18.7 Cartan formalism

中文

Cartan’s approach to curvature is expressed entirely in terms of differential forms. Let $e_i(i=1,\dots ,n)$ be a local basis of vector fields, spanning $\mathcal{T}(U)$ over an open set $U\subseteq M$ and $\{{\epsilon }^i\}$ the dual basis of ${\mathcal{T}}^{\ast }(U)$ , such that $⟨e_i,{\epsilon }^j⟩={\delta }_i^j$ . For example, in a coordinate chart $(U;x^i)$ we may set $e_i={\partial }_{x^i}$ and ${\epsilon }^j=\mathrm{d}{x}^j$ , but such a coordinate system will exist for an arbitrary basis $\{e_i\}$ if and only if $[e_i,e_j]=0$ for all $i,j=1,\dots ,n$ . We define the connection 1-forms ${\omega }_{\text{ j }}^i:\mathcal{T}(U)\to \mathcal{F}(U)$ by

$$D_X{e}_j={\omega }_j^i(X)e_i.\text{\hspace{1em}(18.63)}$$

These maps are differential 1-forms on $U$ since they are clearly -linear,

$$D_{X+fY}{e}_j=D_X{e}_j+f{D}_Y{e}_j⟹{\omega }_j^i(X+fY)={\omega }_j^i(X)+f{\omega }_j^i(Y).$$

$\mathrm{If}\tau :T(U)\times T(U)\to T(U)$ is the torsion operator defined in Eq. (18.19), set

$$\tau (X,Y)={\tau }^i(X,Y)e_i.$$

The maps ${\tau }^i:\mathcal{T}(U)\times \mathcal{T}(U)\mathcal{F}(U)$ are $\mathcal{F}$-linear in both arguments by the $\mathcal{F}$-linearity of $\tau$, and are antisymmetric ${\tau }^i(X,Y)=-{\tau }^i(Y,X)$ . They are therefore differential 2-forms on $U$ , known as the torsion 2-forms.

Exercise

Show that ${\tau }^i=T_{ik}^i{\epsilon }^j\wedge {\epsilon }^k$ where $T_{jk}^i=⟨\tau (e_j,e_k),{\epsilon }^i⟩$

From the identity $Z=⟨{\epsilon }^i,Z⟩e_i$ for any vector field $Z$ on $U_{:}$ , we have

$$\begin{array}{rl}& \tau (X,Y)=D_X(⟨{\epsilon }^i,Y⟩e_i)-D_Y(⟨{\epsilon }^i,X⟩e_i)-⟨{\epsilon }^i,[X,Y]⟩e_i\\ & =X(⟨{\epsilon }^i,Y⟩)e_i+⟨{\epsilon }^i,Y⟩{\omega }_i^k(X)e_k-Y(⟨{\epsilon }^i,X⟩)e_i-⟨{\epsilon }^i,X⟩{\omega }_i^k(X)e_k\\ & -⟨{\epsilon }^i,[X,Y]⟩e_i\\ & =2\mathsf{d}{\epsilon }^i(X,Y)e_i+2{\omega }_i^k\wedge {\epsilon }^i(X,Y)e_k,\end{array}$$

using the Cartan identity (16.14). Thus

$${\tau }^i(X,Y)e_i=2(\mathrm{d}{\epsilon }^i(X,Y)+2{\omega }_k^i\wedge {\epsilon }^k(X,Y))e_i,$$

and by -linear independence of the vector fields $e_i$ we have Cartan’s first structura equation

$$\mathrm{d}{\epsilon }^i=-{\omega }_k^i\wedge {\epsilon }^k+\frac{1}{2}{\tau }^i.\text{\hspace{1em}(18.64)}$$

Define the curvature 2-forms ${\mathbf{\rho }}_j^i$ by

$${\rho }_{X,Y}{e}_j={\rho }_j^i(X,Y)e_i,\text{\hspace{1em}(18.65)}$$

where ${\rho }_{X,Y}:\mathcal{T}(M)\to \mathcal{T}(M)$ is the curvature operator in Eq. (18.22).

