Example of affine locally symmetric space Who can give an example of an affine locally symmetric space that is not a Riemanian locally symmetric space?
 A: Ben's answer to the question is perfectly fine, but one might also want an example that is not even pseudo-Riemannian, i.e., for which the connection $\nabla$ does not admit any nondegenerate symmetric $2$-form that is $\nabla$-parallel.
The simplest such example is in dimension $2$:  Let $M=\mathbb{R}^2$ with coordinates $x$ and $y$, and let $\nabla$ be the torsion-free connection whose geodesics are either of the form
$$
\bigl(x(t),y(t)\bigr) = \bigl(x_0 + at,\ y_0 \cos(at) + (b/a) \sin( at)\bigr)
$$
where $a\not=0$ or of the form $\bigl(x(t),y(t)\bigr) = (x_0,\ y_0 + bt)$.  (Note that, as $a\to 0$, the `generic' formula for the geodesics converges to the special case.)  Note that all of the geodesics leaving $(x_0,y_0)$, other than the special ones with $x(t)$ constant, must also pass through the points $(x_0+k\pi,(-1)^ky_0)$ for $k$ an integer.
The  affine symmetry group (which has 4 components and has dimension $4$) consists of maps of the form
$$
(x,y)\mapsto \bigl({}\pm x+a,\ r\,y+ b_0\cos(x) + b_1\sin(x)\bigr)
$$
with $r\not=0$, $a$, $b_0$, and $b_1$ arbitrary constants.  The geodesic symmetry at $(0,0)$ is $(x,y)\mapsto (-x,-y)$. 
The quadratic form $\mathrm{d}x^2$ is $\nabla$-parallel, but there is no $\nabla$-parallel pseudo-Riemannian metric. The holonomy of $\nabla$ acts indecomposably but not irreducibly, as it preserves the line field $\mathrm{d}x = 0$.
There is a 'dual' symmetric space to this one, got by replacing cosine and sine by their hyperbolic counterparts, with similar properties (except for the geodesic focusing).
A: Having had hard time visualizing the example from the answer by Robert Bryant, I decided to make some graphs for it. They turned out so beautiful that I decided to share them.
Geodesics passing through $(0,1)$:

Geodesics passing through $(1,-0.3)$:

And for the "dual" example, geodesics passing through $(0,-0.1)$:

Geodesics passing through $(1,1)$:

A: Anti de Sitter space is not Riemannian, as the stabilizer of a point is not compact 
