Example of affine locally symmetric space

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

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).

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 $(0,1)$:

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