Unit sphere and Ricci curvature Why is it that on the unit sphere the Ricci curvature Ric = g (where g is the metric defined on the unit sphere) ?
 A: Here's one way to view this:
For any manifold, the Ricci curvature is an invariant of the metric, and so it must be invariant under the isometries (symmetries) of the metric. Now, the group of isometries of the $n$-sphere $\Bbb S^n$ (endowed with the round metric $g$ is $O(n + 1, \Bbb R)$, which acts transtively, and the isotropy group preserving any point $p \in \Bbb S^n$ can be identified with $O(n, \Bbb R)$. Appealing to representation theory tells us that the only bilinear forms on $T_p \Bbb S^n$ invariant under the action of $O(n, \Bbb R)$ are the multiples of the value $g_p$ of the metric $g$ at $p$; in particular, the value $\text{Ric}_p$ of the Ricci curvature at $p$ is thus given by $$\text{Ric}_p = \lambda(p) g_p.$$ Since this holds just as well for any point $p$, this defines a function $\lambda : \Bbb S^n \to \Bbb R$. Then, because the action of the isometry group is transitive, $\lambda$ is constant, and thus,
$$\text{Ric} = \lambda g$$ for some $\lambda$. Taking a trace of both sides gives that the scalar curvature is $\lambda n$. But for $n > 0$ the round metric has positive scalar curvature, and hence $\lambda > 0$. Finally, we can regard the value $\lambda = 1$ pointed out in the question as a normalization in its own right (except when $n = 0, 1$, for which the Ricci curvature is zero): For $n = 2$ it characterizes the round unit sphere metric among its multiples, that is, among the spheres of positive radii; more generally, the unit $n$-sphere has Ricci curvature $(n - 1) g$.
Remark Metrics $h$ that satisfy $$\text{Ric} = \lambda h$$ for some constant $h$ is called Einstein metrics; the study of these is a rich and old topic (at least by the standards of differential geometry), in part because of their relevance to general relativity.