Exercise

Show that the ${{\rho }^i}_j$ are differential 2-forms on $U;$ namely, they are $\mathcal{F}$-linear with respect to $X$ and $Y$ and ${\rho }^i{}_j(X,Y)=-{\rho }^i{}_j(Y,X)$ .

Changing the dummy suffix on the right-hand side of Eq. (18.65) from $i$ to $k$ and applying $⟨{\epsilon }^i,\cdot ⟩$ to both sides of the equation we have, with the help of Eq. (18.23),

$$\begin{array}{l}{\rho }_j^i(X,Y)=⟨{\epsilon }^i,{\rho }_{X,Y}{e}_j⟩\\ =R({\epsilon }^i,e_j,X,Y)\\ =R_{jkl}^i{X}^k{Y}^l\text{where}{R}_{jkl}^i=R({\epsilon }^i,e_j,e_k,e_l)\\ =\frac{1}{2}{R}_{jkl}^i({\epsilon }^k\otimes {\epsilon }^l-{\epsilon }^l\otimes {\epsilon }^k)(X,Y).\end{array}$$

Hence

$${\rho }_j^i=R_{jkl}^i{\epsilon }^k\wedge {\epsilon }^l.\text{\hspace{1em}(18.66)}$$

A similar analysis to that for the torsion operator results in

$$\begin{array}{l}{\rho }_{X,Y}{e}_j=D_X{D}_Y{e}_j-D_Y{D}_X{e}_j-D_{[X,Y]}{e}_j\\ =D_X(⟨{\omega }_j^i,Y⟩e_i)-D_Y(⟨{\omega }_j^i,X⟩e_i)-⟨{\omega }_j^i,[X,Y]⟩e_i\\ =2[\mathrm{d}{\omega }_j^i(X,Y)+{\omega }_k^i\wedge {\omega }_j^k(X,Y)]e_i,\end{array}$$

and Cartan’s second structural equation

$$\mathrm{d}{\omega }_j^i=-{\omega }_k^i\wedge {\omega }_j^k+\frac{1}{2}{\rho }_j^i.\text{\hspace{1em}(18.67)}$$

Example 18.3

With respect to a coordinate basis $e_i={\partial }_{x^i}$ , the Cartan structural equations reduce to formulae found earlier in this chapter. For any vector field $X=X^k{\partial }_{x^k}$ ,

$$D_X{\partial }_{x^j}=X^k{D}_k{\partial }_{x^j}=X^k{\Gamma }_{jk}^i{\partial }_{x^i}.$$

Hence, by Eq. (18.63), we have ${\omega }_j^i(X)=X^k{\Gamma }_{jk}^i$ , so that

$${\omega }_j^i={\Gamma }_{jk}^i\mathrm{d}{x}^k.\text{\hspace{1em}(18.68)}$$

Thus the components of the connection 1-forms with respect to a coordinate basis are precisely the components of the connection.

Setting ${\epsilon }^i=\mathrm{d}{x}^i$ in Cartan’s first structural equation Eq. (18.64), we have

$$\mathrm{d}{\epsilon }^i={\mathrm{d}}^2{x}^i=0=-{\omega }_j^i\wedge \mathrm{d}{x}^j+\frac{1}{2}{\tau }^i.$$

Hence ${\tau }^i=2{\Gamma }_{jk}^i\mathrm{d}{x}^k\wedge \mathrm{d}{x}^j$ , and it follows that the components of the torsion 2-forms are identical with those of the torsion tensor $T$ in Eq. (18.21),

$${\tau }^i=T_{jk}^i\mathrm{d}{x}^j\wedge \mathrm{d}{x}^k\text{ where }{T}_{jk}^i={\Gamma }_{kj}^i-{\Gamma }_{jk}^i.\text{\hspace{1em}(18.69)}$$

Finally, Cartan’s second structural equation Eq. (18.67) reduces in a coordinate basis to

$$\mathrm{d}{\Gamma }_{jk}^i\wedge \mathrm{d}{x}^k=-{\Gamma }_{kl}^i\mathrm{d}{x}^l\wedge {\Gamma }_{jm}^k\mathrm{d}{x}^m+\frac{1}{2}{\rho }_j^i,$$

whence, on using the decomposition Eq. (18.66),

$$R_{jkl}^i\mathrm{d}{x}^k\wedge \mathrm{d}{x}^l=2{\Gamma }_{jk,l}^i\mathrm{d}{x}^l\wedge \mathrm{d}{x}^k+2{\Gamma }_{ml}^i{\Gamma }_{jk}^m\mathrm{d}{x}^l\wedge \mathrm{d}{x}^k.$$

