18.3 Curvature and torsion

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Torsion tensor

In the transformation law of the components ${\Gamma }_{jk}^i$ , Eq. (18.11), the term involving second derivatives of the transformation functions is symmetric in the indices $jk$ . It follows that the antisymmetrized quantity $T_{jk}^i={\Gamma }_{kj}^i-{\Gamma }_{jk}^i$ does transform as a tensor, since the nontensorial parts of the transformation law Eq. (18.11) cancel out.

To express this idea in an invariant non-coordinate way, observe that for any vector field $Y$ the tensor field $DY$ defined in Eq. (18.4) satisfies the identities

$$DY(f\omega ,X)=fDY(\omega ,X),DY(\omega ,fX)=fDY(\omega ,X)$$

for all functions $f\in \mathcal{F}(M)$ . These are called $\mathcal{F}$-linearity in the respective arguments. On the other hand, the map $D^{\prime }:{\mathcal{T}}^{\ast }(M)\times \mathcal{T}(M)\times \mathcal{T}(M)\mathcal{F}(M)$ defined by

$$D^{\prime }(\omega ,X,Y)=⟨\omega ,D_{X}Y⟩$$

is not a tensor field of type (1, 2), since $\mathcal{F}$-linearity fails for the third argument by (Con4),

$$\begin{array}{c}{D}^{\prime }(\omega ,X,fY)=⟨\omega ,D_X(fY)⟩=⟨\omega ,(Xf)Y+f{D}_{X}Y⟩\\ =(Xf)⟨\omega ,Y⟩+f{D}^{\prime }(\omega ,X,Y)\ne f{D}^{\prime }(\omega ,X,Y).\end{array}$$

Exercise

Show that the ‘components’ of $D$ are $D^{\prime }(\mathrm{d}{x}^i,{\partial }_{x^j},{\partial }_{x^k})={\Gamma }_{jk}^i$

Now let the torsion map $\tau$ : $\mathcal{T}(M)\times \mathcal{T}(M)\to \mathcal{T}(M)$ be defined by

$$\tau (X,Y)=D_{X}Y-D_{Y}X-[X,Y]=-\tau (Y,X).\text{\hspace{1em}(18.19)}$$

This map is $\mathcal{F}$-linear in the first argument,

$$\begin{array}{rl}& \tau (fX,Y)=D_{fX}Y-D_Y(fX)-[fX,Y]\\ & =f{D}_{X}Y-(Yf)X-f{D}_{Y}X-f[X,Y]+(Yf)X\\ & =f\tau (X,Y),\end{array}$$

and by antisymmetry it is also -linear in the second argument $Y$. Hence $\tau$ gives rise to a tensor field $T$ of type (1, 2) by

$$T(\omega ,X,Y)=⟨\omega ,\tau (X,Y)⟩,\text{\hspace{1em}(18.20)}$$

known as the torsion tensor of the connection $D$. In a local coordinate chart $(U;x^i)$ its components are precisely the antisymmetrized connection components:

$$T_{jk}^i={\Gamma }_{kj}^i-{\Gamma }_{jk}^i.\text{\hspace{1em}(18.21)}$$

The prooffollows from setting $T_{jk}^i=⟨\mathrm{d}{x}^i,\tau \left({\partial }_{x^j},{\partial }_{x^k}\right)⟩$ and sustituting Eq. (18.19). We cal a connection torsion-free or symmetric if its torsion tensor vanishes, $T=0;$ ; equivalently, its components are symmetric with respect to all coordinates, ${\Gamma }_{jk}^i={\Gamma }_{kj}^i$

Exercise

Prove Eq. (18.21).

Curvature tensor

A similar problem occurs when commuting repeated covariant derivatives on a vector or tensor field. If $X$, $Y$ and $Z$ are any vector fields on $M$ then $D_X{D}_{Y}Z-D_Y{D}_{X}Z$ is obviously a vector field, but the map $P:{\mathcal{T}}^{\ast }(M)\times \mathcal{T}(M)\times \mathcal{T}(M)\times \mathcal{T}(M)\to \mathcal{F}(M)$ defined by $P(\omega ,X,Y,Z)=⟨\omega ,D_X{D}_{Y}Z-D_Y{D}_{X}Z⟩$ fails to be a tensor field oftype (1, 3) as it is not $\mathcal{F}$-linear in the three vector field arguments. The remedy is similar to that for creating the torsion tensor.

