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?


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.


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.

  1. It's a tensor.
  2. 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.

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