We thus find, in agreement with Eq. (18.25),

$$R_{jkl}^i={\Gamma }_{jl,k}^i-{\Gamma }_{jk,l}^i+{\Gamma }_{jl}^m{\Gamma }_{mk}^i-{\Gamma }_{jk}^m{\Gamma }_{ml}^i=-R_{jlk}^i.$$

The big advantage ofCartan’s structural equations over these various coordinate expressions is that they give expressions for torsion and curvature for arbitrary vector field bases.

Bianchi identities

Taking the exterior derivative of Eq. (18.64) gives, with the help of Eq. (18.67),

$$\begin{array}{rl}& {\mathrm{d}}^2{\epsilon }^i=0=-\mathrm{d}{\omega }_k^i\wedge {\epsilon }^k+{\omega }_k^i\wedge \mathrm{d}{\epsilon }^k+\frac{1}{2}\mathrm{d}{\tau }^i\\ & ={\omega }_j^i\wedge {\omega }_k^j\wedge {\epsilon }^k-\frac{1}{2}{\rho }_k^i\wedge {\epsilon }^k-{\omega }_k^i\wedge {\omega }_j^k\wedge {\epsilon }^j+\frac{1}{2}{\omega }_k^i\wedge {\tau }^k+\frac{1}{2}\mathrm{d}{\tau }^i\\ & =\frac{1}{2}(-{\rho }_k^i\wedge {\epsilon }^k+{\omega }_k^i\wedge {\tau }^k+\mathrm{d}{\tau }^i).\end{array}$$

Hence, we obtain thefirst Bianchi identity

$$\mathrm{d}{\tau }^i={\rho }_k^i\wedge {\epsilon }^k-{\omega }_k^i\wedge {\tau }^k.\text{\hspace{1em}(18.70)}$$

Its relation to the earlier identity Eq. (18.24) is left as an exercise (see Problem 18.15).

Similarly

$$\begin{array}{r}{\mathrm{d}}^2{\omega }_j^i=0=-\mathrm{d}{\omega }_k^i\wedge {\omega }_j^k+{\omega }_k^i\wedge \mathrm{d}{\omega }_j^k+\frac{1}{2}\mathrm{d}{\rho }_j^i\\ =\frac{1}{2}(-{\rho }_k^i\wedge {\omega }_j^k+{\omega }_k^i\wedge {\rho }_j^k+\mathrm{d}{\rho }_j^i),\end{array}$$

resulting in the second Bianchi identity

$$\mathrm{d}{\rho }_j^i={\rho }_k^i\wedge {\omega }_j^k-{\omega }_k^i\wedge {\rho }_j^k.\text{\hspace{1em}(18.71)}$$

Pseudo-Riemannian spaces in Cartan formalism

In a pseudo-Riemannian manifold with metric tensor ${}^{g,}$ , set $g_{ij}=g(e_i,e_j)$ . Since $g_{ij}$ is a scalar field for each $i,j=1,\dots ,n$ and $D_{X}g=0$ , we have

$$\begin{array}{rl}⟨X,\mathrm{d}{g}_{ij}⟩=X(g_{i}j)& =D_X(g(e_i,e_j))\\ & =g(D_X{e}_i,e_j)+g(e_i,D_X{e}_j)\\ & =g({\omega }_i^k(X)e_k,ej)+g(e_i,{\omega }_j^k(X)e_k)\\ & =g_{kj}{\omega }_i^k(X)+g_{ik}{\omega }_j^k(X)\\ & =⟨X,{\omega }_{ji}⟩+⟨X,{\omega }_{ij}⟩\end{array}$$