For any pair of vector fields $X$ and $Y$ define the operator ${\rho }_{X,Y}:\mathcal{T}(M)\to \mathcal{T}(M)$ by

$${\rho }_{X,Y}Z=D_X{D}_{Y}Z-D_Y{D}_{X}Z-D_{[X,Y]}Z=-{\rho }_{Y,X}Z.\text{\hspace{1em}(18.22)}$$

This operator is $\mathcal{F}.$ -linear with respect to $X$, and therefore $Y$, since

$$\begin{array}{rl}& {\rho }_{fX,Y}Z=D_{fX}{D}_{Y}Z-D_Y{D}_{fX}Z-D_{[fX,Y]}Z\\ & =f{D}_X{D}_{Y}Z-(Yf)D_{X}Z-f{D}_Y{D}_{X}Z-f{D}_{[X,Y]}Z+(Yf)D_{X}Z\\ & =f{\rho }_{X,Y}Z.\end{array}$$

$\mathcal{F}$-linearity with respect to $Z$ follows from

$$\begin{array}{l}{\rho }_{X,Y}fZ=D_X{D}_Y(fZ)-D_Y{D}_X(fZ)-D_{[X,Y]}(fZ)\\ =D_X((Yf)Z+f{D}_{Y}Z)-D_Y((Xf)Z+f{D}_{X}Z)-([X,Y]f)Z-f{D}_{[X,Y]}Z\\ =(Yf)D_{X}Z+X(Yf)Z+(Xf)D_{Y}Z+f{D}_X{D}_{Y}Z-Y(Xf)Z-(Xf)D_{Y}Z\\ -(Yf)D_{X}Z-f{D}_Y{D}_{X}Z-([X,Y]f)Z-f{D}_{[X,Y]}Z\\ =f{\rho }_{X,Y}Z.\end{array}$$

We can therefore define a tensor field $R$ of type (1, 3), by setting

$$R(\omega ,Z,X,Y)=⟨\omega ,{\rho }_{X,Y}Z⟩=-R(\omega ,Z,Y,X),\text{\hspace{1em}(18.23)}$$

called the curvature tensor of the connection $D$.

For a torsion-free connection, $D_{X}Y-D_{Y}X=[X,Y]$ , there is a cyclic identity:

$$\begin{array}{rl}{\rho }_{X,Y}Z+{\rho }_{Y,Z}X+{\rho }_{Z,X}Y& =D_X{D}_{Y}Z-D_Y{D}_{X}Z-D_{[X,Y]}Z\\ & +D_Y{D}_{Z}X-D_Z{D}_{Y}X-D_{[Y,Z]}X+D_Z{D}_{X}Y\\ & -D_X{D}_{Z}Y-D_{[Z,X]}Y\\ & =D_X[Y,Z]+D_Y[Z,X]+D_Z[X,Y]-D_{[Y,Z]}X\\ & -D_{[Z,X]}Y-D_{[X,Y]}Z\\ & =[X,[Y,Z]]+[Y,[Z,X]]+[Z,[X,Y]]=0\end{array}$$

using $\tau (X,[Y,Z])=0$ , etc. and the Jacobi identity (15.24). For $T=0$ we thus have the so-called first Bianchi identity,

$$R(\omega ,Z,X,Y)+R(\omega ,X,Y,Z)+R(\omega ,Y,Z,X)=0.\text{\hspace{1em}(18.24)}$$

