# Riemannian curvature tensor of hyperbolic space $\mathbb H^3$

Question:

Given metric $$g=\frac{4}{(1-\rho^2)^2}(d\rho^2+\rho^2d\theta^2+\rho^2\cos^2\theta d\varphi^2)$$ and unit orthogonal vector fields $$X_1=\frac{1-\rho^2}{2}\frac{\partial}{\partial \rho}, X_2=\frac{1-\rho^2}{2\rho}\frac{\partial}{\partial \theta}, X_3=\frac{1-\rho^2}{2\rho\cos\theta}\frac{\partial}{\partial \varphi}.$$

In order to check $$\langle R(X_i,X_j)X_k,X_l\rangle=-(\delta_{ik}\delta_{jl}-\delta_{il}\delta_{jk})\ (i,j,k,l=1,2,3)$$, is it necessary to calculate explicitly it's ture for every component?

Or is there an abstract way to derive this?

Background:

I'm going to check hyperbolic space $$\Bbb H^n:=(B^n,g)$$ has constant sectional curvature $$-1$$,

where $$B^n=\{x\in \Bbb R^n ,|x|<1\}$$ is unit ball and $$g=\frac{4}{(1-\sum_i (x^i)^2)^2}\sum_i dx^i\otimes dx^i$$ is hyperbolic metric.

By using a trick in Riemannian Geometry, it remains to show $$\Bbb H^3$$ has constant sectional curvature $$-1$$.

In spherical coordinate $$\{\rho, \varphi, \theta\}$$ on $$\Bbb R^3-\{0\},$$ hyperbolic metric can be written as $$\frac{4}{(1-\rho^2)^2}(d\rho^2+\rho^2d\theta^2+\rho^2\cos^2\theta d\varphi^2)$$.

Vector fields $$X_1=\frac{1-\rho^2}{2}\frac{\partial}{\partial \rho}, X_2=\frac{1-\rho^2}{2\rho}\frac{\partial}{\partial \theta}, X_3=\frac{1-\rho^2}{2\rho\cos\theta}\frac{\partial}{\partial \varphi}$$ are unit orthogonal vector fields under hyoerbolic metric.

$$[X_1,X_2]=-\frac{1+\rho^2}{2\rho}X_2,\ [X_2,X_3]=\frac{1-\rho^2}{2\rho}\tan \theta\cdot X_3,\ [X_1,X_3]=-\frac{1+\rho^2}{2\rho} X_3.$$

For unit orthogonal vector fields, Koszul formula becomes

$$2\langle\nabla_XY,Z\rangle =\langle[X,Y],Z\rangle-\langle[X,Z],Y\rangle-\langle[Y,Z],X\rangle$$ and we have

$$\nabla_{X_1}X_1=\nabla_{X_1}X_2=\nabla_{X_1}X_3=0.$$ $$\nabla_{X_2}X_2=-\frac{1+\rho^2}{2\rho} X_1,\nabla_{X_2}X_3=0,\nabla_{X_3}X_3=-\frac{1+\rho^2}{2\rho} X_1+\frac{1-\rho^2}{2\rho}\tan \theta\cdot X_2.$$

Sectional curvature is determined by all its 2-dim subspace, and by torsion-free property and calculation, $$K(X_1,X_2):=\langle R(X_1,X_2)X_2,X_1\rangle=-1,$$ $$K(X_2,X_3):=\langle R(X_2,X_3)X_3,X_2\rangle=-1,$$ $$K(X_1,X_3):=\langle R(X_1,X_3)X_3,X_1\rangle=-1.$$ So $$\Bbb H^3$$ has constant sectional curvature $$-1$$.

My textbook also says $$\langle R(X_i,X_j)X_k,X_l\rangle=-(\delta_{ik}\delta_{jl}-\delta_{il}\delta_{jk})\ (i,j,k,l=1,2,3)$$, it's an abstract equation while calculating sectional curvature requires a lot of explicit calculation, so is there an abstract way(without calculating all its components) to derive this equation for Riemmanian curvature?

Thanks for your time and patience.

Lemma(3.4): Let $$M$$ be a Riemannian manifold and $$p$$ a point of $$M$$. Define a tri-linear mapping $$R':T_p M \times T_p M \times T_p M \to T_p M$$ by: $$\left\langle R^{\prime}(X, Y, W), Z\right\rangle=\langle X, W\rangle\langle Y, Z\rangle-\langle Y, W\rangle\langle X, Z\rangle$$ for all $$X, Y, W, Z \in T_{p} M$$. Then $$M$$ has constant sectional curvature equal to $$K_0$$ if and only if $$R = K_0 R'$$, where $$R$$ is the curvature of $$M$$.
If we write $$\langle R(X, Y) Z, T\rangle=(X, Y, Z, T)$$ those will let us deduce: $$(a)(X, Y, Z, T)+(Y, Z, X, T)+(Z, X, Y, T)=0$$ $$\begin{array}{l}{\text { (b) }(X, Y, Z, T)=-(Y, X, Z, T)} \\ {\text { (c) }(X, Y, Z, T)=-(X, Y, T, Z)} \\ {\text { (d) }(X, Y, Z, T)=(Z, T, X, Y)}\end{array}$$ (a) is the Bianchi identity. Those identities will give you the power to use linear algebra to show that if you have data on all 2 dimensional subspaces (sectional curvature) you can deduce stuff about the curvature.