Does 1+1D Gravity Contract to a point if the spatial dimension has the topology of a circle? I image if I put a point mass on a circle, a test particle anywhere on the circle will move toward it.
I am now trying to imagine the shape of space-time for this by embedding it into a 3D space.  I imagine if the area of the circle in the 3D space didn't change over time, we would have a cylinder, but then the time-like geodesics would curl around toward the mass.  But a cylinder doesn't have these geodesics.  It seems like the only way to give it that, is as time increases, the circle contracts toward the mass point.  I'm having trouble imaging this though.
Is it wrong to try embedding 1+1 gravity in a 3D space?
If not, does anyone have a picture of this?
Is it the right picture?
 A: Unfortunately, the curvature tensor vanishes for 1-dimensional gravity. Therefore, according to Einstein's equation, the stress-energy tensor has to be zero, which means NO GRAVITY EXISTS.
EDIT: oops! I mistook it as 0+1 dimensional... I will update in a minute.
UPDATE: I will try to simplify things. First assume the initial space is completely 'circular', which means the curvature everywhere the same on a space-like slice. Then use coordinates $(\theta, t)$, where $\theta\in [0,2\pi)$. The metric will not depend on $\theta$, because of the symmetry. Suppose 
$$
g_{\mu\nu}=\left(\begin{matrix}g_{tt}&g_{t\theta}\\g_{\theta t}&g_{\theta\theta}\end{matrix}\right)=\left(\begin{matrix}g_{1}&g_{2}\\g_{2}&g_{3}\end{matrix}\right)
$$
Where I've used the symmetry of the metric.
The inverse metric would be:
$$
g^{\mu\nu}=\frac1{g_1g_3-g_2^2}\left(\begin{matrix}g_3&-g_2\\-g_2&g_1\end{matrix}\right)
$$
We proceed by calculating the Christoffel coefficients:
$$
\begin{align}
\Gamma^t_{\mu\nu}=\frac1{g_1g_3-g_2^2}\left(
\begin{matrix}\frac12(g_3\dot g_1-g_2\dot g_2)&-\frac12g_2\dot g_3\\-\frac12g_2\dot g_3&\frac12 g_2\dot g_3
\end{matrix}
\right)
\end{align}
$$ Where we denote the derivative with respect to $t$ by a dot (and all derivatives with respect to $\theta$ is zero by symmetry).
If I haven't made any mistake, we can go on to the next four:
$$
\begin{align}
\Gamma^\theta_{\mu\nu}=\frac1{g_1g_3-g_2^2}\left(
\begin{matrix}\frac12(-g_2\dot g_1+g_1\dot g_2)&\frac12g_1\dot g_3\\ \frac12g_1\dot g_3&\frac12 g_2\dot g_3
\end{matrix}
\right)
\end{align}
$$
After that comes the Ricci curvature tensor, which fortunately contains only 1 independent component... Ahhhh I think I need a computer algebra system, back in a minute.

Another question of yours is Is it wrong to try embedding 1+1 gravity in a 3D space?
It is not completely wrong. However, embedding is only possible in 3+1D spacetime, because 1) there exists 2-dimensional (pseudo-)Riemannian manifold that requires 4-dimensional flat space to be embedded in, and 2) it requires the flat space metric to also be pseudo-Riemannian. Therefore, most likely your visualization would be a animated curve.
