18.6 The Riemann tensor and its symmetries

中文

In a pseudo-Riemannian manifold $(M,g)$ it is possible to lower the contravariant index of the curvature tensor to form a tensor $\overline{R}$ of type (0, 4),

$$\overline{R}(W,Z,X,Y)=R(\omega ,Z,X,Y)\text{ where }g(W,A)=⟨\omega ,A⟩.$$

This tensor will be referred to as the Riemann curvature tensor or simply the Riemann tensor. Setting $W={\partial }_{x^i},Z={\partial }_{x^j}$ , etc. then $\omega =g_{ia}\mathrm{d}{x}^a$ , whence

$${\overline{R}}_{ijkl}=g_{ia}{R}_{jkl}^a.$$

In line with the standard index lowering convention, we denote the components of $\overline{R}$ by $R_{ijkl}$

The following symmetries apply to the Riemann tensor:

$$R_{ijkl}=-R_{ijlk},\text{\hspace{1em}(18.49)}$$

$$R_{ijkl}=-R_{jikl},\text{\hspace{1em}(18.50)}$$

$$R_{ijkl}+R_{iklj}+R_{iljk}=0,\text{\hspace{1em}(18.51)}$$

$$R_{ijkl}=R_{klij}.\text{\hspace{1em}(18.52)}$$

Proof

Antisymmetry in the second pair of indices, Eq. (18.49), follows immediately from the definition of the curvature tensor Eq. (18.23) – it is not changed by the act of lowering the first index. Similarly, Eq. (18.51) follows immediately from Eq. (18.26). The remaining symmetries Eq. (18.50) and Eq. (18.52) may be proved by adopting geodesic coordinates at any given point $p$ and using the expression Eq. (18.43) for the components of ${R^i}_{jkl}$

$$R_{ijkl}(p)=\frac{1}{2}(g_{ml,jk}+g_{jk,ml}-g_{mk,jl}-g_{jl,mk}){|}_p.\text{\hspace{1em}(18.53)}$$

A more ‘invariant’ proof of Eq. (18.50) is to apply the generalized Ricci identities, Problem 18.9, to the metric tensor $g_{:}$ ,

$$0=g_{ij;kl}-g_{ij;lk}=g_{aj}{R}_{ikl}^a+g_{ia}{R}_{jkl}^a=R_{jikl}+R_{ijkl}.$$

The symmetry Eq. (18.52) is actually a consequence of the first three symmetries, as may be shown by performing cyclic permutations on all four indices of Eq. (18.51):

$$R_{jkli}+R_{jlik}+R_{jikl}=0,\text{\hspace{1em}(18.51a)}$$ $$R_{klij}+R_{kijl}+R_{kjli}=0,\text{\hspace{1em}(18.51b)}$$ $$R_{lijk}+R_{ljki}+R_{lkij}=0.\text{\hspace{1em}(18.51c)}$$

The combination Eq. (18.51) (18.51$a$) (18.51$b$) (18.51$c$) gives, after several cancellations using the symmetries Eq. (18.49) and Eq. (18.50),

$$2R_{ijkl}-2R_{klij}=0.$$

This is obviously equivalent to $Eq.$ . Eq. (18.52).

Exercise

Prove from these symmetries that the cyclic symmetry also holds for any three indices; for example

$$R_{ijkl}+R_{jkil}+R_{kijl}=0.$$

These symmetries permit us to count the number of independent components of the Riemann tensor. Since a skew symmetric tensor of type (0, 2) on an $n$-dimensional vec tor space has $\frac{1}{2}n(n-1)$ independent components, a tensor of type (0, 4) subject to sym metries Eq. (18.49) and Eq. (18.50) will have $\frac{1}{4}{n}^2(n-1{)}^2$ independent components. For fixed $i=1,2,\dots ,n$ only unequal triples $j\ne k\ne l$ need be considered in the symmetry Eq. (18.52), for if some pair are equal nothing new is added, by the cyclic identity: for example, $i$ $k=l=1$ then $R_{ij11}+R_{i11j}+R_{i1j1}=0$ merely reiterates the skew symmetry on the sec ond pair ofindices, $R_{i11j}=-R_{i1j1}$ . For every triple $j\ne k\ne l$ the total number ofrelations generated by Eq. (18.52) that are independent of Eq. (18.49) and Eq. (18.50) is therefore the number of such triples ofnumbers $j\ne k\ne l$ in the range 1, … , $n$, namely $n$ $\left(\genfrac{}{}{0ex}{}{n}{3}\right)=n^2(n-1)(n-2)/6.$ By the above proof we need not consider the symmetry Eq. (18.52), and the total number of independent components of the Riemann tensor is

$$\begin{array}{l}N=\frac{n^2(n-1{)}^2}{4}-\frac{n^2(n-1)(n-2)}{6}\\ =\frac{n^2(n^2-1)}{12}.\end{array}\text{\hspace{1em}(18.54)}$$

