How can isotropic manifolds be of non-constant curvature? Wikipedia states that an isotropic manifold is homogeneous, that is, all points are the same.
But later on, Wikipedia (or this answer) states that not all isotropic manifolds are of constant curvature (e.g. $\Bbb{CP}^n$). How can this be? If two points have distinct curvature, how can they be the same?
 A: An isotropic manifold $M$ is one for which the isometry group $G$ acts transitively on $M$, and for which the isotropy group $G_p$ acts transitively on spheres in $T_p M$ (for some and hence all $p\in M$).
While the curvature at any two points is "the same" in the sense that we may write the Riemann tensor $R$ as $R(p)=g^*R(q)$ for any $p,q\in M$, this doesn't fully constrain the sectional curvature, since different planes in $T_pM$ may have different sectional curvatures. The isotropy condition may not either, since while $G_p$ acts transitively on lines in $T_pM$, it won't necessarily act transitively on planes in $T_pM$.
As an example, consider the complex projective space $\mathbb{CP}(n)$ with the Fubini-Study metric. The isometry group is the projective unitary group $G=PU(n+1)=U(n+1)/U(1)$ (since the FS metric is $U(n+1)$ invariant by definition and the center $U(1)$ is the kernel of the action on $\mathbb{CP}(n)$); one can verify this action is transitive. Fixing the line $l=\{(z,0,\dots,0)\in\mathbb{C}^{n+1},z\in\mathbb{C}\}$, one can show that the isotropy group is isomorphic to $U(n)$, and thus the isotropy representation $G_p\to GL(T_pM)$ is just the defining representation of $U(n)$ (though we can forget the complex structure and consider it an action on $\mathbb{R}^{2n}$). This representation is transitive on lines, but not on planes for $n\ge 2$. The second claim is evident since the sectional curvature is invariant under this representation, but is not constant on the space of plains $Gr(2,T_pM)$ (en explicit formula is given here).
As for why this is, it seems related to the fact that it is "harder" for the isotropy representation to act transitively on $Gr(2,T_pM)$ than on $Gr(1,T_pM)$, by dimension counting if nothing else. It will if $G_p\cong SO(\dim M,\mathbb{R})$, but in general $G_p$ is only a subgroup of the orthogonal group.
I'm not familiar with the classification of istropic spaces, but it seems that those with constant sectional curvature form only a fraction of the full picture. For instance, if we restrict to the compact, complete, and simply connected case, the only isotropic spaces of constant sectional curvature are round $S^n$, while the other classes $\mathbb{CP}(n)$, $\mathbb{HP}(n)$, and $\mathbb{OP}(2)$ have no constant curvature metrics.
