Who can give an example of an affine locally symmetric space that is not a Riemanian locally symmetric space?



Example of affine locally symmetric space
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).


