I'm aware of Bianchi's (local) classification of homogenous 3-manifolds into the Bianchi types I through IX, and I can follow the algebra for classifying the Lie algebras. However, I still can't visualize the different spaces, except for the simpler ones. For example, I can see that type I is just locally Euclidean $E^3$. Other homogenous spaces that I can visualize are locally spherical $S^3$, which I think corresponds to type IX? I can further imagine $S^2\times\mathbb{R}$ as a homogenous space, but I can't tell if that corresponds to one or more Bianchi types by combining the two spaces in different ways.


How do I go about understanding the (local) geometry of the different Bianchi types and visualizing them?


EDIT: Added (local) above. I am not asking anything topological, I am just interested in the local geometry.



Visualizing Bianchi type/homogenous spaces
For a different viewpoint from the excellent treatments by Scott and Thurston of 3-dimensional geometries, if you are trying to get a feel for the homogeneous Riemannian $3$-manifolds (which, as noted, were first classified by Bianchi), you might want to try looking at them from the point of view of their most basic invariants, their curvatures. This provides a natural classification and made it easy for me to organize the information when I was learning the subject. From this point of view, it doesn't quite divide the way that Bianchi did it, but you shouldn't have any trouble making the comparison. In outline, it goes like this:


Let $(M,g)$ be a connected Riemannian $3$-manifold that is homogeneous, i.e., the group $G$ of isometries of $M$ acts transitively on $M$. Consider the Ricci curvature $\mathrm{Ric}(g)$. This is also a quadratic form on $M$ and, as such, relative to $g$, it has $3$ real eigenvalues $\lambda\_1$, $\lambda\_2$, and $\lambda\_3$, which are constant functions on $M$ because of the homogeneity hypothesis.


Case 1: $\lambda\_1=\lambda\_2=\lambda\_3=\lambda$. In this case, the sectional curvature of $g$ is constant, and so $M$ is a space form, of elliptic, parabolic, or hyperbolic type, depending on whether $\lambda$ is positive, negative, or zero. These are easily visualized, as models for them are straightforward to construct.


Case 2: $\lambda\_1\not=\lambda\_2=\lambda\_3$. In this case, there is a well-defined $G$-invariant line field $L$ on $M$ which is the eigendirection of multiplicity $1$ for $\mathrm{Ric}(g)$. After passing to a double cover, if necessary, there is a unit vector field $Y$ tangent to $L$ and one can compute the Lie derivative of $g$ in the direction $Y$. There are two subcases: 


Case 2a: $L\_Yg=0$ (i.e., $Y$ is a Killing field). In this case, the isometry group of $g$ has dimension $4$, the integral curves of $Y$ are geodesics and the set of these geodesics is a surface $S$. The projection $M\to S$ with fibers tangent to $L$ is a Riemannian submersion, and the induced metric on $S$ has constant Gauss curvature. Then $M$ has the structure of a principal bundle over $S$ (with group action generated by $Y$) with a connection whose curvature $2$-form is a constant multiple of the area form on $S$. There is a $2$-parameter family of such metrics.


Case 2b: $L\_Yg$ is nonzero, in which case, it turns out that there is a coframing $\omega\_1$, $\omega\_2$, and $\omega\_3$ defined up to sign on $M$ so that $g = {\omega\_1}^2+{\omega\_2}^2+{\omega\_3}^2$, while $\mathrm{Ric}(g) = \lambda\_1\ {\omega\_1}^2$ (i.e., $\lambda\_2=0$) and $L\_Yg = \mu\ ({\omega\_2}^2-{\omega\_3}^2)$ for some constant $\mu>0$. This coframe field is invariant under $G$ (at least, up to sign), so $G$ has dimension $3$ and, essentially, the metric $g$ is a left-invariant metric on $G$, which is a covering of $M$. Thus, this subcase, which consists of a $2$-parameter family of metrics (parametrized by $\lambda\_1\not=0$ and $\mu>0$) can be covered as a limit of Case 3, which consists of left invariant metrics on $3$-dimensional Lie groups.


Case 3: The $\lambda\_i$ are all distinct, in which case, after passing to a cover and restricting $G$ to its identity component if necessary, one sees that $G$ leaves fixed a homogeneous coframing on $M$, so $g$ is essentially a left-invariant metric on $G$. By the usual algebra that classifies the $3$-dimensional Lie groups, one knows that there will be a $g$-orthogonal $G$-invariant coframing $\omega\_1$, $\omega\_2$, $\omega\_3$ so that $g = {\omega\_1}^2+{\omega\_2}^2+{\omega\_3}^2$ and so that
\begin{aligned}
d\omega\_1 &= a\_1\ \omega\_2\wedge\omega\_3\\\\
d\omega\_2 &= a\_2\ \omega\_3\wedge\omega\_1 + b\ \omega\_1\wedge\omega\_2\\\\
d\omega\_3 &= a\_3\ \omega\_1\wedge\omega\_2 - b\ \omega\_3\wedge\omega\_1
\end{aligned}
where $a\_1$, $a\_2$, $a\_3$, and $b$ are constants satisfying either $a\_1=0$ or $b=0$ (these latter alternatives are necessary in order for the Jacobi identity to be satisfied). If $b=0$, then the condition that the metric $g$ so defined on the corresponding Lie group $G$ have distinct eigenvalues of its Ricci curvature is just that the $a\_i$ be distinct *and* that $a\_1{+}a\_2{+}a\_3\not=2a\_i$ for any $i$. If $a\_1=0$, then the condition that the metric so defined have all Ricci eigenvalues be distinct is that $a\_2\not=a\_3$ *and* $b^2+a\_2a\_3\not=0$. Thus, there are two $3$-parameter families of these metrics in Case 3. Meanwhile, when one removes some of the inequalities, one gets limiting cases that cover Case 2b.


Thus, in all the homogeneous cases, the curvature (together with, sometimes, derivatives of curvature eigendirections) actually produces a natural coframing and/or foliation of the metric $g$, and, because these invariants are so natural, they are a good place to start to study the local geometry of the homogeneous Riemannian $3$-manifolds.