In a coordinate system $(U;x^i)$ , using Eq. (18.5) and $\left[{\partial }_{x^k},{\partial }_{x^l}\right]=0$ for all $k$, $l$, the components of the curvature tensor are

$$\begin{array}{rl}& R_{jkl}^i=R(\mathrm{d}{x}^i,{\partial }_{x^j},{\partial }_{x^k},{\partial }_{x^l})\\ & =⟨\mathrm{d}{x}^i,D_k{D}_l{\partial }_{x^j}-D_l{D}_k{\partial }_{x^j}-D_{[{\partial }_{x^k},{\partial }_{x^l}]}{\partial }_{x^j}⟩\\ & =⟨\mathrm{d}{x}^i,D_k({\Gamma }_{jl}^m{\partial }_{x^m})-D_l({\Gamma }_{jk}^m{\partial }_{x^m})⟩\\ & =⟨\mathrm{d}{x}^i,{\Gamma }_{jl,k}^m{\partial }_{x^m}+{\Gamma }_{jl}^m{\Gamma }_{mk}^p{\partial }_{x^p}-{\Gamma }_{jk,l}^m{\partial }_{x^m}-{\Gamma }_{jk}^m{\Gamma }_{ml}^p{\partial }_{x^p}⟩\end{array}$$

where ${\Gamma }_{jk,l}^m=\partial {\Gamma }_{jk}^k/\partial x^l$ . Hence

$$R_{jkl}^i={\Gamma }_{jl,k}^i-{\Gamma }_{jk,l}^i+{\Gamma }_{jl}^m{\Gamma }_{mk}^i-{\Gamma }_{jk}^m{\Gamma }_{ml}^i=-R_{jlk}^i.\text{\hspace{1em}(18.25)}$$

Setting $\omega =\mathrm{d}{x}^i,X={\partial }_{x^k}$ , etc. in the first Bianchi identity Eq. (18.24) gives

$$R_{jkl}^i+R_{klj}^i+R_{ljk}^i=0.\text{\hspace{1em}(18.26)}$$

Another class of identities, known as Ricci identities, are sometimes used to define the torsion and curvature tensors in a coordinate region $(U;x^i)$ . For any smooth function $f$ on $U$, set $f_{;ij}\equiv (f_{;i}{)}_{;j}=(f_{,i}{)}_{;j}$ . Then, by Eq. (18.13),

$$f_{;ij}-f_{;ji}=f_{,ij}-{\Gamma }_{ij}^k{f}_{,k}-f_{,ji}+{\Gamma }_{ji}^k{f}_{,k}=T_{ij}^k{f}_{,k}.\text{\hspace{1em}(18.27)}$$

Similarly, for a smooth vector field $X=X^k{\partial }_{x^k}$ , Eq. (18.25) gives rise to

$$X_{;ij}^k-X_{;ji}^k=X^a{R}_{aji}^k+T_{ij}^a{X}_{;a}^k,\text{\hspace{1em}(18.28)}$$

and for a 1-form $\omega =w_i\mathrm{d}{x}^i$

$$w_{k;ij}-w_{k;ji}=w_a{R}_{kij}^a+T_{ij}^a{w}_{k;a}.\text{\hspace{1em}(18.29)}$$

Problems

Problem 18.6

Let $f$ be a smooth function, $X=X^i{\partial }_{x^i}$ a smooth vector field and $\omega =w_i\mathrm{d}{x}^i$ a differential 1-form. Show that

$$\begin{array}{l}(D_j{D}_i-D_i{D}_j)f=0,\\ (D_j{D}_i-D_i{D}_j)X=X^a{R}_{aji}^k{\partial }_{x^k},\\ (D_j{D}_i-D_i{D}_j)\omega =w_a{R}_{kji}^a\mathrm{d}{x}^k.\end{array}$$

Why does the torsion tensor not appear in these formulae, in contrast with the Ricci identities Eq. (18.27)– Eq. (18.29)?

