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 $R(X,Y)Z$.

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    $\begingroup$ Do you know that if two curvature tensors have the same sectional curvature, then they are equal? Can you write down the curvature tensor of a constant curvature metric? $\endgroup$ – Gunnar Þór Magnússon Apr 7 '17 at 15:04
  • $\begingroup$ I'm not sure if I do know that, would I be able to get a reference to this? The book I've been using is Do Carmo. $\endgroup$ – user291678 Apr 7 '17 at 18:03
  • $\begingroup$ It must be somewhere in Do Carmo, it's a very basic fact. It's really just the same as a quadratic form determining a unique bilinear form by polarization. I know it's in Kobayashi-Nomizu vol 1, but probably somewhere in wikipedia as well. (Of course, you could prove it. :) $\endgroup$ – Gunnar Þór Magnússon Apr 7 '17 at 18:20
  • $\begingroup$ Btw, this is a purely linear algebra fact. You don't have to take the derivative of anything to prove it. $\endgroup$ – Gunnar Þór Magnússon Apr 7 '17 at 18:22
  • $\begingroup$ For what it's worth, you're not the first to have problems with this: math.stackexchange.com/questions/1932735/… $\endgroup$ – Gunnar Þór Magnússon Apr 7 '17 at 18:24

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 $w$.


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