Given a $n$-dimensional connected Riemannian manifold $(M,g)$, its symmetry group $G$ can be considered as a subbundle of orthonormal frame bundle of $M$ (which I call $F_OM$), yielding: $$\dim G\le \dim F_OM=\dim M +\dim O(n)={n+n(n-1)\over 2}=\frac{n(n+1)}{2}$$ (Here the embedding $\phi:G \hookrightarrow F_OM$ is defined by singling out an arbitrary point $p \in M$ and orthonormal frame $(v_1,...,v_n) \in F_{O,p} M$ and defining $\phi: g \mapsto (gp,g_*v_1,...,g_*v_n)$. However, it takes some work to verify that this map is indeed injective.)
These considerations made me wonder which manifolds have a maximally large symmetry group, i.e. which $M$ do satisfy $\dim G= \frac{n(n+1)}{2}$. Of, course $\Bbb R^n,\Bbb S^n$ and $\Bbb H^n$ have got this property, but are there any other exotic examples?
(Some thoughts of mine on this problem: One can see that $M$ is homogenuous and isotropic. Each stabiliser of a point has to be isomorphic to either $SO(n)$ or $O(n)$. In particular, $M$ has constant curvature, which implies that $M$ has to be one of $\Bbb R^n,\Bbb S^n$ and $\Bbb H^n$ if $M$ is complete and simply-connected.)
The complete list of such manifolds consists of the three simply connected ones as well as the real-projective space with the constant curvature metric. Here is why. Suppose that $\pi$ is the group of covering transformations of M, identified with the subgroup of the full group G of isometries of the universal cover of M. Then $\pi$ commutes with the connected component $G_o$ of identity in G, otherwise the isometry group of M has dimension less than G. Now, you just have to do case by case analysis. You see that only in the spherical case the group $G_o$ has nontrivial centralizer in G, namely, it is the center of G, generated by the antipodal map. This gives us the projective space.