Problem 18.7

Show that the coordinate expression for the Lie derivative of a vector field may be written

$$({\mathcal{L}}_{X}Y{)}^i=[X,Y{]}^i=Y_{;j}^i{X}^j-X_{;j}^i{Y}^j+T_{jk}^i{X}^k{Y}^j.\text{\hspace{1em}(18.30)}$$

For a torsion-free connection show that the Lie derivative (15.39) of a general tensor field $S$ of type $(r,s)$ may be expressed by

$$\begin{array}{c}({\mathcal{L}}_{X}S{)}_{kl\dots }^{ij\dots }=S_{kl\dots ;m}^{ij\dots }{X}^m-S_{kl\dots }^{mj\dots }{X}_{;m}^i-S_{kl\dots }^{im\dots }{X}_{;m}^j-\dots \\ +S_{ml\dots }^{ij\dots }{X}_{;k}^m+S_{km\dots }^{ij\dots }{X}_{;l}^m+\dots \end{array}\text{\hspace{1em}(18.31)}$$

Write down the full version of this equation for a general connection with torsion.

Problem 18.8

Prove the Ricci identities Eq. (18.28) and Eq. (18.29).

Problem 18.9

For a torsion-free connection prove the generalized Ricci identities

$$\begin{array}{c}{S}_{mn\dots ;ij}^{kl\dots }-S_{mn\dots ;ji}^{kl\dots }=S_{mn\dots }^{al\dots }{R}_{aji}^k+S_{mn\dots }^{ka\dots }{R}_{aji}^l+\dots \\ +S_{an\dots }^{kl\dots }{R}_{mij}^a+S_{ma\dots }^{kl\dots }{R}_{nij}^a+\dots \end{array}$$

How is this equation modified in the case of torsion?

Problem 18.10

For arbitrary vector fields $Y$, $Z$ and $W$ show that the operator ${\Sigma }_{Y,Z,W}:T(M)\to \mathcal{T}(\mathbf{M})$ defined by

$${\Sigma }_{Y,Z,W}X=D_W\left({\rho }_{Y,Z}X\right)-{\rho }_{Z,[Y,W]}X-{\rho }_{Y,Z}\left(D_{W}X\right)$$

has the cyclic symmetry

$${\Sigma }_{Y,Z,W}X+{\Sigma }_{Z,W,Y}X+{\Sigma }_{W,Y,Z}X=0.$$

Express this equation in components with respect to a local coordinate chart and show that it is equivalent to the (second) Bianchi identity

$$R_{jkl;m}^i+R_{jlm;k}^i+R_{jmk;l}^i=R_{jpk}^i{T}_{ml}^p+R_{jpl}^i{T}_{km}^p+R_{jpm}^i{T}_{lk}^p.\text{\hspace{1em}(18.32)}$$

Problem 18.11

Let $Y^i(t)$ be a vector that is parallel propagated along a curve having coordinate representation $x^j={\stackrel{0}{x}}^j+A^{j}t$ , Show that for $t\ll 1$

$$Y^i(t)=\stackrel{0}{Y^i}-\stackrel{0}{{\Gamma }_{ja}^i}\stackrel{0}{Y^j}{A}^{a}t+\frac{t^2}{2}(\stackrel{0}{{\Gamma }_{ka}^i}\stackrel{0}{{\Gamma }_{ja}^k}-\stackrel{0}{{\Gamma }_{ja,b}^j})A^a{A}^b\stackrel{0}{Y^j}+O(t^3)$$

where ${\stackrel{0}{{\Gamma }^i}}_{jk}={\Gamma }_{jk}^i(\stackrel{0}{x^a})$ and ${\zeta }^0{Y}^i=Y^i(0)$ . From the point $P$, having coordinates ${\chi }_{x^i}^0$ , parallel transport the tangent vector $\stackrel{0}{Y^i}$ around a coordinate rectangle $PQRSP$ whose sides are each of paramete

length $t$ and are along the $a$- and $b$-axes successively through these points. For example, the $a$-axis through $P$ is the curve $x^j={\stackrel{0}{x}}^j+{\delta }_a^{j}t$ . Show that to order $t^2,$ , the final vector at $P$ has components

$$Y^i=\stackrel{0}{Y^i}+t^2{\stackrel{0}{R^i}}_{jba}\stackrel{0}{Y^j}$$

where ${\stackrel{0}{R^i}}_{jba}$ are the curvature tensor components at $P$