where

$${\omega }_{ij}=g_{ki}{\omega }_j^k.$$

As $X$ is an arbitrary vector field,

$$\mathrm{d}{g}_{ij}={\omega }_{ij}+{\omega }_{ji}.\text{\hspace{1em}(18.72)}$$

For an orthonormal basis $e_i$ , such that $g_{ij}={\eta }_{ij}$ , we have $\mathrm{d}{g}_{ij}=0$ and

$${\omega }_{ij}=-{\omega }_{ji}.\text{\hspace{1em}(18.73)}$$

In particular, all diagonals vanish, $g_{ii}=0$ for $i=1,\dots ,n$

Lowering the first index on the curvature 2-forms ${\rho }_{ij}=g_{ik}{\rho }_j^k$ , we have from the second Cartan structural equation Eq. (18.67),

$$\begin{array}{rl}& {\rho }_{ij}=2(\mathrm{d}{\omega }_{ij}+{\omega }_{ik}\wedge {\omega }_j^k)\\ & =2(-\mathrm{d}{\omega }_{ji}+{\omega }_j^k\wedge {\omega }_{ki})\\ & =2(-\mathrm{d}{\omega }_{ji}-{\omega }_{jk}\wedge {\omega }_i^k),\end{array}$$

whence

$${\rho }_{ij}=-{\rho }_{ji}.\text{\hspace{1em}(18.74)}$$

Exercise

Show that Eq. (18.74) is equivalent to the symmetry $R_{ijkl}=-R_{jikl}$

Example 18.4

The 3-sphere of radius $a$ is the submanifold of ${\mathbb{R}}^4$ ,

$$S^3(a)=\{(x,y,z,w)|x^2+y^2+z^2+w^2=a^2\}\subset {\mathbb{R}}^4.$$

Spherical polar coordinates $\chi ,\theta ,\varphi$ are defined by

$$\begin{array}{l}x=a\sin \chi \sin \theta \cos \varphi \\ y=a\sin \chi \sin \theta \sin \varphi \\ z=a\sin \chi \cos \theta \\ w=a\cos \chi \end{array}$$

where $0<\chi ,\theta <\pi$ and $0<\varphi <2\pi$ . These coordinates cover all of $S^3(a)$ apart from the points $y=0,x\ge 0$ . The Euclidean metric on ${\mathbb{R}}^4$

$$\mathrm{d}{s}^2=\mathrm{d}{x}^2+\mathrm{d}{y}^2+\mathrm{d}{z}^2+\mathrm{d}{w}^2$$

induces a metric on $S^3(a)$ as for the 2-sphere in Example 18.1:

$$\mathrm{d}{s}^2=a^2\left[\mathrm{d}{\chi }^2+{\sin }^2\chi (\mathrm{d}{\theta }^2+{\sin }^2\theta \mathrm{d}{\varphi }^2)\right].$$

An orthonormal frame is

$$e_1=\frac{1}{a}\frac{\partial }{\partial \chi },e_2=\frac{1}{a\sin \chi }\frac{\partial }{\partial \theta },e_3=\frac{1}{a\sin \chi \sin \theta }\frac{\partial }{\partial \varphi }$$ $${\epsilon }^1=a\mathrm{d}\chi ,{\epsilon }^2=a\sin \chi \mathrm{d}\theta ,{\epsilon }^3=a\sin \chi \sin \theta \mathrm{d}\varphi$$

where

$$g_{ij}=g(e_i,e_j)={\delta }_{ij},g={\epsilon }^1\otimes {\epsilon }^1+{\epsilon }^2\otimes {\epsilon }^2+{\epsilon }^3\otimes {\epsilon }^3.$$

Since the metric connection is torsion-free, ${\tau }^i=0$ , the first structural equation reads

$$\mathsf{d}{\epsilon }^i=-{\omega }_k^i\wedge {\epsilon }^k=-{\omega }_{ik}\wedge {\epsilon }^k$$

setting

$${\omega }_{ij}=A_{ijk}{\epsilon }^k$$

where $A_{ijk}=-A_{jik}$ . By interchanging dummy suffixes $j$ and $k$ we may also write

$$\mathbf{d}{\epsilon }^i=-A_{ikj}{\epsilon }^j\wedge {\epsilon }^k=A_{ikj}{\epsilon }^k\wedge {\epsilon }^j=A_{ijk}{\epsilon }^j\wedge {\epsilon }^k.$$

