# What makes a Riemannian manifold “directionally homogeneous”?

Consider the three Riemannian 2-manifolds: Euclidean plane, 2-sphere and Poincaré disk. Each of them has the following property (D):

For any point $p$ on the surface $S$ and any two unit tangent vectors $v,w\in T_pS$, there exists an isometry $f:S\to S$ such that $f(p)=p$ and the pushforward $(f_*)_p:T_pS\to T_pS$ sends $v$ to $w$.

This property is not seen in arbitrary smooth Riemannian 2-manifolds. For instance, consider the infinite cylinder $S=\{(x,y,z)\in\Bbb R^3:x^2+y^2=1\}$ with metric inherited from $\Bbb R^3$. Even though the space is homogeneous in the sense that for any two points $p_1,p_2\in S$, there exists an isometry $f:S\to S$ sending $p_1$ to $p_2$, there is no isometry whose pushfoward sends a unit tangent vector that goes horizontally (along the direction of $xy$-plane) to a unit tangent vector that goes vertically (parallel the direction of $z$-axis). If there is, such an isometry would send a horizontal circle on $S$ (which is a geodesic generated by a horizontal tangent vector) to a vertical line on $S$ (which is a geodesic generated by a vertical tangent vector), and this is absurd.

On the cylinder, directions at a point are not "equal" to each other, and I say the cylinder is not "directionally homogeneous".

So what are some necessary and/or sufficient conditions for a smooth Riemannian manifold to have property (D)? Please note that I have only taken a course on differential geometry at the level of do Carmo's Differential Geometry of Curves and Surfaces.

Using the terminology of isotropic from the comment of @AnthonyCarapetis, one sufficient condition for an $m$-dimensional Riemannian manifold $M$ to be isotropic is that it be complete, simply connected, and have constant sectional curvature $k$. These manifolds are all known:
• If $k>0$ then $M$ is isometric to an $m$-sphere of radius $1/\sqrt{k}$, i.e. the unit $m$-sphere with metric multiplied by $1/\sqrt{k}$;
• If $k=0$ then $M$ is isometric to Euclidean space $\mathbb{R}^m$;
• And if $k<0$ then $M$ is isometric to hyperbolic $m$-space with metric multiplied by $1/\sqrt{|k|}$.
These are not all possible examples, however. For another example, the quotient of the $m$-sphere of radius $1/\sqrt{k}$ by the antipodal map gives a Riemannian metric on real projective $m$-space which is isotropic.