moment.
(3) Finally there is the case when $G$ is not compact and has a maximal compact subgroup of dimension $\ge 2$. Suppose (3a) that $G$ is not solvable. Then it is locally isomorphic to $\mathrm{SL}_2(\mathbf{R})\times\mathbf{R}$. I haven't looked in details. After replacing $G/K$ by a manifold finitely covered by $G/K$, we can reduce to the case $G=\mathrm{PSL}_2(\mathbf{R})\times\mathrm{SO}(2)$. Again using $i$ as previously, we get families of compact subgroups $K_n$. For $n=0$, this yields $\mathbb{H}^2\times$(circle). For $n\ge 1$ this yields manifolds infinitely covered by the geometry $\widetilde{\mathrm{SL}_2(\mathbf{R})}$, which in this specific case with $\mathrm{PSL}_2(\mathbf{R})\subset G$ have infinite cyclic fundamental group. For $n=1$ this gives $\mathrm{PSL}_2(\mathbf{R})$ itself with left-invariant (right-$\mathrm{PSO}_2$-invariant) metric. Of course before the reduction this yields more examples, such as finite covers of the latter.
(3b) $G$ is solvable, not compact, with maximal compact subgroup of dimension $\ge 2$. Then $G/K$ has infinite fundamental group. We thus see that the universal covering of the manifold is one of the cases in (1b). This yields the possibilities:


*

*$N=\mathbf{R}^2\times\mathbf{R}/\mathbf{Z}$. In that case $G/K$ is product of the Euclidean plane with a circle (this is the only possibility up to homothety, as we can obtain this as quotient of $\mathbb{E}^3$ by a translation);

*$N=$ Heisenberg modulo quotient by a discrete central subgroup.


Sorry for the missing details but I hope this is useful anyway.
A: The simply connected homogeneous 3-folds with 4-dimensional isometry group are well known:
They are


*

*The product spaces $\mathbb S^2\times\mathbb R$ and $\mathbb H^2\times\mathbb R$; 

*The Berger spheres;

*3-folds with $Nil_3$ isometry, e.g. Heisenberg space; and

*3-folds with $SL_2(\mathbb R)$ isometry.
See
F. Bonahon. Geometric structures on 3-manifolds, Handbook of geometric topology, 93-164,
North-Holland, Amsterdam, 2002
P. Scott. The geometries of 3-manifolds, Bull. London Math. Soc., 15 (5): 401-487, 1983
W. Thurston. Three-Dimensional Geometry and Topology, Princeton, 1997
