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.
 A: There is an abstract way to deduce this equation from a lemma proved in Do-Carmo Riemannian geometry book:

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
  $M$. 

From this you can easily deduce the equation.
There is work to be done to prove this written in chapter 3 of the book. The proof in the book uses two properties of curvature. 


*

*It's a tensor.

*The Bianchi identity.


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. 