For $i=1$ , using $A_{11k}=0$

$$\mathrm{d}{\epsilon }^1=a{\mathrm{d}}^2\chi =0=A_{121}{\epsilon }^2\wedge {\epsilon }^1+A_{131}{\epsilon }^3\wedge {\epsilon }^1+(A_{123}-A_{132}){\epsilon }^2\wedge {\epsilon }^3,$$

whence

$$A_{121}=A_{131}=0,A_{123}=A_{132}.$$

For $i=2$

$$\begin{array}{rl}& \mathrm{d}{\epsilon }^2=a\cos \chi \mathrm{d}\chi \wedge \mathrm{d}\theta =a^{-1}\cot \chi {\epsilon }^1\wedge {\epsilon }^2\\ & =A_{212}{\epsilon }^1\wedge {\epsilon }^2+A_{232}{\epsilon }^3\wedge {\epsilon }^2+(A_{213}-A_{231}){\epsilon }^1\wedge {\epsilon }^3,\end{array}$$

which implies

$$A_{212}=a^{-1}\cot \chi ,A_{213}=A_{231},A_{232}=0.$$

Similarly the $i$ 3 equation gives

$$A_{312}=A_{321},A_{313}=a^{-1}\cot \chi ,A_{323}=a^{-1}\frac{\cot \theta }{\sin \chi }.$$

All coefficients having all three indices different, such as ${\mathbf{A}}_{123}$ , must vanish since

$$A_{123}=A_{132}=-A_{312}=-A_{321}=A_{231}=A_{213}=-A_{123}⟹A_{123}=0.$$

There is enough information now to write out the connection 1-forms:

$${\omega }_{12}=-{\omega }_{21}=-a^{-1}\cot \chi {\epsilon }^2,$$ $${\omega }_{13}=-{\omega }_{31}=-a^{-1}\cot \chi {\epsilon }^3,$$ $${\omega }_{23}=-{\omega }_{32}=-a^{-1}\frac{\cot \theta }{\sin \chi }{\epsilon }^3.$$

The second structural relations Eq. (18.67) can now be used to calculate the curvature 2- forms:

$$\begin{array}{l}{\rho }_{12}=2(\mathrm{d}{\omega }_{12}+{\omega }_{1k}\wedge {\omega }_2^k)\\ =2(a^{-1}{\mathrm{cosec}}^2\chi \mathrm{d}\chi \wedge {\epsilon }^2-a^{-1}\cot \chi \mathrm{d}{\epsilon }^2+{\omega }_{13}\wedge {\omega }_{32})\\ =2a^{-2}({\mathrm{cosec}}^2\chi {\epsilon }^1\wedge {\epsilon }^2-{\cot }^2\chi {\epsilon }^1\wedge {\epsilon }^2)\\ =2a^{-2}{\epsilon }^1\wedge {\epsilon }^2,\end{array}$$

and similarly

$$\begin{array}{l}{\rho }_{13}=2(\mathrm{d}{\omega }_{13}+{\omega }_{12}\wedge {\omega }_{23})=2a^{-2}{\epsilon }^1\wedge {\epsilon }^3,\\ {\rho }_{23}=2(\mathrm{d}{\omega }_{23}+{\omega }_{21}\wedge {\omega }_{13})=2a^{-2}{\epsilon }^2\wedge {\epsilon }^3.\end{array}$$

The components of the Riemann curvature tensor can be read off using Eq. (18.66):

$$R_{1212}=R_{1313}=R_{2323}=a^{-2},$$

and all other components of the Riemann tensor are simply related to these components by symmetries; for example, $R_{2121}=-R_{1221}=a^{-2},R_{1223}=0$ , etc. It is straightforward to verify the relation

$$R_{ijkl}=\frac{1}{a^2}(g_{ik}{g}_{jl}-g_{il}{g}_{jk}).$$

For any Riemannian space $(M,g)$ the sectional curvature of the vector 2-space spanned by a pair of tangent vectors $X$ and $Y$ at any point $p$ is defined to be

