Compute curvature tensor from constant sectional curvature Given sectional curvature as a constant, i.e. $\dfrac{R_m(X,Y,Y,X)}{|X|^2|Y|^2-<X,Y>^2} = C$, I want to compute the curvature tensors $R(X,Y)Z$ and $R_m(X,Y,Z,W)$.
I believe I need to use the identity
$$-6R_m(X,Y,Z,W) = \partial_t\partial_s\{R_m(X+sZ, Y+tW, Y+tW, X+sZ) - R_m(X+sW, Y+tZ, Y+tZ, X+sW)\}$$
I can almost see what to do for $R_m(X,Y,Z,W)$ and how this identity and the defintion of sectional curvature are related but can't really see how to finish it off. I don't really even know where to begin with $R(X,Y)Z$.
 A: To fix notation, we define the Riemann tensor as
$$
\operatorname{Rm}(x,y,z,w) = \langle R(x,y)w, z \rangle,
$$
and the sectional curvature as $K(x,y) = \operatorname{Rm}(x,y,x,y) / (|x|^2|y|^2 - \langle x,y \rangle^2)$.
The symmetries of the curvature tensor $\operatorname{Rm}$ mean that the bilinear form $b$ on $\bigwedge^2 V$ defined by
$$
b(x \wedge y, z \wedge w) = \operatorname{Rm}(x,y,z,w)
$$
is actually well defined, and we note that the sectional curvature of the metric is just the quadratic form defined by $b$ (divided by the square of the norm of $x \wedge y$). Thus the sectional curvature determines the curvature tensor, because a quadratic form determines a unique bilinear form by polarization.
We now pull out of our hat the tensor
$$
\operatorname{Rm}'(x,y,z,w) = \langle x, z \rangle \langle y, w \rangle
- \langle x, w \rangle \langle y, z \rangle
$$
(or recall where the idea of constant sectional curvature comes from and calculate the curvature tensor of the sphere). This tensor has sectional curvature
$$
K'(x,y)
= \frac{|x|^2 |y|^2 - \langle x, y \rangle^2}{|x|^2|y|^2 - \langle x, y \rangle^2} = 1.
$$
It thus follows that if our curvature tensor $R$ has constant sectional curvature $C$, then $R = C R'$.
We can now stare at $\operatorname{Rm}'$ for a couple of minutes and see that we must have $R'(x,y)z = \langle y, z \rangle \, x - \langle x, z \rangle \, y$, because that is a tensor such that
$$
\langle R'(x,y)w, z \rangle
= \operatorname{Rm}'(x,y,z,w)
$$
for all $w$.
A: For $Rm(X, Y, Z, W)$, you can use equations (5) and (6) in Karcher's 1970 paper "A short proof of Berger's curvature tensor estimates". These formulas give the result for $X, Y, Z, W$ orthonormal. Then, some work is needed to generalize to arbitrary vectors, along the same lines as in the proof of the second theorem in that (3 page) paper. Theorems in this paper provide bounds on $Rm$ in terms of sectional curvature in the general case, which may be useful to go beyond constant sectional curvature. 
