Compute curvature tensor from constant sectional curvature

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 $w$.