$$K(X,Y)=\frac{R(X,Y,X,Y)}{A(X,Y)}$$

where $A(X,Y)$ is the ‘area’ of the parallelogram spanned by $X$ and $Y$,

$$A(X,Y)=g(X,Y)g(X,Y)-g(X,X)g(Y,Y).$$

For the 3-sphere

$$K(X,Y)=\frac{1}{a^2}\frac{X_i{Y}^i{X}_k{Y}^k-X_i{X}^i{Y}_k{Y}^k}{X_i{Y}^i{X}_k{Y}^k-X_i{X}^i{Y}_k{Y}^k}=\frac{1}{a^2}$$

independent of the point $p\in S^3(a)$ and the choice of tangent vectors $X$, $Y$. For this reason the 3-sphere is said to be a space of constant curvature.

Locally flat spaces

A manifold $M$ with affine connection is said to be locally flat if for every point $p\in M$ there is a chart $(U;x^i)$ such that all components of the connection vanish throughout $U$. This implies of course that both torsion tensor and curvature tensor vanish throughout $U$, but more interesting is that these conditions are both necessary and sufficient. This result is most easily proved in Cartan’s formalism, and requires the transformation of the connection 1-forms under a change of basis

$${\epsilon }^{\prime i}=A_j^i{\epsilon }^j\text{ where }{A}_j^i=A_j^i(x^1,\dots ,x^n).$$

Evaluating $\mathrm{d}{\epsilon }^{\prime i}$ , using Eq. (18.64), gives

$$\mathrm{d}{\epsilon }^{\prime i}=\mathrm{d}{A}_j^i\wedge {\epsilon }^j+A_j^i\mathrm{d}{\epsilon }^j=-{\omega }_k^{\prime i}\wedge {\epsilon }^{\prime k}+\frac{1}{2}{\tau }^{\prime i}$$

where ${\tau }^{\prime i}={A^i}_j{\tau }^j$ and

$$A_j^k{\omega }_k^{\prime i}=\mathrm{d}{A}_j^i-A_k^i{\omega }_j^k.\text{\hspace{1em}(18.75)}$$

Exercise

Show from this equation and Eq. (18.67) that if a transformation exists such that ${\omega }_k^{\prime i}=0$ then the curvature 2-forms vanish, ${\rho }_j^k=0$

Theorem 18.1 · Criterion for local flatness of a torsion-free connection

A manifold with symmetric connection is locally flat everywhere if and only ifthe curvature 2-forms vanish, ${{\rho }^k}_j=0$

Proof

The only ifpart follows from the above comments. For the converse, we suppose ${\tau }^i={\rho }_j^i=0$ everywhere. If $(U;x^i)$ is any chart on $M$, le $N=U\times {\mathbb{R}}^{n^2}$ and denote coor dinates on ${\mathbb{R}}^{n^2}$ by ${\mathbf{z}}_{\mathbf{k}}^j$ . Using the second structural formula Eq. (18.67) with ${{\rho }^i}_j=0$ , the 1-forms ${\mathbf{\alpha }}_j^i=\mathrm{d}{z}_j^i-z_k^i{\mathbf{\omega }}_j^k$ on $N$ satisfy

$$\begin{array}{rl}\mathrm{d}{\alpha }_j^i& =-\mathrm{d}{z}_k^i\wedge {\omega }_j^k-z_k^i\mathrm{d}{\omega }_j^k\\ & =-({\alpha }_k^i+z_m^i\mathrm{d}{\omega }_k^m)\wedge {\omega }_j^k+z_k^i{\omega }_m^k\wedge {\omega }_j^m\\ & ={\omega }_j^k\wedge {\alpha }_k^i.\end{array}\text{\hspace{1em}(18.76)}$$

By the Frobenius theorem 16.4, $\mathrm{d}{\alpha }_j^i=0$ is an integrable system on $N_z$ , and has a local integral submanifold through any point $x_0^i,A_{0k}^j$ where $\det$[ $A_{0k}^j]\ne 0$ , that may be assumed to be of the form

$$z_i^j=A_i^j(x^1,\dots ,x^n)\text{ where }{A}_i^j(x_0^1,\dots ,x_0^n)=A_{0i}^j.$$