For low dimensions the number of independent components of the Riemann tensor is

$$\begin{array}{ll}n=1\text{:}& N=0,\\ n=2\text{:}& N=1,\\ n=3\text{:}& N=6,\\ n=4\text{:}& N=20.\end{array}$$

A tensor of great interest in general relativity is the Ricci tensor, defined by

$$\mathrm{Ric}=C_2^{1}R.$$

It is common to write the components of $\mathrm{Ric}({\partial }_{x^i},{\partial }_{x^j})$ in any chart $(U;x^i)$ as $R_{ij}$ :

$$R_{ij}=R_{iaj}^a=g^{ab}{R}_{aibj}.\text{\hspace{1em}(18.55)}$$

This tensor is symmetric since, by symmetry Eq. (18.52),

$$R_{ij}=g^{ab}{R}_{aibj}=g^{ab}{R}_{bjai}=R_{jai}^a=R_{ji}.$$

Contracting again gives the quantity known as the Ricci scalar,

$$R=R_i^i=g^{ij}{R}_{ij}.\text{\hspace{1em}(18.56)}$$

Bianchi identities

For a torsion-free connection we have, on setting $T_{km}^p=0$ in Eq. (18.32) of Problem 18.10, the second Bianchi identity

$$R_{jkl;m}^i+R_{jlm;k}^i+R_{jmk;l}^i=0.\text{\hspace{1em}(18.57)}$$

These are often referred to simply as the Bianchi identities. An alternative demonstration is to use normal coordinates at any point $p\in M$ , such that ${\Gamma }_{jk}^i(p)=0$ . Making use of Eqs. Eq. (18.14) and Eq. (18.25) we have

$$\begin{array}{rl}& R_{jkl;m}^i(p)=R_{jkl,m}^i(p)+({\Gamma }_{am}^i{R}_{jkl}^a-{\Gamma }_{jm}^a{R}_{akl}^i-{\Gamma }_{km}^a{R}_{jal}^i-{\Gamma }_{lm}^a{R}_{jka}^i)(p)\\ & =R_{jkl,m}^i(p)\\ & ={\Gamma }_{jl,km}^i(p)-{\Gamma }_{jk,lm}^i(p)+{\Gamma }_{jl,m}^a{\Gamma }_{ak}^i(p)+{\Gamma }_{jl}^a(p){\Gamma }_{ak,m}^i-{\Gamma }_{jk,m}^a{\Gamma }_{al}^i(p)\\ & -{\Gamma }_{jk}^a(p){\Gamma }_{al,m}^i\\ & ={\Gamma }_{jl,km}^i(p)-{\Gamma }_{jk,lm}^i(p).\end{array}$$

If we substitute this expression in the left-hand side of Eq. (18.57) and use ${\Gamma }_{jk}^i={\Gamma }_{kj}^i$ , all terms cancel out.

Contracting Eq. (18.57) over $i$ and $m$ gives

$$R_{jkl;i}^i-R_{jl;k}+R_{jk;l}=0.\text{\hspace{1em}(18.58)}$$

Contracting Eq. (18.58) again by multiplying through by $g^{jl}$ and using Eq. (18.41) we find

$$R_{k;i}^i-R_{;k}+R_{k;j}^j=0,$$

or equivalently, the contracted Bianchi identities

$$R_{k;j}^j-\frac{1}{2}{R}_{,k}=0.\text{\hspace{1em}(18.59)}$$

A useful way of writing Eq. (18.59) is

$$G_{k;j}^j=0,\text{\hspace{1em}(18.60)}$$

where $G^i{}_j$ is the Einstein tensor,

$$G_j^i=R_j^i-\frac{1}{2}R{\delta }_j^i.\text{\hspace{1em}(18.61)}$$

This tensor is symmetric when its indices are lowered,

$$G_{ij}=g_{ia}{G}_j^a=R_{ij}-\frac{1}{2}R{g}_{ij}=G_{ji}.\text{\hspace{1em}(18.62)}$$