We may assume that de $[A_k^j]$ is non-singular in a neighbourhood of ${\mathbf{X}}_0$ . Hence

$${\alpha }_j^i=0⟹\mathrm{d}{A}_j^i-A_k^i{\omega }_j^k=0$$

and substituting in Eq. (18.75) results in

$${\omega }_k^{\prime i}=0\text{ if }{\epsilon }^{\prime i}=A_j^i{\epsilon }^j.$$

Finally, the structural equation Eq. (18.64) gives $\mathbf{d}{\epsilon }^{\prime i}=0$ , and the Poincaré lemma 17.5 implies that there exist local coordinates $y^i$ such that ${\epsilon }^{\prime i}=\mathrm{d}{\mathbf{y}}^i$ -

In the case of a pseudo-Riemannian space, locally flat coordinates such that ${\Gamma }_{jk}^i=0$ imply that $g_{ij,k}=0$ by Eq. (18.37). Hence $g_{ij}=\mathrm{const}.$ . throughout the coordinate region, and a linear transformation can be used to diagonalize the metric into standard diagonal form $g_{ij}={\eta }_{ij}$ with 1 along the diagonal.

Problems

Problem 18.15

Let $e_i={\partial }_{x^i}$ $i$ be a coordinate basis.

(a) Show that the first Bianchi identity reads

$$R_{[jkl]}^i=T_{[jk;l]}^i-T_{[jk}^a{T}_{l]a}^i,$$

and reduces to the cyclic identity Eq. (18.26) in the case of a torsion-free connection

(b) Show that the second Bianchi identity becomes

$$R_{j[kl;m]}^i=R_{ja[k}^i{T}_{ml]}^a,$$

which is identical with Eq. (18.32) of Problem 18.10

Problem 18.16

In a Riemannian manifold $(M,g)$ show that the sectional curvature $K(X,Y)$ at a point $p_{:}$ defined in Example ${}^{18.4,}$ is independent of the choice of basis of the 2-space; i.e., $K(X^{\prime },Y^{\prime })=K(X,Y)\mathrm{if}{X}^{\prime }=aX+bY,Y^{\prime }=cX+\mathrm{d}Y$ where $ad-bc\ne 0$

The space is said to be isotropic at $p\in M$ if $K(X,Y)$ is independent of the choice of tangen vectors $X$ and $Y\mathrm{at}p$ . If the space is isotropic at each point $p$ show that

$$R_{ijkl}=f(g_{ik}{g}_{jl}-g_{il}{g}_{jk})$$

where $f$ is a scalar field on $M$. Ifthe dimension of the manifold is greater than 2, show Schur’s theorem: a Riemannian manifold that is everywhere isotropic is a space ofconstant curvature, $f=const.[Hint;$ Use the contracted Bianchi identity Eq. (18.59).]

Problem 18.17

Show that a space is locally flat if and only if there exists a local basis of vector fields $\{e_i\}$ that are absolutely parallel, $D{e}_i=0$

Problem 18.18

Let $(M,\phi )$ be a surface ofrevolution defined as a submanifold of ${\mathbb{E}}^3$ of the form

$$x=g(u)\cos \theta ,y=g(u)\sin \theta ,z=h(u).$$

Show that the induced metric (see Example 18.1) is

$$\mathrm{d}{s}^2=\left(g^{\prime }(u{)}^2+h^{\prime }(u{)}^2\right)\mathrm{d}{u}^2+g^2(u)\mathrm{d}{\theta }^2.$$

Picking the parameter $u$ such that $g^{\prime }(u{)}^2+h^{\prime }(u{)}^2=1$ (interpret this choice!), and setting the basis 1-forms to be $\begin{array}{r}{\epsilon }^1=\mathrm{d}u,{\epsilon }^2=g\mathrm{d}\theta \end{array}$ , calculate the connection 1-forms ${\omega }_j^i$ , the curvature 1-forms ${{\rho }^i}_j$ and the curvature tensor component $R_{1212}$

Problem 18.19

For the ellipsoid

$$\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1$$

show that the sectional curvature is given by

$$K={\left(\frac{x^{2}bc}{a^3}+\frac{y^{2}ac}{b^3}+\frac{z^{2}ab}{c^3}\right)}^{-2}.$$